Chapter 1
INTRODUCTION
Conceptual Questions
1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry,
biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of
the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for
example, physics is essential. Also, many study physics for the sense of fulfillment that comes with learning about
the world we inhabit.
2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would
be impossible.
3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features
of a physical system. Often a problem would become too complicated if one attempted to match the real
conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful.
4. (a) 3
(b) 9
5. Scientific notation eliminates the need to write many zeros in very large or small numbers. Also, the appropriate
number of significant digits is unambiguous when written this way.
6. In scientific notation the decimal point is placed after the first (leftmost) numeral. The number of digits written
equals the number of significant figures.
7. Not all of the significant digits are precisely known. The least significant digit (rightmost) is an estimate and is
less precisely known than the others.
8. It is important to list the correct number of significant figures so that we can indicate how precisely a quantity is
known and not mislead the reader by writing digits that are not at all known to be correct.
9. The kilogram, meter, and second are three of the base units used in the SI system.
10. The SI system uses a well-defined set of internationally agreed upon standard units and makes measurements in
terms of these units and their powers of ten. The U.S. Customary system contains units that are primarily of
historical origin and are not based upon powers of ten. As a result of this international acceptance and the ease of
manipulation that comes from dealing with powers of ten, scientists around the world prefer to use the SI system.
11. Fathoms, kilometers, miles, and inches are units with dimensions of length. Grams and kilograms are units with
dimensions of mass. Years, months, and seconds are units with dimensions of time.
12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and
obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick
sketch to outline the details of the situation and the known parameters.
13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more
apparent when data is plotted graphically rather than listed in numerical tables.
14. The statement gives a numerical value for the speed of sound in air, but fails to indicate the units used for the
measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s.
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15. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. It may
also be useful to explore other possible methods of solution as a check on the validity of the first.
Problems
1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence.
Solution Find the height of the new fence.
1.37 × 1.8 m = 2.5 m
2. Strategy There are
60 s 60 min 24 h
×
×
= 86, 400 seconds in one day and 24 hours in one day.
1 min
1h
1d
Solution Find the ratio of the number of seconds in a day to the number of hours in a day.
86, 400 24 × 3600
=
= 3600 1
24
24
3. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 1.160S1 = 1.160(4π r12 ).
4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 1.160
⎝ r1 ⎠
r2
= 1.160 = 1.077
r1
The radius of the balloon increases by 7.7%.
4. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 2.0 S1 = 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 2.0
⎝ r1 ⎠
r2
= 2.0 = 1.4
r1
The radius of the balloon increases by a factor of 1.4.
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Chapter 1: Introduction
5. Strategy The surface area S and the volume V are given by S = 6s 2 and V = s3 , respectively.
Solution Find the ratio of the surface area to the volume.
S 6s 2
6
=
=
3
V
s
s
6. Strategy To find the factor Samantha’s height increased, divide her new height by her old height. Subtract 1 from
this value and multiply by 100 to find the percent increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .
7. Strategy Recall that area has dimensions of length squared.
Solution Find the ratio of the area of the park as represented on the map to the area of the actual park.
map length
1
map area
=
= 10−4 , so
= (10−4 ) 2 = 10−8 .
actual length 10, 000
actual area
8. Strategy Let X be the original value of the index.
Solution Find the net percentage change in the index for the two days.
(first day change) × (second day change) = [ X × (1 + 0.0500)] × (1 − 0.0500) = 0.9975 X
The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% .
9. Strategy Use a proportion.
Solution Find Jupiter’s orbital period.
T 2 R3
T 2 ∝ R3 , so J = J = 5.193. Thus, TJ = 5.193/2 TE = 11.8 yr .
TE2 RE3
10. Strategy The area of the circular garden is given by A = π r 2 . Let the original and final areas be A1 = π r12 and
A2 = π r22 , respectively.
Solution Calculate the percentage increase of the area of the garden plot.
π r 2 − π r12
r2 − r2
1.252 r12 − r12
∆A
1.252 − 1
× 100% = 2
× 100% = 2 1 × 100% =
× 100% =
× 100% = 56%
A
1
π r12
r12
r12
11. Strategy The area of the poster is given by A = w. Let the original and final areas be A1 = 1w1 and
A2 = 2 w2 , respectively.
Solution Calculate the percentage reduction of the area.
A2 = 2 w2 = (0.800 1 )(0.800w1 ) = 0.640 1w1 = 0.640 A1
A1 − A2
A − 0.640 A1
× 100% = 1
× 100% = 36.0%
A1
A1
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Chapter 1: Introduction
Physics
12. Strategy The volume of the rectangular room is given by V = wh. Let the original and final volumes be
V1 = 1w1h1 and V2 = 2 w2 h2 , respectively.
Solution Find the factor by which the volume of the room increased.
V2
w h
(1.50 1 )(2.00 w1 )(1.20h1 )
= 2 2 2 =
= 3.60
V1
1w1h1
1w1h1
13. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then add and give the answer
with the number of significant figures determined by the less precise of the two numbers.
Solution Perform the operation with the appropriate number of significant figures.
3.783 × 106 kg + 1.25 × 108 kg = 0.03783 × 108 kg + 1.25 × 108 kg = 1.29 × 108 kg
(b) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Perform the operation with the appropriate number of significant figures.
(3.783 × 106 m) ÷ (3.0 × 10−2 s) = 1.3 × 108 m s
14. (a) Strategy Move the decimal point eight places to the left and multiply by 108.
Solution Write the number in scientific notation.
290,000,000 people = 2.9 × 108 people
(b) Strategy Move the decimal point 15 places to the right and multiply by 10−15.
Solution Write the number in scientific notation.
0.000 000 000 000 003 8 m = 3.8 × 10−15 m
15. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then subtract and give the
answer with the number of significant figures determined by the less precise of the two numbers.
Solution Perform the calculation using an appropriate number of significant figures.
3.68 × 107 g − 4.759 × 105 g = 3.68 × 107 g − 0.04759 × 107 g = 3.63 × 107 g
(b) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Perform the calculation using an appropriate number of significant figures.
6.497 × 104 m 2
= 1.273 × 102 m
5.1037 × 102 m
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Physics
Chapter 1: Introduction
16. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then add and give the answer
with the number of significant figures determined by the less precise of the two numbers.
Solution Write your answer using the appropriate number of significant figures.
6.85 × 10−5 m + 2.7 × 10−7 m = 6.85 × 10−5 m + 0.027 × 10−5 m = 6.88 × 10−5 m
(b) Strategy Add and give the answer with the number of significant figures determined by the less precise of
the two numbers.
Solution Write your answer using the appropriate number of significant figures.
702.35 km + 1897.648 km = 2600.00 km
(c) Strategy Multiply and give the answer with the number of significant figures determined by the number with
the fewest significant figures.
Solution Write your answer using the appropriate number of significant figures.
5.0 m × 4.3 m = 22 m 2
(d) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Write your answer using the appropriate number of significant figures.
( 0.04 π ) cm =
0.01 cm
(e) Strategy Find the quotient and give the answer with the number of significant figures determined by the
number with the fewest significant figures.
Solution Write your answer using the appropriate number of significant figures.
( 0.040 π ) m =
0.013 m
17. Strategy Multiply and give the answer in scientific notation with the number of significant figures determined by
the number with the fewest significant figures.
Solution Solve the problem.
(3.2 m) × (4.0 × 10−3 m) × (1.3 × 10−8 m) = 1.7 × 10−10 m3
18. Strategy Follow the rules for identifying significant figures.
Solution
(a) All three digits are significant, so 7.68 g has 3 significant figures.
(b) The first zero is not significant, since it is used only to place the decimal point. The digits 4 and 2 are
significant, as is the final zero, so 0.420 kg has 3 significant figures.
(c) The first two zeros are not significant, since they are used only to place the decimal point. The digits 7 and 3
are significant, so 0.073 m has 2 significant figures.
(d) All three digits are significant, so 7.68 × 105 g has 3 significant figures.
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Chapter 1: Introduction
Physics
(e) The zero is significant, since it comes after the decimal point. The digits 4 and 2 are significant as well, so
4.20 × 103 kg has 3 significant figures.
(f) Both 7 and 3 are significant, so 7.3 × 10−2 m has 2 significant figures.
(g) Both 2 and 3 are significant. The two zeros are significant as well, since they come after the decimal point, so
2.300 × 104 s has 4 significant figures.
19. Strategy Divide and give the answer with the number of significant figures determined by the number with the
fewest significant figures.
Solution Solve the problem.
3.21 m
3.21 m
=
= 459 m s
7.00 ms 7.00 × 10−3 s
20. Strategy Convert each length to meters. Then, rewrite the numbers so that the power of 10 is the same for each.
Finally, add and give the answer with the number of significant figures determined by the less precise of the two
numbers.
Solution Solve the problem.
3.08 × 10−1 km + 2.00 × 103 cm = 3.08 × 102 m + 2.00 × 101 m = 3.08 × 102 m + 0.200 × 102 m = 3.28 × 102 m
21. Strategy There are approximately 39.37 inches per meter.
Solution Find the thickness of the cell membrane in inches.
7.0 × 10−9 m × 39.37 inches m = 2.8 × 10−7 inches
22. (a) Strategy There are approximately 3.785 liters per gallon and 128 ounces per gallon.
Solution Find the number of fluid ounces in the bottle.
128 fl oz
1 gal
1L
×
× 355 mL ×
= 12.0 fluid ounces
1 gal
3.785 L
103 mL
(b) Strategy From part (a), we have 355 mL = 12.0 fluid ounces.
Solution Find the number of milliliters in the drink.
355 mL
16.0 fl oz ×
= 473 mL
12.0 fl oz
23. Strategy There are approximately 3.281 feet per meter.
Solution Convert to meters and identify the order of magnitude.
(a) 1595.5 ft ×
(b) 6016 ft ×
1m
= 4.863 × 102 m ; the order of magnitude is 102 .
3.281 ft
1m
= 1.834 × 103 m ; the order of magnitude is 103 .
3.281 ft
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Chapter 1: Introduction
24. Strategy There are 3600 seconds in one hour and 1000 m in one kilometer.
Solution Convert 1.00 kilometers per hour to meters per second.
1.00 km
1h
1000 m
×
×
= 0.278 m s
1h
3600 s 1 km
25. (a) Strategy There are 60 seconds in one minute, 5280 feet in one mile, and 3.28 feet in one meter.
Solution Express 0.32 miles per minute in meters per second.
0.32 mi 1 min 5280 ft
1m
×
×
×
= 8.6 m s
1 min
60 s
1 mi
3.28 ft
(b) Strategy There are 60 minutes in one hour.
Solution Express 0.32 miles per minute in miles per hour.
0.32 mi 60 min
×
= 19 mi h
1 min
1h
26. Strategy There are 0.6214 miles in 1 kilometer.
Solution Find the length of the marathon race in miles.
0.6214 mi
= 26.22 mi
42.195 km ×
1 km
27. Strategy Calculate the change in the exchange rate and divide it by the original price to find the drop.
Solution Find the actual drop in the value of the dollar over the first year.
1.27 − 1.45 −0.18
=
= −0.12
1.45
1.45
The actual drop is 0.12 or 12% .
28. Strategy There are 1000 watts in one kilowatt and 100 centimeters in one meter.
Solution Convert 1.4 kW m 2 to W cm 2 .
2
1.4 kW 1000 W ⎛ 1 m ⎞
2
×
×⎜
⎟ = 0.14 W cm
1 kW ⎝ 100 cm ⎠
1 m2
29. Strategy There are 1000 grams in one kilogram and 100 centimeters in one meter.
Solution Find the density of mercury in units of g cm3 .
3
1.36 × 104 kg 1000 g ⎛ 1 m ⎞
3
×
×⎜
⎟ = 13.6 g cm
1 kg ⎝ 100 cm ⎠
1 m3
30. Strategy The distance traveled d is equal to the rate of travel r times the time of travel t. There are 1000
milliseconds in one second.
Solution Find the distance the molecule would move.
459 m
1s
d = rt =
× 7.00 ms ×
= 3.21 m
1s
1000 ms
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Chapter 1: Introduction
Physics
31. Strategy There are 1000 meters in a kilometer and 1,000,000 millimeters in a kilometer.
Solution Find the product and express the answer in km3 with the appropriate number of significant figures.
1 km
1 km
(3.2 km) × (4.0 m) × (13 × 10−3 mm) ×
×
= 1.7 × 10−10 km3
1000 m 1,000,000 mm
32. (a) Strategy There are 12 inches in one foot and 2.54 centimeters in one inch.
Solution Find the number of square centimeters in one square foot.
2
2
⎛ 12 in ⎞ ⎛ 2.54 cm ⎞
2
1 ft 2 × ⎜
⎟ ×⎜
⎟ = 929 cm
1
ft
1
in
⎝
⎠ ⎝
⎠
(b) Strategy There are 100 centimeters in one meter.
Solution Find the number of square centimeters in one square meter.
2
⎛ 100 cm ⎞
4
2
1 m2 × ⎜
⎟ = 1× 10 cm
⎝ 1m ⎠
(c) Strategy Divide one square meter by one square foot. Estimate the quotient.
Solution Find the approximate number of square feet in one square meter.
1 m 2 10, 000 cm 2
=
≈ 11
1 ft 2
929 cm 2
33. (a) Strategy There are 12 inches in one foot, 2.54 centimeters in one inch, and 60 seconds in one minute.
Solution Express the snail’s speed in feet per second.
5.0 cm 1min
1 in
1 ft
×
×
×
= 2.7 × 10−3 ft s
1 min 60 s 2.54 cm 12 in
(b) Strategy There are 5280 feet in one mile, 12 inches in one foot, 2.54 centimeters in one inch, and 60 minutes
in one hour.
Solution Express the snail’s speed in miles per hour.
5.0 cm 60 min
1 in
1 ft
1 mi
×
×
×
×
= 1.9 × 10−3 mi h
1 min
1h
2.54 cm 12 in 5280 ft
34. Strategy A micrometer is 10−6 m and a millimeter is 10−3 m; therefore, a micrometer is 10−6 10−3 = 10−3 mm.
Solution Find the area in square millimeters.
2
⎛ 10−3 mm ⎞
150 µm 2 × ⎜
= 1.5 × 10−4 mm 2
⎜ 1 µm ⎟⎟
⎝
⎠
35. Strategy Replace each quantity in U = mgh with its SI base units.
Solution Find the combination of SI base units that are equivalent to joules.
U = mgh ⇒ J = kg × m s 2 × m = kg ⋅ m 2 ⋅ s −2
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Chapter 1: Introduction
36. (a) Strategy Replace each quantity in ma and kx with its dimensions.
Solution Show that the dimensions of ma and kx are equivalent.
[L]
[M]
[L]
ma has dimensions [M] ×
× [L] = [M] ×
.
and kx has dimensions
2
2
[T]
[T]
[T]2
Since [M][L][T]−2 = [M][L][T]−2 , the dimensions are equivalent.
(b) Strategy Use the results of part (a).
Solution Since F = ma and F = − kx, the dimensions of the force unit are [M][L][T]−2 .
37. Strategy Replace each quantity in T 2 = 4π 2 r 3 (GM ) with its dimensions.
Solution Show that the equation is dimensionally correct.
4π 2 r 3
[L]3
[L]3 [M][T]2
has dimensions
=
×
= [T]2 .
T 2 has dimensions [T]2 and
3
3
[L]
[M]
GM
[L]
× [M]
[M][T]2
Since [T]2 = [T]2 , the equation is dimensionally correct.
38. Strategy Determine the SI unit of momentum using a process of elimination.
Solution Find the SI unit of momentum.
p2
kg ⋅ m 2
kg 2 ⋅ m 2
. Since the SI unit for m is kg, the SI unit for p 2 is
. Taking the square
has units of
K=
2
2m
s
s2
root, we find that the SI unit for momentum is kg ⋅ m ⋅ s −1 .
39. (a) Strategy Replace each quantity (except for V) in FB = ρ gV with its dimensions.
Solution Find the dimensions of V.
F
[MLT −2 ]
V = B has dimensions
= [L3 ] .
−3
−2
ρg
[ML ] × [LT ]
(b) Strategy and Solution Since velocity has dimensions [LT −1 ] and volume has dimensions [L3 ], the correct
interpretation of V is that is represents volume .
40. Strategy Replace v, r, ω , and m with their dimensions. Then use dimensional analysis to determine how v
depends upon some or all of the other quantities.
[L]
1
, [L],
, and [M], respectively. No combination of r, ω , and m
[T]
[T]
1 [L]
=
and there is no dimensionless
gives dimensions without [M], so v does not depend upon m. Since [L] ×
[T] [T]
Solution v, r , ω , and m have dimensions
constant involved in the relation, v is equal to the product of ω and r , or v = ω r .
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Physics
41. Strategy Approximate the distance from your eyes to a book held at your normal reading distance.
Solution The normal reading distance is about 30-40 cm, so the approximate distance from your eyes to a book
you are reading is 30-40 cm.
42. Strategy Estimate the length, width, and height of your textbook. Then use V = wh to estimate its volume.
Solution Find the approximate volume of your physics textbook in cm3 .
The length, width, and height of your physics textbook are approximately 30 cm, 20 cm, and 4.0 cm, respectively.
V = wh = (30 cm)(20 cm)(4.0 cm) = 2400 cm3
43. (a) Strategy and Solution The mass of the lower leg is about 5 kg and that of the upper leg is about 7 kg, so an
order of magnitude estimate of the mass of a person’s leg is 10 kg.
(b) Strategy and Solution The length of a full size school bus is greater than 1 m and less than 100 m, so an
order of magnitude estimate of the length of a full size school bus is 10 m.
44. Strategy and Solution A normal heart rate is about 70 beats per minute and a person lives for about 70 years, so
70 beats
70 y
5.26 × 105 min
×
×
= 2.6 × 109 times per lifetime, or about 3 × 109 .
the heart beats about
1 min
lifetime
1y
45. Strategy (Answers will vary.) In this case, we use San Francisco, CA for the city. The population of San
Francisco is approximately 750,000. Assume that there is one automobile for every two residents of San
Francisco, that an average automobile needs three repairs or services per year, and that the average shop can
service 10 automobiles per day.
Solution Estimate the number of automobile repair shops in San Francisco.
3 repairs 1 y
0.01 repairs
.
×
≈
If an automobile needs three repairs or services per year, then it needs
auto ⋅ y 365 d
auto ⋅ d
1 auto
× 750, 000 residents ≈ 4 × 105 autos.
2 residents
If a shop requires one day to service 10 autos, then the number of shops-days per repair is
1d
0.1 shop ⋅ d
=
1 shop ×
.
10 repairs
repair
If there is one auto for every two residents, then there are
0.01 repairs 0.1 shop ⋅ d
×
= 400 shops .
auto ⋅ d
repair
Checking the phone directory, we find that there are approximately 463 automobile repair and service shops in
400 − 463
× 100% = −16% . The estimate was 16% too low, but in the ball
San Francisco. The estimate is off by
400
park!
The estimated number of auto shops is 4 × 105 autos ×
46. Strategy Estimate the appropriate orders of magnitude.
Solution Find the order of magnitude of the number of seconds in one year.
seconds/minute ~ 102
minutes/hour ~ 102
hours/day ~ 101
102 ⋅102 ⋅101 ⋅102 = 107 s
174
days/year ~ 102
Physics
Chapter 1: Introduction
47. Strategy One story is about 3 m high.
Solution Find the order of magnitude of the height in meters of a 40-story building.
(3 m)(40) ~ 100 m
48. Strategy To determine if c and A0 are correct, graph A versus B3 .
49. Strategy The plot of temperature versus
elapsed time is shown. Use the graph to
answer the questions.
A on the vertical axis and B3 on the horizontal axis .
103.00
Temperature (°F)
Solution To graph A versus B3 , graph
102.00
101.00
100.00
10 A.M.
11 A.M. 12 P.M.
Time
1 P.M.
Solution
(a) By inspection of the graph, it appears that the temperature at noon was 101.8°F.
(b) Estimate the slope of the line.
102.6°F − 100.0°F
2.6°F
=
= 0.9 °F h
m=
1:00 P.M. − 10:00 A.M.
3h
(c) In twelve hours, the temperature would, according to the trend, be approximately
T = (0.9 °F h)(12 h) + 102.5°F = 113°F.
The patient would be dead before the temperature reached this level. So, the answer is no.
50. Strategy Use the slope-intercept form, y = mx + b.
Solution Since x is on the vertical axis, it corresponds to y. Since t 4 is on the horizontal axis, it corresponds to x
(in y = mx + b) . So, the equation for x as a function of t is x = (25 m s 4 )t 4 + 3 m .
51. Strategy Use the two temperatures and their corresponding times to find the rate of temperature change with
respect to time (the slope of the graph of temperature vs. time). Then, write the linear equation for the temperature
with respect to time and find the temperature at 3:35 P.M.
Solution Find the rate of temperature change.
∆T 101.0°F − 97.0°F
=
= 1.0 °F h
m=
∆t
4.0 h
Use the slope-intercept form of a graph of temperature vs. time to find the temperature at 3:35 P.M.
T = mt + T0 = (1.0 °F h)(3.5 h) + 101.0°F = 104.5°F
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Chapter 1: Introduction
Strategy Plot the weights and ages on a
weight versus age graph.
Solution See the graph.
20.0
Weight (lb)
52. (a)
Physics
15.0
10.0
5.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0
Age in months
(b) Strategy Find the slope of the best-fit line between age 0.0 and age 5.0 months.
Solution Find the slope.
13.6 lb − 6.6 lb
7.0 lb
=
= 1.4 lb mo
m=
5.0 mo − 0.0 mo 5.0 mo
(c) Strategy Find the slope of the best-fit line between age 5.0 and age 10.0 months.
Solution Find the slope.
17.5 lb − 13.6 lb
3.9 lb
=
= 0.78 lb mo
m=
10.0 mo − 5.0 mo 5.0 mo
(d) Strategy Write a linear equation for the weight of the baby as a function of time. The slope is that found in
part (b), 1.4 lb mo. The intercept is the weight of the baby at five months of age.
Solution Find the projected weight of the child at age 12.
W = (1.4 lb mo)(144 mo − 5 mo) + 13.6 lb = 210 lb
53. Strategy Put the equation that describes the line in slope-intercept form, y = mx + b.
at = v − v0
v = at + v0
Solution
(a) v is the dependent variable and t is the independent variable, so a is the slope of the line.
(b) The slope-intercept form is y = mx + b. Find the vertical-axis intercept.
v ↔ y, t ↔ x, a ↔ m, so v0 ↔ b.
Thus, +v0
is the vertical-axis intercept of the line.
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54. (a) Strategy The equation of the speed versus time is given by v = at + v0 , where a = 6.0 m s 2 and
v0 = 3.0 m s.
Solution Find the change in speed.
v2 = at2 + v0
− (v1 = at1 + v0 )
v2 − v1 = a (t2 − t1 )
v2 − v1 = (6.0 m s 2 )(6.0 s − 4.0 s) = 12 m s
(b) Strategy Use the equation found in part (a).
Solution Find the speed when the elapsed time is equal to 5.0 seconds.
v = (6.0 m s 2 )(5.0 s) + 3.0 m s = 33 m s
55. (a) Strategy Plot the decay rate on the vertical axis and the time on the horizontal axis.
Solution The plot is shown.
Decay Rate (decays/s)
450
300
150
0
0 10 20 30 40 50 60 70 80 90 100
Time (min)
(b) Strategy Plot the natural logarithm of the decay rate on the vertical axis and the time on the horizontal axis.
Natural Logarithm of the
Decay Rate
Solution The plot is shown.
Presentation of the data in this form—as the
natural logarithm of the decay rate—might be
useful because the graph is linear.
6.0
4.0
2.0
0
0 10 20 30 40 50 60 70 80 90 100
Time (min)
56. (a) Strategy Refer to the figure. Use the definition of the slope of a line and the fact that the vertical axis
intercept is the x-value corresponding to t = 0.
Solution Compute the slope.
∆x 17.0 km − 3.0 km
=
= 1.6 km h .
∆t
9.0 h − 0.0 h
When t = 0, x = 3.0 km; therefore, the vertical axis intercept is 3.0 km.
(b) Strategy and Solution The physical significance of the slope of the graph is that it represents the speed of
the object. The physical significance of the vertical axis intercept is that it represents the starting position of
the object (position at time zero).
177
Chapter 1: Introduction
Physics
57. Strategy For parts (a) through (d), perform the calculations.
Solution
(a) 186.300 + 0.0030 = 186.303
(b) 186.300 − 0.0030 = 186.297
(c) 186.300 × 0.0030 = 0.56
(d) 186.300 0.0030 = 62, 000
(e) Strategy For cases (a) and (b), the percent error is given by
0.0030
× 100%.
Actual Value
Solution Find the percent error.
0.0030
× 100% = 0.0016%
Case (a):
186.303
0.0030
× 100% = 0.0016%
Case (b):
186.297
For case (c), ignoring 0.0030 causes you to multiply by zero and get a zero result. For case (d), ignoring
0.0030 causes you to divide by zero.
(f) Strategy Make a rule about neglecting small values using the results obtained above.
Solution
You can neglect small values when they are added to or subtracted from sufficiently large values.
The term “sufficiently large” is determined by the number of significant figures required.
58. Strategy The weight is proportional to the mass and inversely proportional to the square of the radius, so
W ∝ m r 2 . Thus, for Earth and Jupiter, we have WE ∝ mE rE2 and WJ ∝ mJ rJ2 .
Solution Form a proportion.
WJ
WE
=
mJ rJ2
mE rE2
=
mJ ⎛ rE
⎜
mE ⎜⎝ rJ
2
⎞
320mE
⎟⎟ =
mE
⎠
On Jupiter, the apple would weigh
⎛ rE
⎜⎜
⎝ 11rE
320 (1.0
121
2
⎞
320
⎟⎟ =
121
⎠
N) = 2.6 N .
59. Strategy Assuming that the cross section of the artery is a circle, we use the area of a circle, A = π r 2.
Solution
A1 = π r12 and A2 = π r22 = π (2.0r1 )2 = 4.0π r12 .
Form a proportion.
A2 4.0π r12
=
= 4.0
A1
π r12
The cross-sectional area of the artery increases by a factor of 4.0.
178
Physics
Chapter 1: Introduction
60. (a) Strategy The diameter of the xylem vessel is one six-hundredth of the magnified image.
Solution Find the diameter of the vessel.
d magnified 3.0 cm
d actual =
=
= 5.0 × 10−3 cm
600
600
(b) Strategy The area of the cross section is given by A = π r 2 = π (d 2)2 = (1 4)π d 2 .
Solution Find by what factor the cross-sectional area has been increased in the micrograph.
Amagnified 14 π d magnified 2 ⎛ 3.0 cm ⎞2
=
=⎜
= 360,000 .
2
−3 cm ⎟
1πd
Aactual
×
5.0
10
⎝
⎠
actual
4
61. Strategy If s is the speed of the molecule, then s ∝ T where T is the temperature.
Solution Form a proportion.
Tcold
scold
=
swarm
Twarm
Find scold .
scold = swarm
Tcold
250.0 K
= (475 m s)
= 434 m s
Twarm
300.0 K
62. Strategy Use dimensional analysis to convert from furlongs per fortnight to the required units.
Solution
(a) Convert to µm s.
1 furlong
220 yd 1 fortnight
1 day
3 ft
1 m 1, 000, 000 µm
×
×
×
×
×
×
= 166 µm s
1 fortnight 1 furlong 14 days
86,400 s 1 yd 3.28 ft
1m
(b) Convert to km day.
1 furlong
220 yd 1 fortnight 3 ft
1m
1 km
×
×
×
×
×
= 0.0144 km day
1 fortnight 1 furlong 14 days 1 yd 3.28 ft 1000 m
63. Strategy Use the rules for determining significant figures and for writing numbers in scientific notation.
Solution
(a) 0.00574 kg has three significant figures, 5, 7, and 4. The zeros are not significant, since they are used only to
place the decimal point. To write this measurement in scientific notation, we move the decimal point three
places to the right and multiply by 10−3.
(b) 2 m has one significant figure, 2. This measurement is already written in scientific notation
(c) 0.450 × 10−2 m has three significant figures, 4, 5, and the 0 to the right of 5. The zero is significant, since it
comes after the decimal point and is not used to place the decimal point. To write this measurement in
scientific notation, we move the decimal point one place to the right and multiply by 10−1.
179
Chapter 1: Introduction
Physics
(d) 45.0 kg has three significant figures, 4, 5, and 0. The zero is significant, since it comes after the decimal point
and is not used to place the decimal point. To write this measurement in scientific notation, we move the
decimal point one place to the left and multiply by 101.
(e) 10.09 × 104 s has four significant figures, 1, 9, and the two zeros. The zeros are significant, since they are
between two significant figures. To write this measurement in scientific notation, we move the decimal point
one place to the left and multiply by 101.
(f) 0.09500 × 105 mL has four significant figures, 9, 5, and the two zeros to the right of 5. The zeros are
significant, since they come after the decimal point and are not used to place the decimal point. To write this
measurement in scientific notation, we move the decimal point two places to the right and multiply by 10−2.
The results of parts (a) through (f) are shown in the table below.
Measurement
Significant Figures
Scientific Notation
(a)
0.00574 kg
3
5.74 × 10−3 kg
(b)
2m
1
2m
(c)
0.450 × 10−2 m
3
4.50 × 10−3 m
(d)
45.0 kg
3
4.50 × 101 kg
(e)
10.09 × 104 s
4
1.009 × 105 s
(f)
0.09500 × 105 mL
4
9.500 × 103 mL
64. Strategy Use the conversion factors from the inside cover of the book.
Solution
(a)
12.5 US gal 3.785 L 103 mL 0.06102 in 3
×
×
×
= 2890 in 3
1
US gal
L
mL
(b)
2887 in 3 ⎛ 1 cubit ⎞
×⎜
⎟ = 0.495 cubic cubits
1
⎝ 18 in ⎠
3
65. Strategy Use the metric prefixes n (10−9 ), µ (10−6 ), m (10−3 ), or M (106 ).
Solution
(a) M (or mega) is equal to 106 , so 6 × 106 m = 6 Mm .
(b) There are approximately 3.28 feet in one meter, so 6 ft ×
(c) µ (or micro) is equal to 10−6 , so 10−6 m = 1 µm .
180
1m
= 2m .
3.28 ft
Physics
Chapter 1: Introduction
(d) n (or nano) is equal to 10−9 , so 3 × 10−9 m = 3 nm .
(e) n (or nano) is equal to 10−9 , so 3 × 10−10 m = 0.3 nm .
66. Strategy The volume of the spherical virus is given by Vvirus = (4 3)π rvirus3 . The volume of viral particles is one
billionth the volume of the saliva.
Solution Calculate the number of viruses that have landed on you.
10−9 Vsaliva
0.010 cm3
number of viral particles =
=
= 104 viruses
3 10−7 cm 3
Vvirus
9
85
nm
10 43 π
2
1 nm
( )(
)
(
)
67. Strategy The circumference of a viroid is approximately 300 times 0.35 nm. The diameter is given by C = π d ,
or d = C π .
Solution Find the diameter of the viroid in the required units.
(a) d =
300(0.35 nm) 10−9 m
×
= 3.3 × 10−8 m
1 nm
π
(b) d =
300(0.35 nm) 10−3 µm
×
= 3.3 × 10−2 µm
1 nm
π
(c) d =
300(0.35 nm) 10−7 cm
1 in
×
×
= 1.3 × 10−6 in
1 nm
2.54 cm
π
68. (a) Strategy There are 3.28 feet in one meter.
Solution Find the length in meters of the largest recorded blue whale.
1m
1.10 × 102 ft ×
= 33.5 m
3.28 ft
(b) Strategy Divide the length of the largest recorded blue whale by the length of a double-decker London bus.
Solution Find the length of the blue whale in double-decker-bus lengths.
1.10 × 102 ft
1m
×
= 4.2 bus lengths
m
3.28 ft
8.0
bus length
69. Strategy The volume of the blue whale can be found by dividing the mass of the whale by its average density.
Solution Find the volume of the blue whale in cubic meters.
V=
m
ρ
=
1.9 × 105 kg
3
1000 g ⎛ 1 m ⎞
2 3
×⎜
⎟ = 2.2 × 10 m
3
1
kg
100
cm
⎝
⎠
0.85 g cm
×
181
Chapter 1: Introduction
Physics
70. Strategy The shape of a sheet of paper (when not deformed) is a rectangular prism. The volume of a rectangular
prism is equal to the product of its length, width, and height (or thickness).
Solution Find the volume of a sheet of paper in cubic meters.
1m
0.0254 m
1m
×
×
= 6.0 × 10−6 m3
27.95 cm × 8.5 in × 0.10 mm ×
100 cm
1 in
1000 mm
71. (a) Strategy a has dimensions
[L]
[T]2
[L]
; v has dimensions [T] ; r has dimension [L].
2
[L]2 1
[L]
Solution If we square v and divide by r, we have vr , which implies that
⋅ =
, which are the
[T]2 [L] [T]2
dimensions for a. Therefore, we can write a = K
v2
r
, where K is a dimensionless constant.
(b) Strategy Divide the new acceleration by the old, and use the fact that the new speed is 1.100 times the old.
Solution Find the percent increase in the radial acceleration.
v2
2
2
2
⎛ 1.100v1 ⎞
a2 K r ⎛ v2 ⎞
=
=
=
= 1.1002 = 1.210
⎜
⎟
⎜
⎟
a1 K v12 ⎝ v1 ⎠
⎝ v1 ⎠
r
1.210 − 1 = 0.210, so the radial acceleration increases by 21.0%.
72. Strategy Replace each quantity in v = K λ p g q by its units. Then, use the relationships between p and q to
determine their values.
Solution Find the values of p and q.
m
mq m p + q
.
= mp ⋅
=
In units,
s
s 2q
s2q
So, we have the following restrictions on p and q: p + q = 1 and 2q = 1.
Solve for q and p.
p + q =1
2q = 1
1
p + =1
1
q=
2
2
1
p=
2
Thus, v = K λ1 2 g1 2 = K λ g .
73. Strategy There are 2.54 cm in one inch and 3600 seconds in one hour.
Solution Find the conversion factor for changing meters per second to miles per hour.
1 m 100 cm
1 in
1 ft
1 mi
3600 s
×
×
×
×
×
= 2.24 mi h = 1 m s
1s
1m
2.54 cm 12 in 5280 ft
1h
So, for a quick, approximate conversion, multiply by 2.
182
Physics
Chapter 1: Introduction
74. Strategy The order of magnitude of the volume of water required to fill a bathtub is 101 ft 3 . The order of
magnitude of the number of cups in a cubic foot is 102.
Solution Find the order of magnitude of the number of cups of water required to fill a bathtub.
101 ft 3 × 102 cups ft 3 = 103 cups
75. Strategy Since there are hundreds of millions of people in the U.S., a reasonable order-of-magnitude estimate of
the number of automobiles is 108. There are about 365 days per year; that is, about 102. A reasonable estimate of
the gallons used per day per person is greater than one, but less than one hundred; that is, 101.
Solution Calculate the estimate.
108 automobiles × 102 days × 101
gallons
= 1011 gallons
automobile ⋅ day
76. (a) Strategy There are 10,000 (104 ) half dollars in $5000. The mass of a half-dollar coin is about 10 grams, or
10−2 kilograms.
Solution Estimate the mass of the coins.
104 coins × 10−2 kg coin = 102 kg, or 100 kg .
(b) Strategy There are $1, 000, 000 $20 = 50, 000 twenty-dollar bills in $1,000,000. The mass of a twenty-dollar
bill is about 1 gram, or 10−3 kilograms.
Solution Estimate the mass of the bills.
50, 000 bills × 10−3 kg bill = 50 kg .
77. Strategy The SI base unit for mass is kg. Replace each quantity in W = mg with its SI base units.
Solution Find the SI unit for weight.
m
kg ⋅ m
kg ⋅ =
2
s
s2
78. Strategy It is given that T 2 ∝ r 3 . Divide the period of Mars by that of Venus.
Solution Compare the period of Mars to that of Venus.
2
TMars
3
⎛ r
⎞ 2
⎛ 2r
⎞
r3
2
, or TMars = ⎜⎜ Venus ⎟⎟
= Mars , so TMars
= ⎜⎜ Mars ⎟⎟ TVenus
2
3
TVenus
rVenus
⎝ rVenus ⎠
⎝ rVenus ⎠
32
TVenus = 23 2 TVenus ≈ 2.8TVenus .
79. Strategy $59,000,000,000 has a precision of 1 billion dollars; $100 has a precision of 100 dollars, so the net
worth is the same to one significant figure.
Solution Find the net worth.
$59, 000, 000, 000 − $100 = $59, 000, 000, 000
183
Chapter 1: Introduction
Physics
80. Strategy There are about 103 hairs in a one-square-inch area of the average human head. An order-of-magnitude
estimate of the area of the average human head is 102 square inches.
Solution Calculate the estimate.
103 hairs in 2 × 102 in 2 = 105 hairs
81. (a) Strategy There are 7.0 leagues in one pace and 4.8 kilometers in one league.
Solution Find your speed in kilometers per hour.
120 paces 7.0 leagues 4.8 km 60 min
×
×
×
= 2.4 × 105 km h
1 min
1 pace
1 league
1h
(b) Strategy The circumference of the earth is approximately 40,000 km. The time it takes to march around the
Earth is found by dividing the distance by the speed.
Solution Find the time of travel.
1h
60 min
40,000 km ×
×
= 10 min
1h
2.4 × 105 km
82. Strategy Use the fact that RB = 1.42 RA .
Solution Calculate the ratio of PB to PA .
V2
PB RB RA
RA
1
= 2 =
=
=
= 0.704
PA V
RB 1.42 RA 1.42
R
A
83. (a) Strategy Inspect the units of G, c, and h and use trial-and-error to find the correct combination of these
constants.
Solution Through a process of trial and error, we find that the only combination of G, c, and h that has the
hG
c5
dimensions of time is
.
(b) Strategy Substitute the values of the constants into the formula found in part (a).
Solution Find the time in seconds.
2
⎛
−34 kg⋅m ⎞ ⎛ 6.7 × 10−11 m3 ⎞
⎜ 6.6 × 10
⎟⎜
⎟
s
kg⋅s 2 ⎠
⎠⎝
== ⎝
= 1.3 × 10−43 s
5
8
c5
m
3.0 × 10 s
hG
(
)
84. Strategy The dimensions of L, g, and m are length, length per time squared, and mass, respectively. The period
has units of time, so T cannot depend upon m. (There are no other quantities with units of mass with which to
cancel the units of m.) Use a combination of L and g.
Solution The square root of L g has dimensions of time, so
T = C Lg , where C is a constant of proportionality .
184
Physics
Chapter 1: Introduction
85. Strategy The dimensions of k and m are mass per time squared and mass, respectively. Dividing either quantity
by the other will eliminate the mass dimension.
Solution The square root of k m has dimensions of inverse time, which is correct for frequency.
So, f = k m . Find k.
f1 =
k
k
, so f12 =
, or k = m1 f12 .
m1
m1
Find the frequency of the chair with the 75-kg astronaut.
f2 =
k
=
m2
m1 f12
m2
= f1
m1
62 kg + 10.0 kg
= (0.50 s −1 )
= 0.46 s −1
m2
75 kg + 10.0 kg
86. Strategy Solutions will vary. One example follows:
The radius of the Earth is about 106 m. The area of a sphere is 4π r 2 , or about 101 ⋅ r 2 . The average depth of the
oceans is about 4 × 103 m. The oceans cover more than two-thirds of the Earth’s surface, but in this rough
estimation, we assume that oceans cover the entire Earth.
Solution Calculate an order-of-magnitude estimate of the volume of water contained in Earth’s oceans.
The surface area of the Earth is about 101 ⋅ (106 m)2 = 1013 m 2 ; therefore, the volume of water in the oceans is
about area × depth = (1013 m 2 )(4 × 103 m) = 4 × 1016 m3 ∼ 1016 m3 .
Total Mass of
Yeast Cells (g)
87. (a) Strategy Plot the data on a graph with
mass on the vertical axis and time on the
horizontal axis. Then, draw a best-fit
smooth curve.
Solution See the graph.
100.0
90.0
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
0.0
0.0
5.0 10.0 15.0 20.0 25.0
Time (h)
(b) Strategy Answers will vary. Estimate the value of the total mass that the graph appears to be approaching
asymptotically.
Solution The graph appears to be approaching asymptotically a maximum value of 100 g, so the carrying
capacity is about 100 g .
(c) Strategy Plot the data on a graph with the natural
logarithm of m m0 on the vertical axis and time on
the horizontal axis. Draw a line through the points and
find its slope to estimate the intrinsic growth rate.
m
ln m
0
2.0
1.0
Solution See the graph. From the plot of ln
m
m0
vs. t,
the slope r appears to be
1.8 − 0.0
1.8
r=
=
= 0.30 s −1 .
6.0 s − 0.0 s 6.0 s
0.0
0.0
185
2.0
4.0
6.0 t (h)
Chapter 2
MOTION ALONG A LINE
Conceptual Questions
1. Distance traveled is a scalar quantity equal to the total length of the path taken in moving from one point to
another. Displacement is a vector quantity directed from the initial point towards the final point with a magnitude
equal to the straight line distance between the two points. The magnitude of the displacement is always less than
or equal to the total distance traveled.
2. The velocity of an object is a vector quantity equal to the displacement of the object per unit time interval. The
speed of an object is a scalar quantity equal to the distance traveled by the object per unit time interval.
3. The area under the curve of a vx versus time graph is equal to the x-component of the displacement.
4. The slope of a line tangent to a curve on a vx versus time graph is equal to the x-component of the acceleration at
the time corresponding to the point where the tangent line intersects the curve.
5. The area under the curve of an a x versus time graph is equal to the change in the x-component of the velocity.
6. The slope of a line tangent to a curve on a graph plotting the x-component of position versus time is equal to the
magnitude of the x-component of the instantaneous velocity at the time corresponding to the point where the
tangent line intersects the curve.
7. The average velocity of an object is defined as the ratio of the displacement of the object during an interval of
time to the length of the time interval. The instantaneous velocity of an object is obtained from the average
velocity by using a time interval that approaches zero. An object can have different average velocities for different
time intervals. However, the average velocity for a given time interval has a unique value.
8. Yes, the instantaneous velocity of an object can be zero while the acceleration is nonzero. When you toss a ball
straight up in the air, its acceleration is directed downward, with a magnitude of g, the entire time it’s in the air. Its
velocity is zero at the highest point of its path, however.
9. (a) a x > 0 and vx < 0 means you are moving south and slowing down.
(b) a x = 0 and vx < 0 means you are moving south at a constant speed.
(c) a x < 0 and vx = 0 means you are momentarily at rest but speeding up in a southward direction.
(d) a x < 0 and vx < 0 means you are moving south and speeding up.
(e) As can be seen from our answers above, it is not a good idea to use the term “negative acceleration” to mean
slowing down. In parts (c) and (d), the acceleration is negative, but the bicycle is speeding up. Also, in part
(a), the acceleration is positive, but the bicycle is slowing down.
10. At the highest point of the coins motion, it is momentarily at rest, so its velocity is zero. Throughout the coins
motion, its acceleration is due only to gravity (ignoring air resistance).
186
Physics
Chapter 2: Motion Along a Line
Problems
1. Strategy Let east be the +x-direction.
Solution Draw a vector diagram; then compute the sum of the three displacements.
The vector diagram:
32 cm
48 cm
64 cm
N
16 cm
The sum of the three displacements is (32 cm + 48 cm − 64 cm) east = 16 cm east .
2. Strategy Let the positive direction be to the right. Make a vector diagram with the location of the stone as the
starting point.
Solution The vector diagram:
Stone
Add the displacements.
G
∆r = 4.0 m right + 1.0 m left + 6.5 m right + 8.3 m left
= 4.0 m right − 1.0 m right + 6.5 m right − 8.3 m right
= 1.2 m right
The squirrel’s total displacement from his starting point is 1.2 m to the right of the starting point.
3. Strategy Let east be the +x-direction.
Solution Compute the displacements; then find the total distance traveled.
(a) The runner’s displacement from his starting point is
G G G
∆r = rf − ri = 20 m west − 60 m east = 20 m west + 60 m west = 80 m west or − 80 m .
(b) Since the runner is located 20 m west of the milestone, his displacement from the milestone is
20 m west or − 20 m .
(c) The runner’s displacement from his starting point is
G G G
∆r = rf − ri = 140 m east − 60 m east = 80 m east or + 80 m .
(d) The runner first jogs 60 m + 20 m = 80 m; then he jogs 20 m + 140 m = 160 m. The total distance traveled is
80 m + 160 m = 240 m .
4. Strategy Darren’s apartment is due west of Johannes’s dorm, so Johannes’s dorm is due east of Darren’s
apartment. Since the pizza shop is due east of Johannes’s dorm and Johannes’s dorm is due east of Darren’s
apartment, Darren must travel due east.
Solution The distance between Darren’s apartment and the pizza shop is equal to the sum of the distances.
1.50 mi + 3.00 mi = 4.50 mi
Darren must travel 4.50 mi east.
187
Chapter 2: Motion Along a Line
Physics
5. Strategy Let south be the +x-direction.
Solution Draw vector diagrams for each situation; then find the displacements of the car.
12 km
x3
20 km
3 P.M.
(a)
S
South
x1
x2
96 km
4 P.M.
x3 − x1
x3
− x1
x3 − x1 = 12 km − 20 km = −8 km
The displacement of the car between 3 P.M. and 6 P.M. is 8 km north of its position at 3 P.M.
(b)
x1
S
x2
x1 + x2
x1 + x2 = 20 km + 96 km = 116 km
The displacement of the car from the starting point to the location at 4 P.M. is 116 km south of the starting
point.
(c)
S
x3
− (x1 + x2)
x3 − (x1 + x2)
x3 − ( x1 + x2 ) = 12 km − 116 km = −104 km
The displacement of the car between 4 P.M. and 6 P.M. is 104 km north of its position at 4 P.M.
6. Strategy Use the definition of average velocity.
Solution Find the average velocity of the train.
G
∆r 10 km east − 3 km east 7 km east 60 min
G
v av =
=
=
×
= 30 km h east
∆t
3:28 − 3:14
14 min
1h
7. Strategy Use the definition of average velocity.
Solution Find the average velocity of the cyclist in meters per second.
G
∆r 10.0 km east 10.0 × 103 m east
G
v av =
=
=
= 14.3 m s east
∆t 11 min 40 s
700 s
8. Strategy Use the definition of average speed.
Solution Find the time it took the ball to get to home plate.
∆r
∆r
18.4 m
vav =
, so ∆t =
=
= 0.408 s .
∆t
vav 45.1 m s
188
Physics
Chapter 2: Motion Along a Line
9. Strategy Jason never changes direction, so the direction of the average velocity is due west. Find the average
speed by dividing the total distance traveled by the total time.
Solution The distance traveled during each leg of the trip is given by ∆x = vav ∆t.
vav =
(35.0 mi h )(0.500 h) + (60.0 mi h )(2.00 h) + (25.0 mi h )(10.0 60.0 h)
= 53.1 mi h
0.500 h + 2.00 h + 10.0 60.0 h
So, the average velocity is 53.1 mi h due west .
10. Strategy When the Boxster catches the Scion, the displacement of the Boxster will be ∆r + 186 m and the
displacement for the Scion will be ∆r. ∆t for both cars will be the same.
Solution Find the time it takes for the Boxster to catch the Scion.
∆r
∆r + 186 m
or ∆r = vav,B ∆t − 186 m.
vav = car , so for the Boxster, vav,B =
∆t
∆t
For the Scion,
∆r
vav,S =
, so ∆r = vav,S ∆t.
∆t
Equate the two expressions for ∆r.
vav,S ∆t = vav,B ∆t − 186 m
(vav,S − vav,B )∆t = −186 m
∆t =
186 m
186 m
186 m
=
=
= 32 s
vav,B − vav,S 24.4 m s − 18.6 m s 5.8 m s
11. Strategy Use the area under the curve to find the displacement of the car.
Solution The displacement of the car is given by the area under the vx vs. t curve. Under the curve, there are
16 squares and each square represents (5 m/s)(2 s) = 10 m. Therefore, the car moves 16(10 m) = 160 m .
12. (a) Strategy Use the area between the v y vs. t curve and the x-axis to find the displacement of the elevator.
Solution From t = 0 s to t = 10 s, there are 8 squares. From t = 14 s to t = 20 s, there are 4 squares. Each
square represents (1 m/s)(2 s) = 2 m. The displacement from t = 14 s to t = 20 s is negative (v y < 0). So, the
total displacement is ∆y = 8(2 m) + (−4)(2 m) = 8 m, and the elevator is 8 m above its starting point.
(b) Strategy Use the slope of the curve to determine when the elevator reaches its highest point.
Solution The vertical velocity is positive for t = 0 s to t = 10 s. It is negative for t = 14 s to t = 20 s. It is zero
for t = 10 s to t = 14 s. So, the elevator reaches its highest location at t = 10 s and remains there until t = 14 s
before it goes down. Thus, the elevator is at its highest location from t = 10 s to t = 14 s .
13. Strategy Use the graph to answer the questions. The slope of the graph at any instant represents the speed at that
instant.
Solution
(a) The section of the graph with the largest magnitude (steepest) slope represents the highest speed, DE.
(b) The slope changes from positive to negative at D, and from negative to positive at E, so the object reverses its
direction of motion at times 4 s and 5 s.
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Chapter 2: Motion Along a Line
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(c) During the time interval t = 0 s to t = 2 s, the speed of the object is
20 m − 0
20 m − 20 m
= 10 m s , and from t = 2 s to t = 3 s it is v = vav =
= 0.
v = vav =
2s
3 s−2 s
Therefore, the distance traveled is (10 m s )(2 s) = 20 m .
14. Strategy Set up ratios of speeds to distances.
Solution Find the speed of the baseball v.
v
65.0 mph
65.0 mph
, so v =
(60.5 ft) = 91.5 mph .
=
60.5 ft
43.0 ft
43.0 ft
15. Strategy Use vector subtraction to find the change in velocity.
Solution Find the change in velocity of the scooter.
G G
G
∆v = v f − vi = 15 m s west − 12 m s east = 15 m s west − (−12 m s west) = 27 m s west
16. Strategy Determine the maximum time allowed to complete the run.
Solution Massimo must take no more than 1000 m ÷ 4.0 m s = 250 s to complete the run. Since he ran the first
900 m is 250 s, he cannot pass the test because he would have to run the last 100 m in 0 s.
17. Strategy Use the area under the curve to find the displacement of the skateboard.
Solution The displacement of the skateboard is given by the area under the v vs. t curve. Under the curve for
t = 3.00 s to t = 8.00 s, there are 16.5 squares and each square represents (1.0 m/s)(1.0 s) = 1.0 m; so the board
moves 16.5 m.
18. Strategy Use the definition of average speed.
Solution Find the average speeds of the skater.
(a) vav, x =
∆x 6.0 m − 0
=
= 1.5 m s
∆t
4.0 s
(b) vav, x =
6.0 m − 0
= 1.2 m s
5.0 s
19. Strategy The slope of the x vs. t curve is equal to vx . Use the definition of average speed.
Solution Compute the average speed at t = 2.0 s.
6.0 m − 4.0 m
= 1.0 m s
vx =
3.0 s − 1.0 s
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Physics
Chapter 2: Motion Along a Line
20. Strategy The slope of each segment (between changes in the slope) of the graph is equal to the speed during that
time period.
Solution Find the speed during each time period.
Then, plot vx as a function of time.
vx (m/s)
5
4m
0 < t < 1 s: vx =
=4 m s
1s
6 m−4 m
1 < t < 3 s: vx =
=1 m s
3 s −1 s
6 m−6 m
3 < t < 5 s: vx =
=0 m s
5 s−3 s
1 m−6 m
5 < t < 6 s: vx =
= −5 m s
6 s−5 s
0 m −1 m
6 < t < 8 s: vx =
= −0.5 m s
8 s−6 s
0
5
10 t (s)
5
21. (a) Strategy Let the positive direction be to the right. Draw a diagram.
Solution Find the chipmunk’s total displacement.
80 cm right
30 cm left
90 cm right
310 cm left
170 cm left
80 cm − 30 cm + 90 cm − 310 cm = −170 cm
The total displacement is 170 cm to the left.
(b) Strategy The average speed is found by the dividing the total distance traveled by the elapsed time.
Solution Find the total distance traveled.
80 cm + 30 cm + 90 cm + 310 cm = 510 cm
Find the average speed.
510 cm
= 28 cm s
18 s
(c) Strategy The average velocity is found by dividing the displacement by the elapsed time.
Solution Find the average velocity.
G
∆r 170 cm to the left
G
v av =
=
= 9.4 cm s to the left
∆t
18 s
22. Strategy The average speed is found by dividing the total distance traveled by the elapsed time.
(a) Solution Find the average speed for the first 10-km segment of the race.
10.0 × 103 m 10.0 × 103 m
=
= 4.88 m s
v1 =
0.5689 h
2048 s
(b) Solution Find the average speed for the entire race.
42,195 m 42,195 m
vav =
=
= 4.90 m s
2.3939 h
8618 s
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Chapter 2: Motion Along a Line
Physics
23. Strategy Use the definition of average velocity. Find the time spent by each runner in completing her portion of
the race.
Solution The times for each runner are 300.0 m ÷ 7.30 m s = 41.1 s, 300.0 m ÷ 7.20 m s = 41.7 s, and
100.0 m ÷ 7.80 m s = 12.8 s. The net displacement of the baton is 100.0 m to the north, so the average velocity of
G
∆r
100.0 m north
G
=
= 1.05 m s to the north .
the baton for the entire race is v av =
∆t 41.1 s + 41.7 s + 12.8 s
24. Strategy Use ∆v = a∆t and solve for ∆t.
Solution
∆v
22 m s
∆t =
=
= 13 s
a 1.7 m s 2
25. Strategy Use the definition of average acceleration.
Solution
G
G G
G
∆v v f − vi
=
aav =
∆t
∆t
0 − 28 m s in the direction of the car’s travel
=
= 7.0 m s 2 in the direction opposite the car’s velocity
4.0 s
26. Strategy Use Newton’s second law of motion.
Solution Find the average acceleration of the airplane.
G
∆v 35 m s
∑ F = ma, so aav =
=
= 4.4 m s 2 ; thus aav = 4.4 m s 2 forward .
∆t
8.0 s
27. Strategy Use the definition of average acceleration.
Solution Find the change in velocity.
G G
G
∆v = v f − v i
= 3.0 m s toward the paddle − 4.0 m s away from the paddle
= 3.0 m s toward the paddle + (− 4.0 m s away from the paddle)
= 3.0 m s toward the paddle + 4.0 m s toward the paddle
= 7.0 m s toward the paddle
Calculate the average acceleration.
G
G
∆v 7.0 m s toward the paddle
=
= 28 m s 2 toward the paddle
aav =
0.25 s
∆t
192
Physics
Chapter 2: Motion Along a Line
28. Strategy Use the definitions of instantaneous acceleration, displacement, and average velocity.
Solution
(a) We draw a line tangent to the point at (14.0 s, 55.0 m s ) .
vx (m/s)
60.0
50.0
40.0
30.0
55.0 m/s
20.0
10.0
7.0 s
0
0
2.0
4.0
6.0
8.0 10.0 12.0 14.0 16.0 18.0 20.0 t (s)
It looks as if the tangent line passes through the point ( 7.0 s, 45.0 m s ) .
G
∆v x (55.0 m s − 45.0 m s) in the +x-direction
G
=
= 1.4 m s 2 in the +x-direction
ax =
14.0 s − 7.0 s
∆t
(b) The area under the vx vs. t curve from t = 12.0 s to t = 16.0 s represents the displacement of the body. Each
grid square represents (10.0 m/s)(1.0 s) = 1.0 × 101 m, and there are approximately 22 squares under the curve
for t = 12.0 s to t = 16.0 s, so the car travels 220 m in the +x-direction.
G
∆r 220 m in the +x-direction
G
(c) v av =
=
= 55 m s in the +x-direction
4.0 s
∆t
29. Strategy The magnitude of the acceleration is the absolute value of the slope of the graph at t = 7.0 s.
Solution
∆vx
0 − 20.0 m s
ax =
=
= 2.5 m s 2
12.0 s − 4.0 s
∆t
30. (a) Strategy The acceleration a x is equal to the slope of the vx versus t graph. Use the definition of average
acceleration.
Solution
∆v
14 m s − 4 m s
aav, x = x =
= 2 m s2
11 s − 6 s
∆t
(b) Strategy For constant acceleration, vav, x = (vf + vi ) 2.
Solution
14 m s + 4 m s
vav, x =
= 9.0 m s
2
193
Chapter 2: Motion Along a Line
Physics
(c) Strategy vav, x = ∆x ∆t and ∆x is the area under the graph in the figure. Find the area.
Solution Each square represents (1.0 m s)(1.0 s) = 1.0 m and there are 195 squares under the graph. So,
195(1.0 m)
vav, x =
= 9.8 m s .
20.0 s
(d) Strategy At t = 10 s, vx = 12 m/s, and at t = 15 s, vx = 14 m/s.
Solution
∆vx = 14 m s − 12 m s = 2 m s
(e) Strategy The area under the vx vs. t graph for t = 10 s to t = 15 s represents the displacement of the car.
Solution Each square represents 1.0 m. There are 69 of these squares, so the car has traveled (1.0 m)(69) =
69 m.
31. Strategy The acceleration a x is equal to the slope of the vx versus t graph. The displacement is equal to the area
under the curve.
Solution
(a) a x is the slope of the graph at t = 11 s.
ax =
∆vx 10.0 m s − 30.0 m s
=
= −10 m s 2
12.0 s − 10.0 s
∆t
(b) Since vx is constant, a x = 0 at t = 3 s.
(c) The area under the vx vs. t curve from t = 12 s to t = 14 s represents the displacement of the body. Each
square represents (10.0 m/s)(1.0 s) = 1.0 × 101 m, and there is 1 2 square under the curve for
t = 12 s to t = 14 s, so the car travels 5.0 m.
32. (a) Strategy Plot the data on a v versus t graph. Draw a best-fit line.
Solution
v (m/s)
8.0
6.0
4.0
2.0
0
0
1.00
2.00
3.00 t (s)
(b) Strategy The slope of the graph gives the acceleration.
Solution The points appear to lie upon a line, so yes, it is plausible that the acceleration is constant. Compute
the magnitude of the acceleration.
∆v 7.0 m s − 0
a=
=
= 2.43 m s 2
2.88 s − 0
∆t
The acceleration is 2.43 m s 2 in the direction of motion .
194
Physics
Chapter 2: Motion Along a Line
33. (a) Strategy The graph will be a line with a slope of 1.20 m s 2 .
vx (m/s)
Solution vx = 0 when t = 0. The graph is shown.
(b) Strategy Use Eq. (2-15).
Solution Find the distance the train traveled.
1
1
1
∆x = vix ∆t + a x (∆t ) 2 = (0)∆t + a x (∆t ) 2 = a x (∆t ) 2
2
2
2
1
2
2
= (1.20 m s )(12.0 s) = 86.4 m
2
16
14
12
10
8
6
4
2
0
0 2 4 6 8 10 12 t (s)
(c) Strategy Use Eq. (2-12).
Solution Find the final speed of the train.
vfx − vix = vfx − 0 = ax ∆t , so vfx = ax ∆t = (1.20 m s 2 )(12.0 s) = 14.4 m s .
(d) Strategy Refer to Figure 2.16, which shows a motion diagram.
Solution The motion diagram is shown.
t = 4.0 s
t = 2.0 s
t=0
t = 6.0 s t = 8.0 s
0.0
20.0
40.0
t = 10.0 s
60.0
t = 12.0 s
80.0
x (m)
34. Strategy Relate the acceleration, speed, and distance using Eq. (2-16). Let southwest be the positive direction.
Solution Find the constant acceleration required to stop the airplane. The acceleration must be opposite to the
direction of motion of the airplane, so the direction of the acceleration is − southwest = northeast.
v 2 − vix 2 0 − (55 m s) 2
vfx 2 − vix 2 = 2a x ∆x, so a x = fx
=
= −1.5 m s 2 .
2∆x
2(1.0 × 103 m)
Thus, the acceleration is 1.5 m s 2 northeast .
35. (a) Strategy Between 0 to 2 s the speed is 24.0 m s , and between 11 and 12 s the speed is 6.0 m s. Since the
acceleration is constant between 2 and 11 s, draw a straight line between the two horizontal lines of constant
speed.
Solution Draw the graph.
vx (m/s)
24
16
8
0
0
4
8
12 t (s)
195
Chapter 2: Motion Along a Line
Physics
(b) Strategy Let south be the +x-direction. Use Eq. (2-12) to find the acceleration of the train between 2 and 11
s. Before 2 s and after 11 s the acceleration is zero.
Solution
vfx − vix = a x ∆t , so ax = (vfx − vix ) ∆t = (6.00 m s − 24.0 m s) (9.00 s) = − 2.00 m s 2 .
The acceleration is 2.00 m s 2 north .
(c) Strategy Use Eq. (2-15).
Solution Find the distance the train traveled up the incline.
1
1
∆x = vix ∆t + a x (∆t ) 2 = (24.0 m s)(9.00 s) + (−2.00 m s 2 )(9.00 s) 2 = 135 m
2
2
36. (a) Strategy For motion in a straight line, the magnitude of a constant acceleration is equal to the change in
speed divided by the time elapsed.
Solution Find how long the airplane accelerated.
∆v
∆v 46.0 m s
a=
, so ∆t =
=
= 9.20 s .
∆t
a 5.00 m s 2
(b) Strategy Use Eq. (2-16).
Solution Find the distance the plane traveled along the runway.
v2
(46.0 m s)2
vf2x − vi2x = vf2x − 0 = 2a x ∆x, so ∆x = fx =
= 212 m .
2a x 2(5.00 m s 2 )
37. Strategy Use Eq. (2-12).
Solution
(a) Since the acceleration is constant, we have ∆vx = a x ∆t = (2.0 m s 2 )(12.0 s − 10.0 s) = 4.0 m s .
(b) The speed when the stopwatch reads 12.0 s is the sum of the speed at 10.0 s plus the change in speed, so
v(t = 10.0 s) + ∆vx = 1.0 m s + 4.0 m s = 5.0 m s .
38. Strategy Find the time it takes for the car to collide with the tractor (assuming it does) by setting the distance
the car travels equal to that of the tractor plus the distance between them and solving for t. Use Eq. (2-15). Then,
use the essential relationships for constant acceleration to answer the remaining questions.
Solution Solve for the time.
1
(27.0 m s)∆t + (−7.00 m s 2 )(∆t )2 = (10.0 m s)∆t + 25.0 m, so 0 = (3.50 m s 2 )(∆t ) 2 − (17.0 m s)∆t + 25.0 m
2
and solving for t we get an imaginary value for the time. Therefore, you won’t hit the tractor.
Find the distance the car requires to stop.
2
∆xc = (vfc
− vic2 ) (2a ) = [0 − (27.0 m s) 2 ] [2(−7.00 m s 2 )] = 52.1 m . Since the acceleration of the car is
constant, the average speed of the car as it attempts to stop is vc, av = (vfc + vic ) 2 = (0 + 27.0 m s) 2 = 13.5 m s.
Thus, the time required for the car to stop is ∆t = ∆xc vc, av = (52.1 m) (13.5 m s) = 3.86 s. The distance the
tractor travels in this time is ∆xt = vt ∆t = (10.0 m s)(3.86 s) = 38.6 m. Now, 38.6 m + 25.0 m = 63.6 m, which is
63.6 m − 52.1 m = 11.5 m beyond the stopping point of the car.
196
Physics
Chapter 2: Motion Along a Line
39. Strategy Use Eq. (2-16).
Solution Find the distance the train travels.
∆x =
vfx 2 − vix 2
2
= 236 m
2(−1.52 m s 2 )
236 m > 184 m, so the answer is no; it takes 236 m for the train to stop.
2a x
=
0 − ( 26.8 m s )
40. (a) Strategy Use Eq. (2-16) and Newton’s second law.
Solution Find the final speed of the electrons.
⎛F ⎞
vf2x − vi2x = vf2x − 0 = 2a x ∆x = 2 ⎜ x ⎟ ∆x, so
⎝m⎠
vfx = ±
2 Fx ∆x
m
2(6.4 × 10−17 N)(0.020 m)
=
9.109 × 10
−31
kg
= 1.7 × 106 m s . (Speed is always positive.)
(b) Strategy Use Eq. (2-15).
Solution Find the time it takes the electrons to travel the length of the tube.
x = 0 to 2.0 cm:
1
1
∆x1 = vix ∆t1 + a x (∆t1 ) 2 = 0 + a x (∆t1 ) 2 , so
2
2
2∆x1 2∆x1
2m∆x1
(∆t1 ) 2 =
=
or ∆t1 =
.
Fx
ax
Fx
m
x = 2.0 cm to 47 cm:
∆x2 = vfx ∆t2 , so ∆t2 =
∆x2
.
vfx
Find the total time.
∆t = ∆t1 + ∆t2 =
2m∆x1 ∆x2
+
=
Fx
vfx
2(9.109 × 10−31 kg)(0.020 m)
6.4 × 10−17
N
+
0.45 m
1.7 × 106 m s
= 290 ns
41. Strategy Refer to the figure. Analyze graphically and algebraically.
Solution Graphical analysis:
The displacement of the object is given by the area under the vx vs. t curve between t = 9.0 s and
t = 13.0 s. The area is a triangle, A = 12 bh.
1
(13.0 s − 9.0 s)(40 m s) = 80 m
2
Algebraic solution:
Use the definition of average velocity.
v +v
40 m s + 0
∆x = vav, x ∆t = ix fx ∆t =
(13.0 s − 9.0 s) = 80 m
2
2
∆x =
The object goes 80 m .
197
Chapter 2: Motion Along a Line
Physics
42. Strategy Refer to the figure. Analyze graphically and algebraically.
Solution Graphical analysis: Find the slope of the graph.
40 m s − 20 m s
aav, x =
= 5.0 m s 2
9.0 s − 5.0 s
v −v
40 m s − 20 m s
Algebraic solution: vfx − vix = aav, x ∆t , so aav, x = fx ix =
= 5.0 m s 2 .
∆t
9.0 s − 5.0 s
The average acceleration is 5.0 m s 2 in the +x-direction .
43. (a) Strategy The graph will be a line with a slope of −1.40 m s 2 .
Solution vx = 22 m s when t = 0.
vx (m/s)
22
18
14
10
0
2
6
4
8 t (s)
(b) Strategy Since the train slows down, the acceleration is negative. Use Eq. (2-12).
Solution
vfx − vix = a x ∆t , so vfx = vix + ax ∆t = 22 m s + (−1.4 m s 2 )(8.0 s) = 11 m s .
(c) Strategy Use Eq. (2-15) to find the distance the train traveled up the incline.
Solution
1
1
xf = xi + vix ∆t + a x (∆t ) 2 = 0 + (22 m s)(8.0 s) + (−1.4 m s 2 )(8.0 s) 2 = 130 m
2
2
(d) Strategy Refer to Figure 2.16, which shows a motion diagram.
Solution The motion diagram is shown.
t=0
0
t = 2.0 s
40
t = 4.0 s t = 6.0 s t = 8.0 s
80
120
x (m)
44. Strategy Use Eq. (2-15).
Solution Solve for ∆t using the quadratic formula.
1
1
∆y = viy ∆t + a y (∆t ) 2 , so a y (∆t ) 2 + viy ∆t − ∆y = 0.
2
2
2
−viy ± viy + 2a y ∆y −3.00 m s ± (3.00 m s)2 + 2(−9.80 m s 2 )(−78.4 m)
∆t =
=
= −3.71 s or 4.32 s
ay
−9.80 m s 2
Since ∆t > 0, the brick lands on the ground 4.32 s after it is thrown from the roof.
198
Physics
Chapter 2: Motion Along a Line
45. Strategy Use Eq. (2-16).
Solution Find the final speed of the penny.
vfy 2 − viy 2 = vfy 2 − 0 = −2 g ∆y, so vfy = −2 g ∆y = −2(9.80 m s 2 )(0 m − 369 m) = 85.0 m s.
G
Therefore, v = 85.0 m s down .
46. Strategy Ignoring air resistance, the golf ball is in free fall. Use Eq. (2-15).
Solution
(a) Find the time it takes the golf ball to fall 12.0 m.
1
2∆y
−2(0 − 12.0 m)
=
= 1.6 s .
viy = 0, so ∆y = − g (∆t )2 and ∆t = −
2
g
9.80 m s 2
(b) Find how far the golf ball would fall in 2
−2(0 − 12.0 m)
9.80 m s 2
= 3.13 s.
1
1
∆y = − g (∆t ) 2 = − (9.80 m s 2 )(3.13 s)2 = − 48 m, so the golf ball would fall 48 m.
2
2
47. Strategy The final speed is zero. Use Eq. (2-16).
Solution Find the initial speed.
vfy 2 − viy 2 = 0 − viy 2 = −2 g ∆y, so viy = 2 g ∆y = 2(9.80 m s 2 )(1.3 m − 0) = 5.0 m s .
48. Strategy The acceleration of the camera is given by v y1 / ∆t1 , where v y1 = 3.3 m s and ∆t1 = 2.0 s. Use Eq.
(2-15).
Solution After 4.0 s, the camera has fallen
1
1 ⎛ v y1 ⎞
3.3 m s
∆y = a y ( ∆ t ) 2 = ⎜
(4.0 s)2 = 13 m .
⎟ ( ∆t ) 2 =
2
2 ⎜⎝ ∆t1 ⎟⎠
2(2.0 s)
49. Strategy Use Eq. (2-15) to find the time it takes for the coin to reach the water. Then, find the time it takes the
sound to reach Glenda’s ear. Add these two times. Let h = 7.00 m.
Solution Find the time elapsed between the release of the coin and the hearing of the splash.
1
1
2h
h
h = viy ∆t + a y (∆t ) 2 = 0 + g (∆t1 ) 2 , so ∆t1 =
. h = vs ∆t2 , so ∆t2 = .
2
2
g
vs
Therefore, the time elapsed is ∆t = ∆t1 + ∆t2 =
2h h
2(7.00 m)
7.00 m
+ =
+
= 1.22 s .
2
g vs
343
ms
9.80 m s
50. (a) Strategy The stone is instantaneously at rest at its maximum height. Use Eq. (2-16).
Solution Find the maximum height of the stone.
vfy 2 − viy 2 = 0 − viy 2 = −2 g ∆y = −2 g ( yf − yi ), so yf = yi +
199
viy 2
2g
= 1.50 m +
(19.6
m s)
2
2(9.80 m s 2 )
= 21.1 m .
Chapter 2: Motion Along a Line
Physics
(b) Strategy The stone is instantaneously at rest at its maximum height of 21.1 m. Use Eq. (2-15).
Solution Going up 21.1 m − 1.50 m = 19.6 m (to rest) takes the same time as falling down 19.6 m (from rest),
so ∆yup = 12 g (∆tup ) 2 , or ∆tup = 2(19.6 m) (9.80 m s 2 ) = 2.00 s. Falling 21.1 m from rest takes
∆tdown = 2(21.1 m) (9.80 m s 2 ) = 2.08 s. The total time elapsed is 2.00 s + 2.08 s = 4.08 s .
51. Strategy Use Eqs. (2-15), (2-16), and (2-12). Let the +y-direction be down.
Solution
(a) Ignoring air resistance, the lead ball falls
1
1
∆y = g (∆t ) 2 = (9.80 m s 2 )(3.0 s)2 = 44 m .
2
2
(b) The lead ball is initially at rest. Find the speed of the ball after it has fallen 2.5 m.
vfy 2 = 2 g ∆y, so vfy = 2 g ∆y = 2(9.80 m s 2 )(2.5 m) = 7.0 m s .
(c) After 3.0 s, the lead ball is falling at a speed of v y = g ∆t = (9.80 m s 2 )(3.0 s) = 29 m s .
(d) Find the change in height of the ball when ∆t = 2.42 s.
1
1
∆y = viy ∆t − g (∆t ) 2 = (4.80 m s)(2.42 s) − (9.80 m s 2 )(2.42 s)2 = −17.1 m
2
2
The ball will be 17.1 m below the top of the tower.
52. Strategy Use Eq. (2-16).
Solution Find the sandbag’s speed when it hits the ground.
vfy 2 − viy 2 = −2 g ∆y, so vfy = viy 2 − 2 g ∆y =
(10.0
2
m s ) − 2(9.80 m s 2 )(−40.8 m) = 30.0 m s .
53. (a) Strategy Use Eq. (2-15) and the quadratic formula to find the time it takes the rock to reach Lois. Then, use
Eq. (2-15) again to find Superman’s required constant acceleration.
Solution Solve for ∆t using the quadratic formula.
1
1
∆y = viy ∆t + a y (∆t )2 , so a y (∆t ) 2 + viy ∆t − ∆y = 0.
2
2
2
−viy ± viy + 2a y ∆y −(−2.8 m s) ± (−2.8 m s)2 + 2(−9.80 m s 2 )(−14.0 m)
∆t =
=
= −2 s or 1.4286 s
ay
−9.80 m s 2
Since ∆t > 0, it takes about 1.43 s for the rock to reach Lois. Find Superman’s required acceleration.
1
1
1
2∆x
2(120 m)
∆x = vix ∆t + a x (∆t ) 2 = 0 + a x (∆t ) 2 = a x (∆t ) 2 , so a x =
=
= 120 m s 2 .
2
2
2
(∆t ) 2 (1.43 s)2
Superman must accelerate at 120 m s 2 toward Lois to save her.
(b) Strategy Use Eq. (2-12).
Solution Find Superman’s speed when he reaches Lois.
vx = ax ∆t = (120 m s 2 )(1.43 s) = 170 m s
200
Physics
Chapter 2: Motion Along a Line
54. Strategy The average speed of the flower pot as it passes the student’s window is approximately equal to its
instantaneous speed, so vav, y = ∆y ∆t ≈ v y .
Solution Determine the distance the flower pot fell to reach the speed v y .
v 2y = 2 gh, so h =
23.6 m
m
4.0 floor
v 2y
2g
=
2
1 ⎛ ∆y ⎞
(2.0 m)2
= 23.6 m.
⎜ ⎟ =
2 g ⎝ ∆t ⎠
2(9.80 m s 2 )(0.093 s)2
= 5.9 floors, so the pot fell from the 4th floor + 5.9 floors = 10th floor.
55. Strategy Find and subtract the time it took for the sound of the rock hitting the bottom of the well to reach your
ears from the total time (3.20 s) to find the time it took the rock to reach the bottom. Then, use Eq. (2-15) to
determine the depth of the well.
Solution The time it took for the sound of the rock hitting the bottom to reach you is ∆tsound = d vs , where d is
the depth of the well. So, ∆tfall = ∆ttotal − d vs . Find d.
d=
1
1 ⎛
d
g (∆t )2 = g ⎜⎜ ∆ttotal −
2
2 ⎝
vs
2
⎞
1 ⎡
1 2⎤
2 2∆t total
d+
d ⎥ , so
⎟⎟ = g ⎢ (∆ttotal ) −
2 ⎢⎣
vs
vs 2
⎠
⎦⎥
2∆ttotal
2
1 2
d = (∆ttotal )2 −
d+
d
g
vs
vs 2
2vs 2
g
d = (∆ttotal ) 2 vs 2 − 2vs ∆ttotal d + d 2
v
⎛
0 = d 2 − 2vs ⎜ ∆ttotal + s
g
⎝
⎞
2 2
⎟ d + (∆ttotal ) vs
⎠
v ⎞
⎛
Use the quadratic formula with a = 1, b = −2vs ⎜ ∆ttotal + s ⎟ , and c = (∆ttotal )2 vs 2 to solve for d.
g⎠
⎝
d=
v
⎛
2vs ⎜ ∆ttotal + s
g
⎝
⎡
vs
⎞
⎛
⎟ ± ⎢ −2vs ⎜ ∆ttotal +
g
⎠
⎝
⎣
2(1)
2
⎞⎤
2 2
⎟ ⎥ − 4(1)(∆ttotal ) vs
⎠⎦
2
⎡
⎛
⎛
343 m s ⎞
343 m s ⎞ ⎤
± ⎢ −2(343 m s) ⎜ 3.20 s +
2(343 m s) ⎜ 3.20 s +
⎟
⎟⎟ ⎥ − 4(3.20 s) 2 (343 m s) 2
2
2
⎜
⎟
⎜
9.80 m s ⎠
9.80 m s ⎠ ⎦⎥
⎢⎣
⎝
⎝
=
2
= 46 m (26,000 m is extraneous.)
56. (a) Strategy Use the definition of average speed.
Solution Find the average speed of the swimmer.
∆r
1500 m
vav =
=
= 1.7 m s
60 s
∆t
14 min ×
+ 53 s
1 min
(b) Strategy and Solution The swimmer pushes off from each end of the pool and he goes faster during the
push-off than when swimming.
201
Chapter 2: Motion Along a Line
Physics
57. (a) Strategy Use Eqs. (2-3) and (2-13) since the acceleration is constant.
Solution Find the distance traveled.
v +v
27.3 m s + 17.4 m s
∆x = vav, x ∆t = fx ix ∆t =
(10.0 s) = 224 m .
2
2
(b) Strategy Use the definition of average acceleration.
Solution Find the magnitude of the acceleration.
∆v 27.3 m s − 17.4 m s
a=
=
= 0.99 m s 2
10.0 s
∆t
58. (a) Strategy Use the definition of average acceleration.
Solution Find the magnitude of the acceleration.
∆v 24 m s
a=
=
= 12 m s 2
2.0 s
∆t
(b) Strategy Relate the distance traveled, acceleration, and time using Eq. (2-15).
Solution Find the distance traveled.
1
1
∆x = a (∆t )2 = (12 m s 2 )(2.0 s)2 = 24 m
2
2
(c) Strategy Refer to part (a).
Solution The magnitude of the acceleration of the runner is
∆v 6.0 m s
a=
=
= 3.0 m s 2 .
2.0 s
∆t
Find ac / ar .
ac 12 m s 2
=
= 4.0
ar 3.0 m s 2
59. Strategy Use the definitions of average velocity and average acceleration.
Solution
G
∆y
160 × 103 m up
G
=
= 330 m s up
(a) v av =
∆t (8.0 min) 60 s
1 min
(
)
G
G
∆v 7600 m s up − 0
=
= 16 m s 2 up
(b) aav =
∆t (8.0 min) 60 s
1 min
(
)
202
Physics
Chapter 2: Motion Along a Line
60. Strategy Use the essential relationships for constant acceleration problems, Eqs. (2-12) through (2-16).
∆x1
Solution Find vx1 , the speed after 10.0 s.
∆vx1 = a x1∆t1 = (1.0 m s 2 )(10.0 s) = 10 m s
∆x2
∆x3
0.60 km
Find ∆x1 , ∆x3 , and ∆t3 .
v 2 − vix32 0 − vx12
1
1
−(10 m s) 2
=
=
= 25 m
a x1 (∆t1 )2 = (1.0 m s 2 )(10.0 s)2 = 50 m; ∆x3 = x3
2
2
2a3
2a3
2(−2.0 m s 2 )
∆x
25 m
∆x3 = vav, x ∆t3 , so ∆t3 = 3 =
= 5.0 s. Find ∆x2 .
vav3 (10 m s) 2
∆x1 =
∆x
∆x2 = vx 2 ∆t2 = vx1∆t2 , so ∆t2 = 2 .
vx1
Find the total time.
∆x
0.60 × 103 m − 50 m − 25 m
∆t = ∆t1 + ∆t2 + ∆t3 = ∆t1 + 2 + ∆t3 = 10.0 s +
+ 5.0 s = 68 s
10 m s
vx1
61. Strategy Each car has traveled the same distance ∆x in the same time ∆t when they meet.
Solution Using Eq. (2-15), we have
1
1
2v
∆x = vi ∆t + a (∆t )2 = 0 + a (∆t )2 = v∆t , so ∆t = . The speed of the police car is vp = a∆t = a(2v a) = 2v .
a
2
2
62. (a) Strategy Use Eq. (2-16).
Solution Find the initial velocity of the stone.
vfy 2 − viy 2 = 2a y ∆y, so viy = ± vfy 2 − 2a y ∆y = ± (−25.0 m s)2 − 2(−9.80 m s 2 )(−16.0 m) = ±17.6 m s.
Since the stone was thrown vertically downward, its initial velocity was 17.6 m s downward .
(b) Strategy Use Eq. (2-15).
Solution Find the change in height of the stone.
1
1
∆y = viy ∆t + a y (∆t )2 = (−17.647 m s)(3.00 s) + (−9.80 m s 2 )(3.00 s)2 = −97.0 m
2
2
The height of the building is 97.0 m.
203
Chapter 2: Motion Along a Line
Physics
63. Strategy The direction of the acceleration is opposite the direction of motion. Use Eq. (2-16).
Solution
(a) Find the magnitude of the acceleration.
2
2
vfx − vix = 2a x ∆x, so a x =
vfx 2 − vix 2
2∆x
=
0 − ( 29 m s )
2
2(1.0 m)
= −420 m s 2 .
G
So, a = 420 m s 2 opposite the direction of motion .
(b) Find the magnitude of the acceleration.
ax =
vfx 2 − vix 2
2∆x
=
0 − ( 29 m s )
2(0.100 m)
2
= −4200 m s 2
G
So, a = 4200 m s 2 opposite the direction of motion .
64. Strategy Use the fact that distance traveled equals average rate times time elapsed.
Solution
(a) Marcella must run the distance in t =
1000 m
500 m
= 250 s. She ran the first 500 m in t1 =
= 119 s,
4.00 m s
4.20 m s
so she must run the last 500 m in t2 = t − t1 =
1000 m
500 m
−
= 131 s .
4.00 m s 4.20 m s
(b) Marcella’s average speed for the last 500 m must be v =
d 500 m
=
= 3.8 m s .
t2 131 s
65. Strategy Use the definitions of displacement, average velocity, and average acceleration.
Solution
G G G
∆r = rf − ri = 185 mi north − 126 mi north = 59 mi north
G
∆r 59 mi north ⎛ 60 min ⎞
G
v av =
=
⎜
⎟ = 96 mi h north
∆t
37 min ⎝ 1 h ⎠
G
G
∆v 105.0 mi h north − 112.0 mi h north −7.0 mi h north ⎛ 60 min ⎞
2
aav =
=
=
⎜
⎟ = 11 mi h south
∆t
37 min
37 min
1
h
⎝
⎠
66. Strategy If the rocket moves with constant acceleration, its average acceleration is equal to its instantaneous
(constant) acceleration.
G
Solution From Problem 58, aav = 16 m s 2 up. Use ∆y = viy ∆t + 12 a y (∆t )2 to find a y .
∆y = viy ∆t + 12 a y (∆t )2 = (0)∆t + 12 a y (∆t ) 2 , so a y =
2∆y
(∆t ) 2
G G
a ≠ aav , so the acceleration is not constant.
204
=
2(160 × 103 m)
(8.0
min)2
(
)
60 s 2
1 min
= 1.4 m s 2 .
Physics
Chapter 2: Motion Along a Line
67. Strategy Analyze the graph to answer each question about the motion of the elevator.
Solution
(a) The area under the curve represents the change in velocity. Assume each tick mark along the t-axis represents
1 s; for example, t1 = 3 s.
1
1
⎛1
⎞ 1
bh = (4 s) ⎜ m s 2 ⎟ = m s
2
2
4
⎝
⎠ 2
The elevator accelerates (a y > 0) to 1 2 m s during the first 4 s. Then, the elevator travels at
A1 =
1 2 m s (a y = 0) for the next 4 s.
1
1
1
⎛ 1
⎞
bh = (2 s) ⎜ − m s 2 ⎟ = − m s
2
2
2
⎝ 2
⎠
The elevator slows down (a y < 0) until it comes to rest. Then, it sits for the next 2 s. So, the passenger has
A2 =
gone to a higher floor.
(b) Sketch the graph of v y vs. t by plotting points at one-second intervals with v y determined from the a y vs. t
graph. Each rectangle represents (1 s)(1/4 m/s 2 ) = 1/4 m/s.
t (s)
vy
t (s)
vy
0
The elevator is at rest, so v y = 0.
6 (t2 )
12 ms
1
1 4 × (1 4 m s) = 1 16 m s
7
12 ms
2
1(1 4 m s) = 1 4 m s
8
12 ms
3 (t1)
(1 3 4)(1 4 m s) = 7 16 m s
9 (t3 )
−1(1 4 m s) + 1 2 m s = 1 4 m s
4
2(1 4 m s) = 1 2 m s
10
−1(1 4 m s) + 1 4 m s = 0
5
1 2 m s , since a y = 0.
11
0, since a y = 0
vy
t1
t2
t3
t
205
Chapter 2: Motion Along a Line
Physics
(c) The graph from part (b) shows that the velocity is nonnegative for t ≥ 0, so the position of the elevator is
always increasing in height until it stops.
Break the region under the graph from part (b) into six sections (blocks of time):
( 23 t1 )
Section 4: ( 89 t3, t3 )
( 23 t1, 43 t1 )
Section 5: ( t3, 10
t
9 3)
Section 1: 0,
Section 2:
( 43 t1, 89 t3 )
Section 6: ( 10
t , 4t
9 3 3 3)
Section 3:
Find the relative distance the elevator travels in each section by counting squares under the curve.
1: 2.7 at
2t
3 1
2: 13.5 at
4t
3 1
3: 32 at
8
t3
9
4: 6.7 at t3
5: 1.2 at
10 t
9 3
6: 0 at
4t
3 3
Plot points and draw a smooth curve.
y
t1
t2
t3
t
68. Strategy Use the graph to answer the questions. The slope of the graph represents the acceleration of the ball.
Solution
(a) The ball reaches its maximum height the first time v y = 0, or at t = 0.30 s .
(b) The time it takes for the ball to make the transition from its negative-most velocity to its positive-most
velocity is the time that the ball is in contact with the floor.
0.65 s − 0.60 s = 0.05 s
(c) Using Eq. (2-15) and the definition of average acceleration, we find that the maximum height of the ball is
∆v y
1
0 − 3.0 m s
(∆t ) 2 = (3.0 m s)(0.30 s) +
(0.30 s) = 0.45 m .
∆y = viy ∆t + a y (∆t ) 2 = viy ∆t +
2
2∆t
2
∆v y 0 − 3.0 m s
(d) a y =
=
= −10 m s 2 , so the acceleration is 10 m s 2 down .
0.30 s
∆t
(e) aav =
∆v 3.0 m s − ( −3.0 m s )
=
= 120 m s 2 , so the acceleration is 120 m s 2 up .
∆t
0.05 s
206
Physics
Chapter 2: Motion Along a Line
69. (a) Strategy Use Eq. (2-15).
Solution Find the rocket’s altitude when the engine fails.
1
1
∆y = a (∆t1 ) 2 = (20.0 m s 2 )(50.0 s)2 = 25.0 km
2
2
(b) Strategy viy = the speed when the engine fails = a∆t1; v y = viy − g ∆t = 0 at maximum height.
Solution Find the time elapsed from the engine failure to maximum height.
a
20.0 m s 2
0 = viy − g ∆t = a∆t1 − g ∆t , so ∆t = ∆t1 =
(50.0 s) = 102 s.
g
9.80 m s 2
The time to maximum height from lift off is ∆t + ∆t1 = 102 s + 50.0 s = 152 s .
(c) Strategy Use Eq. (2-15).
Solution Find the maximum height reached by the rocket.
2
a 2 (∆t1 ) 2 a 2 (∆t1 ) 2
a 2 (∆t1 ) 2
⎛a
⎞ 1 ⎛a
⎞
1
yf = yi + viy ∆t − g (∆t ) 2 = yi + (a∆t1 ) ⎜ ∆t1 ⎟ − g ⎜ ∆t1 ⎟ = yi +
−
= yi +
2
g
2g
2g
⎝g
⎠ 2 ⎝g
⎠
= 25.0 km +
(20.0 m s 2 ) 2 (50.0 s) 2
= 76.0 km
2(9.80 N kg)
(d) Strategy Use Eq. (2-16). viy = 0 at the maximum height.
Solution Find the final velocity.
vfy 2 − viy 2 = vfy 2 − 0 = 2a y ∆y = −2 g ∆y, so vfy = −2 g ∆y = −2(9.80 N kg)(0 − 76.0 × 103 m) = 1220 m s.
G
Thus, v = 1220 m s downward .
70. Strategy Analyze the graph to answer each question about the motion of the Engine. The x-component of the
engine’s velocity is represented by the slope.
Solution
(a) a x < 0 when the engine is moving in the +x-direction and slowing down or when it is moving in the
− x-direction and speeding up. So, at t3 and t4 a x < 0.
(b) a x = 0 when the engine’s speed is constant or zero. So, at t0 , t2 , t5 , and t7 a x = 0.
(c) a x > 0 when the engine is moving in the +x-direction and speeding up and when it is moving in the
− x-direction and slowing down. So, at t1 and t6 a x > 0.
(d) vx = 0 when the slope of the graph is zero. So, at t0 , t3 , and t7 vx = 0.
(e) The speed is decreasing when a x and vx have opposite directions. So, at t6 the speed is decreasing.
207
Chapter 2: Motion Along a Line
Physics
71. Strategy and Solution Since the acceleration is constant, the speed of the glider at the gate is its average speed as
it passes through.
8.0 cm
vf = vgate, av =
= 24 cm s
0.333 s
Also, since the acceleration is constant, the displacement of the glider equals the average speed times the time of
travel; vav = vf 2 = 12 cm s. Find ∆t.
∆x = vav ∆t , so ∆t =
∆x
.
vav
Find aav from its definition; aav = a.
v ∆v (12 cm s )( 24 cm s − 0 )
∆v
∆v
=
= av =
= 3.0 cm s 2
∆t ∆x vav
∆x
96 cm
G
So, a = 3.0 cm s 2 parallel to the velocity.
a=
72. Strategy v1fx 2 − v1ix 2 = 2a1d1 , where a1 = 10.0 ft s 2 and d1 is the distance to the point of no return.
v2fx 2 − v2ix 2 = 2a2 d 2 , where a2 = −7.00 ft s 2 and d 2 is the distance from the point of no return to the end of the
runway. The initial speed v1ix and the final speed v2fx are zero. The speed at the point of no return is v1fx = v2ix .
Let v1fx = v2ix = v for simplicity. Also, d = d1 + d 2 is the length of the runway.
Solution From the setup, we have v 2 = 2a1d1 and −v 2 = 2a2 d 2 = 2a2 (d − d1 ).
2a1d1 = 2a2 (d1 − d )
a1d1 = a2 d1 − a2 d
(a1 − a2 )d1 = −a2 d
a2
a2 − a1
−7.00 ft s 2
d=
s2
−7.00 ft
Find the time to d1 using Eq. (2-15).
d1 =
d2
d = 1.50 mi
Eliminate v 2 .
d1 =
d1
1
a (∆t ) 2 , so ∆t =
2 1
2d1
a1
=
− 10.0 ft
2(3260 ft)
10.0 ft s 2
s2
⎛ 5280 ft ⎞
(1.50 mi) ⎜
⎟ = 3260 ft
⎝ 1 mi ⎠
= 25.5 s .
208
Physics
Chapter 2: Motion Along a Line
73. (a) Strategy Use the definition of average speed.
Solution
∆x 100 × 10−9 m
vav =
=
= 1.0 mm s
∆t 0.10 × 10−3 s
(b) Strategy Find the time it takes the pain signal to travel the length of a 1.0-m long neuron. Then, add the
times of travel across synapses and neurons.
Solution
x
1.0 m
tn = =
= 10 ms
v 100 m s
Find the total time to reach the brain.
tn + tsyn + tn + tsyn = 2tn + 2tsyn = 2(tn + tsyn ) = 2(10 ms + 0.10 ms) = 20 ms
(c) Strategy Use the definition of average speed.
Solution
∆x 2.0 m + 2(100 × 10−9 m)
vav =
=
= 100 m s
∆t
20 × 10−3 s
209
Chapter 3
MOTION IN A PLANE
Conceptual Questions
1. No; to be equal they must also have the same direction. If the magnitudes are different, they cannot be equal.
2. (a) Yes, since the direction matters. See Fig. 3.1c.
(b) No. The largest possible magnitude occurs when the two vectors point in the same direction. Then the
magnitude of the sum equals the sum of the magnitudes.
3. A vector is a quantity that has both a magnitude and a direction associated with it. Velocity and displacement are
both examples of vector quantities. A scalar is a quantity that is only defined by a magnitude—it has no direction
associated with it. Scalar quantities include speed and distance traveled.
4. Yes, it is possible for two different projectiles with identical initial speeds but different angles of elevation to land
in the same spot. An object’s range is proportional to its horizontal velocity and also its time of flight. Roughly,
with a small angle of elevation the horizontal component of the velocity is greater than it is with a large angle of
elevation, but the time of flight is correspondingly less. The figure below shows the trajectories of two projectiles
launched with an initial speed of 100 m s at angles 30° and 60°.
y (m)
400
200
0
0
200
400
600
800
1000 x (m)
5. The trajectory is sometimes parabolic in another reference frame that moves with a constant velocity with respect
to the first. The only possibility other than parabolic is a straight-line trajectory.
6. After the first ball has reached its highest point and fallen back down to where it started, it will be moving
downward with speed vi just as it passes the height at which it was first thrown. From that point on, its trajectory
will be identical to the initial trajectory of the second ball. The balls therefore reach the ground with the same
speed, albeit at different times.
7. The two balls will cross at a height greater than half h. Prior to crossing, the bottom ball will be moving faster
than the top ball, and so will cover more distance.
8. The path of an object may only diverge from a straight line if the object accelerates. An object moving with
constant velocity is not accelerating and therefore must travel in a straight line.
9. The average speed and the magnitude of the average velocity of an object are equal if and only if the object travels
along a straight line path without changing direction. In all other cases, the average speed is greater than the
magnitude of the average velocity because the total distance traveled must be greater than the straight-line
distance between the starting and ending points.
210
Physics
Chapter 3: Motion in a Plane
10. An object traveling vertically downward in the gravitational field of the Earth is accelerating in the same direction
as its velocity. An object traveling vertically upward is accelerating in the direction opposite its velocity. A
projectile traveling in a parabolic path under the influence of gravity accelerates in a direction perpendicular to its
velocity at the apex of its flight.
11. A car driving around a curve at a constant speed does not have a constant velocity, because the direction is
changing.
12. That an object is in a state of equilibrium is to say that the sum of all forces acting on that object is zero and
therefore that the object is not accelerating.
(a) A car driving around a curve at constant speed is not in equilibrium because it is accelerating to maintain a
curved trajectory.
(b) A car driving straight up an incline at constant speed is not accelerating and is therefore in equilibrium. In this
case, the car is in dynamic equilibrium because the vehicle is not at rest.
(c) The Moon moves in a circular path around the Earth and is therefore an example of an object that is not in
equilibrium.
13. (a) A vector of magnitude 1L may be obtained by adding two vectors with lengths 3L and 4L aligned in opposite
directions.
(b) A vector of magnitude 7L may be obtained by adding two vectors with lengths 3L and 4L aligned in the same
direction.
(c) A vector of magnitude 5L may be obtained as the hypotenuse of a right triangle with sides composed of
vectors with magnitudes 3L and 4L.
14. The primary benefit of graphical vector addition is its use in providing a visual understanding of the problem—a
feature that is often obscured in algebraic vector addition. Despite this benefit, adding vectors graphically is a
cumbersome and imprecise process whereas the algebraic method is relatively easy to perform and provides much
greater accuracy. These benefits make the algebraic method the favored choice in most situations.
15. No single component of a vector can ever be greater than the magnitude of the vector. This is equivalent to the
statement that each side of a right triangle must be shorter than its hypotenuse—a statement that can be verified
using the Pythagorean Theorem.
16. Neglecting air resistance, the trajectory of a bullet that has exited the muzzle of a rifle is solely influenced by
gravity. The force due to gravity causes the bullet to accelerate downward toward the Earth but does not influence
its horizontal motion. Thus, to hit a target, the muzzle must be aimed above the target by a distance equal to the
amount that the bullet will fall in the course of its travel. If aimed at the target instead of above it, the bullet will
miss low.
17. The demonstration works when the hunter is aiming either up or down at the monkey and coconut. In the absence
of gravity, either case will result in the arrival of a bullet at the position occupied by the coconut. Gravity alters
the vertical motion of the coconut and the bullet identically—in a given time interval, both objects will fall an
equal distance from the trajectory they would have followed in the absence of gravity. Thus, either case concludes
with the result that the bullet and coconut arrive at the same position.
211
Chapter 3: Motion in a Plane
Physics
Problems
1. Strategy Let east be the +x-direction.
Solution Draw vector diagrams; then find the magnitudes and directions of the vectors.
(a)
B
N
A
A+B
(2.56 km + 7.44 km) west = 10.00 km west
(b)
−B
A
A−B
N
(2.56 km − 7.44 km) west = − 4.88 km west = 4.88 km east
(c)
B
−A
B−A
N
(7.44 km − 2.56 km) west = 4.88 km west
2. Strategy Let the +x-direction be to the right.
Solution Draw vector diagrams; then find the magnitudes and directions of the vectors.
(a)
A
B
A+B
G G
A + B = (1.73 − 1.00) units in the + x-direction = 0.73 units in the + x -direction
(b)
−B
A
A−B
G G
A − B = (1.73 + 1.00) units in the + x-direction = 2.73 units in the + x-direction
(c)
−A
B
B−A
G G
B − A = (1.00 + 1.73) units in the − x-direction = 2.73 units in the − x -direction
3. Strategy Use the properties of vectors to answer the questions.
Solution
(a) The only way for the sum to have a magnitude of 7.0 N is if the vectors are in the same direction.
(b) Recognizing that the three vectors form a 3-4-5 right triangle, we know that the vectors are perpendicular.
(c) The smallest magnitude sum can only be obtained if the two vectors are in opposite directions; the magnitude
of the smallest vector is 1.0.
212
Physics
Chapter 3: Motion in a Plane
4. Strategy Use the definitions of distance and displacement, and the properties of circles.
Solution
(a) Halfway around the track is a distance equal to half of the circumference of the circular track.
300 m
= 150 m
2
(b) The displacement is the shortest distance between the starting and ending points of the run. Since the runner
ran halfway around the track, the magnitude of the displacement is equal to the diameter of the circular track.
The runner ran from west to east, so the direction of the displacement is east. Find the diameter of the track.
C = π d , so d = C π = (300 m) π = 95 m.
The runner’s displacement is 95 m east.
5. Strategy Sketch the displacement vectors using graph paper, ruler, and protractor. Then find the vector sum by
sketching a graphical addition of the displacement vectors.
Solution The combined sketches are shown.
60°
20 km
20 km
60°
20 km in the +x-direction
x
6. Strategy Draw a diagram of Orville’s trip. Use it to find how far he walked during the second portion of the trip.
Solution The diagram is shown.
200 m
N
45°
320 m
Using a ruler and a protractor (and the scale of the diagram), we find that Orville walked about 230 m at 40° north
of west.
G G G G
G G G
G G G
7. Strategy Find graphically the vectors D = A + B and E = A + C. Then, show graphically that A + B = B + A.
Solution Use graph paper, ruler, and protractor to find the magnitude and direction of the vector sum of the two
forces in each case.
(a)
(b)
E=A+C
B
C
A
B
A
A+B
D
B+A
E
A
B
D=A+B
A
A+B=B+A
213
Chapter 3: Motion in a Plane
Physics
G
G
G
G
8. (a) Strategy Use the fact that | A | = | B | and symmetry to determine the direction of C; then sketch C.
G
G
Solution By symmetry, we find that C points downward; the horizontal components cancel when A and
G
B are added. The downward components of each vector have the same magnitude, about 0.7 cm. So, the
G
magnitude of C is about 1.4 cm. The sketch is shown:
α
α
4.0 cm
4.0 cm
C
A
B
1.4 cm
G
G
G
G
(b) Strategy Use the fact that | A | = | B | and symmetry to determine the direction of D; then sketch D.
G
G
Solution The vertical components cancel when B is subtracted from A. The direction of the horizontal
G
component of B is reversed due to the subtraction, and so the vector resulting from the subtraction is in the
G
direction of the horizontal component of A; that is, to the left. The horizontal components of each vector
G
have the same magnitude, which is about 3.95 cm; so the magnitude of D is about 7.9 cm. The sketch is
shown:
D
α
4.0 cm
A
9. Strategy Draw the displacement vectors. Use the diagram to answer the questions.
N
39 km
W
Killarney
Mallow
E
22 km
S
Cork
Solution
(a) The magnitude of the displacement from Killarney to Cork is the hypotenuse of a right triangle with legs 22
km and 39 km.
∆r = (22 km) 2 + (39 km) 2 = 45 km
(b) The distance along Michaela’s chosen route is 39 km + 22 km = 61 km, so the additional distance traveled is
61 km − 45 km = 16 km .
214
Physics
Chapter 3: Motion in a Plane
10. Strategy Use graph paper, ruler, and protractor to draw a diagram.
Solution
N
2.7 km, 45° W of N
Return Trip:
2.0 km, 20° E of S
1.2 km E
The scout troop must walk 2.0 km at 20° east of south to return to their starting point.
11. Strategy Write an expression for each displacement as the difference of two position vectors and then add them.
Solution Prove that the total displacement of a trip is equal to the vector sum of the displacements of each leg.
G G G
G G G
G G
G G
G G
We have, ∆r = (r2 − r1 ) + (r3 − r2 ) + (r4 − r3 ) + " + (rn − rn−1 ), so ∆r = rn − r1 , since all position vectors are
G
G
added and subtracted exactly once except for the first, r1 , which is only subtracted, and the last, rn , which is only
added.
12. Strategy Sketch the displacement vectors using graph paper, ruler, and protractor. Then estimate the position by
sketching a graphical addition of the displacement vectors.
Solution The combined sketches are shown.
Marblehead
Harbor
N
~30 nautical miles,
15°-20° south of east
The position is about 30 nautical miles at about 15° to 20° south of east.
13. Strategy The vector makes an angle of 60.0° counterclockwise from the y-axis. So, the angle from the positive xaxis is 90.0° + 60.0° = 150.0°.
Solution Find the components of the vector.
y
x-comp = (20.0 m) cos (150.0°) = −17.3 m and y-comp = (20.0 m) sin (150.0°) = 10.0 m .
20.0 m
60.0°
x
G
G
G
14. Strategy Let A be directed along the +x-axis and let B be 60.0° CCW from A.
G G
Solution Find the magnitude of A + B .
( A + B) x = Ax + Bx = 4.0 + 6.0 cos 60.0° = 7.0 and ( A + B) y = Ay + B y = 0 + 6.0sin 60.0° = 5.2, so
G G
A + B = 7.02 + 5.22 = 8.7 units .
215
y
6.0 B
60.0° A
x
4.0
Chapter 3: Motion in a Plane
Physics
15. (a) Strategy Since each vector is directed along a different axis, each component of the vector sum is equal to
the vector that lies along that component’s axis. Use the Pythagorean theorem.
G G
Solution Find the magnitude of A + B .
G G
A + B = [( A + B ) x ]2 + [( A + B ) y ]2 = (−1.0)2 + ( 3.0)2 = 2.0 units
y
A
B
Find the direction.
3.0
= 30° CCW from the +y -axis, so
θ = tan −1
−
1.0
G G
A + B = 2.0 units at 30° CCW from the +y -axis .
3.0
1.0
x
(b) Strategy Subtract a vector by adding its opposite. Use the Pythagorean theorem.
Solution Find the magnitude.
G G
A − B = [( A − B ) x ]2 + [( A − B ) y ]2 = 1.02 +
(
3.0
)
2
= 2.0 units
Find the direction.
G G
3.0
= 60° CW from the − x -axis, so A − B = 2.0 units at 30° CW from the +y -axis .
θ = tan −1
1.0
(c) Strategy The vectors lie on the axes.
G G
Solution Find the components of B − A.
x-comp = Bx = −1.0 unit
and y-comp = − Ay = − 3.0 units .
G
16. Strategy The components of a are given. Since the x-component is negative and the y-component is positive, the
vector lies in the second quadrant. Give the angle with respect to the axis to which it lies closest.
G
Solution Find the magnitude and direction of a.
y
(a) a = a x 2 + a y 2 = (−3.0 m s 2 )2 + (4.0 m s 2 ) 2 = 5.0 m s 2
a
ax
(b) θ = tan −1
ay
x
4.0
= 37° CCW from the +y -axis
−3.0
G
G
G
17. Strategy Use the fact that | A | = | B |, symmetry, and the component method to find the magnitude of C.
G
G
G
Solution By symmetry, C points downward, since the horizontal components cancel when A and B are added.
The downward components of each vector have the same magnitude, Ay = B y = (4.0 cm) sin10° = 0.69 cm. So,
G
the magnitude of C is 1.4 cm.
G
G
G
18. Strategy Use the fact that | A | = | B |, symmetry, and the component method to find the magnitude of D.
G
G
Solution The vertical components cancel when B is subtracted from A. The direction of the horizontal component
G
of B is reversed due to the subtraction, and so the vector resulting from the subtraction is in the direction of the
G
horizontal component of A; that is, to the left. The horizontal components of each vector have the same
G
magnitude, which is Ax = Bx = (4.0 cm) cos10° = 3.94 cm. so the magnitude of D is 7.9 cm.
216
Physics
Chapter 3: Motion in a Plane
19. Strategy Determine the angle each vector makes with the positive x-axis.
Solution Find the components of each vector.
G
Vector A :
G
Vector B :
Ax = (7.0 m) cos 20.0° = 6.6 m
Bx = (7.0 m s) cos(−20.0°) = 6.6 m s
Ay = (7.0 m) sin 20.0° = 2.4 m
B y = (7.0 m s) sin(−20.0°) = −2.4 m s
G
Vector C :
G
Vector D :
C x = (7.0 m) cos110.0° = −2.4 m
Dx = (7.0 m s) cos(−110.0°) = −2.4 m s
C y = (7.0 m) sin110.0° = 6.6 m
D y = (7.0 m s) sin(−110.0°) = −6.6 m s
G
20. Strategy The components of v are given. Since the x-component is positive and the y-component is negative, the
vector lies in the fourth quadrant. Give the angle with respect to the axes.
G
Solution Find the magnitude and direction of v.
y
vx
2
2
(a) v = vx + v y = (16.4 m
(b) θ = tan −1
s)2
+ (−26.3 m
s)2
= 31.0 m s
x
vy
v
−26.3
= 58.1° with the +x-axis and 31.9° with the − y -axis
16.4
21. Strategy Use the Pythagorean theorem to find the magnitude of each vector. Give the angle with respect to the
axis to which it lies closest.
Solution Find the magnitude and direction of each vector.
(a) A = (−5.0 m s) 2 + (8.0 m s) 2 = 9.4 m s and θ = tan −1
(b) B = (120 m) 2 + (− 60.0 m)2 = 130 m and θ = tan −1
8.0
= 32° CCW from the +y -axis .
−5.0
−60.0
= 27° CW from the +x-axis .
120
(c) C = (−13.7 m s) 2 + (−8.8 m s) 2 = 16.3 m s and θ = tan −1
(d) D = (2.3 m s 2 )2 + (6.5 × 10−2 m s 2 )2 = 2.3 m s 2
θ = tan −1
−8.8
= 33° CCW from the − x-axis .
−13.7
and
0.065
= 1.6° CCW from the +x-axis .
2.3
22. Strategy The vector makes an angle of 50.0° (50.0° + 130.0° = 180.0°) clockwise from the − x-axis. This is the
angle to use in determining the components.
Solution Find the components of the vector.
Ax = − A cos θ = −(22.2 cm) cos 50° = −14.3 cm and
Ay = A sin θ = (22.2 cm) sin 50° = 17.0 cm .
22.2 cm y
50.0°
x
217
Chapter 3: Motion in a Plane
Physics
23. (a) Strategy Since the angle is below the +x-axis, it is negative.
Solution Compute the components.
Bx = 7.1cos(−14°) = 6.9 and B y = 7.1sin(−14°) = −1.7 .
Cx = −1.8
y
14°
7.1
x
B
Cy = − 6.7
G
(b) Strategy The components of C are given. Use the Pythagorean theorem.
G
Solution Compute the magnitude and direction of C.
C = C x 2 + C y 2 = (−1.8)2 + (− 6.7)2 = 6.9
θ = tan −1
− 6.7
= 15° CW from the − y -axis
−1.8
(c) Strategy Add the components of the vectors to find the components of the vector sum. Use the Pythagorean
theorem. Give the angle with respect to the axis to which it lies closest.
G G
Solution Find the magnitude and direction of C + B.
G G
C + B = (C x + Bx )2 + (C y + B y )2 = (−1.8 + 6.9)2 + (−6.7 − 1.7)2 = 9.8 and
θ = tan −1
−8.4
= 31° CCW from the − y -axis .
5.1
G G
G
G
(d) Strategy Use the components of C and B to find those of C − B.
Solution Compute the magnitude and direction.
G G
C − B = (C x − Bx )2 + (C y − B y )2 = (−1.8 − 6.9) 2 + [− 6.7 − (−1.7)]2 = 10
θ = tan −1
−5.0
= 30° CCW from the −x-axis
−8.7
G G
G
G
(e) Strategy Use the components of C and B to find those of C − B.
Solution Compute the components.
x-comp = C x − Bx = −1.8 − 6.9 = −8.7 and y-comp = C y − B y = − 6.7 + 1.7 = −5.0 .
24. Strategy Margaret’s total displacement is the vector sum of the three displacements along the path.
Solution Add the displacements.
0.500 mi west + 0.200 mi north + 0.300 mi east = 0.500 mi west + 0.200 mi north − 0.300 mi west
= 0.200 mi west + 0.200 mi north
Let north be along the +y-axis and west be along the –x-axis. Then, the components of the total displacement are
∆x = −0.200 mi and ∆y = 0.200 mi. Find the magnitude.
∆r = (∆x)2 + (∆y )2 = (−0.200 mi)2 + (0.200 mi)2 = 0.283 mi
Find the direction.
0.200
θ = tan −1
= 45.0° CCW from the + y -axis = 45.0° N of W or NW
− 0.200
218
Physics
Chapter 3: Motion in a Plane
25. Strategy Add the displacement from Jerry’s dorm to the fitness center to the displacement from Cindy’s
apartment to Jerry’s dorm to find the total displacement from Cindy’s apartment to the fitness center.
Solution Add the displacements.
1.50 mi east + 2.00 mi north + 3.00 mi east = 4.50 mi east + 2.00 mi north
Let north be along the +y-axis and east be along the +x-axis. Then, the
components of the total displacement are ∆x = 4.50 mi and ∆y = 2.00 mi.
Find the magnitude.
∆r = (∆x)2 + (∆y )2 = (4.50 mi)2 + (2.00 mi) 2 = 4.92 mi
y
N
Apartment
1.50 mi
Dorm
Find the direction.
2.00
θ = tan −1
= 24.0° north of east
4.50
26. Strategy Let east be the +x-direction and north be the +y-direction.
Solution Compute the direction of travel and the distance walked.
∆r1x = 1.2 km
∆r1y = 0
∆r2 x = (2.7 km) cos135° = −1.9 km
∆r2 y = (2.7 km) sin135° = 1.9 km
∆rx = ∆r1x + ∆r2 x = 1.2 km − 1.9 km = −0.7 km
∆ry = ∆r1 y + ∆r2 y = 1.9 km
G
∆r = (∆rx )2 + (∆ry )2 = (− 0.7 km) 2 + (1.9 km)2 = 2.0 km
The direction of the return trip is opposite the displacement vector found.
∆rx′ = 0.7 km and ∆ry′ = −1.9 km.
−1.9
= 20° east of south
θ = tan −1
0.7
So, they must travel 2.0 km at 20° east of south.
27. Strategy Draw a diagram and use the component method. Let north be +y and east be +x.
Solution The diagram is shown (with the answer included).
Marblehead
Harbor
N
29 nautical miles,
17° south of east
Find the position of the sailboat.
∆x = 45 n.m. + (20.0 n.m.) cos 300° + 30.0 n.m. + (10.0 n.m.) cos 60° − 62 n.m. = 28 n.m.
∆y = (20.0 n.m.) sin 300° + (10.0 n.m.) sin 60° = −8.7 n.m.
∆r = (∆x)2 + (∆y )2 = (28 n.m.)2 + (−8.7 n.m.) 2 = 29 n.m.
−8.7
= 17° south of east
θ = tan −1
28
G
So, ∆r = 29 nautical miles at 17° south of east .
219
∆r
θ
3.00 mi
Fitness
center
2.00 mi
x
Chapter 3: Motion in a Plane
Physics
28. Strategy Use the component method to find the displacement from the starting point.
Solution Compute the total displacement. Let north be +y and east be +x.
∆x = (2.2 mi) cos 55° + (1.1 mi) cos15° = 2.3 mi
∆y = 1.6 mi + (2.2 mi) sin 55° + (1.1 mi) sin15° = 3.7 mi
∆r =
θ=
(∆x)2
3.7
tan −1
+ ( ∆y ) 2
= (2.3
mi)2
+ (3.7
mi) 2
y
2.2 mi
= 4.4 mi
1.6 mi
15°
1.1 mi
35°
N
∆r
θ
x
= 58° north of east
2.3
G
So, ∆r = 4.4 miles at 58° north of east .
29. Strategy Draw diagrams of the situation. Use the definitions of average speed and average velocity.
Solution
C = 0.478 mi
y
N
(a) Find the runner’s average speed.
∆x 0.750 mi 1609 m 1 min
=
×
×
= 5.03 m s
vav =
∆t 4.00 min
mi
60 s
(b) Find the location of the runner on the track.
0.750
= 1.569, so the runner has gone around once plus 0.569 times.
0.478
Find the angle θ shown in the diagram.
0.569 × 360° − 180° = 24.84°
Find the radius of the track.
C
C = 2π r , so r =
.
2π
G
Find ∆r.
x
vi
rf
y
θ
x
∆r
ri
∆r = (ri + rf cos θ )2 + (rf sin θ )2 = (r + r cos θ ) 2 + (r sin θ ) 2 = r (1 + cos θ )2 + (sin θ ) 2
C
0.478 mi
2(1 + cos θ ) =
2(1 + cos 24.84°)
= r 1 + 2 cos θ + cos 2 θ + sin 2 θ = r 1 + 2 cos θ + 1 =
2π
2π
= 0.1486 mi
r sin θ
sin 24.84°
= tan −1
= 12.4°
φ = tan −1
r + r cos θ
1 + cos 24.84°
Find the runner’s average velocity.
∆r 0.1486 mi 1609 m 1 min
G
G
=
×
×
= 0.996 m s , so v av = 0.996 m s at 12.4° west of north .
v av =
∆t
4.00 min
mi
60 s
30. Strategy Use the definitions of average speed and average velocity.
Solution
(a) Find the runner’s average speed.
∆x 1.00 mi 1609 m 1 min
=
×
×
= 6.7 m s
vav =
∆t 4.0 min
mi
60 s
(b) Find the location of the runner on the track.
1.00
= 2.0, so the runner has gone around 2.0 times.
0.50
G
The runner has gone around two complete times, so his displacement is zero; therefore, v av = 0 .
220
Physics
Chapter 3: Motion in a Plane
31. (a) Strategy Draw the position vectors with respect to Illium.
Solution
N
W
Atkins Glen
Illium
15°
km
27.2
E
73.
6k
m
25°
Cornwall
S
(b) Strategy Use the component method.
Solution
G G G
∆r = rf − ri ; ∆x = xf − xi = rf cos θf − ri cos θi ; ∆y = yf − yi = rf sin θ f − ri sin θi
Find the magnitude of the displacement.
G
∆ r = ( ∆ x ) 2 + ( ∆y ) 2
= [(27.2 km) cos195° − (73.6 km) cos 245°]2 + [(27.2 km) sin195° − (73.6 km) sin 245°]2
= 59.9 km
Find the direction of the displacement.
(27.2 km) sin195° − (73.6 km) sin 245°
= 85° north of east
θ = tan −1
(27.2 km) cos195° − (73.6 km) cos 245°
G
So, ∆r = 59.9 km at 85° north of east .
(c) Strategy Use the definition of average velocity.
Solution
G
∆r 59.9 km at 85° north of east
G
v av =
=
= 80 km h at 85° north of east
∆t
(45 min) 1 h
( 60 min )
32. (a) Strategy Find the distance traveled during the first part of the trip; then compute the speed required to travel
the remaining distance in the time required.
Solution There is 48.0 min = 0.800 h left to complete the trip. The harpsichordist has traveled
(55.0 mi h)(1.20 h) = 66.0 mi, so he has 122 mi − 66 mi = 56 mi to go. To get to the concert on time, he
must travel at a speed of (56 mi) (0.800 h) = 70 mi h .
221
Chapter 3: Motion in a Plane
Physics
(b) Strategy Use the definition of average velocity. Use components for the displacements.
Solution Let the +x-direction be west and the +y-direction be south. Then,
∆x = 66.0 mi + (56 mi) cos 30.0° = 114.5 mi and ∆y = (56 mi) sin 30.0° = 28 mi.
(114.5 mi)2 + (28 mi)2
= 59 mi h .
2.00 h
28
The direction of the average velocity is θ = tan −1
= 14°.
114.5
The magnitude of the average velocity is v =
So, the average velocity for the entire trip was 59 mi h at 14° south of west .
33. Strategy Use the definition of average velocity. Draw a diagram.
3.2 km
Solution Let east be in the +x-direction and north be in the +y-direction.
G
Find the magnitude of ∆r.
N
G
4.8 km
∆r = [3.2 km + (4.8 km) cos 75.0° + 3.2 km]2 + [(4.8 km) sin 75.0°]2 = 8.9 km
15.0°
G
75.0°
Find the direction of ∆r.
3.2 km
4.6 km
θ = tan −1
= 31° north of east
7.6 km
G
∆r
8.94 km
∆r
G
G
=
= 26 km h and v av =
= 26 km h at 31° north of east .
So, v av =
∆t 0.10 h + 0.15 h + 0.10 h
∆t
34. (a) Strategy Find the average speed by dividing the total distance traveled by the total time.
Solution Since the time traveled at each speed is the same, we can simply add the speeds and divide by 2.0.
96 km h + 128 km h 224 km h
=
= 110 km h
vav =
2.0
2.0
(b) Strategy Use the definition of average velocity. Draw a diagram.
Solution Let east be the +x-direction and north be the +y-direction.
N
30° 128 km
60°
96 km
G
Find the magnitude of ∆r.
G
2
2
∆r = [96 km h + (128 km h) cos 60°] (1.0 h) 2 + [ (128 km h) sin 60°] (1.0 h) 2 = 195 km
G
Find the direction of ∆r.
110.85 km
θ = tan −1
= 35° north of east
160 km
G
∆r 194.65 km
∆r
G
G
=
= 97 km h and v av =
= 97 km h at 35° north of east .
So, v av =
∆t
2.0 h
∆t
222
Physics
Chapter 3: Motion in a Plane
35. (a) Strategy Find the average speed by dividing the total distance traveled by the total time.
Solution Each distance is given by the product of the speed and time.
distance (108 km h)(20.0 min) + (90.0 km h)(10.0 min)
vav =
=
= 102 km h
time
20.0 min + 10.0 min
(b) Strategy Use the definition of average velocity. Draw a diagram.
Solution Compute the distance of each leg of the trip, then draw the diagram.
⎛ 1h ⎞
⎛ 1h ⎞
(108 km h)(20.0 min) ⎜
⎟ = 36.0 km and (90.0 km h)(10.0 min) ⎜
⎟ = 15.0 km.
60
min
⎝
⎠
⎝ 60 min ⎠
N
36.0 km
60.0°
15.0 km
G
Find ∆r. Let east be in the +x-direction and north be in the +y-direction.
G
∆r = [−36.0 km + (15.0 km) cos 240.0°]2 + [(15.0 km) sin 240.0°]2
= (−43.5 km)2 + (−13.0 km)2 = 45.4 km
−13.0
= 16.6° south of west
−43.5
∆r
45.4 km
G
So, v av =
=
= 90.8 km h and
∆t
⎛ 1h ⎞
(10.0 min + 20.0 min) ⎜
⎟
⎝ 60 min ⎠
G
∆r
G
v av =
= 90.8 km h at 16.6° south of west .
∆t
θ = tan −1
36. (a) Strategy From Problem 9, the distance Michaela traveled between Killarney to Cork via Mallow was 61 km.
Solution Find Michaela’s average speed.
distance traveled 61 km 1000 m 1 min
v=
=
×
×
= 21 m s
total time
48 min 1 km
60 s
(b) Strategy From Problem 4, the magnitude of Michaela’s displacement was 45 km.
Solution Find the magnitude of Michaela’s average velocity.
∆r 45 km 1000 m 1 min
G
=
×
×
= 16 m s
v av =
∆t 48 min 1 km
60 s
223
Chapter 3: Motion in a Plane
Physics
37. Strategy Draw diagrams. Use the definitions of average speed and average velocity.
Solution
90.0 km/h
30.0°
76.0 km/h
N
(a) Find the displacement.
G G G
∆r = rf − ri
= (90.0 km h)(80.0 min)[1 h (60 min)] east − (76.0 km h)(45.0 min)[1 h (60 min)] south of west
= 120 km east − 57.0 km at 30.0° south of west
120 km
30.0°
N
∆r
57.0 km
Find the distance.
∆r = [120 km − (57.0 km) cos 30.0°]2 + [−(57.0 km) sin 30.0°]2 = 76.2 km
(b) Find the magnitude of the average velocity.
∆r 76.2 km 60 min
=
×
= 102 km h
vav =
∆t 45.0 min
1h
Find the direction.
(57.0 km) sin 30.0°
= 22.0° north of west
θ = tan −1
−120 km + (57.0 km) cos 30.0°
The average velocity on the third leg is 102 km h at 22.0° north of west .
(c) Compute the time.
80.0 min + 15.0 min + 45.0 min = 140.0 min
Find the magnitude of the average velocity.
∆r 76.17 km 60 min
=
×
= 32.6 km h
vav =
∆t 140.0 min
1h
The direction is opposite that found in part (b), so the average velocity during the first two legs is
32.6 km h at 22.0° south of east .
(d) Since the displacement is zero, the average velocity over the entire trip is 0 .
(e) Compute the total time.
80.0 min + 15.0 min + 45.0 min + 45.0 min + 55.0 min = 240.0 min, or 4.000 h
Compute the total distance.
120 km + 57.0 km + 76.2 km = 253.2 km
253.2 km
= 63.3 km h .
The average speed during the entire trip is
4.000 h
224
Physics
Chapter 3: Motion in a Plane
38. Strategy Use the definition of average acceleration.
Solution South is the positive direction.
G
G G
G ∆v v f − vi
=
a=
∆t
∆t
5.0 m s south − 2.0 m s north 5.0 m s south + 2.0 m s south
=
=
= 0.70 m s 2 south
10.0 s
10.0 s
39. Strategy Use the definition of average acceleration.
Solution Up is the positive direction.
G
G G
G ∆v v f − vi 8.3 m s down − 55 m s down
=
=
= −13 m s 2 down = 13 m s 2 up .
a=
3.5 s
∆t
∆t
40. (a) Strategy Use the definition of average velocity. Draw a diagram.
Solution Let the center of the circle be the origin, then
G G G
∆r = rf − ri = 20.0 m east − 20.0 m south.
G
∆r = (20.0 m) 2 + (−20.0 m) 2 = 28.3 m
Let east be the +x-direction and north the +y-direction.
20.0
θ = tan −1 =
= 45.0° north of east
20.0
G
∆r 28.3 m
∆r
G
G
=
= 9.4 m s and v av =
= 9.4 m s at 45° north of east .
So, v av =
∆t
∆t
3.0 s
y
N
x
20.0 m
vf
vi
(b) Strategy Use the definition of average acceleration and the fact that C = 2π r.
Solution Find the average acceleration of the car.
G
2π r (3 4) 3π r
3π r
G
G
=
, so aav =
(south − west).
v f = vi =
∆t
2∆t
2(∆t )2
G
3π r
3π r
12 + (−1)2 =
aav =
2
2(∆t )
2(∆t )2
−1
θ = tan −1 = 45° south of east
1
G
G
∆v 3π (20.0 m)
aav =
=
at 45° south of east = 15 m s2 at 45° south of east
∆t
2(3.0 s) 2
(c) Strategy Consider Newton’s first law of motion.
Solution Although the magnitude of the velocity is constant, its direction must change continuously for the
car to travel in a circle; changing the direction of the velocity requires an acceleration .
225
Chapter 3: Motion in a Plane
Physics
41. Strategy The magnitude of the velocity is constant, but the direction changes. Recall that the circumference of a
circle is given by 2π r.
Solution
(a) Find the car’s speed.
C 4
C
2π r π r π (10.0 m)
v=
=
=
=
=
= 9.82 m s
4∆t 4∆t 2∆t 2(1.60 s)
∆t
vf
10.0 m
vi
(b) Let east be the +x-direction and north the +y-direction.
G G
G
∆v = v f − vi = v east − v north
G
∆v = v 2 + (− v) 2 = 2v 2 = v 2
−v
= tan −1 (−1) = 45° south of east (SE)
θ = tan −1
v
G
∆v = 2(9.82 m s) southeast = 13.9 m s southeast
N
10.0 m
G
G
∆v 13.88 m s southeast
=
= 8.68 m s 2 southeast
(c) aav =
∆t
1.60 s
42. (a) Strategy Let east be in the +x-direction and north be
in the +y-direction.
N
vi
Solution See the figure.
vf
192 km/h
240 km/h
45°
W
E
S
(b) Strategy Use the component method to subtract the initial velocity vector from the final velocity vector.
Solution
G G
G
∆v = v f − vi = 240 km h NW − 192 km h N, so
G
2
2
∆v = [ (240 km h) cos135°] + [ −192 km h + (240 km h) sin135°] = 170 km h .
θ = tan −1
−192 km h + (240 km h) sin135°
= 7° south of west
(240 km h) cos135°
G
So, ∆v = 170 km h at 7° south of west .
(c) Strategy Use the definition of average acceleration.
Solution
G
G
∆v 170 km h at 7° south of west
aav =
=
= 57 km h 2 at 7° south of west
∆t
3.0 h
226
Physics
Chapter 3: Motion in a Plane
43. (a) Strategy Draw a diagram. Use the component method to subtract the initial velocity vector from the final
velocity vector. Let east be in the +x-direction and north be in the +y-direction.
Solution
N
90 km/h
W
E
45°
100 km/h
S
G G
G
∆v = v f − vi = 100 km h SE − 90 km h W, so
G
2
2
∆v = [ (100 km h) cos 315° + 90 km h ] + [ (100 km h) sin 315°] = 180 km h .
θ = tan −1
(100 km h) sin 315°
G
= 24° south of east, so ∆v = 180 km h at 24° south of east .
(100 km h) cos 315° + 90 km h
(b) Strategy Find the total time of travel. Use the definition of average acceleration.
Solution
16 km
8.0 km
34 km
∆t =
+
+
= 0.62 h, so
90 km h 80 km h 100 km h
G
G
∆v 176 km h at 24° south of east
aav =
=
= 280 km h 2 at 24° south of east .
∆t
0.62 h
44. Strategy Since the particle is moving to the east and is accelerated to the south, its velocity in 8.00 s will be
between east and south. Use the component method. Let north be in the +y-direction and east be in the +xdirection.
Solution
vx = 40.0 m s and v y = a y ∆t = (−2.50 m s 2 )(8.00 s) = −20.0 m s.
G
v = vx 2 + v y 2 = (40.0 m s)2 + (−20.0 m s) 2 = 44.7 m s
θ = tan −1
vy
= tan −1
−20.0 m s
= 26.6° south of east
40.0 m s
vx
G
So, v = 44.7 m s at 26.6° south of east .
45. Strategy Use the component method. Solve for the time. Let north be in the +y-direction and east be in the +xdirection.
Solution
vx = 60 m s and v y = a y ∆t.
Use the Pythagorean theorem.
v 2 = vx 2 + v y 2 = vx 2 + (a y ∆t )2 , so ∆t =
v 2 − vx 2
ay
=
(100 m s) 2 − (60 m s)2
227
100 m s 2
= 0.8 s .
Chapter 3: Motion in a Plane
Physics
46. Strategy Use equations of motion with constant acceleration to determine the vertical and horizontal positions of
the baseball after 1.40 s have elapsed.
Solution Find the position of the ball.
xf = vix ∆t = (30.0 m s)(1.40 s) = 42.0 m
1
1
yf = yi + viy ∆t − g (∆t )2 = 9.60 m + 0 − (9.80 m s 2 )(1.40 s)2 = 0.00 m
2
2
After 1.40 s have elapsed, the ball is on the ground at a horizontal distance of 42.0 m from the launch point.
47. Strategy Use equations of motion with constant acceleration to determine the vertical and horizontal positions of
the clay after 1.50 s have elapsed.
Solution Find the position of the clay.
xf = vix ∆t = (20.0 m s)(1.50 s) = 30.0 m
1
1
yf = yi + viy ∆t − g (∆t )2 = 8.50 m + 0 − (9.80 m s 2 )(1.50 s)2 = −2.53 m
2
2
The clay cannot pass through the ground, so it hit and stuck prior to 1.50 s. Find the time it took for the clay to
land.
1
1
yf = 0 = yi + viy ∆t − g (∆t )2 = yi − g (∆t )2 , so
2
2
2 yi
2(8.50 m)
∆t =
=
= 1.317 s, and xf = vix ∆t = (20.0 m s)(1.317 s) = 26.3 m.
g
9.80 m s 2
The clay hits and it is on the ground after 1.32 s, so the horizontal distance along the ground is 26.3 m.
48. (a) Strategy Use equations of motion with constant acceleration to determine the vertical and horizontal
positions of the tennis ball after 1.60 s have elapsed.
Solution Find the position of the ball.
xf = vix ∆t = (20.0 m s)(1.60 s) = 32.0 m
1
1
yf = yi + viy ∆t − g (∆t )2 = 14.0 m + 0 − (9.80 m s 2 )(1.60 s)2 = 14.0 m − 12.5 m = 1.5 m
2
2
After 1.60 s have elapsed, the ball has fallen 12.5 m vertically and has traveled 32.0 m horizontally.
(b) Strategy The ball is still in the air. Set the final position equal to zero in Eq. (4-9) to find the time when the
ball hits the ground.
Solution Find the elapsed time.
1
1
yf = 0 = yi + viy ∆t − g (∆t ) 2 = yi − g (∆t ) 2 , so ∆t =
2
2
2 yi
g
=
2(14.0 m)
9.80 m s 2
= 1.69 s.
Find the landing position.
xf = vix ∆t = (20.0 m s)(1.69 s) = 33.8 m
The ball will land after another 0.09 s and will then be at a horizontal distance of 33.8 m.
228
Physics
Chapter 3: Motion in a Plane
49. Strategy Use Eqs. (3-13) and (3-14). Set vfy = 0, since the vertical component of the velocity is zero at the
maximum height.
Solution
(a) Find the maximum height.
19.6 m/s
vfy 2 − viy 2 = 0 − viy 2 = −2 g ∆y, so ∆y =
yf =
viy
viy
2
2g
= yf − yi and
30.0°
vix
vi 2 sin 2 θ
(19.6 m s) 2 sin 2 30.0°
+ yi =
+ 1.0 m = 5.9 m .
2g
2(9.80 m s 2 )
(b) At the ball’s highest point, vfy = 0, so the speed v equals vx .
v = vx = vix = vi cos θ = (19.6 m s ) cos 30.0° = 17.0 m s
50. (a) Strategy Use Eqs. (3-10) and (3-14).
Solution Find the components of the velocity.
vx = vix = vi cos θ = ( 20.0 m s ) cos 60.0° = 10.0 m s
viy
v y = vi sin θ − g ∆t = (20.0 m s) sin 60.0° − (9.80 m s 2 )(3.0 s) = −12 m s
20.0 m/s
60.0°
vix
(b) Strategy Use Eqs. (3-5) and (3-12).
Solution Calculate the x-component of the displacement.
∆x = vx ∆t = (10.0 m s ) (3.0 s) = 30 m
Calculate the y-component of the displacement.
1
1
∆y = viy ∆t − g (∆t )2 = ( 20.0 m s ) sin 60.0°(3.0 s) − (9.80 m s 2 )(3.0 s) 2 = 8 m
2
2
51. (a) Strategy Ignoring air resistance, the bomb will travel a distance equal to the speed of the plane times the
time is takes the bomb to reach ground level.
Solution Use Eq. (3-12) to find the time it takes for the bomb to reach ground level.
1
1
−2 ∆ y
∆y = viy ∆t + a y (∆t ) 2 = 0 − g (∆t ) 2 , so ∆t =
. So, the bomb should be released
2
2
g
∆x = v x ∆ t = v x
−2∆y
−2(−125 m)
= (40.0 m s)
= 202 m horizontally from the target.
g
9.80 m s 2
(b) Strategy The horizontal component of the velocity of the bomb is 40.0 m s. Find the vertical component of
the velocity and use the components to find the direction the bomb is traveling just before it hits the target.
Solution Use Eq. (3-13) to find the vertical component of the bomb’s velocity.
2
2
vfx
2
vfy − viy = 2a y ∆y = −2 g ∆y = vfy − 0, so vfy = −2 g ∆y . Find the angle.
θ = tan −1
vfy
vfx
= tan −1
−2(9.80 m s 2 )(−125 m)
−2 g ∆y
= tan −1
vfx
40.0 m s
= 51.1° below the horizontal
229
θ
vfy
vf
Chapter 3: Motion in a Plane
Physics
52. Strategy Solve ∆x = vx ∆t for the time and substitute the result into Eq. (3-12). Then, solve for ∆x to find the
required distance from the cannon.
Solution ∆x = vx ∆t = (vi cos θ )∆t , so ∆t = ∆x (vi cos θ ). Substitute.
∆x
1
∆y = yf − yi = viy ∆t + a y (∆t )2
2
∆x
g (∆x)2
1
(∆x)2
= (vi sin θ )
− g
= ∆x tan θ −
, so
vi cos θ 2 vi 2 cos 2 θ
2vi 2 cos 2 θ
0=
g
2
2
2vi cos θ
18.0 m/s
35.0°
Cannon
5.0 m
Net
(∆x) 2 − (tan θ )∆x + ∆y.
Use the quadratic formula.
2
⎛
⎞
g
tan θ ± tan 2 θ − 4 ⎜ 2 2 ⎟ ∆y tan 35.0° ± tan 2 35.0° − 2(9.80 m s2 )( −25.0 m)
v
θ
2
cos
(18.0
m
s)
cos 35.0°
⎝ i
⎠
∆x =
=
= 37.1 m or − 6.0 m
2
9.80 m s
⎛
⎞
g
2⎜ 2 2 ⎟
(18.0 m s) 2 cos 2 35.0°
⎝ 2vi cos θ ⎠
Since the cannon won’t fire backward, −6.0 m is extraneous. So, you tell the ringmaster to place the net such that
its center is 37.1 m in front of the cannon.
53. (a) Strategy At the maximum height of the cannonball’s trajectory, vfy = 0. Use Eq. (3-13).
Solution Find the maximum height reached by the cannonball.
vfy 2 − viy 2 = 0 − (vi sin θ )2 = 2a y ∆y = −2 g ( yf − yi ), so
yf = yi +
vi sin 37°
vi 2 sin 2 θ
(40 m s)2 sin 2 37°
= 7.0 m +
= 37 m .
2g
2(9.80 m s 2 )
vi
37°
vi cos 37°
(b) Strategy Solve ∆x = vx ∆t for the time and substitute the result into Eq. (3-12). Then, solve for ∆x to find
the horizontal distance from the release point.
Solution When the cannonball hits the ground, ∆y = −7.0 m.
∆x = vx ∆t = (vi cos θ )∆t , so ∆t =
∆x
. Substitute.
vi cos θ
1
∆x
1
(∆x)2
g ( ∆x ) 2
∆y = yf − yi = viy ∆t + a y (∆t )2 = (vi sin θ )
− g
= ∆x tan θ −
, so
2
2
2
vi cos θ 2 vi cos θ
2vi 2 cos 2 θ
g
0=
(∆x)2 − (tan θ )∆x + ∆y. Use the quadratic formula.
2
2
2vi cos θ
tan θ ± tan 2 θ −
∆x =
2g
2vi 2 cos 2 θ
4 g ∆y
2vi 2 cos 2 θ
tan 37° ± tan 2 37° −
=
2(9.80 m s 2 )( −7.0 m)
(40 m s) 2 cos 2 37°
9.80 m s 2
= 170 m or − 9 m
(40 m s)2 cos 2 37°
Since the catapult doesn’t fire backward, −9 m is extraneous. So, the cannonball lands 170 m
release point.
230
from its
Physics
Chapter 3: Motion in a Plane
(c) Strategy The x-component is the same as the initial value. Find the y-component using Eq. (3-13).
Solution The x-component of the velocity is vfx = vix = vi cos θ = (40 m s) cos 37° = 32 m s .
Find the y-component of the velocity.
vfy 2 − viy 2 = vfy 2 − vi 2 sin 2 θ = 2a y ∆y = −2 g ∆y, so
vfy = ± vi 2 sin 2 θ − 2 g ∆y = ± (40 m s) 2 sin 2 37° − 2(9.80 m s 2 )(−7.0 m) = −27 m s ,
where the negative sign was chosen because the cannonball is on its way down.
54. Strategy Solve ∆x = vx ∆t for the time and substitute the result into Eq. (3-12). Then, solve for vi .
Solution ∆x = vx ∆t = (vi cos θ )∆t , so ∆t =
∆x
. Substitute.
vi cos θ
vi sin 53°
53°
vi cos 53°
1
∆x
1
(∆x) 2
g (∆x) 2
∆y = viy ∆t + a y (∆t )2 = (vi sin θ )
− g
= ∆x tan θ −
, so
2
vi cos θ 2 vi 2 cos 2 θ
2vi 2 cos 2 θ
g (∆x)2
vi =
2 cos 2 θ (∆x tan θ − ∆y )
=
(9.80 m s 2 )(50 m) 2
2 cos 2 53°[(50 m) tan 53° − (−12 m)]
vi
= 21 m s .
55. (a) Strategy Consider each quantity’s dependence on time.
Solution
∆x = vx ∆t , so x increases linearly with time.
According to Eq. (3-12), y is parabolic.
y
x
t
t
Since the net acceleration of the stone in the
horizontal direction is zero, vx is constant.
v y starts positive and decreases linearly.
vy
vx
t
t
(b) Strategy Find vi in terms of ∆x, ∆t , and θ .
Solution Solve for the initial speed.
∆x
105 m
∆x = (vi cos θ )∆t , so vi =
=
= 27.6 m s.
∆t cos θ (4.20 s) cos 25.0°
So, the initial velocity is 27.6 m s at 25.0° above the horizontal .
231
vi sin 25.0°
vi
25.0°
vi cos 25.0°
Chapter 3: Motion in a Plane
Physics
(c) Strategy Find h using the result for vi found in part (b).
Solution Use Eq. (3-12).
1
1
1
1
⎛ ∆x
⎞
∆y = viy ∆t + a y (∆t )2 = (vi sin θ )∆t − g (∆t )2 = ⎜
sin θ ⎟ ∆t − g (∆t )2 = ∆x tan θ − g (∆t )2
2
2
2
2
⎝ ∆t cos θ
⎠
1
2
2
= (105 m) tan 25.0° − (9.80 m s )(4.20 s) = −37.5 m
2
So, h = 37.5 m .
(d) Strategy Set v y = 0 to find the time when the stone reaches its maximum height.
Solution Use Eq. (3-10) to find the time.
v sin θ
vfy − viy = a y ∆t = − g ∆t , so vfy = vi sin θ − g ∆t = 0, or ∆t = i
.
g
Find H.
2
vi 2 sin 2 θ vi 2 sin 2 θ
⎞ 1 ⎛ vi sin θ ⎞
−
⎟− g⎜
⎟ = h+
2g
g
⎠ 2 ⎝ g ⎠
2
2
2
2
v sin θ
(27.6 m s) sin 25.0°
= h+ i
= 37.5 m +
= 44.4 m above the ground
2g
2(9.80 m s 2 )
H = h + (vi sin θ )∆t −
⎛ v sin θ
1
g (∆t )2 = h + vi sin θ ⎜ i
2
⎝ g
56. Strategy Use the result for the range given in Problem 57, part (b).
Solution Compute the ranges.
v 2 sin 2θ (36.2 m s) 2 sin 2(36.0°)
(a) R = i
=
= 127 m
g
9.80 m s 2
v 2 sin 2θ (36.2 m s) 2 sin 2(54.0°)
=
= 127 m
R= i
g
9.80 m s 2
(b) R =
vi 2 sin 2θ
g
=
2
(36.2 m s) sin 2(23.0°)
9.80 m s 2
θ2
θ1
= 96.2 m
R
v 2 sin 2θ (36.2 m s)2 sin 2(67.0°)
=
= 96.2 m
R= i
g
9.80 m s 2
v 2 sin 2θ (36.2 m s)2 sin 2(45.0°)
(c) R = i
=
= 134 m
g
9.80 m s 2
(d)
The ranges are the same for each pair of complementary angles. The largest range occurred for an angle
of 45.0° above the horizontal.
232
Physics
Chapter 3: Motion in a Plane
57. (a) Strategy Solve for the time using Eq. (3-12).
Solution
vi sin θ
vi
θ
vi cos θ
R
2viy
2vi sin θ
1
1
1
∆y = 0 = viy ∆t + a y (∆t )2 = viy ∆t − g (∆t )2 = viy − g ∆t , so ∆t =
=
.
2
2
2
g
g
1
(b) Strategy Use ∆x = vx ∆t = vix ∆t and ∆y = viy ∆t − g (∆t )2 to find the range. Use the trigonometric identity
2
sin 2θ = 2sin θ cos θ .
Solution Solve for the time.
∆x
.
∆x = vix ∆t , so ∆t =
vix
∆x = R, so ∆t =
R
. Find ∆t in terms of viy .
vix
∆y = 0 = viy ∆t −
2viy
1
1
R
=
g (∆t )2 = viy − g ∆t , so ∆t =
. Therefore, the range is
2
2
g
vix
R=
2viy vix
g
=
2vi 2 sin θ cos θ vi 2 sin 2θ
=
.
g
g
(c) Strategy and Solution The maximum value of sin 2θ occurs when 2θ = 90° or θ = 45° .
Therefore, Rmax =
vi 2 sin 90°
g
=
vi 2
g
.
58. Strategy Use the expression for the range.
Solution
(a) The maximum value of sin 2θ occurs when 2θ = 90°.
Therefore, Rmax =
vi 2 sin 90° vi 2 (1)
v2
=
= i .
g
g
g
(b) The maximum value of sin 2θ occurs when 2θ = 90° or θ = 45° .
59. (a) Strategy At a projectile’s highest point, the vertical component of its velocity is zero.
Solution Using Eqs. (3-14), we have vx = vi cos θ and v y = 0 .
(b) Strategy Use Eq. (3-10).
Solution v y = viy − g ∆t = vi sin θ − g ∆t = 0, so ∆t = vi sin θ g .
233
Chapter 3: Motion in a Plane
Physics
(c) Strategy Use Eq. (3-12) and the result from part (b).
Solution Find H.
1
1
∆y = H = viy ∆t + a y (∆t ) 2 = viy ∆t − g (∆t )2 , so
2
2
2
vi 2 sin 2 θ vi 2 sin 2 θ (vi sin θ )2
⎛ v sin θ ⎞ 1 ⎛ vi sin θ ⎞
−
=
H = vi sin θ ⎜ i
.
⎟− g⎜
⎟ =
g
2g
2g
⎝ g ⎠ 2 ⎝ g ⎠
60. (a) Strategy At the maximum height, vfy = 0.
Solution Use Eq. (3-10) to find the time it takes the ball to reach its maximum height.
v sin θ
∆v y = vfy − viy = 0 − vi sin θ = − g ∆t , so ∆t = i
.
g
Use Eq. (3-12) to find how much the ball rises.
viy
60.0°
vix
2
vi 2 sin 2 θ vi 2 sin 2 θ
⎛ v sin θ ⎞ 1 ⎛ vi sin θ ⎞
1
∆y = (vi sin θ )∆t − g (∆t )2 = vi sin θ ⎜ i
−
⎟− g⎜
⎟ =
2
g
2g
⎝ g ⎠ 2 ⎝ g ⎠
2
2
v sin θ (22.0 m s)2 sin 2 60.0°
= i
=
= 18.5 m higher than where it was hit
2g
2(9.80 m s 2 )
(b) Strategy The elapsed time is twice that found in part (a).
Solution
∆t = 2vi sin θ g = 2 ( 22.0 m s ) sin 60.0° (9.80 m s 2 ) = 3.89 s
(c) Strategy Use Eq. (3-5).
Solution Find the horizontal displacement of the ball.
∆x = vix ∆t = (vi cos θ )∆t = ( 22.0 m s ) cos 60.0°(3.89 s) = 42.8 m
61. Strategy The skater must be up the ramp far enough for their speed at the end of the horizontal section to be just
great enough so that the skater travels a horizontal distance of 7.00 m while falling 3.00 m. Draw a diagram of the
skater on the ramp to find the acceleration of the skater caused by the force of gravity.
Solution According to Newton’s second law, ∑ F = mg sin15.0° = ma, so
a = g sin15.0° along the surface of the ramp. Use Eq. (2-16) to relate the distance
up the ramp to the speed of the skater at the end of the ramp.
d
m
15.0°
mg
15.0°
mg cos 15.0°
mg sin 15.0°
vf 2 − vi 2 = vf 2 − 0 = 2a∆x = 2( g sin15.0°)d , so d = vf 2 (2 g sin15.0°).
Since the ramp is frictionless, the velocity of the skater at the end of the horizontal part of the ramp is in the xdirection with magnitude equal to vf . So, the components of the displacement are ∆x = vf ∆t (1) and
1
1
1
∆y = viy ∆t + a y (∆t ) 2 = 0 − g (∆t ) 2 = − g (∆t )2 (2). Solving for ∆t in (1) and substituting into (2) gives
2
2
2
2
1 ⎛ ∆x ⎞
g (∆x)2
∆y = − g ⎜⎜ ⎟⎟ , or vf 2 = −
.
2 ⎝ vf ⎠
2∆y
Substitute this result into the equation for d.
d=
vf 2
2 g sin15.0°
=
− g (∆x)2 (2∆y )
( ∆x ) 2
(7.00 m) 2
=−
=−
= 15.8 m
2 g sin15.0°
4∆y sin15.0°
4(−3.00 m) sin15.0°
234
Physics
Chapter 3: Motion in a Plane
62. Strategy In each case, use Eq. (3-12) to find the time it takes for the stone to reach the base of the gorge.
Solution
1
1
1
2∆y
2(0 − 60.0 m)
= −
= 3.49 s .
(a) ∆y = viy ∆t + a y (∆t ) 2 = 0 − g (∆t ) 2 = − g (∆t ) 2 , so ∆t = −
2
2
2
g
9.83 m s 2
1
1
(b) ∆y = viy ∆t + a y (∆t )2 = viy ∆t − g (∆t ) 2 , so
2
2
1
1
0 = g (∆t ) 2 − viy ∆t + ∆y = (9.83 m s 2 )(∆t )2 − ( −20.0 m s ) ∆t − 60.0 m.
2
2
Solve for ∆t using the quadratic formula.
∆t =
−20.0 m s ± (20.0 m s) 2 − 2(9.83 m s 2 )(−60.0 m)
9.83 m s 2
= 2.01 s or − 6.08 s
Since ∆t > 0, ∆t = 2.01 s .
1
1
g (∆t ) 2 , so 0 = g (∆t ) 2 − (vi sin θ )∆t + ∆y.
2
2
Solve for ∆t using the quadratic formula.
(c) ∆y = (vi sin θ )∆t −
∆t =
=
vi sin θ ± vi 2 sin 2 θ − 4
vi = 20.0 m/s
vi sin 30.0°
( 12 g ) ∆y
30.0°
vi cos 30.0°
g
( 20.0
m s ) sin 30.0° ±
( 20.0
2
m s ) sin 2 30.0° − 2(9.83 m s 2 )(−60.0 m)
9.83 m s 2
= 4.66 s or − 2.62 s
Since ∆t > 0, ∆t = 4.66 s.
Find the horizontal distance.
x = (vi cos θ )∆t = ( 20.0 m s ) cos 30.0°(4.656 s) = 80.6 m
63. Strategy The circus performer is moving horizontally as he clears the net, so at that moment, his vertical
component of velocity is zero. His horizontal component of velocity is constant. Use Eqs. (3-10) and (3-11).
Solution Find the time that it takes the performer to reach the net.
∆x
∆x = vx ∆t = vi cos θ∆t , so ∆t =
.
vi cos θ
Find the muzzle speed of the cannon.
vi sin 40°
vi
40°
vi cos 40°
g ∆x
g ∆x
(9.80 m s 2 )(6.0 m)
vfy = 0 = viy − g ∆t = vi sin θ −
, so vi =
=
= 11 m s .
vi cos θ
sin θ cos θ
sin 40° cos 40°
Find the height of the net.
1
1
1
∆x
1
1
∆y = h = (vfy + viy )∆t = vi sin θ∆t = vi sin θ
= tan θ∆x = tan 40°(6.0 m) = 2.5 m
2
2
2
vi cos θ 2
2
235
Chapter 3: Motion in a Plane
Physics
64. Strategy vfy = 0 at the maximum height. Use Eqs. (3-10) and (3-12).
Solution
vfy = vi sin θ − g ∆t = vi sin 45° − g ∆t =
vi
2
− g ∆t = 0. Thus, ∆t =
vi
g 2
.
Find H max , the maximum height of the projectile’s trajectory.
2
v ⎛ v ⎞ 1 ⎛ v ⎞
v2 v2 v2
1
g (∆t ) 2 = i ⎜⎜ i ⎟⎟ − g ⎜⎜ i ⎟⎟ = i − i = i
2
2g 4g 4g
2⎝g 2⎠ 2 ⎝g 2⎠
Find R, the range of the projectile.
1
∆y = (vi sin θ )∆t − g (∆t )2 and ∆x = (vi cos θ )∆t.
2
When the projectile returns to its original height, ∆y = 0.
2v sin θ
1
1
.
∆y = (vi sin θ )∆t − g (∆t )2 = vi sin θ − g ∆t = 0, so ∆t = i
2
2
g
H max = (vi sin θ )∆t −
Substitute this value for ∆t into R = ∆x = (vi cos θ )∆t.
⎛ 2v sin θ
R = (vi cos θ )∆t = vi cos θ ⎜ i
g
⎝
Substitute for θ .
R=
2vi 2 sin 45° cos 45°
g
Therefore, H max =
=
1 ⎛ vi 2
⎜
4 ⎜⎝ g
⎞ 2vi 2 sin θ cos θ
⎟=
g
⎠
vi 2
g
⎞ R
⎟= .
⎟ 4
⎠
65. Strategy Consider the relative motion of the two vehicles.
Solution Let north be in the +x-direction.
vJRx = the velocity of the Jeep relative to the road = 82 km h
vRFx = the velocity of the road relative to the Ford = −vFRx = 48 km h
vJFx = the velocity of the Jeep relative to the (observer in the) Ford = vJRx + vRFx = 82 km h + 48 km h
= 130 km h
G
So, v JFx = 130 km h north .
G
66. Strategy Consider the relative motion of the two vehicles. Find v BV = the velocity of the BMW relative to the
VW.
Solution Let north be in the +x-direction.
vBRx = the velocity of the BMW relative to the road = 100.0 km h
vRVx = the velocity of the road relative to the VW = −vVR = 42 km h
vBVx = vBRx + vRVx = 100.0 km h + 42 km h
Find ∆t.
∆x
10.0 km
⎛ 3600 s ⎞
∆t =
=
⎜
⎟ = 254 s
vBVx 100.0 km h + 42 km h ⎝ 1 h ⎠
236
Physics
Chapter 3: Motion in a Plane
67. Strategy Consider the relative motion of the two vehicles. Draw a diagram and use the component method.
Solution Let north be in the +y-direction and east the +x-direction.
G
v tc = the velocity of the truck relative to the car
G
v tr = the velocity of the truck relative to the road
G
v cr = the velocity of the car relative to the road
Find the velocity of the truck relative to the car.
G
G
G
G
G
G
G
v tc = v tr + v rc = v tr + (− v cr ) = v tr − v cr
N
35°
110 km/h
85 km/h
vtcx = vtrx − vcrx = vtr cos125° − 0 = (85 km h) cos125°
vtcy = vtry − vcry = vtr sin125° − vcr = (85 km h) sin125° − 110 km h
G
v tc =
[(85
θ = tan −1
km h) cos125°] + [ (85 km h) sin125° − 110 km h ] = 63 km h
2
2
(85 km h) sin125° − 110 km h
= 40° south of west
(85 km h) cos125°
So, the relative velocity is 63 km h at 40° south of west .
68. Strategy Consider the relative motion of the ship and the water.
Solution The relative speeds are:
vupstream = vship − vwater = vup = vs − vw
vdownstream = vship + vwater = vd = vs + vw
Find the speed of the current, vw .
∆x
∆x
∆x = vup ∆tup = (vs − vw )∆tup , so
= vs − vw (1). ∆x = vd ∆td = (vs + vw )∆td , so
= vs + vw (2).
∆tup
∆td
Subtract (1) from (2).
∆x
∆x
∆x ⎛ 1
1 ⎞ 208 km ⎛ 1
1 ⎞
⎜
⎟=
−
= 2vw , so vw =
−
−
⎜
⎟ = 0.42 km h .
2 ⎜ ∆td ∆tup ⎟
2 ⎝ 19.2 h 20.8 h ⎠
∆td ∆tup
⎝
⎠
69. Strategy The minimum air velocity is in the same direction as the airplane’s.
Solution
210 m s east − 160 m s east = 50 m s east
70. (a) Strategy The ground speed of the small plane will be the magnitude of the vector sum of the two velocities.
Draw a diagram and use the component method.
Solution Let north be in the +y-direction and east the +x-direction.
G
2
2
v = vx 2 + v y 2 = [ −30.0 m s + (10.0 m s) cos 210°] + [ (10.0 m s) sin 210°]
= 39.0 m s
N
30.0 m/s
30° 10.0 m/s
(b) Strategy The new directional heading relative to the ground is in the direction of the velocity relative to the
ground.
Solution
θ = tan −1
(10.0 m s) sin 210°
= 7.4° south of west
−30.0 m s + (10.0 m s) cos 210°
237
Chapter 3: Motion in a Plane
Physics
71. (a) Strategy To compensate for the wind, the plane’s new heading will be north of west, and the north
component of the plane’s velocity relative to the air must be equal in magnitude to that of the south
component of the velocity of the wind. Draw a diagram and use the component method.
Solution The north component of the velocity relative to the air is
equal to (30.0 m s) sin θ . The south component of the wind is
equal to (10.0 m s) sin 30°. We set these equal and solve for θ .
(30.0 m s) sin θ = (10.0 m s) sin 30°, so
(10.0 m s) sin 30°
θ = sin −1
= 9.6° north of west .
30.0 m s
N
θ
30.0 m/s
30° 10.0 m/s
(b) Strategy The new ground speed is equal to the sum of the west components of the two velocities in part (a).
Solution (30.0 m s) cos 9.6° + (10.0 m s) cos 30° = 38 m s
72. Strategy The upstream component of the velocity of the boat must be equal in magnitude to that of the current.
Solution Let +y-direction be toward the opposite shore.
1.8
= 27°.
(4.0 km h) sin θ = 1.8 km h , so θ = sin −1
4.0
The direction of the velocity of the boat relative to the water is 27° upstream.
Upstream
y
4.0 km/h
θ
Opposite
shore
1.8 km/h
73. Strategy Consider the relative motion of the two vehicles. Use the component method.
Solution Let the +y-direction be north and the +x-direction be east.
G
v ps = the velocity of the Pierce Arrow relative to the Stanley Steamer
G
v pg = the velocity of the Pierce Arrow relative to the ground
G
vsg = the velocity of the Stanley Steamer relative to the ground
Compute the components of the velocity of the Pierce Arrow relative to the observer riding
in the Stanley Steamer.
vpsx = vpgx + vgsx = 50 km h + 0, so vx = 50 km h east .
vpsy = vpgy + vgsy = vpgy − vsgy = 0 − 40 km h and − 40 km h north = 40 km h south, so
v y = 40 km h south .
238
y
x
40 km/h
50 km/h
Physics
Chapter 3: Motion in a Plane
74. Strategy Consider the relative motion of the water (w) and Sheena (s). Let the +y-direction be upstream and the
+x-direction be toward the opposite bank (b).
Solution
(a) Find the x-component.
vx = (3.00 mi h) cos 60.0° = 1.50 mi h
3.00 mi/h
y
The y-component is v y = vSby = vSwy + vwby .
60.0°
v y = (3.00 mi h) sin 60.0° − 1.60 mi h = 1.00 mi h
1.60 mi/h
x
Use the Pythagorean theorem.
vSb = (1.50 mi h) 2 + (1.00 mi h)2 = 1.80 mi h
(b) ∆t =
∆x
1.20 mi
⎛ 60 min ⎞
=
= (0.800 h) ⎜
⎟ = 48.0 min
vx 1.50 mi h
⎝ h ⎠
(c) ∆y = v y ∆t = (1.00 mi h)(0.800 h) = 0.800 mi upstream
(d) The upstream component of her velocity relative to the water must be equal in magnitude to the velocity of
the current relative to the bank, or v y = 0.
(3.00 mi h) sin θ − 1.60 mi h = 0, so θ = sin −1
1.60
= 32.2° upstream .
3.00
75. Strategy Consider the relative motion of the dolphin with respect to the bay and the uniform water current. Draw
a diagram.
Solution Let the +x-direction be west and the +y-direction be north.
d = dolphin
b = bay
w = water current
G
G
G
v db = v dw + v wb ; calculate the components.
v
vdbx = vdwx + vwbx = vdw cos θ − vwb cos 45° = vdw cos θ − wb
2
vwb
vdby = vdwy + vwby = vdw sin θ − vwb sin 45° = vdw sin θ −
2
N
Home
Bay
θ
(a) Set vdby = 0 to find θ .
v
v
2.83 m s
vdw sin θ − wb = 0, so θ = sin −1 wb = sin −1
= 30.0° N of W .
vdw 2
2
(4.00 m s) 2
(b) ∆t =
∆x
0.80 × 103 m
=
vdbx (4.00 m s) cos 30.0° − 2.83 m
2
s
⎛ 1 min ⎞
⎜
⎟ = 9.1 min
⎝ 60 s ⎠
239
4.00 m/s
45°
2.83 m/s
Chapter 3: Motion in a Plane
Physics
76. Strategy and Solution In the figure below we see two different reference frames, labeled O and O′, that are at
rest with respect to each other. The points labeled ‘start’ and ‘finish’ represent the beginning and ending points of
an object’s path, such as a bee flying from one point to another. The bee’s displacement is denoted by the vector
G
G G G
G G G
∆r. It is evident from the figure that in frame O, ∆r = rf − ri , while in frame O′, ∆r = rf′ − ri′. Thus, the
displacement is the same in the two different reference frames.
y′
y
Finish
r′f
rf
∆r
ri
x′
Start
r′i O′
x
O
77. (a) Strategy Consider the relative motion of the boy and the water.
Solution
d
d
∆t = across and vwater = downstream , so
vboy
∆t
vwater =
ddownstream = 50.0 m
0.500 m/s
ddownstream ddownstream
50.0 m
=
vboy =
(0.500 m s) = 1.00 m s
dacross vboy
dacross
25.0 m
(b) Strategy Use the Pythagorean theorem.
vwater
dacross= 25.0 m
Solution Find the speed of the boy relative to the friend.
vbf = (0.500 m s) 2 + (1.00 m s)2 = 1.12 m s
78. (a) Strategy The east-west components of the airplane’s and the wind’s velocities must be equal in magnitude
for the plane to travel north.
Solution Let the +y-direction be north and the +x-direction be east.
vpx = vp cos θ = vair, x = vair cos 45.00°, so
vpg
v
100.0 km h
θ = cos −1 air = cos −1
= 76.37° north of east .
(300.0 km h) 2
vp 2
100.0 km/h
Solution Find the time.
∆y
∆y
∆t =
=
v y vp sin θ − vair sin 45.00°
600.0 km
⎡
100.0 km h ⎤
− (100.0 km h) sin 45.00°
(300.0 km h) sin ⎢cos −1
300.0
km h ) 2 ⎥⎦
(
⎣
240
θ
45°
(b) Strategy The northern or y-component of the plane’s velocity relative to the ground
is the y-component of its velocity relative to the air minus the y-component of the
air’s velocity relative to the ground.
=
300.0 km/h
N
= 2.717 h
Physics
Chapter 3: Motion in a Plane
79. (a) Strategy Draw a diagram. Determine how long it takes for the ball to reach the wall. Then, use Eq. (3-12) to
find the height of the ball.
Solution Find the time it takes for the ball to reach the wall.
∆x
.
∆x = vx ∆t = (vi cos θ )∆t , so ∆t =
vi cos θ
Find the change in height of the ball.
80° 20 m/s
60 cm
yf
θ = 10°
10 m
1
1
∆x
1
( ∆x ) 2
∆y = yf − yi = viy ∆t + a y (∆t ) 2 = (vi sin θ )∆t − g (∆t )2 = (vi sin θ )
− g
2
2
vi cos θ 2 vi 2 cos 2 θ
= ∆x tan θ −
g (∆x) 2
2vi 2 cos 2 θ
The height is yf = yi + ∆x tan θ −
g (∆x)2
2vi 2 cos 2 θ
= 0.60 m + (10 m) tan10° −
(9.80 m s 2 )(10 m) 2
2(20 m s)2 cos 2 10°
= 1.1 m .
(b) Strategy Determine the sign of the y-component of the velocity to find if the ball is going up or down.
Solution
vfy − viy = vfy − vi sin θ = a y ∆t = − g ∆t = − g
∆x
, so
vi cos θ
g ∆x
(9.80 m s 2 )(10 m)
vfy = vi sin θ −
= (20 m s) sin10° −
= −1.5 m s.
vi cos θ
(20 m s) cos10°
Since vfy < 0, the ball is on its way down.
80. (a) Strategy Compute the displacements. Use the definition of average velocity. East is the positive x-direction.
Solution The first displacement is (80.0 km h)(45.0 min)(1 h 60 min)
= 60.0 km east, and the second is (60.0 km h)(30.0 min)(1 h 60 min)
= 30.0 km at 38.0° north of east. Find the average velocity.
G
∆r = (∆rx )2 + (∆ry ) 2
=
[60.0 km + (30.0 km) cos 38.0°]2 + [(30.0 km) sin 38.0°]2
θ = tan −1
N
30.0 km
38.0°
60.0 km
= 85.7 km
(30.0 km) sin 38.0°
= 12.5° north of east
60.0 km + (30.0 km) cos 38.0°
G
∆r
85.66 km
G
⎛ 60 min ⎞
=
v av =
⎜
⎟ at 12.5° north of east
∆t 45.0 min + 30.0 min ⎝ 1 h ⎠
= 68.5 km h at 12.5° north of east
(b) Strategy and Solution The return trip is the exact opposite journey, so the average velocity has the same
magnitude but the opposite direction. The average velocity is 68.5 km h at 12.5° south of west .
241
Chapter 3: Motion in a Plane
Physics
81. (a) Strategy Draw a diagram and use vector addition.
Solution Find the magnitude of the displacement.
G
∆r = [600.0 km + (300.0 km) cos(−30.0°)]2 + [(300.0 km) sin(−30.0°)]2
= 873 km
N
600.0 km
Start
θ
30.0°
∆r
300.0 km
G
(b) Strategy Refer to the diagram in part (a). Find the angle between the initial displacement vector and ∆r.
Solution Find the direction of the displacement.
(300.0 km) sin(−30.0°)
= 9.90° south of east
θ = tan −1
600.0 km + (300.0 km) cos(−30.0°)
(c) Strategy The flight time is given by the quotient of the distance traveled and the speed of the jetliner.
Solution
d 600.0 km + 300.0 km
∆t = =
= 2.250 h
v
400.0 km h
(d) Strategy The direct flight time is given by the quotient of the magnitude of the displacement and the speed
of the jetliner.
Solution
G
∆r
873 km
∆t =
=
= 2.18 h
v
400.0 km h
82. (a) Strategy Find the time it takes the coconut to strike the ground using Eq. (3-12).
Solution Solve for ∆t. The initial velocity in the vertical direction is zero.
1
1
∆y = viy ∆t + a y (∆t )2 = (0)∆t − g (∆t ) 2 , so
2
2
2∆y
2(−100 m)
∆t = −
= −
= 4.5 s .
g
9.80 m s 2
18 m/s
100 m
∆x
(b) Strategy Use the result of part (a) and ∆x = vx ∆t to find the horizontal distance the coconut travels, ∆x.
Solution The horizontal distance the coconut travels from the release point is
∆x = vx ∆t = (18 m s)(4.5 s) = 81 m .
242
Physics
Chapter 3: Motion in a Plane
83. (a) Strategy and Solution The pilot flew assuming there was no wind, so the velocity of the plane relative to the
air was 160 km h at 20° N of E .
(b) Strategy Let the +x-direction be east and the +y-direction be north. Find the components of the
displacement.
Solution
x = (320 km) cos 20.0° − 20 km
y = (320 km) sin 20.0°
Find the magnitude of the velocity.
x2 + y 2
[(320 km)cos20.0° − 20 km]2 + (320 km)2 sin 2 20.0°
=
= 150 km h
2.0 h
∆t
Find the direction of the velocity.
y
(320 km) sin 20.0°
= 21° N of E
θ = tan −1 = tan −1
x
(320 km) cos 20.0° − 20 km
G
So, v = 150 km h at 21° N of E .
v=
(c) Strategy Since the airplane’s component of velocity in the y-direction was unaffected, and it’s x-component
was reduced, the wind’s velocity must be from the east (west).
Solution Compute the wind speed.
20 km
G
vwind =
= 10 km h , so v wind = 10 km h west .
2.0 h
84. Strategy Use ∆y = (vi sin θ )∆t −
1
g (∆t ) 2 and ∆x = (vi cos θ )∆t for the change in the projectile’s position.
2
Solution When a projectile returns to its original height, ∆y = 0.
0 = (vi sin θ )∆t −
2v sin θ
1
1
1
g (∆t ) 2 = vi sin θ − g ∆t , so g ∆t = vi sin θ or ∆t = i
.
2
2
2
g
Substitute this value for ∆t into R = ∆x = (vi cos θ )∆t.
⎛ 2v sin θ ⎞ 2vi 2 sin θ cos θ
R = (vi cos θ )∆t = (vi cos θ ) ⎜ i
⎟=
g
g
⎝
⎠
Substitute the given values.
2(1.46 × 103 m s)2 sin 55° cos 55°
R=
= 200 km
9.80 m s 2
243
1.46 km/s
55°
Big Bertha
Chapter 3: Motion in a Plane
Physics
85. Strategy The projectile must be displaced 75.0 m vertically in the same amount of time that it travels 350 m
horizontally. The projectile may hit the headquarters on its way up, on its way down, or at its maximum height.
Use ∆x = vx ∆t and Eq. (3-12).
Solution Solve for the initial speed, vi .
∆x = vx ∆t , so ∆t =
∆x
∆x
=
.
vx vi cos θ
1
1
∆x
1 ⎛ ∆x
∆y = viy ∆t + a y (∆t ) 2 = (vi sin θ )∆t − g (∆t )2 = (vi sin θ )
− g ⎜⎜
2
2
vi cos θ 2 ⎝ vi cos θ
vi =
g (∆x)2
2(∆x tan θ − ∆y ) cos 2 θ
=
(9.80 m s 2 )(350 m)2
2[(350 m) tan 40.0° − 75.0 m]cos 2 40.0°
2
⎞
g (∆x)2
, so
⎟⎟ = ∆x tan θ − 2
2vi cos 2 θ
⎠
= 68 m s .
86. (a) Strategy To return home, the pilot must travel opposite his previous day’s displacement.
Solution Use the component method to find the displacement. Let north be +y and
east be +x.
∆x = −55 mi + (25 mi) cos 285° = −49 mi
∆y = (25 mi) sin 285° = −24 mi
∆r = (∆x)2 + (∆y )2 = (−48.5 mi) 2 + (−24.1 mi)2 = 54 mi
−24.1
= 26° south of west
θ = tan −1
−48.5
So, to return to his original destination, the pilot must travel
54 mi at 26° north of east .
(b) Strategy Add each distance traveled.
Solution The pilot has flown 55 mi + 25 mi + 54 mi = 134 mi extra.
87. Strategy Use Eqs. (2-15) and (3-12). Let +x be east and +y be north.
G
Solution Find the components of the position vector r.
1
1
xf = xi + vix ∆t + a x (∆t )2 = 2.0 m + (0)(2.0 s) + (5.0 m s 2 )(2.0 s) 2 = 12 m
2
2
and
1
1
yf = yi + viy ∆t + a y (∆t )2 = 0 + (20 m s)(2.0 s) + (0)(2.0 s) 2 = 40 m.
2
2
G
So, r = 12 m east and 40 m north .
244
y
55 mi
15°
25 mi
x
N
Physics
Chapter 3: Motion in a Plane
88. Strategy Assume that the outfielder catches the ball at the same height at which it was hit, and that he begins
running at the same time that it was hit. θ = 45° for the maximum range.
Solution Find vi . At the maximum height H, v y = 0. Use Eq. (3-13).
vfy 2 − viy 2 = 0 − viy 2 = 2a y ∆y = −2 gH , so viy 2 = vi 2 sin 2 45° =
vi 2
= 2 gH , or vi = 2 gH .
2
Find the elapsed time using Eq. (3-12).
1
1
1
1
∆y = 0 = viy ∆t + a y (∆t )2 = viy ∆t − g (∆t )2 = (vi sin θ )∆t − g (∆t ) 2 = vi sin θ − g ∆t , so
2
2
2
2
2v sin θ 2(2 gH ) sin 45°
2H
.
∆t = i
=
=2
g
g
g
⎛
2(44 m)
Thus, the farthest distance is d = v∆t = (7.6 m s) ⎜ 2
⎜ 9.80 m s 2
⎝
⎞
⎟ = 46 m .
⎟
⎠
89. Strategy Use the results from Problems 57 and 59.
Solution
(a) At the maximum height, vfy = 0 = vi sin θ − g ∆t , so ∆t = vi sin θ g .
Also, ∆x = (vi cos θ )∆t , so ∆t =
Equate the expressions for ∆t.
∆x
. (∆x is half of 0.800 m.)
vi cos θ
v 2 sin 2 θ
vi sin θ
∆x
g ∆x
=
, so vi 2 =
. From Problem 50, H = i
, so
g
vi cos θ
sin θ cos θ
2g
H=
g ∆x sin 2 θ
∆x
0.400 m
=
tan θ =
tan 55.0° = 28.6 cm .
2 g sin θ cos θ
2
2
(b) Since H ∝ tan θ , and since tan θ increases if θ increases (0 ≤ θ ≤ 90), the maximum height would be
smaller (45.0° < 55.0°).
(c) The range would be larger, since the range is maximized for θ = 45°.
(d) Calculate vi 2 .
(9.80 m s 2 )(0.400 m)
= 8.34 m 2 s 2
sin 55.0° cos 55.0°
Calculate the maximum height and range for 45.0°.
(8.34 m 2 s 2 ) sin 2 45.0°
H=
= 21.3 cm
2(9.80 m s 2 )
vi 2 =
From Problem 48, R =
vi 2 sin 2θ
(8.34 m 2 s 2 ) sin 90.0°
= 85.1 cm .
, so R =
g
9.80 m s 2
245
Chapter 3: Motion in a Plane
Physics
90. Strategy Use Eq. (3-12) to find the time of flight for the package. The distance separating the package and the
helicopter is given by ∆x = vx ∆t , where vx is the speed of the package relative to the helicopter, 12 m s.
Solution Find the time the package takes to reach the ground.
1
1
1
yf = yi + viy ∆t + a y (∆t ) 2 = yi + (0)∆t − g (∆t ) 2 = 0, so g (∆t )2 = yi , or ∆t =
2
2
2
Compute the horizontal distance.
2 yi
2(18 m)
∆x = vx ∆t = vx
= (12 m s )
= 23 m
g
9.80 m s 2
2 yi
g
.
91. (a) Strategy The time required for the round-trip is equal to the round-trip distance divided by the cruising
speed.
Solution
round-trip distance 2(5.80 × 103 km)
∆t =
=
= 33.1 h
cruising speed
350.0 km h
(b) Strategy Consider the relative motion of the plane, air, and the ground.
Solution Subscripts: airplane = p; air = a; ground = g.
∆t = ∆ttailwind + ∆theadwind
=
d
d
5.80 × 103 km
5.80 × 103 km
+
=
+
= 34.1 h
vpa + vag vpa − vag 350.0 km h + 60.0 km h 350.0 km h − 60.0 km h
(c) Strategy Let +y be antiparallel to the crosswind. Set vpgy = 0, so the plane can travel in a straight line.
vpax = vpa cos θ is the speed of the plane along the straight line between the cities.
Solution
y
vpgy = vpay + vagy = vpa sin θ − vag = 0, so θ = sin
Thus, ∆t =
v
−1 ag
3
vpa
= sin
2 ∆x
2(5.80 × 10 km)
=
= 33.6 h .
vpa cos θ ( 350.0 km h ) cos 9.87°
246
−1
60.0
= 9.87°.
350.0
vpa
vpay
x
θ
New York
vag
vpax
Paris
Physics
Chapter 3: Motion in a Plane
92. (a) Strategy Use Eqs. (3-5) and (3-12).
Solution ∆x = vix ∆t = vi ∆t , so ∆t = ∆x vi . Relate the distance to the initial height.
6.00 m/s
2
1
1
1 ⎛ ∆x ⎞
yf = yi + viy ∆t + a y (∆t )2 = yi + (0)∆t − g (∆t )2 = yi − g ⎜⎜ ⎟⎟ = 0, so
2
2
2 ⎝ vi ⎠
∆x = vi
2 yi
g
= ( 6.00 m s )
2(8.00 m)
9.80 m s 2
8.00 m
= 7.67 m .
∆x
(b) Strategy Use Eq. (3-10) and the Pythagorean theorem.
Solution ∆v y = v y − 0 = − g ∆t = − g ∆x vi and vx = vi . Calculate the speed.
v = vx 2 + v y 2 = vi 2 +
g 2 (∆x)2
vi 2
=
( 6.00
2
m s) +
(9.80 m s 2 )2 (7.67 m)2
(6.00 m s) 2
= 13.9 m s
(c) Strategy vx relative to the seagull is zero.
Solution
vy = g
∆x
7.67 m
= (9.80 m s 2 )
= 12.5 m s
vi
6.00 m s
93. Strategy Use Eqs. (3-5), (2-12), and (3-12).
Solution Find the time of flight in terms of h and vi .
∆x = vi ∆t = h, so ∆t =
h
.
vi
h
vx
Find the time of flight in terms of vi and g.
h
vy
2
⎞
2v 2
h 1 ⎛ 2v 2 ⎞ 2v
gh 2
, so h = i and ∆t = = ⎜ i ⎟ = i .
⎟⎟ =
g
vi vi ⎜⎝ g ⎟⎠ g
2vi 2
⎠
G
Find the components of v.
vy
−2vi
⎛ 2v ⎞
= tan −1
= tan −1 (−2) = −63°,
vx = vi and v y = − g ∆t = − g ⎜ i ⎟ = −2vi , so θ = tan −1
vx
vi
⎝ g ⎠
θ
v
1
1 ⎛h
h = g (∆t )2 = g ⎜⎜
2
2 ⎝ vi
or 63° below the horizontal .
94. Strategy The cutter must move with the moving glass to cut perpendicularly to the direction of motion of the
conveyor belt. Thus, the cutter must be set at some angle with respect to the width of the belt and toward the
direction of motion of the belt. Draw a diagram.
Solution Let d be the distance that the sheet of glass travels in the time t that it takes
the cutter to cut it. Then, d = (15.0 cm s)t. Since the cutter moves across the width at
15.0 cm/s
θ
a speed of 24.0 cm s and the width is 72.0 cm, the time t is given by
t = (72.0 cm) (24.0 cm s) = 3.00 s. Solve for θ .
d
(15.0 cm s)t (15.0 cm s)(3.00 s)
=
=
= 0.625, so
72.0 cm
72.0 cm
72.0 cm
θ = tan −1 0.625 = 32.0° .
tan θ =
247
72.0 cm
d
Chapter 3: Motion in a Plane
Physics
95. (a) Strategy Use the definition of average velocity. Draw a diagram.
Solution Since the pilot has traveled 15 km due west in 1.0 h, the average
velocity of the wind must be 15 km h due west.
Oklahoma City
330 km
15 km
10.0°
N
200 km
Dallas
(b) Strategy The angle θ is the angle with respect to the vertical (north) in which the pilot should have headed
his plane to get directly to Oklahoma city from Dallas without being blown off course. Draw a vector
diagram. Use the Law of Sines and Eq. (3-15).
Solution Let the velocities of the plane with respect to the ground and to the air be
G
G
v pg and v pa , respectively. Let the velocity of the air with respect to the ground be
G
G
G
G
v ag . Then, we have v pg = v pa + v ag .
From the diagram, we see that θ = 10.0° − α . We need to find α to find θ . The
vectors form a triangle. We know two sides of the triangle, and the angles α and β
are opposite those sides. If we can find the angle β , we can use the Law of Sines to
G
G
find α . Since v pg is 10.0° from the vertical (west of north) and v ag is horizontal
15 km/h
β
200 km/h
y
θ
10.0°
(due west), we see that β = 90.0° − 10.0° = 80.0°. Find α .
sin α
sin β
sin 80.0°
15sin 80.0°
=
=
, so α = sin −1
= 4.2°.
15 km h 200 km h 200 km h
200
Thus, the direction the pilot should have headed his plane is
θ = 10.0° − 4.2° = 5.8° west of north.
α
N
vpg
x
96. (a) Strategy Find the time it takes the ball to reach the ground using Eq. (3-12).
Solution Solve for ∆t. The initial velocity in the vertical direction is zero.
1
1
2∆y
2(−20.0 m)
∆y = viy ∆t + a y (∆t ) 2 = (0)∆t − g (∆t ) 2 , so ∆t = −
= −
= 2.02 s .
2
2
g
9.80 m s 2
(b) Strategy and Solution It would still take 2.02 s for the ball to fall to the ground, since v y = 0 for both
cases.
248
Physics
Chapter 3: Motion in a Plane
(c) Strategy Use Eq. (3-12) and the quadratic formula to find the time.
Solution
∆y = (vi sin θ )∆t −
y
1
g ( ∆t ) 2
2
x
1
g (∆t )2 − (vi sin θ )∆t + ∆y
2
0 = (4.90 m s 2 )(∆t )2 + (20.0 m s) sin18°∆t − 20.0 m
Solve for ∆t using the quadratic formula.
0=
∆t =
vi sin 18°
−(20.0 m s) sin18° ± (20.0 m s) 2 sin 2 18° − 4(4.90 m s 2 )(−20.0 m)
2(4.90 m s 2 )
18°
vi
= 1.5 s or − 2.7 s
∆t > 0, so ∆t = 1.5 s .
1
97. Strategy Let +y be downward. vix = vx = vi and viy = 0, so ∆x = vi ∆t and ∆y = g (∆t )2 .
2
Let the step number n =
x
0.30 m
=
y
,
0.18 m
such that n = 0 to 1 represents step 1, n = 1 to 2 represents step 2, etc.
Solution Find the step the marble strikes first.
2(0.18 m)vi
v
v ∆t
g (∆t ) 2
=
n= i
= 1.2 i .
, so ∆t =
(0.30 m)g
g
0.30 m 2(0.18 m)
Therefore,
3.0 m/s
0.18 m
0.30 m
2
2
vi ⎛ vi ⎞
−1 vi = (4.0 m −1 ) ( 3.0 m s ) = 3.7.
⎜ 1.2 ⎟ = (4.0 m )
0.30 m ⎝
g⎠
g
9.80 m s 2
The value of n is between 3 and 4, so the marble first strikes step 4.
y
n=
98. Strategy Draw a diagram. Use the definition of average acceleration and the formula for arc length.
Solution Compute the travel time.
s r ∆θ (150 m)(36°× 2π 360°)
∆t = =
=
= 4.7 s
v
v
20.0 m s
Find the magnitude of the change in velocity.
y
N
x
150 m
∆v = (v cos θ − v)2 + (v sin θ − 0) 2 = (20.0 m s) (cos 36° − 1)2 + (sin 36°) 2 = 12.4 m s
Find the direction.
sin 36°
= 18° west of north
φ = tan −1
cos 36° − 1
Compute the average acceleration.
G
G
∆v 12.4 m s
=
at 18° west of south = 2.6 m s 2 at 18° west of north
aav =
∆t
4.7 s
249
v
Chapter 3: Motion in a Plane
Physics
99. Strategy Use ∆x = vx ∆t and Eq. (3-12).
Solution Find ∆t in terms of ∆x and vix .
∆x
∆x = vx ∆t = vix ∆t , so ∆t =
.
vix
Substitute the expression for ∆t into Eq. (3-12), where yi = 0, since the projectile is launched from the origin.
2
⎛ −g
⎛ viy ⎞
⎛ ∆x ⎞ 1 ⎛ ∆x ⎞
1
yf = yi + viy ∆t + a y (∆t )2 = 0 + viy ⎜⎜
⎟⎟ ∆x + ⎜
⎟⎟ − g ⎜⎜
⎟⎟ = ⎜⎜
⎜ 2v 2
2
⎝ vix ⎠ 2 ⎝ vix ⎠
⎝ vix ⎠
⎝ ix
⎞
⎟ ( ∆x ) 2
⎟
⎠
100. Strategy Consider the motion of the person relative to the escalator.
Solution
vw = the speed of the person walking on the stalled escalator
vr = the speed of the person riding on the escalator without walking
vwr = the speed of the person walking while riding = vw + vr and x = the distance traveled = vw tw = vr tr = vwr t ,
vw t w
x
x
94 s
so t =
=
=
=
= 39 s .
t
vwr vw + vr v + v w 1 + 94 s
w
w t
r
66 s
250
Chapter 4
FORCE AND NEWTON’S LAWS OF MOTION
Conceptual Questions
1. In an automobile accident the force due to the collision changes the motion of the car, but the driver and
passengers continue to move in accordance with Newton’s first law. Seat belts supply the force necessary to
change their motion and slow them down. Without seat belts people would collide with the steering wheel or
windshield, for example, and stand a greater risk of injury.
2. When the person strikes the rug with the carpet beater, the rug begins to move forward. The carpet frame supplies
the force to overcome the inertia of the rug and hold it in place, while the inertia of the dust causes it to continue
moving forward. Similarly, when someone throws a baseball the inertia of the ball causes it to continue moving
after it has left the person’s hand.
3. There are a number of forces still acting on you, such as the normal force from the ground pushing up and the
force of gravity pulling down. The forces largely cancel each other though, so the net force acting on you is
essentially zero.
4. When the dog shakes, his wet fur changes velocity back and forth. The water will only remain on the fur if it has
the same velocity as the fur, so there must be a sufficiently large force holding it on as the dog shakes. When the
force is not sufficient, drops of water lose contact with the fur and experience no more force from the dog’s
motion. From the principle of inertia (Newton’s first law) these drops resist changes in velocity, so they fly off the
dog’s body with whatever velocity they had when they lost contact. The drops in the air will then fall to the
ground due to the force of gravity. As the dog continues shaking himself, more drops are shaken loose.
5. When the handle hits the board and stops abruptly, Newton’s first law says that the steel head will continue to
move for a short distance, resisting changes in velocity, until the force of friction between the head and the handle
has brought it to rest. It will have then moved down some to where the handle is a little wider, resulting in a
tightening of the head onto the handle.
6. The road pushes on the tires causing the car to move forward. The engine facilitates this process by rotating the
wheels so they push backwards on the road. In accordance with Newton’s third law the road exerts an equal and
opposite force on the tires.
7. Because of the principle of inertia, the cars continue moving until a sufficient force has caused them to stop. The
contact force between the cars at the moment of collision starts to slow them down. Before this force has stopped
the cars completely they will have moved a small distance, crumpling the front ends. The rear end of the car
continues to move while the front end is being crumpled, until it too comes to rest.
8. It is impossible to have zero net force on an object if the object is influenced exclusively by a single non-zero
force. Accordingly, because an object in free fall is influenced solely by gravity, it must have a non-zero net force.
For an object to be in equilibrium, the net external force acting on it must equal zero. Thus, an object in free fall
cannot be in a state of equilibrium.
9. (a) The reading of the scale is the magnitude of the normal force pushing up on you. This equals your weight as
long as the normal force and the force of gravity are the only forces acting on you, and you are at rest or
moving with a constant velocity.
251
Chapter 4: Force and Newton’s Laws of Motion
Physics
(b) If you were standing on the scale in a swimming pool for example, there would be a buoyant force from the
water pushing up on you, and the scale would read a smaller apparent weight. Also, the scale would not read
your weight if you were accelerating—for example standing on the scale in an elevator as it was moving
upward with increasing speed.
10. (a) False. Moving at constant speed, the engine must be pulling with a force equal to the force of friction, which
under ordinary conditions is much less than the train’s weight.
(b) False. By Newton’s third law, the engine’s pull on the first car and that car’s pull on the engine must always
be equal in magnitude and opposite in direction.
(c) False. Its inertia would cause it to keep coasting at a constant speed. The force of friction would cause the
train to slow down and eventually stop.
11. The weight of a person is the force of gravitational attraction on that person due to the Earth. This force is
inversely proportional to the square of the distance between the person and the center of the Earth.
(a) The rotation of the Earth causes a flattening of the planet such that the radius along the equator is greater than
the radius from pole to pole. The man would therefore weigh more at the North Pole where his distance to the
center of the Earth is less.
(b) The man would weigh more at the base of the mountain because, once again, this location is closer to the
center of the Earth, thus increasing the force of gravitational attraction.
G
12. The acceleration of an object thrown straight up into the air is equal to g at the top of its flight. This is
independent of whether air resistance is negligible, because the force of air resistance depends on how fast the
object moves through the air. At the highest point of the object’s motion, it has an instantaneous velocity of zero,
and therefore there is no force from air resistance.
13. The key is that the equal and opposite forces of Newton’s third law are acting on two different objects—one on
the wagon and the other on you. The wagon can therefore experience a non-zero net force, which causes it to
accelerate forward.
14. (a) Constant speed implies zero acceleration; thus, the scale must read the same.
(b) Increasing speed implies acceleration; thus, the scale must not read the same.
(c) Free fall implies acceleration; thus, the scale must not read the same.
15. The top string would be the first to break, since the tension it experiences is larger by an amount equal to the
weight of the ball it is holding up.
16. The forces are equal in magnitude, in accordance with Newton’s third law. The resulting changes in velocity will
not be equal though, because the masses are different.
17. Newton’s third law tells us that if the person on the raft walks away from the pier, the raft will in turn move
toward the pier. Thus, after walking the length of the raft, it should be possible for the person with the hook on the
pier to grab the raft and reel it in. Without the person on the pier to hold the raft, this technique would be of no
use, as the raft would move back away from the pier on the return walk.
18. It means that the cord and the pulley may be assumed to have no mass and the pulley exerts no opposing frictional
force.
252
Physics
Chapter 4: Force and Newton’s Laws of Motion
19. The terminal velocity of an object depends linearly on its mass and inversely on a parameter determined by its
size and shape. The mass of a feather and brick differ by several orders of magnitude. Therefore, if the difference
in the drag parameters of the two objects is much smaller than their mass differences, the brick must have a higher
terminal velocity. Because the brick is traveling at a higher velocity for the majority of its journey, it arrives at the
ground first. The density of the atmosphere on the surface of the Moon is much less than the density of the
atmosphere on Earth. Thus, the effects of drag are reduced on the Moon and both objects will hit the ground at
nearly the same instant—however, only in a perfect vacuum would the objects fall at exactly identical rates.
20. Since the force of air resistance always opposes an objects motion, the drag force on the baseball will be directed
downward on the way up and upward on the way down. There will be no drag force at the top of the flight, since
the velocity at the very top is zero. Thus the magnitude of the ball’s acceleration will be greater than g on the way
up, equal to g at the top of the flight, and less than g on the way down.
21. The apparent weight of the load is increased by an amount equal to the force required to accelerate it.
22. The only force acting is gravity.
23. If the initial velocity and net force are along the same line, the object moves along a straight line. It may reverse
direction along that line if the initial velocity and net force are in opposite directions. If the initial velocity and net
force are not along the same line, then the object moves along a curved path (a parabola).
24. Simple machines allow work to be done at a slower rate; thus, the power required is reduced.
25. Yes, as long as the y-axis is perpendicular to the chosen x-axis. This will often simplify a problem.
26. The tension is the same everywhere along the line.
27. Neglecting air resistance, the barrel (with you inside) will be free-falling, so the ball will hover apparently
motionless. With air resistance taken into account, the barrel will be somewhat slowed as it falls, so from within
the ball will be seen to fall a little faster.
Problems
1. Strategy Determine the forces not acting on the scale.
Solution The scale is in contact with the floor, so a contact force due to the floor is exerted on the scale. The
scale is in contact with the person’s feet, so a contact force due to the person’s feet is exerted on the scale. The
scale is in the proximity of a very large mass (Earth), so the weight of the scale is a force exerted on the scale. The
weight of the person is a force exerted on the person due to the very large mass, so it is not a force exerted on the
scale.
2. Strategy There are 0.2248 pounds per newton.
Solution Find the weight of the sack of flour in pounds.
0.2248 lb
19.8 N ×
= 4.45 lb
1N
3. Strategy There are 0.2248 pounds per newton.
Solution Find the weight of the astronaut in newtons.
1N
175 lb ×
= 778 N
0.2248 lb
253
Chapter 4: Force and Newton’s Laws of Motion
Physics
4. Strategy Consider the concept of contact on the atomic scale. Is the concept of contact force valid only for the
macroscopic scale?
Solution Does the concept of contact force apply to the atomic scale? No; the idea of contact breaks down at the
atomic scale. There is no way to define contact between atoms.
5. Strategy Use graph paper to draw a diagram.
Solution Find the vector sum of the vectors.
y
20
20 N
60.0
60.0
20
x
20 N
Because of symmetry, the y-components of the vectors cancel. The x-components look to be about 10 N, so the
vector sum is 10 N + 10 N = 20 N, or 20 N in the positive x -direction .
6. Strategy Represent 10 N as a length of 1 cm. Then, 30 N is represented by 3 cm and 40 N is represented by 4 cm.
Solution Use graph paper, ruler, and protractor to find the magnitude and direction of the vector sum of the two
forces.
30 N
40 N
Using the ruler, we find that the magnitude of the vector sum of the forces is about 70 N. Using the protractor, we
find that the direction of the vector sum of the forces is about 5° below the horizontal.
7. Strategy Graph the vectors and their sum. Use the scale of the graph to find the magnitude of the vector sum.
Solution The length of the vector sum is equal to one side of
a grid square, so the magnitude is 2 N. The vector points east,
so the vector sum of the forces is 2 N to the east.
N
A
A+B+C
254
B
C
Physics
Chapter 4: Force and Newton’s Laws of Motion
8. Strategy Graph the vectors and their sum. Use the scale of the graph to find the magnitude of the vector sum.
Solution The length of the vector sum is approximately equal to
seven sides of a grid square, so the magnitude is 14 N. The vector
points east, so the vector sum of the forces is 14 N to the east.
G
(Note that F and the vector sum overlap.)
N
D
E
F
D+E+F
9. Strategy Graph the vectors and their sum. Use the scale of the graph to find the magnitude of the vector sum.
Solution Use graph paper, ruler, and protractor to find the
magnitude and direction of the vector sum of the two forces.
The vector sum points due north. Each side of a grid square
represents 10 N, so the magnitude of the net force on the
sledge is about 120 N north.
62 N
N
62 N
10. Strategy Use Newton’s laws of motion. Let the y-direction be perpendicular to the canal and the +x-direction be
parallel to the center line in the direction of motion.
Solution Find the net force on the barge.
∑ Fy = T sin15° − T sin15° = 0 and
Centerline
∑ Fx = T cos15° + T cos15° = 2T cos15° = 2(560 N) cos15° = 1.1 kN.
G
So, Fnet = 1.1 kN forward (along the center line) .
560 N
15°
560 N
x
y
11. Strategy Draw a free-body diagram. Then, solve for the angle Red Riding Hood was pulling such that the net
force on the basket is straight up.
Solution Since we don’t know all quantities, the vectors in the diagram are not to scale.
For the net force to be straight up, the net force in the horizontal direction must be zero.
Only the forces due to Red and the wolf have components in the horizontal direction, so
these components must be equal in magnitude but opposite in direction. Set these
components equal and solve for the angle. Since the angle is measured from the vertical,
the horizontal component of the wolf’s force is the magnitude times the sine of the angle
(and similarly for Red’s force).
(12 N) sin θ = (6.4 N) sin 25°
(6.4 N) sin 25°
θ = sin −1
= 13° from the vertical
12 N
Fwolf
FRed
mg
12. Strategy Assume that the (now crashed) car has completely come to rest against the wall. There are two bodies
exerting forces on the car, the wall and the Earth. Let the subscripts be the following:
c = car
e = Earth
w = wall
Solution The free-body diagram is shown.
Fcw
Fce
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Chapter 4: Force and Newton’s Laws of Motion
Physics
13. Strategy Make a scale drawing to determine which net force is greater.
Solution The scale drawing is shown.
A
B
45°
4N
45°
2N
2N
45°
2N
4N
2N
The net force magnitude on object B is greater than that on object A because two of the forces acting
on B are directed at an angle greater than 45° with respect to the horizontal and contribute more to the
downward directed net force.
14. Strategy For each object, add the forces to find the net force.
Solution
(a) 10 N left + 40 N right = −10 N right + 40 N right = 30 N to the right
(b) The forces balance, so the net force is 0 .
(c) The horizontal forces balance, so the net force is due only to the downward force. The net force is
18 N downward.
15. Strategy Draw a free-body diagram and add the force to find the net force on the truck.
Solution The vertically directed forces balance, so the net
force is due to the difference in the east-west forces.
7 kN east + 5 kN west = 7 kN east − 5 kN east = 2 kN east
N
52 kN
5 kN
7 kN
52 kN
16. Strategy The force of the lake on the boat must be equal in magnitude and opposite in direction to the weight of
the boat. The force of the wind on the boat must be equal in magnitude and opposite in direction to that of the line.
Let the subscripts be the following:
s = sailboat e = Earth w = wind l = lake m = mooring line
Solution The free-body diagram is shown.
N
|Fsl| = 820 N
|Fsm| = 110 N
|Fsw| = 110 N
|Wse| = 820 N
256
Physics
Chapter 4: Force and Newton’s Laws of Motion
17. Strategy Since the hummingbird is hovering motionless, there is no net force on the hummingbird.
Solution The weight (downward force exerted by the Earth) is equal to the upward force exerted by the air,
0.30 N.
18. Strategy Since the suitcase is moving at a constant speed, the net force on it must be zero. The force of friction
must oppose the force of the pull. So, the force of friction must be equal in magnitude and opposite in direction to
the horizontal component of the force of the pull. Draw a free-body diagram to illustrate the situation.
Solution Find the force of friction.
N
The horizontal component of the pull force is (5.0 N) cos 60° = 2.5 N. Since the
horizontal component of the pull force is equal and opposite to the friction force, the
force of friction acting on the suitcase is 2.5 N, opposite the direction of motion.
5.0 N
60°
fk
mg
19. (a) Strategy Use Newton’s second law.
Solution Find the mass.
F
0.375 N
m= =
= 1.3 kg
a 0.30 m s 2
(b) Strategy Draw a diagram. Use the component method to find the new velocity.
Solution The wind gives the boat an additional velocity of
s 2 )(2.0
v = a∆t = (0.30 m
Find the new velocity.
G
v = vx 2 + v y 2
y
s) = 0.60 m s at 28° south of west.
x
N
0.33 m/s
28°
vwind
= [ − 0.33 m s + (0.60 m s) cos 208°] + [ (0.60 m s) sin 208°]
= 0.90 m s
vy
(0.60 m s) sin 208°
θ = tan −1 = tan −1
= 18° south of west
vx
− 0.33 m s + (0.60 m s) cos 208°
G
So, v = 0.90 m s at 18° south of west .
2
2
20. Strategy Since the man and mattress are neither moving upward nor downward, the net force must be zero in the
vertical direction.
Solution So that the net force is zero, the upward force of the water must be equal to the combined weight of the
man and the air mattress, or 806 N.
21. Strategy Use Newton’s second law for the vertical direction.
Solution Draw a free-body diagram. Find the mass of the potatoes.
T − mg 46.8 N − 39.2 N
∑ Fy = T − mg = ma y , so m =
=
= 4.0 kg .
ay
1.90 m s 2
y
T
a
mg
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Chapter 4: Force and Newton’s Laws of Motion
Physics
22. Strategy Use Newton’s second law for the vertical direction.
Solution Draw a free-body diagram. Find the tension.
∑ Fy = T − mg = ma y , so
y
T
T = m(a y + g ) = (2010 kg)(1.50 m s 2 + 9.80 m s 2 ) = 22.7 kN.
a
The tension in the cable is 22.7 kN upward.
mg
23. (a) Strategy Use Newton’s second law for the vertical direction. Let +y be in the upward direction.
Solution Draw a free-body diagram.
T − mg 33.6 kN − 24.8 kN
∑ Fy = T − mg = ma y , so a y =
=
= +3.5 m s 2 .
m
2530 kg
y
T
a
So, the acceleration of the elevator is 3.5 m s 2 up .
mg
(b) Strategy Use the definition of average acceleration and solve for the final speed.
Solution
∆v
ay =
, so ∆v = vf − vi = a y ∆t , or vf = vi + a y ∆t = 1.20 m s + (3.5 m s 2 )(4.00 s) = 15 m s.
∆t
So, the velocity of the elevator 4.00 s later will be 15 m s up .
24. (a) Strategy and Solution The weight of the glider is equal and opposite to the force due to the air, 3.0 kN
downward. The force on the Earth due to the glider is equal and opposite to the weight of the glider,
3.0 kN upward .
(b) Strategy Use Newton’s second law and solve for the acceleration.
Solution The net force is 3.0 kN downward + 2.0 kN upward = 1.0 kN downward.
G
G
G
G F
F
Fg (1.0 kN downward)(9.80 m s 2 )
a= =
=
=
= 3.3 m s 2 downward
m Fg g Fg
3.0 kN
25. Strategy Use Newton’s laws of motion.
Solution The stone is lifted with constant velocity, so the net force on the stone in the vertical direction is zero.
The force of the man’s hand on the stone is equal and opposite to the force of gravity on the stone, which is mg
downward. The magnitude of the total force of the man’s hand on the stone is
F = mg = (2.0 kg)(9.80 m s 2 ) = 20 N .
26. Strategy Use Newton’s laws of motion.
Solution Find the magnitude of the total force of the man’s hand on the stone.
∑ Fy = Fhs − mg = ma y , so Fhs = ma y + mg = m(a y + g ) = (2.0 kg)(1.5 m s 2 + 9.80 m s 2 ) = 23 N .
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Physics
Chapter 4: Force and Newton’s Laws of Motion
27. Strategy Use Newton’s second law.
Solution Compute the acceleration of the automobile.
G
G ΣF 3.36 × 103 N forward
a=
=
= 2.40 m s 2 forward
m
1.40 × 103 kg
28. Strategy Use Newton’s second law.
Solution Compute the mass of the crate.
ΣF
100 N
ΣFx = ma x , so m = x =
= 40 kg .
ax
2.5 m s 2
29. Strategy Find the net force on the airplane; and from it determine the acceleration. Then use Eq. (4-4) to find the
distance traveled.
Solution Find the net force on the airplane.
∑ Fx = 1.800 kN − 1.400 kN = 0.400 kN = ma x , so a x =
0.400 kN 0.400 kN
=
= 0.3448 m s 2 .
m
1160 kg
∑ Fy = 16.000 kN − 16.000 kN = 0 = ma y , so a y = 0.
Find the distance traveled.
1
1
∆x = vix ∆t + a x (∆t ) 2 = (60.0 m s)(60.0 s) + (0.3448 m s 2 )(60.0 s) 2 = 4.22 km
2
2
30. Strategy Find the net force on the elevator; and from it determine the acceleration. Then use Eq. (4-5) to find the
speed.
Solution Find the net force on the elevator.
T − mg 7730 N − (832 kg)(9.80 m s 2 )
∑ Fy = T − mg = ma y , so a y =
=
= −0.5091 m s 2 .
m
832 kg
Find the speed.
vf2x − vi2x = vf2x − 0 = 2a x ∆x, so vfx = 2a x ∆x = 2(−0.5091 m s 2 )(−5.00 m) = 2.26 m s .
31. Strategy Draw vector arrows representing all of the forces acting on the object. Make sure that the directions of
the arrows correctly illustrate the directions of the forces and that their lengths are proportional to the magnitudes
of the forces. Let the subscripts be the following:
p = system of plant, soil, pot
h = hook
c = cord
C = ceiling
e = Earth
s = system of plant, soil, pot, cord, hook
Solution The free-body diagrams are shown.
(a)
(b)
(c)
Fpc
(d)
Fch
FsC
FhC
Fpe
Fse
Fce
Fcp
Fhe
259
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Chapter 4: Force and Newton’s Laws of Motion
Physics
32. Strategy We are concerned with the interactions of pairs of objects that exert forces on each other. Analyze the
given forces in light of Newton’s first and third laws.
Solution Forces (a) and (b) are third law pairs. This is an interaction between two objects, the bike and the Earth.
Each body exerts a gravitational force on the other body; and these forces are equal in magnitude and opposite in
direction. Forces (a) and (c) are equal and opposite due to the first law. They act not on each other, but on the
same body.
33. Strategy Analyze the forces due to and on the three interacting objects: the woman, the chair, and the floor.
Solution
(a) The weight of the woman is directed downward. The forces on the woman due to the seat and armrests are
directed upward and total 25 N + 25 N + 500 N = 550 N. The chair and floor must support her entire weight,
so the balance of her weight to support is 600 N − 2(25 N) − 500 N = 50 N. Thus, the floor exerts a force on
the woman’s feet of 50.0 N upward.
(b) The force exerted by the floor on the chair must be equal to the weight of the chair plus the weight of the
woman supported by the chair, or 600.0 N + 100.0 N − 50.0 N = 650.0 N. Thus, the floor exerts a force on the
chair of 650.0 N upward.
(c) The two forces acting on the woman and chair system are the upward force due to the
floor and the downward gravitational force due to the Earth. Let the subscripts be the
following: s = woman and chair system, e = Earth, f = floor.
Fsf
Wse
34. Strategy Consider forces acting on the rod.
Solution
One force acting on the rod is the downward force on the rod by the line; its interaction partner is the
upward force on the line by the rod. Another force acting on the rod is the downward gravitational force
on the rod by the Earth; its interaction partner is the upward gravitational force on the Earth by the rod.
35. Strategy Consider forces acting on the fish suspended by the line.
Solution
One force acting on the fish is an upward force on the fish by the line; its interaction partner is a
downward force on the line by the fish. A second force acting on the fish is the downward gravitational
force on the fish; its interaction partner is the upward gravitational force on the Earth by the fish.
36. (a) Strategy Determine the force of air resistance using the force of the parachute and the weight of the
skydiver.
Solution Both the upward force of the parachute and the upward force of air resistance act to oppose the
force due to gravity on the skydiver (the weight). Since the skydiver is falling at constant speed, the net force
on the skydiver is zero. Thus, the sum of the magnitudes of the forces of air resistance and the parachute must
be equal to that of the diver’s weight. So, Fair + 620 N = 650 N, or Fair = 30 N .
260
Physics
Chapter 4: Force and Newton’s Laws of Motion
(b) Strategy Identify each force acting on the skydiver and use Newton’s laws to identify the interaction
partners of each.
Solution Refer to part (a).
The parachute pulls up on the skydiver with a force of 620 N and the skydiver pulls down on the
parachute with a force of 620 N; the Earth pulls down on the skydiver with a force of 650 N and
the skydiver pulls up on the Earth with a force of 650 N; air resistance pushes upward on the
skydiver with a force of 30 N and the skydiver pushes downward on the air with a force of 30 N.
(c) Strategy Identify each force acting on the parachute and use Newton’s laws to identify the interaction
partners of each.
Solution
The skydiver pulls downward on the parachute with a force of 620 N and the parachute pulls up on
the skydiver with a force of 620 N; air resistance pushes up on the parachute and the parachute pushes
downward on the air; Earth pulls downward on the parachute and the parachute pulls upward on the Earth.
37. Strategy Use Newton’s first and third laws.
Solution
(a) Margie exerts a downward force on the scale equal to her weight, 543 N. According to the third law, the scale
exerts an upward force on Margie equal in magnitude to the magnitude of the force exerted by Margie on it,
or 543 N.
(b) Refer to part (a). The interaction partner of the force exerted on Margie by the scale is the contact force of
Margie’s feet on the scale.
(c) The Earth must hold up both the scale and (indirectly) Margie, since Margie is standing on the scale. So, the
Earth must push up on the scale with a force equal to the combined weight of Margie and the scale, or
543 N + 45 N = 588 N.
(d) Refer to part (c). The interaction partner of the force exerted on the scale by the Earth is the contact force on
the Earth due to the scale.
38. Strategy Treat the skydiver and parachute as one object.
Solution The two external forces acting on the skydiver and parachute system are the upward directed drag force
due to the air and the downward directed force due to gravity.
39. Strategy Use the conversion factor for pounds to newtons, 1 lb = 4.448 N, and the Earth’s average gravitational
field strength, g = 9.80 N kg.
Solution
(a) Answers will vary. For a 150-lb person, (150 lb)(4.448 N lb ) = 670 N .
(b) Weight of 250 g of cheese = mg = (0.25 kg)(9.80 N kg) = 2.5 N
(c) Answers will vary. A stick of butter weighs about 0.25 lb.
(0.25 lb)(4.448 N lb ) = 1.1 N
So, a stick of butter weighs about 1 N.
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Chapter 4: Force and Newton’s Laws of Motion
Physics
40. Strategy Use the conversion factor for pounds to newtons, 0.2248 lb = 1 N.
Solution
(a) Find the weight of the girl in newtons.
W = mg = (40.0 kg)(9.80 N kg) = 392 N
(b) Find the weight of the girl in pounds.
(392 N)(0.2248 lb N) = 88.1 lb
41. Strategy Find the mass using the weight of the man and the Earth’s average gravitational field strength,
g = 9.80 N kg.
Solution Find the mass of the man.
W 0.80 × 103 N
m=
=
= 82 kg
g
9.80 N kg
42. (a) Strategy Compare the strengths of the forces at the location of the rock.
Solution The gravitational force between to bodies is inversely proportional to the square of the distance
between them. The Moon is much closer to the rock than is the Earth, so (even though the Earth is much
more massive than the Moon) the rock will fall toward the Moon’s surface.
(b) Strategy The Moon’s gravitational field strength is 1.62 N kg. The force of the Moon on the rock is equal
to the weight of the rock on the Moon.
Solution Find the weight of the rock to find the gravitational force exerted by the Moon on it.
F = W = mg = (1.0 kg)(1.62 N kg) = 1.6 N
The force on the rock due to the Moon is 1.6 N toward the Moon.
(c) Strategy Use Newton’s law of universal gravitation. The average Earth-Moon distance is 3.845 × 108 m.
The mass of the Earth is 5.974 × 1024 kg.
Solution Find the gravitational force exerted by the Earth on the rock.
Gm1m2 (6.674 × 10−11 N ⋅ m 2 kg 2 )(1.0 kg)(5.974 × 1024 kg)
=
= 2.7 mN
F=
(3.845 × 108 m)2
r2
Since gravitational force is attractive, the force exerted by the Earth on the rock is 2.7 mN toward Earth.
(d) Strategy Recognize that the force due to the Moon is much greater than that due to the Earth.
Solution Find the net gravitational force.
1.6 N toward the Moon + 0.0027 N toward Earth = 1.6 N toward the Moon − 0.0027 N toward the Moon
= 1.6 N toward the Moon
43. Strategy Use Newton’s universal law of gravitation.
Solution Estimate the magnitude of the gravitational attraction between Alex and Pat.
Gm1m2 (6.674 × 10−11 N ⋅ m 2 kg 2 )(55 kg)(40 kg)
=
= 2 nN
F=
(8 m)2
r2
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Chapter 4: Force and Newton’s Laws of Motion
44. Strategy Use Newton’s universal law of gravitation.
Solution Find the ratio.
F1
F2
=
Gm1m2
r12
÷
Gm1m2
r22
2
2
2
⎛ r + 6.00 × 106 m ⎞
⎛r ⎞
⎛ 6.371× 106 m + 6.00 × 106 m ⎞
⎟ =⎜
=⎜ 2 ⎟ =⎜ 1
⎟⎟ = 3.770
⎜r ⎟
⎜
⎜
⎟
r1
6.371× 106 m
⎝ 1⎠
⎝
⎠
⎝
⎠
45. Strategy The gravitational field strength is given by g = GM R 2 . Use the mass of the Earth and the
gravitational field strength of the Moon and solve for R, which, in this case, is the distance from the center of the
Earth. Then, subtract the radius of the Earth to find the height above the surface.
Solution Solving for R, we have
R=
GM
=
g
(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
1.62 m s 2
= 1.57 × 104 km.
So, the height above the surface is 1.57 × 104 km − 6.371× 103 km = 9.3 × 103 km .
46. Strategy On Earth, g = 9.80 N kg.
Solution Find the man’s weight on Earth, Mars, Venus, and Earth’s moon.
mg = (65 kg)(9.80 N kg) = 640 N
(a) Find the man’s weight on Mars.
mg = (65 kg)(3.7 N kg) = 240 N
(b) Find the man’s weight on Venus.
mg = (65 kg)(8.9 N kg) = 580 N
(c) Find the man’s weight on Earth’s moon.
mg = (65 kg)(1.6 N kg) = 100 N
47. Strategy Gravitational field strength is given by g = GM R 2 , so let the new field strength be g ′ = ng = GM r 2 ,
where n = 2 3 for part (a) and 1 3 for part (b).
Solution Determine r in terms of R.
g ′ ng
=
=n=
g
g
GM
r2
GM
R2
=
R2
r
2
, so r =
R
.
n
Find an expression for the altitude, h.
⎛ 1
⎞
R
h=r−R=
− R = R⎜
− 1⎟
n
n
⎝
⎠
⎛ 1
⎞
− 1⎟ = 1432 km
(a) h = (6.371× 103 km) ⎜
⎜ 23 ⎟
⎝
⎠
⎛ 1
⎞
(b) h = (6.371× 103 km) ⎜
− 1⎟ = 4664 km
⎜ 13 ⎟
⎝
⎠
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Chapter 4: Force and Newton’s Laws of Motion
Physics
48. Strategy Gravitational field strength is given by g = GM R 2 . Find H = R2 − R1 , where R1 and R2 are the
distances from the center of the Earth to the surface of the Earth and the location of the balloon, respectively.
Solution Find the height above sea level of the balloon, H.
GM
GM
−
= GM g 2−1/ 2 − g1−1/ 2
H = R2 − R1 =
g2
g1
(
)
= (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg) ⎡ (9.792 N kg)−1/ 2 − (9.803 N kg)−1/ 2 ⎤ = 4 km
⎣
⎦
49. Strategy This is the same as asking, “At what altitude is the gravitational field strength half of its value at the
surface of the Earth?” g = GM R 2 , so let the new field strength be g ′ = ng = GM r 2 where n = 1 2.
Solution Determine r in terms of R.
GM
2
g ′ ng
r 2 = R , so r = R .
=
= n = GM
g
g
n
r2
2
R
Find an expression for the altitude, h.
h=r−R=
⎛ 1
⎞
⎛ 1
⎞
R
− R = R⎜
− 1⎟ , so h = (6.371× 103 km) ⎜
− 1⎟ = 2639 km .
⎜ 12 ⎟
n
⎝ n ⎠
⎝
⎠
50. (a) Strategy Use Newton’s universal law of gravitation and r = 3.845 × 108 m for the distance between Earth
and the Moon.
Solution Find the magnitude of the gravitational force that Earth exerts on the Moon.
GM E M M (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)(7.35 × 1022 kg)
=
= 1.98 × 1020 N
F=
2
8
2
(3.845 × 10 m)
r
(b) Strategy Use Newton’s third law.
Solution According to Newton’s third law, the magnitude of the gravitational force that the moon exerts on
the Earth is the same as the force that the Earth exerts on the moon.
51. Strategy Use Newton’s universal law of gravitation. The Voyager spacecraft are approximately 15 billion
kilometers from the Sun. Use this value for the distance between the Earth and the spacecraft.
Solution Find the approximate magnitude of the gravitational force between Earth and the spacecraft.
Gm1m2 (6.674 × 10−11 N ⋅ m 2 kg 2 )(825 kg)(5.974 × 1024 kg)
=
= 1.5 × 10−9 N
F=
(15 × 1012 m)2
r2
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Chapter 4: Force and Newton’s Laws of Motion
52. Strategy Use Newton’s law of universal gravitation.
Solution Find h such that ∆g = g − g ′ < 0.01000 g .
GM E ( RE + h) 2
RE 2
1
∆g
g − g′
g′
= 0.01000 =
= 1− = 1−
= 1−
= 1−
2
2
g
g
g
( RE + h)
(1 + h RE )2
GM E RE
Solve for h.
0.01000 = 1 −
1
(1 + h RE ) 2
⎛
1
h
= ⎜1 +
0.99000 ⎜⎝ RE
1
h
±
= 1+
RE
0.99000
⎞
⎟⎟
⎠
2
h
1
=
−1
RE
0.99000
⎛
⎞
⎛
⎞
1
1
h = RE ⎜⎜
− 1⎟⎟ = (6.371× 103 km) ⎜⎜
− 1⎟⎟ = 32 km
⎝ 0.99000 ⎠
⎝ 0.99000 ⎠
The positive root was chosen because h > 0 and RE > 0.
53. Strategy Consider each of the four forces and any possible relationships between them.
Solution (a) The force of the Earth pulling on the book and (d) the force of the book pulling on the Earth are an
interaction pair; they are equal and opposite. (b) The force of the table pushing on the book and (c) the force of the
book pushing on the table are an interaction pair; they are equal and opposite. There are two forces acting on the
book: the gravitational force of Earth pulling on it and the contact force of the table pushing on it. Since the book
is in equilibrium, the net force on it must be zero; therefore, the forces due to Earth and the table on the book are
equal and opposite, so the pair of forces given in (a) and (b) are equal in magnitude and opposite in direction even
though they are not an interaction pair.
54. Strategy The direction of the normal force is always perpendicular to the surface of the ramp. The friction force
is in whatever direction necessary to oppose the motion of the object.
Solution The results are shown in the table.
G
f
G
N
(a)
perpendicular to and away from
along the ramp upward
(b)
perpendicular to and away from
along the ramp downward
(c)
perpendicular to and away from
along the ramp upward
55. Strategy Use Newton’s laws of motion.
Solution Without a machine, the force is equal to the weight of the object mg. According to Newton’s laws of
mg
d
h
= .
motion and Fig. 2.32, with a frictionless plane, the force is equal to mg sin φ = mg . So,
h
h
d
mg
d
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Chapter 4: Force and Newton’s Laws of Motion
Physics
56. Strategy Use Newton’s laws of motion. Draw a diagram.
Solution
N
y
x
20.0°
20.0°
mg
fs
mg cos 20.0°
mg sin 20.0°
(a) Compute the magnitude of the normal force.
∑ Fy = N − mg cos θ = 0, so N = mg cos θ = (80.0 N) cos 20.0° = 75.2 N.
The normal force is 75.2 N perpendicular to and above the surface of the ramp.
(b)
The interaction partner is equal in magnitude to the component of the apple crate’s weight perpendicular to
the ramp, 75.2 N perpendicular to the ramp and opposite in direction to the normal force; it is exerted by
the crate on the ramp; it is a contact force.
(c) Compute the magnitude of the force of static friction on the crate.
∑ Fx = fs − mg sin θ = 0, so fs = mg sin θ = (80.0 N) sin 20.0° = 27.4 N.
The force of static friction exerted on the crate by the ramp is 27.4 N up the incline.
(d) The minimum possible value of the coefficient of static friction is the value that just makes the force of static
friction oppose the component of the crate’s weight that is directed down the incline.
mg sin θ mg sin θ
fs = µs, min N = mg sin θ , so µs, min =
=
= tan θ = tan 20.0° = 0.364 .
N
mg cos θ
(e) Find the magnitude.
F = fs 2 + N 2 = (27.4 N) 2 + (75.2 N) 2 = 80.0 N
Find the direction.
N
mg cos θ
1
θ = tan −1 = tan −1
= tan −1
= 70.0° or upward
f
mg sin θ
tan 20.0°
G
So, F = 80.0 N upward .
266
Physics
Chapter 4: Force and Newton’s Laws of Motion
57. (a) Strategy Draw a diagram and use Newton’s laws of motion.
Solution According to the first law, since the skier is moving
with constant velocity, the net force on the skier is zero.
Calculate the force of kinetic friction.
∑ Fx = f k − mg sin θ = 0, so
f k = mg sin θ = (85 kg)(9.80 N kg) sin11° = 160 N.
The force of kinetic friction is 160 N up the slope.
N
y
x
fk
mg
θ
mg cos θ
mg sin θ
(b) Strategy Use the diagram and results from part (a).
Solution Find the normal force.
∑ Fy = N − mg cos θ = 0, so N = mg cos θ . Since f k = µk N ,
f
mg sin θ
µk = k =
= tan θ = tan11° = 0.19 .
N mg cos θ
58. Strategy While the crate is at rest, the force of static friction must be equal in magnitude and opposite in
direction to the component of the crate’s weight along and down the incline.
Solution Find the frictional force on the crate.
∑ Fx = fs − mg sin θ = 0, so fs = mg sin θ = (18.0 kg)(9.80 N kg) sin 30° = 88 N.
The frictional force is 88 N up the ramp.
59. Strategy While the crate is sliding down the ramp, the force of kinetic friction has its maximum magnitude and is
opposite in direction to the crate’s motion down the incline.
Solution Find the frictional force on the crate.
∑ Fy = N − mg cos θ = 0, so N = mg cos θ .
∑ Fx = f k = µk N = µk mg cos θ = 0.40(18.0 kg)(9.80 N kg) cos 30° = 61 N
The frictional force is 61 N up the ramp.
60. Strategy While the crate is sliding up the ramp, the force of kinetic friction has its maximum magnitude and is
opposite in direction to the crate’s motion up the incline.
Solution Find the frictional force on the crate.
∑ Fy = N − mg cos θ = 0, so N = mg cos θ .
∑ Fx = f k = µk N = µk mg cos θ = 0.40(18.0 kg)(9.80 N kg) cos 30° = 61 N
The frictional force is 61 N down the ramp.
61. Strategy Draw free-body diagrams for each situation. Let the subscripts be the following:
b = book
t = table
e = Earth
h = hand
Solution The diagrams are shown.
(a)
(b)
Nbt
Fbh
fbt
Wbe
(c)
Nbt
Nbt
fbt
Wbe
267
Wbe
Chapter 4: Force and Newton’s Laws of Motion
Physics
(d) Strategy and Solution In cases (a) and (b), the book is accelerating; so in these cases, the net force is not
zero.
(e) Strategy and Solution The normal force on the book is equal to its weight, (0.50 kg)(9.80 m s 2 ) = 4.9 N.
The net force acting on the book in part (b) is equal to the force of kinetic friction. The force of kinetic
friction is opposite the direction of motion. The magnitude is µk N = 0.40(4.9 N) = 2.0 N. Thus, the net force
on the book is 2.0 N opposite the direction of motion.
(f) Strategy and Solution The free-body diagram would look just like the diagram for part (c) and the book
would not slow down because there is no net force on the book (friction is zero).
62. (a) Strategy Refer to Example 4.10. The maximum static friction force must be greater than the x-component of
the weight. Use Newton’s laws of motion.
Solution
∑ Fx = fs − mg sin θ ≥ 0, so fs ≥ mg sin θ and ∑ Fy = N − mg cos θ = 0, so N = mg cos θ .
Compare the forces of friction and gravity.
µs N ≥ mg sin θ
µs mg cosθ ≥ mg sin θ
µs ≥ tan θ
0.42 ≥ tan15°
0.42 ≥ 0.27 True
Yes; the static friction force can hold the safe in place.
(b) Part (b) is unnecessary, since the answer to part (a) is yes.
63. (a) Strategy To just get the block to move, the force must be equal to the maximum force of static friction.
Solution Solve for µs .
F = f max
F
12.0 N
= µs N = µs mg , so µs =
=
= 0.41 .
mg (3.0 kg)(9.80 N kg)
N
f
F
mg
(b) Strategy The maximum static frictional force is now proportional to the total mass of the two blocks. The
free-body diagram is the same as before, except that the mass m is now the sum of the masses of both blocks.
Solution Find the magnitude F of the force required to make the two blocks start to move.
F = µs mg = 0.41(3.0 kg + 7.0 kg)(9.80 N kg) = 40 N
64. (a) Strategy Use Newton’s first law of motion.
Solution Since the sleigh is moving with constant speed, the net force acting on the sleigh is zero.
(b) Strategy Since Fnet = 0, the force of magnitude T must be equal to the force of kinetic friction.
Solution Find the coefficient of kinetic friction.
T
T = f k = µk mg , so µk =
.
mg
268
Physics
Chapter 4: Force and Newton’s Laws of Motion
65. Strategy Since the block moves with constant speed, there is no net force on the block. Draw the free-body
diagram using this information. Let the subscripts be the following:
b = block
B = Brenda
w = wall
e = Earth
Solution Find the coefficient of kinetic friction between the wall and the block.
∑ Fx = N bw − FbB sin θ = 0, so N bw = FbB sin θ .
y
x
∑ Fy = FbB cos θ − Fbe − f bw = 0, so f bw = FbB cos θ − Fbe .
Since f bw = µk N bw , we have
F cos θ − Fbe FbB cos θ − Fbe
F
=
= cot θ − be csc θ
µk = bB
N bw
FbB sin θ
FbB
2.0 N
= cot 30.0° −
csc 30.0° = 0.4
3.0 N
FbB
fbw
Nbw
Fbe
66. Strategy Let the subscripts be the following:
t = table
e = Earth
1 = block 1
2 = block 2
3 = block 3
4 = block 4
h = horizontal force
1234 = system of blocks
(The blocks are numbered from left to right.)
Solution The diagrams are shown.
(a)
(b)
N2t
F23
F2l
f2t
N1234t
F1234h
f1234t
W2e
W1234e
67. Strategy To just get the block to move, the magnitude of the gravitational force acting on the box must be equal
to the magnitude of the maximum force of static friction. See the free-body diagram.
Solution Find the angle of the ramp.
mg sin θ = f max
= µs N
= µs mg cos θ
tan θ = µs
θ = tan
−1
µs = tan
−1
N
f
mg
0.30 = 17°
θ
269
mg cos θ
mg sin θ
Chapter 4: Force and Newton’s Laws of Motion
Physics
68. Strategy For each slide, the +x-direction is up the slide, and the +y-direction is away from the slides’ top
surfaces. Use Newton’s second law.
Solution
N
f
y
x
θ
mg cos θ
mg
θ
mg sin θ
(a) Slide 1:
∑ Fx = f − mg sin θ1 = 0, so f = mg sin θ1 and ∑ Fy = N − mg cos θ1 = 0, so N = mg cos θ1.
Thus, f = µ N = µ mg cos θ1 = mg sin θ1 and µ = tan θ1.
Slide 2:
∑ Fx = f − mg sin θ 2 = − ma, so f = − ma + mg sin θ 2 and ∑ Fy = N − mg cos θ 2 = 0, so N = mg cos θ 2 .
Thus, f = µ N = µ mg cos θ 2 = −ma + mg sin θ 2 and µ = −
a
+ tan θ 2 .
g cos θ 2
Set the two expressions for µ equal and solve for a.
a
−
+ tan θ 2 = tan θ1 , so a = − g (cos θ 2 tan θ1 − cos θ 2 tan θ 2 ) = g (sin θ 2 − cos θ 2 tan θ1 ) .
g cos θ 2
(b) Compute the magnitude of the acceleration.
a = (9.80 m s 2 )(sin 61° − cos 61° tan 45°) = 3.8 m s 2
69. Strategy Identify each force acting on the sailboat. Draw a free-body diagram.
Solution The forces acting on the sailboat are:
1) the force of gravity
2) the vertical force of the water opposing gravity
3) the force of the wind
4) the force of the line tied to the mooring
Nsw
Fsl
FsW
s = sailboat
l = line
w = water
W = wind
g = gravity
Wsg
70. Strategy and Solution The towline and glider are interaction partners; thus, according to Newton’s third law of
motion, the forces they exert on each other are equal in magnitude and opposite in direction. Therefore, the force
exerted by the glider on the towline is 850 N, due west.
71. Strategy Count the number of couplings behind car 2, and multiply that number by T5 = 459 N, the tension in
the fifth coupling (between cars 4 and 5).
Solution There are three couplings behind car 2, so tension is 3 × 459 N = 1.38 kN .
270
Physics
Chapter 4: Force and Newton’s Laws of Motion
72. Strategy Identify all forces acting on the strut. Decompose the tension into its x- and y-components.
Solution Use Newton’s laws of motion. See the diagram.
∑ Fy = T sin 30.0° − 200.0 N = 0, so
T=
200.0 N
= 400 N .
sin 30.0°
T sin 30.0°
T cos 30.0°
Fsw
Fsign
73. Strategy Use Newton’s laws of motion. The lower cord supports only the lower box, whereas the upper cord
supports both boxes. Draw a diagram.
Solution Find the tension in each cord.
Lower cord
∑ Fx = Tl − ml g sin θ = 0, so
y
x
Tl = ml g sin θ = (2.0 kg)(9.80 N kg) sin 25° = 8.3 N .
Nl
Tl
Upper cord
∑ Fx = Tu − mu g sin θ − Tl = 0, so
Nu
Tl
mug cosθ
mug
Tu = mu g sin θ + Tl = (1.0 kg)(9.80 N kg) sin 25° + 8.3 N = 12.4 N .
mug sin θ
mlg cos θ
θ
mlg
Tu
mlg sin θ
74. Strategy Recall that the tension in the rope is the same along its length.
Solution The tension is equal to the weight at the end of the rope, 120 N. Therefore, both scales read 120 N.
75. Strategy Recall that the tension in the rope is the same along its length.
Solution The tension is equal to the weight at the end of the rope, 120 N. Therefore, scale A reads 120 N.
There are two forces pulling downward on the pulley due to the tension of 120 N in each part of the rope.
Therefore, TB = TA + TA = 2TA = 240 N. Scale B reads 240 N, since it supports the pulley.
76. Strategy Use Newton’s laws of motion and analyze each scale separately.
Solution Scale B reads 120 N due to the apples hanging from it. According to Newton’s third law, scale A also
reads 120 N, since B is attached directly below it, which is attached to the weight.
77. Strategy Use Newton’s laws of motion. Draw a free-body diagram.
Solution
(a) Find the tension in the rope from which the pulley hangs.
∑ Fy = T1 sin θ − Mg = 0 and ∑ Fx = T1 cos θ − T2 = 0.
The tension in T2 is due to the mass M, so T2 = Mg .
Thus, T1 cos θ = Mg and T1 sin θ = Mg .
According to these equations, cos θ = sin θ , which is true only if
θ = 45° for 0° ≤ θ ≤ 90°.
Mg
= 2 Mg .
Therefore, T1 =
cos 45°
(b) As found in part (a), θ = 45° .
271
y
x
θ
T2
Mg
T1
Chapter 4: Force and Newton’s Laws of Motion
Physics
78. Strategy Use Newton’s law of motion. Draw a free-body diagram.
Solution
y
T
30.0°
x
mg
F
(a) ∑ Fx = F − T cos θ = 0, so F = T cos θ . ∑ Fy = T sin θ − mg = 0, so T sin θ = mg , or T =
Thus, F =
(b) T =
mg
.
sin θ
mg
mg
(2.0 kg)(9.80 N kg)
=
= 34 N .
cos θ =
sin θ
tan θ
tan 30.0°
(2.0 kg)(9.80 N kg)
= 39 N
sin 30.0°
79. Strategy Use Newton’s laws of motion. Draw a free-body diagram.
Solution Find the tension in each wire.
∑ Fx = T25 sin 25° − T15 sin15° = 0, so T25 sin 25° = T15 sin15°, or
25°
sin15°
T .
sin 25° 15
∑ Fy = T25 cos 25° + T15 cos15° − F = 0, so T15 cos15° + T25 cos 25° = F .
Substitute for T25 .
T25 =
⎛ sin15°
⎞
T15 cos15° + ⎜
T15 ⎟ cos 25° = F
⎝ sin 25°
⎠
sin15° ⎞
⎛
T15 ⎜ cos15° +
⎟=F
tan
25° ⎠
⎝
T15 =
T25 =
45 N
°
cos15°+ sin15
tan25°
y
15°
T15
T25
x
45 N
Fg
= 30 N
sin15°
(30 N) = 18 N
sin 25°
80. Strategy Use Newton’s laws of motion. Draw a diagram.
Solution Find the tension.
∑ Fx = T cos θ − T cos θ = 0 and
T
∑ Fy = T sin θ + T sin θ − W = 0. So, 2T sin θ = W , or T =
W
.
2sin θ
θ
θ
T
W
81. Strategy Use Newton’s laws of motion. Let +y be down and +x to the right.
G
Solution Find the force F applied to the front tooth.
∑ Fx = T sin θ − T sin θ = 0 and ∑ Fy = T cos θ + T cos θ − F = 0. So, we have
F = 2T cos θ = 2(1.2 N) cos 33° = 2.0 N. By symmetry, the force is directed toward the back of the mouth, so
G
F = 2.0 N toward the back of the mouth .
272
Physics
Chapter 4: Force and Newton’s Laws of Motion
82. Strategy Use Newton’s laws of motion.
Solution The Earth exerts a force on the mass, which then exerts a force on the scale, which then exerts a force
on the hook, which then exerts a force on the ceiling. All these forces are equal (assuming that the masses of the
spring scale and hook are negligible). In addition, each body, on which a force is exerted, exerts an equal and
opposite force on the other object. So, the ceiling exerts a force on the hook, the hook on the scale, etc. One
person replaces the force on the mass due to the Earth, and the other person replaces the force on the scale due to
the hook. So, each person must exert a force of 98 N.
83. Strategy Use Newton’s second law.
Solution
For m2 : ∑ Fx = T2 − T1 = m2 a
For m1: ∑ Fx = T1 = m1a
Find T2 .
T2 − T1 = m2 a, so T2 = T1 + m2 a = m1a + m2 a = (m1 + m2 )a.
m1
x
T1
T1
m2
T2
Find T1 T2 .
T1
=
T2
m1a
(m1 + m2 )a
=
m1
m1 + m2
84. Strategy Use Eq. (2-16) to find the distance of the block from the top of the incline.
Solution
vf
= vf − 0 = vf = 2ad and ( 0.50vf ) = 2a∆x = 0.25vf
Form a proportion.
2
− vi 2
vf 2
0.25vf 2
2
2
2
2.0 m
2.
∆x
d
2ad
2.0 m
=
=
= 0.50 m .
, so ∆x =
2a∆x
4.0
4.0
85. (a) Strategy The force required to start the block moving is that needed to overcome the maximum force of
static friction. Draw a diagram.
Solution Find the applied horizontal force.
∑ Fx = F − fs = 0, so F = fs = µs N .
∑ Fy = N − Fg = N − mg = 0, so N = mg .
N
F
fs
So, the value of the applied force at the instant that the block
starts to slide is F = µs mg = 0.40(5.0 kg)(9.80 N kg) = 20 N .
Fg
(b) Strategy The force required to keep the block moving is that needed to overcome kinetic friction. At the
instant the block starts to slide, the net force on the block is the difference between the forces required to
overcome static and kinetic friction.
Solution Calculate the net force.
∑ F = fs − f k = µs mg − µk mg = ( µs − µk )mg = (0.40 − 0.15)(5.0 kg)(9.80 N kg) = 12 N
273
Chapter 4: Force and Newton’s Laws of Motion
Physics
86. Strategy Let the direction of motion be +x. Let F be the magnitude of the force exerted by the locomotive on the
caboose. Use Newton’s second law.
Solution Draw a free-body diagram.
∑ Fy = 0, since the vertical component of the acceleration is zero.
x
a
f
∑ Fx = F − f = ma x , so F = ma x + f = (1.0 kg)(3.0 m s 2 ) + 0.50 N = 3.5 N .
F
87. Strategy Choose the +x-axis to the right and +y-axis up. Use Newton’s second law and the essential relationships
for constant acceleration problems.
Solution
y
(a) For m1:
∑ F1 y = N − m1 g = 0, so N = m1 g .
∑ F1x = T = m1a1x
x
N
T
m1
For m2 :
T
∑ F2 x = 0 and ∑ F2 y = T − m2 g = m2 a2 y .
m1g
Now, a1x and a2 y must be equal in magnitude, otherwise the cord will
compress or expand. a1x is in the +x-direction and a2 y is in the − y -direction.
So, let a = a1x = − a2 y . Then, T = m1a and T − m2 g = − m2 a.
Eliminate T and solve for a.
m2 g
(2.0 kg)(9.80 m s 2 )
=
= 3.9 m s 2 .
T − m2 g = m1a − m2 g = − m2 a, so a =
m1 + m2
3.0 kg + 2.0 kg
The acceleration of block 1 is 3.9 m s 2 to the right
and the acceleration of block 2 is
3.9 m s 2 downward .
(b) Compute the velocity of the first block.
G G
v = a∆t = (3.92 m s 2 to the right)(1.2 s) = 4.7 m s to the right
(c) Compute the distance moved by the first block.
1
1
∆x = a (∆t ) 2 = (3.92 m s 2 )(1.2 s) 2 = 2.8 m
2
2
(d) Compute the displacement of the first block.
G 1G
1
∆r1 = a(∆t ) 2 = (3.92 m s 2 to the right)(0.40 s)2 = 0.31 m to the right
2
2
Compute the displacement of the second block.
G
1G
1
∆r2 = a(∆t ) 2 = (3.92 m s 2 down)(0.40 s)2 = 0.31 m down
2
2
274
m2
m2g
Physics
Chapter 4: Force and Newton’s Laws of Motion
88. Strategy Use Newton’s second law. Treat the last 10 freight cars as a system.
Solution The vertical forces cancel.
∑ Fy = N − mg = ma y = 0
N
Let the direction of motion be +x.
The force exerted on the eleventh car by the tenth is the tension at the coupler, T11.
The force is
1.0 × 105
T11
∆vx
4.0 m s
= 10(5.0 × 104 kg)
= 1.0 × 105 N
20.0 s
∆t
T11 = ∑ Fx = ma x = m
mg
N in the direction of motion .
89. Strategy Use the expressions for a y and T found in Example 4.15.
Solution
(m2 − m1 ) g (5.0 kg − 3.0 kg)(9.80 m s 2 )
=
= 2.5 m s 2
m2 + m1
5.0 kg + 3.0 kg
G
G
Since m2 > m1 , a1 = 2.5 m s 2 up and a 2 = 2.5 m s 2 down .
(a) a y =
(b) T =
m2
m1
2m1m2
2(3.0 kg)(5.0 kg)
g=
(9.80 m s 2 ) = 37 N
m1 + m2
3.0 kg + 5.0 kg
90. Strategy Use Newton’s second law to evaluate the situation.
Solution Since F = ma, the maximum acceleration is amax =
F
T
2500 N
=
=
= 1.8 m s 2 . With this
m mcar 1400 kg
v
30 mph
=
= 7.5 s. This is certainly possible. So, yes,
a 1.8 m s 2
the truck driver should be concerned about the rope breaking, particularly when friction is also considered.
acceleration, the truck could reach 30 mph in about ∆t =
91. Strategy From Example 4.15, the acceleration is a y = g (m2 − m1 ) (m2 + m1 ). Use Eq. (2-15).
Solution Find the time it takes block 2 to reach the floor.
1
2∆y
2∆y
∆y = − a y (∆t ) 2 , so (∆t ) 2 = −
=−
. Thus,
m
(
2 − m1 ) g
2
ay
m2
m2 + m1
∆t = −
2∆y (m2 + m1 )
(m2 − m1 ) g
= −
2(0 − 1.4 m)(9.2 kg + 3.6 kg)
(9.2 kg − 3.6 kg)(9.80 m s 2 )
275
= 0.81 s .
m1
Chapter 4: Force and Newton’s Laws of Motion
Physics
92. (a) Strategy Use Newton’s second law. Since the pumpkin and the watermelon are attached by the cord, they
must have the same magnitude of acceleration.
T
T
x
Solution For the pumpkin:
∑ Fx = mp g sin 53.0° − T = mp a (1)
x
m pg
For the watermelon:
∑ Fx = T − mw g sin 30.0° = mw a (2)
Adding (1) and (2) gives
g (mp sin 53.0° − mw sin 30.0°) = (mp + mw )a.
mwg
30°
53°
Solving for a, we have
g (mp sin 53.0° − mw sin 30.0°) (9.80 m s 2 )[(7.00 kg) sin 53.0° − (10.0 kg) sin 30.0°]
=
= 0.34 m s 2 .
a=
7.00 kg + 10.0 kg
mp + mw
The acceleration is positive, so the watermelon slides up the ramp and the pumpkin slides down. Therefore,
G
the acceleration is a = 0.34 m s 2 , where the watermelon moves up and to the left .
(b) Strategy The pumpkin will travel down the ramp with the acceleration found in part (a).
Solution Use Eq. (2-15).
1
1
∆x = a (∆t ) 2 = (0.34 m s 2 )(0.30 s)2 = 1.5 cm
2
2
(c) Strategy Use Eq. (2-12).
Solution
v = a∆t = (0.34 m s 2 )(0.20 s) = 6.8 cm s
93. Strategy Let the +x-direction be down the incline. Use Newton’s second law and Eq. (2-16).
Solution Find the acceleration of the glider.
∑ Fx = mg sin θ = ma x , so a x = g sin θ .
Find the angle of inclination.
vfx 2 − vix 2 = vfx 2 − 0 = 2a x ∆x = 2 g sin θ∆x, so
θ = sin −1
vfx 2
2 g ∆x
= sin −1
(0.250 m s) 2
2(9.80 m
s 2 )(0.500
m)
x
θ
θ
mg
mg sinθ
= 0.365° .
The slope is a x = g sin θ = 0.0625 m s 2 = 6.25 cm s 2 .
The positions and times are shown in the graph.
vx (cm/s)
30
20
10
0
276
0
1
2
3
4 t (s)
Physics
Chapter 4: Force and Newton’s Laws of Motion
94. Strategy Let the +x-direction be down the incline. Use Newton’s second law.
Solution
∑ Fx = mg sin θ = ma x
55°
x
mg
mg sin 55°
55°
G
(a) Fnet = mg sin θ = (10.0 kg)(9.80 m s 2 ) sin 55° = 80 N, so Fnet = 80 N directed down the incline .
(b) ma x = mg sin θ , so a x = g sin θ = (9.80 m s 2 ) sin 55° = 8.0 m s 2 , so
G
a = 8.0 m s 2 directed down the incline .
(c) ∆vx = a x ∆t , so ∆t =
∆vx
ax
=
10.0 m s
(9.80 m s 2 ) sin 55°
= 1.2 s .
(d) The motion diagram shows the distance traveled for equal time intervals.
t = 0 s 1/4 s 1/2 s
x=0m
1.0 m
3/4 s
4/4 s
5/4 s
2.25 m
4.0 m
6.25 m
(e) The slope of the graph is a x = 8.0 m s 2 . The vx vs. t graph is shown.
vx (m/s)
12
8
4
0
0
1/4
1/2
3/4
4/4
5/4 t (s)
95. (a) Strategy The force of static friction is greater than the applied force. Draw a diagram.
Solution Find the possible values for the coefficient of static
friction.
F 120 N
= 0.48.
fs = µs N > F , so µs > =
N 250 N
Therefore, µs > 0.48 .
N
F
fs
Fg
(b) Strategy Refer to part (a).
Solution Compute the coefficient of static friction.
F 150 N
µs = =
= 0.60
N 250 N
(c) Strategy Refer to part (a), but with the coefficient of kinetic friction instead of that for static friction.
Solution Compute the coefficient of kinetic friction.
µk = F N = (120 N) (250 N) = 0.48
277
Chapter 4: Force and Newton’s Laws of Motion
Physics
96. Strategy Draw free-body diagrams for each crate. Then use Newton’s second law to find the tensions.
Solution Upper crate (1):
∑ Fy = T1 − T2 − m1 g = m1a y , so T1 = T2 + m1 g + m1a y .
y
T1
Lower crate (2):
∑ Fy = T2 − m2 g = m2 a y , so T2 = m2 g + m2 a y .
T2
200 kg
Calculate T1.
T2
T1 = T2 + m1 g + m1a y = m2 g + m2 a y + m1 g + m1a y
100 kg
m2g
m1g
= m2 ( g + a y ) + m1 ( g + a y ) = (m2 + m1 )( g + a y )
= (100 kg + 200 kg)(9.80 m s 2 + 1.0 m s 2 ) = 3.2 kN
Calculate T2 .
T2 = m2 g + m2 a y = m2 ( g + a y ) = (100 kg)(9.80 m s 2 + 1.0 m s 2 ) = 1.1 kN
97. Strategy Draw a free-body diagram and use Newton’s second law.
Solution The elevator floor pushes upward on Oliver with a force equal to the
normal force.
∑ Fy = N − W = N − mg = ma y , so N = m( g + a y ).
y
N
a
Therefore, the magnitude of the force exerted by the floor is
F = m( g + a y ) = (76.2 kg)(9.80 m s 2 − 1.37 m s 2 ) = 642 N .
W
98. Strategy When Jaden is on the ground, his weight is equal to mg = 600 N. While on the accelerating elevator,
his apparent weight is 550 N. Since 550 N < 600 N, the acceleration must be downward.
Solution According to Newton’s second law, ∑ Fy = N − W = ma y , so
N −W W ′ −W
⎛W′ ⎞
⎛ 550 N ⎞
=
= g⎜
− 1⎟ = (9.80 m s 2 ) ⎜
− 1⎟ = − 0.8 m s 2 , or
ay =
m
W g
⎝W
⎠
⎝ 600 N ⎠
G
a = 0.8 m s 2 downward .
N
y
a
W
99. Strategy When Ian is on the ground, his weight is equal to mg = 640 N. While on the elevator, his apparent
weight is 700 N. Since 700 N > 640 N, the elevator and Ian must be accelerating upward.
Solution According to Newton’s second law, ∑ Fy = N − W = ma, so
N −W W ′ −W
⎛W′ ⎞
a=
=
= g⎜
− 1⎟ .
m
W g
⎝W
⎠
The magnitude of the net force on the system is
⎛W′ ⎞
⎛ 700 N ⎞
− 1⎟ = (1050 kg)(9.80 m s 2 ) ⎜
− 1⎟ = 1 kN.
F = mcombined a = mcombined g ⎜
⎝W
⎠
⎝ 640 N ⎠
Therefore, the net force is 1 kN upward.
278
y
N
a
W
Physics
Chapter 4: Force and Newton’s Laws of Motion
100. Strategy Refer to Example 4.19.
Solution
(a) The elevator is accelerating downward, so a y = − 0.50 m s 2 .
⎛ ay ⎞
⎛ −0.50 m s 2 ⎞
W
W ′ = ( g + a y ) = W ⎜1 +
= (598 N) ⎜ 1 +
⎟
⎟ = 567 N
⎜
⎜
g
g ⎟⎠
9.80 m s 2 ⎟⎠
⎝
⎝
(b) Since the elevator is moving downward and slowing down, it is accelerating upward, so a y = 0.50 m s 2 .
⎛ ay
W ′ = W ⎜1 +
⎜
g
⎝
⎞
⎛ 0.50 m s 2 ⎞
⎟⎟ = (598 N) ⎜⎜ 1 +
⎟ = 629 N
2⎟
⎝ 9.80 m s ⎠
⎠
101. Strategy The apparent weight is given by W ′ = m( g + a y ).
Solution
(a) Up is the positive direction. Solve for a y .
⎛ ay ⎞
⎛ ay
= W ⎜1 +
W ′ = m( g + a y ) = mg ⎜1 +
⎟
⎜
⎜
g ⎟⎠
g
⎝
⎝
G
So, a is 1.4 m s 2 downward .
⎞
⎛W′ ⎞
⎛ 120 lb ⎞
− 1⎟ = (9.80 m s 2 ) ⎜
− 1⎟ = −1.4 m s 2 .
⎟⎟ , so a y = g ⎜
W
140
lb
⎝
⎠
⎝
⎠
⎠
(b) With a downward acceleration, the elevator could be going up and slowing down, or going down and
speeding up, so the answer is no; one cannot tell whether the elevator is speeding up or slowing down.
102. Strategy Draw a free-body diagram and use Newton’s second law.
Solution The force Yolanda exerts on the floor of the elevator is equal and opposite to the
normal force exerted on her feet by the floor of the elevator.
∑ Fy = N − W = N − mg = ma y , so N = m( g + a y ). Therefore, the magnitude of the force
she exerts is F = m( g + a y ) = (64.2 kg)(9.80 m
s2
+ 2.13 m
s2 )
= 766 N. The force is
766 N downward.
103. Strategy The apparent weight is given by W ′ = W (1 + a y g ).
Solution Up is the positive direction. Find Felipe’s actual weight.
⎛ ay ⎞
W′
750 N
W ′ = W ⎜1 +
=
= 620 N .
⎟⎟ , so W =
a
⎜
2.0 m s 2
y
g ⎠
⎝
1+
1+
2
g
9.80 m s
279
y
N
a
W
Chapter 4: Force and Newton’s Laws of Motion
Physics
104. Strategy The apparent weight is given by W ′ = m( g + a y ).
Solution
(a) Up is the positive direction. L is for Luke and b is for the box.
WL′ = mL ( g + a y ) and WL′ + b = (mL + mb )( g + a y ).
Solve for mL (Luke’s mass) in the first equation, substitute for mL in the second, and solve for a y .
mL =
WL′
g + ay
Substitute.
⎛ W′
⎞
L + m ⎟ ( g + a ) = W ′ + m ( g + a ), so
WL′ + b = ⎜
b
L
b
y
y
⎜ g + ay
⎟
⎝
⎠
W ′ − WL′
1.200 × 103 N − 0.960 × 103 N
a y = L+ b
−g =
− 9.80 m s 2 = 2.2 m s 2 .
mb
20.0 kg
So, the acceleration of the elevator is 2.2 m s 2 up .
(b) Find Luke’s weight.
⎛ ay ⎞
WL′
0.960 × 103 N
WL′ = WL ⎜1 +
, so WL =
=
= 784 N .
⎟
⎜
g ⎟⎠
1 + a y g 1 + (2.2 m s 2 ) (9.80 m s 2 )
⎝
105. Strategy Consider the ranges and natures of the fundamental forces.
Solution Of all of the fundamental forces, the weak force has the shortest range (about 10−17 m). In the Sun, the
weak interaction enables thermonuclear reactions to occur, without which there would be no sunlight.
106. Strategy Consider the fundamental forces and their relative strengths.
Solution The gravitational force is the fundamental force that governs the motion of planets in the solar system.
Gravity is by far the weakest of the fundamental forces, though it dominates interactions at large scales primarily
because planets and larger bodies are extremely massive. The electromagnetic force is ineffective for such bodies
because they are electrically neutral. The final two fundamental forces, the weak and strong nuclear forces,
dominate interactions at very small distance scales, but have no effect on the large distance scales associated with
the motion of large bodies.
107. Strategy Consider the ranges of the forces given.
Solution The range of the strong force is about 10−15 m, so it certainly does not have unlimited range. Contact
forces are not unlimited, as well, since they are limited to the contact region between objects (and there are no
known objects of unlimited size). Both electromagnetic and gravitational forces have unlimited ranges.
108. Strategy Consider the nature of the forces given.
Solution The strong force holds protons and neutrons together in the atomic nucleus, and its range is much
smaller than the radius of an atom; thus, it is not the force that binds electrons to nuclei to form atoms. When
atoms on the surfaces of two objects come very close together, they interact via the electromagnetic force. These
are contact forces. So, contact force is an interaction between atoms, not within atoms; thus, it is not the force that
binds electrons to nuclei. Nuclei and electrons have masses so small that the gravitational forces between them are
vanishingly small; so, gravitational force is not the force that binds electrons to nuclei. So, we are left with
electromagnetism as the force that is the fundamental interaction that binds electrons to nuclei to form atoms.
280
Physics
Chapter 4: Force and Newton’s Laws of Motion
109. Strategy Consider the natures of the fundamental forces.
Solution Of the fundamental forces, the strong force is the strongest, hence its name. It is strong, but has a very
short range. But the range is just the right size (about 10−15 m) to be the fundamental interaction that binds
quarks together to form protons, neutrons, and many exotic subatomic particles.
110. Strategy The car is moving straight with constant speed, so the horizontal pair of forces and the vertical pair of
forces are equal in magnitude and opposite in direction. Let the subscripts be the following:
c = car
e = Earth
r = road
a = air
Solution Since the car is moving with constant velocity, the net force on the car is
zero. The free-body diagram is shown.
Ncr
Fca
Fcr
Wce
111. (a) Strategy Draw a diagram and use Newton’s second law to find the acceleration of the skier. Then, relate the
acceleration, speed, and distance using Eq. (2-16).
Solution Find the acceleration.
∑ Fx = mg sin θ = ma x , so a x = g sin θ .
Find the speed at the bottom of the slope.
vf2x − vi2x = 2a x ∆x
N
32°
x
vf2x − 0 = 2 g sin θ∆x
mg
32°
2
vfx = 2 g sin θ∆x = 2(9.80 m s )(sin 32°)(50 m) = 23 m s
(b) Strategy Find the acceleration using Eq. (2-16). Then, use Newton’s second law.
Solution Find a x .
vf2x
− vi2x
ax =
N
= 2a x ∆x, so
vf2x − vi2x
2 ∆x
Find µk .
=
0−
(
2 g sin θ∆xslope
2∆x
)
2
=−
2 g sin θ∆xslope
2∆x
=−
g sin θ∆xslope
∆x
ma
∑ Fx = − f k = ma x , so f k = µk N = − ma x and µk = − x .
N
∑ Fy = N − mg = 0, so N = mg . Thus, we have
µk = −
ma x
N
=−
sin θ∆xslope (sin 32°)(50 m)
a
=− x =
=
= 0.19 .
mg
g
∆x
140 m
ma x
281
y
x
v
fk
.
mg
Chapter 4: Force and Newton’s Laws of Motion
Physics
112. Strategy Draw a diagram and use Newton’s laws of motion.
Solution According to the first law, for the box to move with
constant speed, the net force on the box must by zero. Calculate the
magnitude of the force of the push required.
∑ Fy = N − mg cos θ = 0, so N = mg cos θ .
∑ Fx = F − f k − mg sin θ = 0, so
F = f k + mg sin θ = µk N + mg sin θ
= µk mg cos θ + mg sin θ = mg ( µk cos θ + sin θ )
N
y
x
F
fk
mg
θ
mg cos θ
mg sin θ
= (65 kg)(9.80 N kg)(0.30 cos 25° + sin 25°) = 440 N .
113. Strategy Use Newton’s laws of motion.
Solution
(a) Since the airplane is cruising in a horizontal level flight (straight line) at constant velocity, it is in equilibrium
and the net force is zero.
(b) The air pushes upward with a force equal to the weight of the airplane: 2.6 × 104 N .
114. Strategy Use Newton’s laws of motion. Let the +y-direction be up and the +x-direction be to the right.
Solution
(a) The tension T is the same along the length of the cord. Its magnitude is equal to the weight of the leg (and the
weight of the hanging weight), 22 N. The only vertical force is due to this tension, so Fy = 22 N.
Find the magnitude of the total force of the traction apparatus applied to the leg.
∑ Fx = T cos θ + T cos θ − Fx = 0, so Fx = 2T cos θ = 2(22 N) cos 30.0° = 38 N.
Thus, F = (38 N) 2 + (22 N) 2 = 44 N .
(b) The horizontal force is Fx = 38 N .
(c) The magnitude of the horizontal force acting on the femur is equal to the horizontal component of the traction
force acting on the leg, Fx = 38 N .
115. Strategy The forces on the forearm in addition to that of the biceps are gravity due to the Earth, a downward
force due to the upper arm bone, and the downward force due to the 100-N weight. Let the subscripts be the
following:
f = forearm
e = Earth
b = biceps
w = weight
u = upper arm bone
Solution The free-body diagram is shown.
Ffb
Ffw
Ffu
Ffe
282
Physics
Chapter 4: Force and Newton’s Laws of Motion
116. Strategy Let the subscripts be the following:
i = ice
e = Earth
s = stone
o = opponent’s stone
Solution
(a) The only forces on the stone are gravity due to the
Earth and the normal force due to the ice.
Nsi
Wse
(b) As the stone slides down the rink, it experiences a force
of kinetic friction opposite to its motion.
Nsi
fsi
Wse
(c) The additional force is that due to the opponent’s stone.
Nsi
fsi
Fso
Wse
117. (a) Strategy Neglect frictional forces. Identify all of the forces acting on the car; then draw a free-body diagram.
Solution The forces are the normal force due to the road,
the gravitational force due to the Earth, and the tension due
to the rope. The diagram is shown including the angles.
y
N
5.0°
x
T
10.0°
Fg
(b) Strategy Use Newton’s laws of motion and the free-body diagram.
Solution Find the tension.
sin 5.0°
N.
cos10.0°
Fg
T sin10.0°
mg
T sin10.0°
.
∑ Fy = N cos 5.0° + T sin10.0° − Fg = 0, so N =
−
=
−
cos 5.0°
cos 5.0°
cos 5.0°
cos 5.0°
Solve for T.
T sin10.0° ⎞ tan 5.0°
sin 5.0°
sin 5.0° ⎛ mg
T=
N=
mg − T tan 5.0° tan10.0°, so
−
⎜
⎟=
cos10.0°
cos10.0° ⎝ cos 5.0°
cos 5.0° ⎠ cos10.0°
tan 5.0°
tan 5.0°(1000 kg)(9.80 N kg)
T (1 + tan 5.0° tan10.0° ) =
mg and T =
= 860 N .
cos10.0°
cos10.0° (1 + tan 5.0° tan10.0° )
∑ Fx = T cos10.0° − N sin 5.0° = 0, so T =
283
Chapter 4: Force and Newton’s Laws of Motion
Physics
118. Strategy Use Newton’s laws of motion. Draw diagrams.
Solution
(a) Below are two diagrams for each of the two cases.
(1)
T1
T2
W1
W2
(2)
T1
T2
(3)
T3
(4)
T4
T3
W3
Sum the forces in (1).
Left weight: ∑ Fy = T1 − W1 = 0, so T1 = W1 = 550 N.
Right weight: ∑ Fy = T2 − W2 = 0, so T2 = W2 = 550 N.
Since W1 = W2 , T1 = T2 = 550 N; the tensions are the same, and the scale in (2) has forces exerted on it of
magnitude 550 N, which are opposite in direction
Sum the forces in (3).
Single weight: ∑ Fy = T3 − W3 = 0, so T3 = W3 = 550 N. The scale in (4) is in equilibrium, so
T4 = T3 = 550 N.
Both scales are in equilibrium and each has two forces which are equal in magnitude and opposite in direction
exerted on it. The magnitudes of the forces are equal, so in each case, the two ropes pull on the scale with
forces of 550 N in opposite directions, so the scales give the same reading.
(b) The reading on each scale is equal to the tension in the rope, 550 N.
119. Strategy Use the method of Example 4.6.
Solution Find the change in the gravitational field strength.
−2
−2
⎛
⎛
h ⎞
h ⎞
= ⎜⎜1 +
⎟⎟ , so g = gsurface ⎜⎜1 +
⎟⎟ .
Wsurface gsurface ⎝ RE ⎠
⎝ RE ⎠
Compute the change in the gravitational field strength.
−2 ⎤
−2
⎡ ⎛
⎡ ⎛
h ⎞ ⎥
8850 m ⎞ ⎤
∆g = gsurface − g = gsurface ⎢1 − ⎜⎜ 1 +
= (9.80 N kg) ⎢1 − ⎜ 1 +
⎟⎟
⎟ ⎥ = 0.027 N kg
⎢ ⎝ RE ⎠ ⎥
⎢⎣ ⎝ 6.37 × 106 m ⎠ ⎥⎦
⎣
⎦
W
=
g
120. Strategy The gravitational field strengths at sea level at the equator and at the North Pole are 9.784 N/kg and
9.832 N/kg, respectively. Weight is directly proportional to g.
Solution Find the percentage by which the weight of an object changes when moved from the equator to the
North Pole.
9.832 − 9.784
× 100% = 0.49%
9.784
284
Physics
Chapter 4: Force and Newton’s Laws of Motion
121. (a) Strategy For the sum of the two forces to be in the forward (+y) direction, the net force in the x-direction
must be zero. Draw a diagram and use Newton’s laws of motion.
Solution Compute the magnitude of the force.
∑ Fx = F sin 38° − (105 N) sin 28° = 0, so
(105 N) sin 28°
= 80 N .
F=
sin 38°
28°
38°
y
105 N
F
x
(b) Strategy Find the sum of the y-components of the two forces to find the magnitude of the net force on the
barge from the two tow ropes.
Solution Find the magnitude of the force.
∑ Fy = F cos 38° + (105 N) cos 28° = (80 N) cos 38° + (105 N) cos 28° = 160 N
122. Strategy Use Newton’s laws of motion and draw a free-body diagram.
Solution Find the tension in the cable.
mg
∑ Fy = T cos θ − mg = 0, so T =
.
cos θ
y
T θ
mg
Fwall
123. (a) Strategy The force on the patella bisects the angle between the directions of the tension of the tendons. So,
the angle is (37.0° + 80.0°) 2 = 58.5°. Use Newton’s laws of motion.
Solution Find the magnitude of the contact force exerted on the patella by the femur.
ΣF = F − T cos θ − T cosθ = 0, so F = 2T cos θ = 2(1.30 kN) cos 58.5° = 1360 N .
(b) Strategy Refer to part (a).
Solution Find the direction of the contact force.
θ = 58.5° − 37.0° = 80.0° − 58.5° = 21.5°
124. Strategy The static and kinetic friction forces always oppose the force applied to the block. Use Newton’s laws
of motion to analyze the forces on the block and the motion of the block in each case.
Solution
(a) The applied force must overcome the maximum force of static friction. So, the minimum horizontal applied
force required to make the block start to slide is F = fs = µs N = µs mg = 0.35(4.6 kg)(9.80 m s 2 ) = 16 N .
(b) Once the block is sliding, the force required to keep it sliding at constant velocity is equal and opposite to the
maximum force of kinetic friction. Since the maximum force of kinetic friction is less than the maximum
force of static friction, the applied force is greater than necessary for constant velocity; therefore, the block
will accelerate.
(c) Use Newton’s second law to find the acceleration of the block.
F − f k µs mg − µk mg
∑ Fx = F − f k = ma, so a =
=
= ( µs − µk ) g = (0.35 − 0.22)(9.80 m s 2 ) = 1.3 m s 2 .
m
m
285
Chapter 4: Force and Newton’s Laws of Motion
Physics
125. Strategy Let left be positive. Both blocks move with acceleration a (to the left). Use Newton’s second law.
Solution For the two-block system:
Fnet = F = (2m + m)a = 3ma
m
F
2m
Let F12 be the force of the smaller block on the larger block and F21 be the force
of the larger block on the smaller block. Also, by Newton’s third law, F21 = − F12.
For the smaller block:
Fnet = F21 = ma
Find F12.
F21 ma 1
F
F
=
= , so
= F21 = − F12 , or F12 = − .
F
3ma 3
3
3
So, the force of the smaller block on the larger block is F 3 to the right .
126. Strategy and Solution Since the 10 cars are identical, each contributes one tenth of the force opposing the force
of the locomotive. The first pulls with nine tenths of the force, the second pulls with eight tenths, etc. So, the last
car pulls with one tenth of the magnitude of the force of the locomotive and is opposite the force’s direction.
2.0 × 106 N east
−
= 2.0 × 105 N west
10
127. (a) Strategy The force of gravity on the brick must be equal and opposite to the maximum force of static friction
on the brick for it to just begin to slide.
Solution Draw a diagram. Use Newton’s second law.
f
µ N
∑ Fx = fs − mg sin θ = 0, so sin θ = s = s .
mg
mg
∑ Fy = N − mg cos θ = 0, so N = mg cos θ . Find θ .
µ N µ mg cos θ
sin θ = s = s
, so tan θ = µs or
mg
mg
θ = tan −1 µs = tan −1 0.40 = 22° with respect to the horizontal .
N
y
x
f
m
θ
θ
mg
(b) Strategy After the brick starts to slide, the net force on it is the difference between the force of gravity and
the maximum force of kinetic friction. Use Newton’s second law.
Solution
∑ Fx = f k − mg sin θ = ma, so
f k − mg sin θ µk mg cos θ − mg sin θ
=
= − g (sin θ − µk cos θ )
m
m
= − (9.80 m s 2 )[sin(tan −1 0.40) − 0.30 cos(tan −1 0.40)] = − 0.9 m s 2 .
a=
The acceleration of the brick is 0.9 m s 2 down the incline .
286
Physics
Chapter 4: Force and Newton’s Laws of Motion
128. (a) Strategy Since mg = (51 kg)(9.80 m s 2 ) = 500 N > 408 N, the woman feels less than her normal weight, so
the elevator is accelerating downward. Use Newton’s second law.
Solution Let the +y-direction be up.
∑ Fy = 408 N − mg = ma y , so a y =
408 N − mg 408 N
=
− 9.80 m s 2 = −1.8 m s 2 .
m
51 kg
G
Thus, a = 1.8 m s 2 down .
(b) Strategy Find the change in speed of the elevator after 4.0 s at the acceleration found in part (a).
Solution Let down be positive. Find the speed of the elevator.
∆v = vf − vi = a y ∆t , so vf = vi + a y ∆t = 1.5 m s + (1.8 m s 2 )(4.0 s) = 8.7 m s .
129. (a) Strategy Determine the maximum force of static friction and compare it to the force of the push, 5.0 N.
Solution Find the maximum force of static friction.
fs,max = µs N = µs mg Moon = 0.35(2.0 kg)(9.80 m s 2 ) = 6.9 N > 5.0 N
No, the puck does not move, since the maximum force of static friction is greater than the force of the push.
(b) Strategy and Solution Since 7.5 N > 6.9 N, the maximum force of static friction, yes, the puck does move.
(c) Strategy Use Newton’s second law.
Solution Find the acceleration of the puck.
∑ F = Fpush − f k = Fpush − µk mg = ma, so
a=
Fpush
m
− µk g =
6.0 N
− 0.25(9.80 m s 2 ) = 0.6 m s 2 .
2.0 kg
(d) Strategy Use Newton’s second law and the fact that gravity is weaker on the Moon.
Solution The acceleration of the puck is
Fpush
a=
− µk g.
m
Since gravity is weaker on the Moon, the second term on the right side of the equation is smaller than it
would be for Earth. This is the same thing as saying that the force of friction will be less on the Moon.
Therefore, the acceleration of the puck is more on the Moon than on Earth.
287
Chapter 4: Force and Newton’s Laws of Motion
Physics
130. Strategy Analyze the forces for each situation. The tension must never exceed 12 N.
Solution Find the tension in each wire. Refer to the diagram.
Figure (a)
∑ Fx = T50 cos 50° − T30 cos 30° = 0, so T50 cos 50° = T30 cos 30°, or
y
x
cos 50°
T .
cos 30° 50
∑ Fy = T30 sin 30° + T50 sin 50° − Fg = 0, so T30 sin 30° + T50 sin 50° = Fg .
30°
T30 =
T50
T30
(a)
Substitute for T30 .
T30 =
cos 50°
(13.2 N) = 9.8 N
cos 30°
T
T
(b)
Fg
Fg
⎛ cos 50°
⎞
T50 ⎟ sin 30° + T50 sin 50° = Fg
⎜
⎝ cos 30°
⎠
T50 ( cos 50° tan 30° + sin 50° ) = Fg
T50 =
50°
15 N
= 13 N
cos 50° tan 30° + sin 50°
Figure (b)
∑ Fy = T + T − Fg = 0, so 2T = Fg , or T = (15 N) 2 = 7.5 N. Since 13 N > 12 N, the arrangement in Fig. (a)
breaks the twine. Since 7.5 N < 12 N, the arrangement in Fig. (b) successfully hangs the picture.
131. Strategy Use Newton’s laws of motion. Neglect friction and draw a diagram. The slope of the incline is equal to
tan θ . Let +x be up the incline.
Solution Find the magnitude of the force exerted on the
rollercoaster by the chain.
∑ Fy = 0
∑ Fx = F − mg sin θ = 0, since the speed is constant.
∆y ⎞
⎛
Thus, F = mg sin θ = mg sin ⎜ tan −1
⎟
∆x ⎠
⎝
3.0 m ⎞
⎛
= (400.0 kg)(9.80 N kg ) sin ⎜ tan −1
⎟ = 120 N .
100.0 m ⎠
⎝
N
F
v
mg
mg cos θ
3.0 m
mg sin θ
θ
100.0 m
132. (a) Strategy Use Newton’s law of universal gravitation.
Solution Find the weight of the satellite when in orbit.
GM E m (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)(320 kg)
Fg =
=
= 250 N
r2
(6.371× 106 m + 16,000 × 103 m)2
(b) Strategy The weight on Earth is equal to the satellite’s mass times g.
Solution Find the weight of the satellite when it was on the launch pad.
Fg = mg = (320 kg)(9.80 N kg) = 3100 N
(c) Strategy Use Newton’s third law of motion.
Solution According to Newton’s third law if motion, the satellite exerts a force on the Earth equal and
opposite to the force the Earth exerts on it; that is, 250 N toward the satellite.
288
Physics
Chapter 4: Force and Newton’s Laws of Motion
133. Strategy Set the magnitudes of the forces on the spaceship due to the Earth and the Moon equal. (The forces are
along the same line.)
Solution Find the distance from the Earth expressed as a percentage of the distance between the centers of the
Earth and the Moon.
GM E m
GM M m
ME
FsE =
, so rE = rM
= FsM =
= 9.02rM .
2
2
0.0123M E
rE
rM
Find the percentage.
9.02rM
rE
9.02
=
=
= 0.900
rE + rM 9.02rM + rM 10.02
The distance from the Earth is 90.0% of the Earth-Moon distance.
134. (a) Strategy Use Eq. (3-12) to find the height and Eq. (3-10) to find the speed of the rocket when it runs out of
fuel. Then, use Eq. (3-13) to find the maximum height of the rocket, where the height and speed of the rocket
when it runs out of fuel are the initial conditions and its final speed is zero.
Solution Find the height h1 and speed v1 of the rocket when it runs out of fuel.
1
1
1
∆y = viy ∆t + a y (∆t ) 2 = 0 + a y (∆t )2 = a y (∆t ) 2 = h1 and ∆v y = a y ∆t = v1.
2
2
2
Find the maximum height.
vf2y − vi2y = −2 g ∆y = 0 − vi2y = −2 g ( yf − yi ), so
vi2y
(a y ∆t ) 2 1
v2 1
[(17.5 m s 2 )(1.5 s)]2
yf = yi +
= h1 + 1 = a y (∆t ) 2 +
= (17.5 m s 2 )(1.5 s) 2 +
= 55 m .
2g
2g 2
2g
2
2(9.80 m s 2 )
(b) Strategy The rocket runs out of fuel after ∆t1 = 1.5 s. Then it travels for a time ∆t2 before it reaches its
maximum height. Finally, it falls freely (from rest) for a time ∆t3 until it reaches the ground.
Solution Find ∆t2 using Eq. (3-10).
a y ∆t1
v
.
vfy − viy = a y ∆t = − g ∆t2 = 0 − v1 , so ∆t2 = 1 =
g
g
Find ∆t3 using Eq. (3-12).
2 yf
1
.
g (∆t3 ) 2 , so ∆t3 =
g
2
Find the total time of flight.
a y ∆t1
2 yf
(17.5 m s 2 )(1.5 s)
2(55 m)
∆t1 + ∆t2 + ∆t3 = ∆t1 +
+
= 1.5 s +
+
= 7.5 s
2
g
g
9.80 m s
9.80 m s 2
yf =
289
Chapter 4: Force and Newton’s Laws of Motion
Physics
135. Strategy Use Newton’s second law. Neglect the mass of the rocket fuel.
Solution
(a) According to Newton’s second law, the net force on the rocket during the first 1.5 s after lift-off was
G
G
F = ma = (0.087 kg)(17.5 m s 2 upward) = 1.5 N upward .
(b) The burning fuel had to overcome the force of gravity on the rocket to give it a net acceleration of
17.5 m s 2 . According to Newton’s second law, ∑ Fy = Ffuel − mg = ma y , so
Ffuel = mg + ma y = m( g + a y ) = (0.087 kg)(9.80 m s 2 + 17.5 m s 2 ) = 2.4 N. Therefore, the force exerted on
the rocket by the burning fuel was 2.4 N upward .
(c) After the fuel was spent, the only force on the rocket was that due to gravity, so the net force on the rocket
G
G
was F = mg = (0.087 kg)(9.80 m s 2 downward) = 0.85 N downward .
(d) The net force was the same as that found in part (c), 0.85 N downward . The acceleration was that due to
the force of gravity, 9.80 m s 2 downward .
136. Strategy Use Newton’s laws of motion.
Solution
(a) Since the train is moving at constant speed, and air resistance and friction are negligible, the readings on the
three scales are all 0.
(b) Air resistance and friction are not considered negligible this time. The engine pulls
the cars against these forces. Since the cars are identical, each car contributes about
one-third of the total frictional and drag forces. Each spring scale will measure the
net force due to the cars behind it, so the relative readings on the three spring scales
are A > B > C. The free-body diagram is shown.
N
FC
FB
f
Fair
W
(c) Spring A measures the forces on all 3 cars. Spring B measures the forces on the latter 2 cars.
Spring C measures the forces on the final 1 car.
A = 5.5 N + 5.5 N + 5.5 N = 16.5 N; B = 5.5 N + 5.5 N = 11.0 N; C = 5.5 N
137. (a) Strategy The tension due to the weight of the potatoes is divided evenly between the two sets of scales.
Solution Find the tension and, thus, the reading of each scale.
2T = mg , so T = mg 2 = (220.0 N) 2 = 110.0 N .
(b) Strategy Scales B and D will read 110.0 N as before. Scales A and C will read an additional 5.0 N due to the
weights of B and D, respectively.
Solution Find the reading of each scale.
TA = 110.0 N + 5.0 N = 115.0 N = TC and TB = 110.0 N = TD .
290
Physics
Chapter 4: Force and Newton’s Laws of Motion
138. (a) Strategy Let the subscripts be the following:
c = computer
d = desk
e = Earth
Solution The only forces on the computer are gravity due to the
Earth and the normal force due to the desk. The free-body diagram
is shown.
Ncd
Fce
(b) Strategy Consider the nature of friction forces.
Solution Since the only forces acting on the computer are in the vertical direction, the friction force is zero.
(c) Strategy Find the maximum force of static friction on the computer due to the desk; this is the horizontal
force necessary to make it begin to slide.
Solution F = fs = µs N = µsW = 0.60(87 N) = 52 N
139. (a) Strategy Identify the interactions between the magnet and other objects.
Solution The interactions are:
1) The gravitational forces between the magnet and the Earth
2) The contact forces, normal and frictional, between the magnet and the photo
3) The magnetic forces between the magnet and the refrigerator
(b) Strategy Refer to part (a). Let the subscripts be the following:
m = magnet
p = photo
e = Earth
r = refrigerator
Solution The magnet is in equilibrium, so the horizontal pair of forces
and the vertical pair of forces are equal in magnitude and opposite in
direction.
fmp
Fmr
Nmp
Wme
(c) Strategy Identify the range of each force and categorize each as long-range or contact.
Solution
The long-range forces are gravity and magnetism. The contact forces are friction and the normal force.
(d) Strategy Refer to part (b). Wme and Fmr are given.
Solution
Wme = 0.14 N, Fmr = 2.10 N, f mp = Wme = 0.14 N, and N mp = Fmr = 2.10 N.
291
Chapter 4: Force and Newton’s Laws of Motion
Physics
140. Strategy The force exerted on the upper scale is the combination of the weight of the crate and the tension of the
lower scale. The forces in the vertical direction sum to zero, since the system is in equilibrium. Draw a diagram.
Solution Find the reading of the upper scale.
0 = Tupper − W − Tlower , so
Tupper
Tupper = W + Tlower = mg + 120 N = (50.0 kg)(9.80 N kg) + 120 N = 610 N .
Tlower
W
141. (a) Strategy Scale A measures the weight of both masses. Scale B only measures the weight of the 4.0-kg mass.
Solution Find the readings of the two scales if the masses of the scales are negligible.
Scale A = (10.0 kg + 4.0 kg)(9.80 N kg) = 137 N and Scale B = (4.0 kg)(9.80 N kg) = 39 N .
(b) Strategy Scale A measures the weight of both masses and scale B. Scale B only measures the weight of the
4.0-kg mass.
Solution Find the readings if each scale has a mass of 1.0 kg.
Scale A = (10.0 kg + 4.0 kg + 1.0 kg)(9.80 N kg) = 147 N and Scale B = 39 N .
142. (a) Strategy and Solution Since the speed is constant, the acceleration is zero; therefore, the force of friction is
zero.
(b) Strategy The force of friction must be in the same direction as the motion of the truck, otherwise the crate
would slide off. The crate’s acceleration is the same as the truck’s.
Solution
f = Fct = ma =
W
1.0 m s 2
a = (180 N)
= 18 N
g
9.80 m s 2
The force of friction is 18 N in the forward direction.
(c) Strategy Set f max = ma and solve for a.
Solution
f
µ N µ mg
a = max = s = s
= µs g = 0.30(9.80 m s 2 ) = 2.9 m s 2
m
m
m
292
Physics
Chapter 4: Force and Newton’s Laws of Motion
143. Strategy Let the +x-direction be up the incline. Use Newton’s second law.
Solution
N
y
x
mg
θ
F cos θ
F
θ
mg cos θ
mg sin θ
(a) Let F be the horizontal force.
∑ Fx = F cos θ − mg sin θ = 0, so F = mg tan θ .
(b) To roll the crate up at constant speed, the net force is zero, so the force is that from part (a), mg tan θ .
(c) ∑ Fx = F cos θ − mg sin θ = ma, so F = mg tan θ +
ma
.
cos θ
144. Strategy Use Newton’s second law. Let T be the maximum tension.
Solution
(a) Find the acceleration of the block plus cart system.
T
∑ Fx = T = (m1 + m2 )a x , so a x =
.
m1 + m2
y
m2
x
Now, switch to the block system.
∑ Fy = N − m2 g = 0, so N = m2 g. ∑ Fx = f max = m2a x , since
the block must not slide.
⎛ T
⎞
= f max = µ N = µ m2 g , so
m2 a x = m2 ⎜
⎜ m + m ⎟⎟
2⎠
⎝ 1
T = (m1 + m2 ) µ g .
(b) Find the acceleration of the block plus cart system.
∑ Fx = T − (m1 + m2 ) g sin θ = (m1 + m2 )a x , so
T
− g sin θ .
ax =
m1 + m2
Now, switch to the block system.
∑ Fy = N − m2 g cos θ = 0, so N = m2 g cos θ .
∑ Fx = f max = m2a x , since the block must not slide.
⎛ T
⎞
− g sin θ ⎟ = f max = µ N = µ m2 g cos θ , so
m2a x = m2 ⎜
⎝ m1 + m2
⎠
T = (m1 + m2 ) g (µ cos θ + sin θ ) .
293
T
m1
N
fmax
m2
m2g
y
T
m2
x
m1
θ
(m1 + m2)g
N
m2
m 2g
fmax
θ
m2g cos θ
m2g sin θ
θ
(m1 + m2)g cos θ
(m1 + m2)g sinθ
Chapter 4: Force and Newton’s Laws of Motion
Physics
145. (a) Strategy Since the downward speed is decreasing at a rate of 0.10g, the acceleration of the truck is 0.10g
upwards. Use Newton’s second law.
Solution
∑ Fx = 0 and ∑ Fy = T − mg = ma y = m(0.10 g ) , so T = 1.10mg .
(b) Strategy and Solution Although the motion of the helicopter has changed, the acceleration of the truck is
the same as in part (a), so the tension is the same, 1.10mg .
146. Strategy Draw a diagram and use Newton’s laws of motion.
Solution
(a) The magnitude of the force of static friction on block A due to
the floor must be equal to the magnitude of the tension in the
cord, so T = fsA = µA N = µA mg .
The magnitude of the applied force must be equal to the
magnitude of the tension in the cord plus the magnitude of the
force of static friction on block B due to the floor. Thus,
F = T + fsB = µA mg + µB mg = mg ( µA + µB )
N
fsA
N
A
T
T
fsB
Fg
F
B
Fg
= (2.0 kg)(9.80 N kg)(0.45 + 0.30) = 15 N .
(b) T = µA mg = 0.45(2.0 kg)(9.80 N kg) = 8.8 N
147. Strategy Let the +y-direction be in the direction of the 360.0-N force (F). Draw a diagram and use Newton’s
laws of motion.
Solution Find the force exerted on the poplar tree.
∑ Fy = F − 2T sin θ = 0 before the poplar is cut through.
So, F = 2T sin θ . The force exerted on the poplar is the tension T, so T =
Refer to the figure in the text to find sin θ .
displacement
2.00 m
=
= 0.0995
sin θ =
half length of rope
(20.0 m) 2 + (2.00 m)2
Thus, T =
T
T
θ
F
.
2sin θ
360.0 N
= 1810 N . Compare the forces.
2(0.0995)
1810 N
≈ 5 times the force with which Yoojin pulls
360.0 N
The values for the two situations are different because the oak tree supplies additional force.
294
θ
360.0 N
y
Physics
Chapter 4: Force and Newton’s Laws of Motion
148. Strategy Use Newton’s laws of motion. Let +y be up and +x be to the right.
Solution
G
(a) Find the magnitude of Fc .
cos θ
.
cos φ
W + Fm sin θ
ΣFy = − Fm sin θ + Fc sin φ − W = 0, so Fc =
.
sin φ
ΣFx = Fm cos θ − Fc cos φ = 0, so Fc = Fm
Eliminate Fc and solve for φ .
cos θ W + Fm sin θ
=
cos φ
sin φ
W + Fm sin θ
W
tan φ =
=
+ tan θ
Fm cos θ
Fm cos θ
⎛ W
⎞
⎡
⎤
50.0 N
φ = tan −1 ⎜
+ tan θ ⎟ = tan −1 ⎢
+ tan 35° ⎥ = 60°
F
cos
θ
(60.0
N)
cos
35
°
⎣
⎦
⎝ m
⎠
cos 35°
So, Fc = (60.0 N)
= 98 N .
cos 59.8°
Fm
(b) As found in part (a), φ = 60° above the horizontal .
149. (a) Strategy Set the magnitudes of the forces on the spacecraft due to the Earth and the Sun equal.
Solution Find the distance of the spacecraft from Earth.
GM S m
GM E m
r
ME
FsS =
, so E =
.
= FsE =
2
2
rS
MS
rS
rE
This is the ratio of the Earth-spacecraft distance to the Sun-spacecraft distance. If this is multiplied by the
Earth-Sun mean distance, the product is the distance of the spacecraft from the Earth.
(1.50 × 1011 m)
5.974 × 1024 kg
1.987 × 1030 kg
= 2.60 × 108 m from Earth
(b) Strategy Imagine the spacecraft is a small distance d closer to the Earth and find out which gravitational
force is stronger, the Earth’s or the Sun’s.
Solution At the equilibrium point the net gravitational force is zero. If the spacecraft is closer to the Earth
than the equilibrium point distance from the Earth, then the force due to the Earth is greater than that due to
the Sun. If the spacecraft is closer to the Sun than the equilibrium point distance from the Sun, then the force
due to the Sun is greater than that due to the Earth. So, if the spacecraft is close to, but not at, the equilibrium
point, the net force tends to pull it away from the equilibrium point.
295
Chapter 4: Force and Newton’s Laws of Motion
Physics
150. Strategy Consider the nature of normal and friction forces. Use Newton’s laws of motion.
Solution
(a) The normal force exerted by the hand is directed toward the wall. (The normal force must be perpendicular to
the plane of the picture.)
(b) The normal force exerted by the wall on the picture is opposite that due to the hand; it is away from the wall.
(c)
The y-component of the force of the hand is (6.0 N) cos 40° = 4.6 N.
This is less than the weight of the painting, so the force of friction
has a magnitude of 5.0 N − 4.6 N = 0.4 N and points in the positive
y-direction. (See the diagram.) The normal force is equal to the xcomponent of the force of the hand, which is (6.0 N) sin 40° = 3.9 N.
fs
0.4 N
=
= 0.1 .
Since fs = µs N , µs =
N 3.9 N
fs
40°
N
y
6.0 N
x
5.0 N
If the frictional force on the picture exerted by the hand is less than the force exerted on the picture
due to gravity, the frictional force on the picture due to the wall is directed upward so that the net
vertical force is zero. If the frictional force exerted by the hand is greater than that due to gravity, the
force due to the wall is directed downward for the same reason.
151. Strategy Use Newton’s second law. Draw a free-body diagram.
Solution The maximum force of static friction must be equal to the force of gravity on
the stuntman for him to just stay on the front of the truck at the minimum acceleration.
Find the minimum acceleration.
∑ Fx = N = ma and ∑ Fy = fs − mg = 0, so fs = µs N = µs ma = mg .
Therefore, the magnitude of the acceleration is a =
g
µs
=
9.80 m s 2
= 15 m s 2 .
0.65
152. (a) Strategy According to Newton’s second law, Fx = ma x . Since v y is constant, Fy = mg.
Solution Measured from the vertical,
F
ma
0.86
θ = tan −1 x = tan −1 x = tan −1
= 5.0° from the vertical .
9.81
Fy
mg
(b) Strategy and Solution Refer to part (a).
F
a x = air sin θ = (0.86 m s 2 ) 2 + (9.81 m s 2 ) 2 sin 3.0° = 0.52 m s 2
m
Fair
cos θ − g = (0.86 m s 2 )2 + (9.81 m s 2 )2 cos 3.0° − 9.81 m s 2 = 0.02 m s 2
ay =
m
296
fs
N
mg
Physics
Chapter 4: Force and Newton’s Laws of Motion
153. Strategy Choose the +x-axis to the right and +y-axis up. Use Newton’s second law.
Solution
y
(a) For m1:
∑ F1 y = N − W1 = N − m1 g = 0, so N = m1 g .
∑ F1x = T − f k = T − µk N = T − µk m1 g = m1a1x
For m2 :
x
fk
N
T
m1
T
∑ F2 x = 0 and ∑ F2 y = T − W2 = T − m2 g = m2 a2 y .
m1g
m2
Now, a1x and a2 y must be equal in magnitude, otherwise the cord will
m2g
compress or expand. a1x is in the +x-direction and a2 y is in the − y -direction.
So, let a = a1x = − a2 y . Then, T − µk m1 g = m1a and T − m2 g = − m2 a.
Subtract the second equation from the first and solve for a.
− µk m1 g + m2 g = m1a + m2 a
g (m2 − µk m1 ) = a (m1 + m2 )
a=
m2 − µk m1
g
m1 + m2
Find T.
T − m2 g = − m2 a
T = m2 g − m2
T = (1 + µ k )
m2 − µk m1
m (m + m2 ) − m2 (m2 − µk m1 )
m m + m22 − m22 + m1m2 µk
g= 2 1
g= 1 2
g
m1 + m2
m1 + m2
m1 + m2
m1m2
g
m1 + m2
(b) For m1 << m2 ,
m − µk (0)
a≈ 2
g = g and
0 + m2
mm
T ≈ (1 + µk ) 1 2 g = (1 + µk )m1 g << m2 g , so the tension is negligible compared to the weight of m2 ; .
0 + m2
it’s essentially in free fall.
If m1 >> m2 , the force of friction between the table and m1 is so large that m1 will not slide, so a = 0 and
T − m2 g = − m2 a = 0, or T = m2 g .
For m1 = m2 = m,
a=
(c)
m − µk m
m(1 − µk )
1
m2
1
g=
g=
(1 − µk ) g and T = (1 + µk )
g=
(1 + µk )mg .
m+m
2m
2
2m
2
a = 0 only for m2 = 0; thus, there is no value at which the two masses slide with constant velocity. For
m2 = 0, there is no tension in the cord.
297
Chapter 5
CIRCULAR MOTION
Conceptual Questions
1. Depressing the gas pedal is not the only way to make the car accelerate. The driver can also apply the brakes or
turn the car to make it accelerate.
2. (a) The child farthest from the axis, the one at 4 m, has the larger linear speed.
(b) Again, the child farthest from the axis, at 4 m, has the larger acceleration.
(c) Both children have the same angular speed.
(d) Both have the same angular displacement.
3. Newton’s law tells us that the gravitational force, which depends on the inverse orbital radius squared, is
proportional to the acceleration. The radial acceleration itself is proportional to the square of the orbital speed and
inversely proportional to the radius. This connects the two quantities so that they are not independent.
Alternatively, by Kepler’s third law the square of the orbital period, which is inversely proportional to the orbital
speed, is proportional to the cube of the orbital radius.
4. In uniform circular motion the velocity is not constant because it is changing direction. The speed however is
constant. The acceleration is also not constant since it too is changing direction, always pointing toward the center
of the circle. The magnitude of the acceleration is, however, constant.
5. In uniform circular motion, the acceleration always points toward the center of the circle. Hence it remains
perpendicular to the velocity the whole time. When a projectile is launched horizontally, the acceleration is
initially perpendicular to the velocity, but does not remain so.
6. The only force acting on the satellite in circular orbit is the gravitational force from the planet. The magnitude of
this force is proportional to both the mass of the satellite and the mass of the planet. The radial acceleration of the
satellite is equal to the gravitational force divided by the satellite’s inertial mass. Because the gravitational and
inertial masses are identical, the magnitude of the satellite’s acceleration is independent of the satellite’s mass—it
does however depend on the mass of the planet. Furthermore, the radial acceleration of the satellite is proportional
to the square of its velocity—the velocity of the satellite therefore depends upon the mass of the planet but not on
the mass of the satellite.
7. The tangential acceleration component at the rim of the flywheel is equal to the product of the flywheel’s angular
acceleration and its radius. Given that the angular acceleration is constant, the tangential acceleration at the rim
must also be constant. The radial acceleration component at the rim is proportional to the square of the tangential
velocity at that radius. Given that the tangential acceleration component is non-zero, the tangential velocity—and
therefore the radial acceleration component—must be changing.
8. The gravitational force on the Moon due to the Earth pulls the Moon toward the Earth just enough to bend its path
into a circle. Equivalently, the Moon is moving just fast enough to stay in a circular orbit. If the Moon was at the
same distance from Earth but moving more slowly, the same gravitational force would pull the Moon on an
inward spiral.
9. When the roller coaster turns hard to the right, the inertia of a rider’s upper body keeps it moving in a straight line
until it runs into the wall of the car. The wall exerts a normal force on the upper body that causes it to accelerate
radially with the car. Thus, no force pushes the rider to the left as they enter a turn—the rider’s inertia simply
carries them forward while the car moves to the right.
298
Physics
Chapter 5: Circular Motion
10. The only two places where a scale would read a person’s true weight are the north and south poles of the Earth’s
rotational axis. A measure of weight by a bathroom scale is a measure of the normal force on a person’s feet at
that location. Using Newton’s second law, the normal force is found to be equal to the true weight of the person
minus a quantity that depends on the straight-line distance from the location of the measurement directly to the
rotation axis. This distance varies from a minimum value of zero at the north and south poles to a maximum value
at the equator. Thus, a scale will read a person’s true weight at the poles and will deviate most from the true
weight along the equator.
11. While the problem could be solved with this choice of axes, it would probably be easier with axes drawn so that
the y-axis is vertical and the x-axis horizontal. A car rounding a banked curve will typically have a horizontal
acceleration toward the center of the curve, but no vertical acceleration. Newton’s second law would then be
easier to set up and solve with the new axes.
12. The groom inverts the snifter over the olive and starts swirling the snifter. As he swirls it, the
olive starts to move in a circle. It pushes against the snifter and the snifter responds by
pushing back (normal force). If the bridegroom moves the snifter fast enough, the force by
the olive on the glass will get stronger, resulting in a stronger normal force. Once the normal
force is great enough, its upward component will overcome the force of gravity, and the olive
will rise in the glass. The clever bridegroom can then quickly invert the glass, keeping the
olive inside.
Problems
1. Strategy Find the arc length swept out by the carnival swing.
Solution Use Eq. (5-4).
⎛ 2π rad ⎞
s = rθ = (8.0 m)(120°) ⎜
⎟ = 17 m
⎝ 360° ⎠
2. Strategy The linear distance traveled divided by the circumference of the ball gives the number of revolutions.
Solution
x
x
18 m
=
=
= 18 rev
C 2π r 2π 0.31 m
2
(
)
3. Strategy During one minute, the second hand of a clock rotates 2π radians. Use the definition of angular speed.
Solution
∆θ
2π rad
=
= 0.105 rad s
ωav =
∆t
60.0 s
4. Strategy Use the conversion factor between degrees and radians, 360° = 2π rad, and the fact that there are
2π radians per revolution.
Solution
⎛ 2π rad ⎞
(a) (30.0°) ⎜
⎟ = 0.524 rad
⎝ 360° ⎠
⎛ 2π rad ⎞
(b) (135°) ⎜
⎟ = 2.36 rad
⎝ 360° ⎠
299
Chapter 5: Circular Motion
Physics
π
⎛1
⎞ ⎛ 2π rad ⎞
(c) ⎜ rev ⎟ ⎜
or 1.57 rad
⎟=
2
⎝4
⎠ ⎝ rev ⎠
⎛ 2π rad ⎞
(d) (33.3 rev) ⎜
⎟ = 209 rad
⎝ rev ⎠
5. Strategy Use Eq. (5-9) to find the angular speed of the bicycle’s tires.
Solution
v 9.0 m s
ω = =
= 26 rad s
r 0.35 m
6. (a) Strategy Use Eq. (5-9) to find the angular speed of the drum.
Solution
v 0.50 m s
ω = =
= 0.56 rad s
r 0.900 m
ω
0.900 m
0.50 m/s
(b) Strategy The linear distance traveled by the drum divided by its circumference gives the number of
revolutions it made.
Solution
y 6.0 m
6.0 m
=
=
= 1.1 rev
2π r
2π (0.900 m)
C
7. (a) Strategy and Solution There are 2π radians per revolution and 60 seconds per minute, so
⎛ 33.3 rev ⎞⎛ 2π rad ⎞ ⎛ 1 min ⎞
⎜
⎟⎜
⎟⎜
⎟ = 3.49 rad s .
⎝ min ⎠⎝ rev ⎠ ⎝ 60 s ⎠
(b) Strategy Use the relationship between linear speed and angular speed.
Solution Find the speed of the doll.
v = r ω = (0.13 m)(3.49 rad s) = 0.45 m s
8. Strategy Use the definition of average angular velocity.
Solution In 1.0 s, the wheel rotates
⎛ 2.0 rev ⎞⎛ 2π rad ⎞
∆θ = ω∆t = ⎜
⎟⎜
⎟ (1.0 s) = 4.2 rad .
⎝ 3.0 s ⎠⎝ rev ⎠
ω
∆θ
9. Strategy Use the conversion factor between degrees and radians and s = rθ , where s = 100.0 ft, θ = 1.5°, and r
is the radius of curvature.
Solution Find the radius of curvature of a “1.5° curve”.
s 100.0 ft ⎛ 360° ⎞
r= =
⎜
⎟ = 3800 ft
1.5° ⎝ 2π rad ⎠
θ
300
Physics
Chapter 5: Circular Motion
10. Strategy Use dimensional analysis.
Solution The dimensions are [L] [T] for v, 1 [T] for ω , and [L] for r.
vω :
[L] 1
⋅
= [L] [T]2
[T] [T]
2
v 2 ⎛ [L] ⎞ 1 [L]2 1
=
⋅
= [L] [T]2
: ⎜
⎟ ⋅
r ⎝ [T] ⎠ [L] [T]2 [L]
2
⎛ 1 ⎞
1
⋅ [L] = [L] [T]2
⎟ ⋅ [L] =
2
[T]
[T]
⎝
⎠
ω 2r : ⎜
Since the dimensions of acceleration are [L] [T]2 , all three expressions are verified.
11. (a) Strategy Use the relationship between linear speed and radial acceleration.
Solution The radius r is half the length of the rod.
v2
ar = , so v = rar = (1.0 m)(980 m s 2 ) = 31 m s .
r
(b) Strategy Use the relationship between angular speed and radial acceleration.
Solution
ar = ω 2 r , so ω =
ar
980 m s 2
=
= 31 rad s .
r
1.0 m
12. (a) Strategy and Solution The force of static friction between the inside wall of the rotor and the people’s
backs keeps them from falling.
(b) Strategy Use Newton’s second law and the relationship between angular speed and radial acceleration. Draw
a free-body diagram.
Solution
y
2
ΣFr = N = mar = mω r and ΣFy = fs − mg = 0. Solve for ω.
fs = µs N = µs mω 2 r = mg , so ω =
g
µs r
=
9.80 m s 2
= 3.1 rad s .
0.40(2.5 m)
fs
N
mg
13. Strategy Use the relationship between angular speed and radial acceleration.
Solution The number of seconds in one day is 86,400, so the angular speed of the Earth (and baobab) is
ω = 2π rad 86, 400 s. Compute the radial acceleration.
2
⎛ 2π rad ⎞
6
2
ar = ω 2 r = ω 2 REarth = ⎜
⎟ (6.371× 10 m) = 3.37 cm s
⎝ 86, 400 s ⎠
301
Chapter 5: Circular Motion
Physics
14. (a) Strategy Use the definition of angular speed and the relationship between the time per revolution and
angular speed.
Solution Find Earth’s daily angular displacement.
∆θ
2π
2π
ω =
∆t =
, so ∆θ = ω ∆t =
(1 d) = 1.72 × 10−2 rad .
∆t
T
365.25 d
G
(b) Strategy Use s = r ∆θ , where for small ∆θ , ∆v ≈ s and r = v ≈ v1 ≈ v2 .
Solution The orbital speed is the circumference divided by the period.
2π r
2π (1.50 × 1011 m)
G
∆v = v∆θ =
∆θ =
(1.72 × 10−2 rad) = 514 m s
T
3.156 × 107 s
G
∆v is perpendicular to the average velocity, so
G
∆v = 514 m s perpendicular to the average velocity .
∆v
v1
∆θ
v2
(c) Strategy Use the definition of average acceleration and the results from parts (a) and (b).
Solution First, find the magnitude of the average acceleration. The average acceleration is radial, so its
direction is perpendicular to the average velocity.
G
∆v 2π r ∆θ 2π (1.50 × 1011 m)(1.72 × 10−2 rad)
aav =
=
=
= 0.00595 m s 2
T ∆t
∆t
86,400 s 2
(365.25 d)(1 d)
d
G
G
∆v
= 0.00595 m s 2 perpendicular to the average velocity .
Thus, aav =
∆t
)
(
(d) Strategy Compute Earth’s instantaneous radial acceleration using the relationship between angular speed
and radial acceleration.
Solution Compare Earth’s average daily acceleration to its instantaneous radial acceleration.
4π 2
4π 2 (1.50 × 1011 m)
ar = ω 2 r =
r=
= 0.00595 m s 2 , so
86,400 s 2
T2
2
(365.25 d)
d
(
)
G
G
a r = 0.00595 m s 2 perpendicular to the velocity, which is the same as aav within 3 significant figures .
15. Strategy The net force must point (horizontally) toward the pole, since the ball is in uniform circular motion.
Draw a free-body diagram for the ball.
Solution According to Newton’s second law,
mg
∑ Fy = T cos 70.0° − mg = 0, so T =
, and ∑ Fx = T sin 70.0° = ma x , so
cos 70.0°
T sin 70.0°
mg ⎛ sin 70.0° ⎞
ax =
=
⎜
⎟ = g tan 70.0°.
m
cos 70.0° ⎝
m
⎠
T
y
70.0°
x
mg
Since the tangential speed is related to the radial acceleration by a = v 2 r and the radius is equal to the length of
the rope times the sine of 70.0°, the tangential speed is
v = a x r = ( g tan 70.0°)r = (9.80 m s 2 ) tan 70.0°(1.30 m) sin 70.0° = 5.74 m s .
302
Physics
Chapter 5: Circular Motion
16. (a) Strategy and Solution Let the tensions in strings A and B be TA and TB , respectively. Draw the diagram.
y
TA
30.0°
30.0°
TB
x
mg
(b) Strategy Use Newton’s second law and the relationship between radial acceleration and angular speed.
Solution The net force must point (horizontally) toward the pole, since the ball is in uniform circular motion.
mg
∑ Fy = TA sin 30.0° − TB sin 30.0° − mg = 0, so TA − TB =
(1).
sin 30.0°
∑ Fx = TA cos 30.0° + TB cos 30.0° = ma x = mω 2 r , so TA + TB =
mω 2 r
(2).
cos 30.0°
Add (1) and (2) to find TA . Note that r = (15.0 cm) cos 30.0°.
mg
mω 2 r
+
, so
sin 30.0° cos 30.0°
m⎛
g
ω 2 r ⎞ 0.100 kg ⎡ 9.80 m s 2 (6.00π rad s)2 (0.150 m) cos 30.0° ⎤
+
+
TA = ⎜
⎟=
⎢
⎥ = 3.64 N .
2 ⎜⎝ sin 30.0° cos 30.0° ⎟⎠
2
cos 30.0°
⎣⎢ sin 30.0°
⎦⎥
2TA =
For TB , we have TB = TA −
mg
(0.100 kg)(9.80 m s 2 )
= 3.64 N −
= 1.68 N .
sin 30.0°
sin 30.0°
17. (a) Strategy Use Newton’s second law and the relationship between linear speed and radial acceleration.
Solution According to Newton’s second law,
ΣFr = T = mar = m
v2
v2
mv 2
= m , thus, T =
.
r
L
L
(b) Strategy Draw a free-body diagram for the rock. Use Newton’s second law.
Solution Decompose the force into vertical (y) and radial (r) components.
mv 2
mv 2
ΣFr = Tr =
=
and ΣFy = Ty − mg = 0.
r
L cos θ
Find the magnitude of the tension.
⎛ mv 2
T = Tr2 + Ty2 = ⎜
⎜ L cos θ
⎝
⎛ v2
= m g2 + ⎜
⎜ L cos θ
⎝
⎞
⎟⎟
⎠
2
⎞
⎟⎟ + (mg )2
⎠
2
303
y
T
θ
mg
Chapter 5: Circular Motion
Physics
18. (a) Strategy Use Newton’s second law. Refer to Fig. 5.11.
Solution
ΣFy = Ty − mg = T cos φ − mg = 0, so T =
mg
.
cos φ
(b) Strategy Use the relationships between angular speed and period and angular speed and radial acceleration.
Solution According the Newton’s second law, we have
ΣFr = Tr = mar = mω 2 r = mω 2 L sin φ = T sin φ , so T = mω 2 L, and from part (a), T =
mg
.
cos φ
Eliminate the tension, T.
mg
mω 2 L =
cos φ
g
2
ω L=
cos φ
2
g
⎛ 2π ⎞
(where T is now the period, not the tension)
⎜
⎟ L=
cos φ
⎝ T ⎠
g
1
=
2
2
T
(2π ) L cos φ
T = 2π
L cos φ
g
19. (a) Strategy Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
ΣFy = N − mg = 0, so N = mg , and ΣFr = fs = mar = m
fs = µs mg = m
v2
, so v =
R
v2
. Find v.
R
µs gR .
(b) Strategy Consider the forces acting on the car.
Solution Initially, the free-body diagram is
N
fs
W
When the force of static friction becomes too small, the tires slip and the free-body diagram is
N
fk
W
So, the static frictional force is not large enough to keep the car in a circular path; the car skids toward the
outside of the curve.
304
Physics
Chapter 5: Circular Motion
20. Strategy Let the x-axis point toward the center of curvature and the y-axis point upward. Draw a free-body
diagram. Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
y
ΣFy = N cos θ − mg = 0, so N cos θ = mg , and ΣFx = N sin θ = mar = m
v2
r
. Solve for θ .
x
θ
N
2
v
(26.8 m s)2
N sin θ m r
v2
v2
=
= tan −1
= 5.08° .
, so tan θ = , or θ = tan −1
N cos θ
mg
rg
rg
(825 m)(9.80 m s 2 )
mg
21. Strategy Let the x-axis point toward the center of curvature and the y-axis point upward. Draw a free-body
diagram. Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
y
ΣFy = N cos θ − mg = 0, so N cos θ = mg , and ΣFx = N sin θ = mar = m
v2
r
.
θ
x
N
Solve for v.
2
v
N sin θ m r
, so v 2 = rg tan θ , or
=
N cos θ
mg
mg
v = rg tan θ = (120 m)(9.80 m s 2 ) tan 3.0° = 7.9 m s .
22. Strategy Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
ΣFx = mar = m
v2
and ΣFy = N − mg = 0, so N = mg .
r
The track exerts an upward force of N = mg and a horizontal force of mv 2 r . Calculate the magnitude of the total
force exerted on the car by the track.
2
⎛ v2 ⎞
v4
(16 m s)4
F = ⎜ m ⎟ + (mg )2 = m
+ g 2 = (320 kg)
+ (9.80 m s 2 )2 = 3900 N
2
2
⎜ r ⎟
r
(35 m)
⎝
⎠
Find the angle.
mg
gr
(9.80 m s 2 )(35 m)
θ = tan −1 2 = tan −1 2 = tan −1
= 53°
v
(16 m s)2
m vr
G
Thus, F = 3900 N at 53° above the horizontal .
23. Strategy Let the x-axis point toward the center of curvature and the y-axis point upward. Draw a free-body
diagram. Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
y
ΣFx = N sin θ = mar = m
v2
r
and ΣFy = N cos θ − mg = 0, so N cos θ = mg . Solve for θ .
x
θ
v2
N sin θ m r
v2
v2
(18 m s) 2
, so tan θ = , or θ = tan −1
=
= tan −1
= 59° .
N cos θ
mg
rg
rg
(20.0 m)(9.80 m s 2 )
305
mg
N
Chapter 5: Circular Motion
Physics
24. Strategy Let the x-axis point toward the center of curvature. Draw a free-body diagram. Use Newton’s second
law and the relationship between radial acceleration and linear speed.
Solution
(a) Since the road is not banked, the force of static friction is the only horizontal force acting on the car.
y
v2
v2
(32 m s)2
ΣFx = fs = mar = m , so fs = m
= (1400 kg)
= 3500 N .
N
r
r
410 m
x
fs
mg
(b) µs N is the maximum frictional force, so fs = µs N only if the radial force (mar = mv 2 r ) is equal to µs N .
Thus, the answer is no.
25. Strategy Let the x-axis point toward the center of curvature and the y-axis point upward. Draw a free-body
diagram. Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
(a) ΣFy = N cos θ − mg − f sin θ = 0 and ΣFx = N sin θ + f cos θ = mv 2 r .
Solve for N in the first equation and substitute into the second.
f sin θ + mg
, so
N=
cos θ
y
θ
x
v2
f sin θ + mg
sin θ + f cos θ = m
r
cos θ
2
v
f sin 2 θ + mg sin θ + f cos 2 θ = m cos θ
r
N
θ
f
mg
v2
cos θ − mg sin θ
r
⎛ v2
⎞
f (1) = m ⎜ cos θ − g sin θ ⎟
⎜ r
⎟
⎝
⎠
⎡ (32 m s) 2
⎤
f = (1400 kg) ⎢
cos 5.0° − (9.80 m s 2 ) sin 5.0° ⎥ = 2300 N
⎣⎢ 410 m
⎦⎥
f (sin 2 θ + cos 2 θ ) = m
(b) Set the expression found for the force of friction equal to zero.
⎛ v2
⎞
v2
f = 0 = m ⎜ cos θ − g sin θ ⎟ , so
cos θ = g sin θ , or
⎜ r
⎟
r
⎝
⎠
v = gr tan θ = (9.80 m s 2 )(410 m) tan 5.0° = 19 m s .
306
Physics
Chapter 5: Circular Motion
26. Strategy Let +y be up and +x be toward the center of curvature. Draw a free-body diagram. Use Newton’s second
law and the relationship between radial acceleration and linear speed.
Solution
y
ΣFy = N y − mg − f y = 0 and ΣFx = N x + f x = mar = m
2
v
, so
R
θ
x
θ
v2
RN
m
= N x + f x = N sin θ + µs N cos θ , or v =
(sin θ + µs cos θ ).
R
m
Find N.
N y − mg − f y = N cos θ − mg − µs N sin θ = 0, so N (cos θ − µs sin θ ) = mg , or
N=
v=
N
fs
mg
mg
. Substitute for N in v.
cos θ − µs sin θ
R⎛
mg
⎜
m ⎜⎝ cos θ − µs sin θ
⎞
⎟⎟ (sin θ + µs cos θ ) =
⎠
gR(sin θ + µs cos θ )
cos θ − µs sin θ
=
gR(tan θ + µs )
1 − µs tan θ
27. Strategy Let the x-axis point toward the radius of curvature and the y-axis point upward. θ is measured from the
vertical axis. Draw a free-body diagram. Use Newton’s second law and the relationship between radial
acceleration and linear speed.
Solution
y
ΣFy = L y − mg = 0 and ΣFx = Lx = m
v2
r
. Thus, L y = L cos θ = mg and
θ
x
L
2
v
v2
L sin θ m r
v2
v2
Lx = L sin θ = m . Solve for θ .
, so tan θ = , or θ = tan −1
.
=
L cos θ
mg
rg
rg
r
mg
28. Strategy Let the x-axis point toward the center of curvature and the y-axis point upward. Draw a free-body
diagram. Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution
y
∑ Fx = N sin θ + fs cos θ = ma x = m
v2
(1) and ∑ Fy = N cos θ − mg − fs sin θ = 0 (2).
r
Now, fs = 0 when the car is traveling at 15.0 m s , since friction is not a factor. Use this
fact and (1) and (2) to solve for θ .
N sin θ + (0) cos θ = N sin θ = m
v152
r
and N cos θ − mg − (0) sin θ = 0, or N =
mg
. Thus,
cos θ
θ
x
N
θ
fs
mg
v152
v152
(15.0 m s)2
⎛ mg ⎞
−1
θ
θ
θ
θ
=
=
=
=
= 17.0°.
sin
tan
,
so
tan
,
or
tan
N sin θ = ⎜
mg
m
⎟
r
gr
⎝ cos θ ⎠
(9.80 m s 2 )(75.0 m)
Now use (1), (2), and the result for θ to find the minimum coefficient of static friction required to keep the car
from slipping at the higher speed.
v2
mv 2
N sin θ + fs cos θ = N sin θ + µs N cos θ = N (sin θ + µs cos θ ) = m , so N =
.
r
r (sin θ + µs cos θ )
N cos θ − mg − fs sin θ = N cos θ − mg − µs N sin θ = 0, so N cos θ − µs N sin θ = N (cos θ − µs sin θ ) = mg , or
mg
N=
.
cos θ − µs sin θ
307
Chapter 5: Circular Motion
Physics
Eliminate N.
mv202
mg
=
cos θ − µs sin θ r (sin θ + µs cos θ )
v 2
sin θ + µs cos θ = 20 (cos θ − µs sin θ )
gr
2
v
v 2
µs cos θ + 20 µs sin θ = 20 cos θ − sin θ
gr
gr
v202
gr
So, µs =
cos θ − sin θ
cos θ +
v202
gr
sin θ
(20.0 m s) 2
=
(9.80 m s 2 )(75.0 m)
cos17.0° +
cos17.0° − sin17.0°
(20.0 m s) 2
(9.80 m s 2 )(75.0 m)
sin17.0°
= 0.204 .
29. Strategy Use v = rω and ω = 2π T , where the radius is the average Earth-Sun distance and the period is one
year.
Solution
1y
⎞
⎛ 2π ⎞ 2π r 2π (1.50 × 1011 m) ⎛
4
v = rω = r ⎜
=
⎟=
⎜
⎟ = 2.99 × 10 m s
7
T
1y
⎝ T ⎠
⎝ 3.156 × 10 s ⎠
30. (a) Strategy Use Newton’s second law and law of universal gravitation.
Solution Note that v = rω = r (2π T ) = 2π r T .
2
GM E
4π 2 r 2
mv 2
⎛ 2π r ⎞
= v2 = ⎜
=
, so
.
⎟
r
r
⎝ T ⎠
r2
T2
Find the height above Earth’s surface, h = r − RE .
∑ Fr = G
GM ET 2
4π 2
h=
mM E
=
= r 3 , so r = 3
GM ET 2
3
2
− RE =
3
GM ET 2
4π 2
= h + RE . Thus,
(
4π
= 16,800 km − 6,371 km = 10, 400 km .
4π
2
(b) Strategy Use the relationship between radial acceleration and linear speed.
Solution
2
ar =
)
s 2
(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)(6.00 h)2 3600
h
v 2 ⎛ 2π r ⎞ 1 4π 2 r 4π 2 (16,800 × 103 m)
=⎜
=
=
= 1.42 m s 2
⎟
2
r ⎝ T ⎠ r
2
3600
s
T2
(6.00 h)
h
(
)
31. Strategy According to Kepler’s third law, r 3 ∝ T 2. Form a proportion.
Solution Find the orbital period of the second satellite.
2
3
T4.02
⎛ T4.0 ⎞
⎛ 4.0r ⎞
2
2
⎟ = 2 , so T4.0 = 64T , or T4.0 = 8.0T = 8.0(16 h) = 130 h .
⎜
⎟ = 64 = ⎜
r
T
⎝
⎠
T
⎝
⎠
308
− 6.371× 106 m
Physics
Chapter 5: Circular Motion
32. Strategy Use Newton’s second law and law of universal gravitation.
Solution Note that v = rω = r (2π T ) = 2π r T .
ΣFr =
GmM E
T = 2π
r2
2
=
mv 2 m ⎛ 2π r ⎞
4π 2 rm
4π 2 r 3
, so T 2 =
, or
= ⎜
⎟ =
r
r⎝ T ⎠
GM E
T2
r3
(613 × 103 m + 6.371× 106 m)3
⎛ 1h ⎞
= 2π
⎜
⎟ = 1.613 h .
2
2
24
−11
GM E
(6.674 × 10
N ⋅ m kg )(5.974 × 10 kg) ⎝ 3600 s ⎠
33. Strategy Use Eq. (5-14) with the mass of Jupiter in place of the mass of the Sun.
Solution Solve for r.
GM J 2
GM J 2
4π 2 3
r = T 2 , so r 3 =
T , or r = 3
T .
2
GM J
4π
4π 2
Compute the distance from the center of Jupiter for each satellite.
rIo = 3
GM J
4π 2
rEuropa = 3
TIo 2 = 3
GM J
4π 2
(6.674 × 10−11 N ⋅ m 2 kg 2 )(1.9 × 1027 kg)
4π 2
TEuropa 2 = 3
2
⎛ 86, 400 s ⎞
(1.77 d)2 ⎜
⎟ = 420, 000 km
⎝ 1d ⎠
(6.674 × 10−11 N ⋅ m 2 kg 2 )(1.9 × 1027 kg)
4π 2
2
⎛ 86, 400 s ⎞
(3.54 d)2 ⎜
⎟ = 670, 000 km
⎝ 1d ⎠
34. Strategy First, use Newton’s law of universal gravitation. Then, use Newton’s second law.
Solution Law of universal gravitation:
GMm (6.674 × 10−11 N ⋅ m 2 kg 2 )(2.0 ×1030 kg)(6.0 × 1024 kg)
Fg =
=
= 3.6 × 1022 N
r2
(1.5 × 1011 m)2
Second law of motion:
v2
(3.0 × 104 m s)2
ΣFr = mar = m
= (6.0 × 1024 kg)
= 3.6 × 1022 N
r
1.5 × 1011 m
Both results are the same, as expected. The magnitude of the force is 3.6 × 1022 N .
35. Strategy The orbital period of the satellite must be equal to that of Mars. Use Kepler’s third law.
Solution Solve for r, the distance from the center of the planet.
GM Mars 2
GM Mars 2
4π 2
r 3 = T 2 , so r 3 =
T , or r = 3
T . Thus, the satellite should be placed
2
GM Mars
4π
4π 2
r=3
(6.674 × 10−11 N ⋅ m 2 kg 2 )(6.42 × 1023 kg)
4π 2
2
⎛ 60 s ⎞
7
(1477 min)2 ⎜
⎟ = 2.04 × 10 m from the center of
⎝ 1 min ⎠
Mars.
309
Chapter 5: Circular Motion
Physics
36. Strategy Refer to the figure. Use v = rω = 2π r T , Newton’s law of universal gravitation, and the definitions of
average velocity and average acceleration.
Solution
(a) v =
2π r 2π (35,800 km + 6371 km)
=
= 3.07 km s , therefore,
T
86, 400 s
G
v = 3.07 km s in the − y -direction at point C.
G
∆r r 2 4r 2 4(35,800 km + 6371 km) 2
G
G
=
=
=
= 2.76 km s , and v av is in the same direction as
(b) v av =
T
∆t
T
86,400 s
4
G
G
∆r , which is 45° above the −x-axis. So, v av = 2.76 km s at 45° above the −x-axis .
G
G
∆v
G G
G
(c) aav =
, so the average acceleration is in the same direction as ∆v = v B − v A , which is 45° below the
∆t
−x-axis.
G
G
G
∆v = [(∆v) x ]2 + [(∆v ) y ]2 = v 2 + v 2 = v 2
G
∆v v 2 4v 2 4 2(3.07 × 103 m s)
=
=
=
= 0.201 m s 2
T
T
86, 400 s
∆t
4
G
So, aav = 0.201 m s 2 at 45° below the − x-axis .
(d) The instantaneous acceleration at point D is in the +y-direction, since the acceleration is always directed
radially inward for uniform circular motion. Its magnitude is ar . Use Newton’s second law and law of
universal gravitation.
GmM E
GM E (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
=
= 0.224 m s 2 .
ΣFr =
= mar , so ar =
2
r
r2
(35,800 × 103 m + 6371× 103 m)2
G
Thus, a = 0.224 m s 2 in the + y -direction .
37. Strategy Use Newton’s second law and law of universal gravitation, as well as the relationship between radial
acceleration and linear speed.
Solution
GmM J
mv 2
ΣFr =
=
(3.0 RJ )2 3.0 RJ
Now, the gravitational field strength of Jupiter is given by g J =
GM J
RJ 2
. Find the period of the spacecraft’s orbit.
2
4π 2 (3.0 RJ )2
mg J
g R
mv 2
⎛ 2π r ⎞
=
=
, so v 2 = J J = ⎜
. Solving for T, we have
⎟
3.0 RJ
9.0
3.0
⎝ T ⎠
T2
T = 2π
27 RJ
gJ
= 2π
27(71,500 × 103 m) ⎛ 1 h ⎞
⎜
⎟ = 16 h .
23 N kg
⎝ 3600 s ⎠
310
Physics
Chapter 5: Circular Motion
38. Stratey Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution Find the normal force.
mvtop 2
mvtop 2
ΣFr = N + mg = mar =
, so N =
− mg .
r
r
The normal force must be greater than or equal to zero so that the passengers do not lose contact with their seats.
The minimum speed is found when N = 0.
0=
mvtop 2
r
− mg , so
vtop 2
r
= g , or vtop = gr = (9.80 m s 2 )(29.5 m) = 17.0 m s .
39. (a) Stratey Draw a diagram. Use Newton’s second law and the relationship between radial acceleration and
linear speed.
Solution Find the tension.
ΣFr = T − mg = mar , so
⎛
⎡
v2 ⎞
(1.6 m s)2 ⎤
T = m( g + ar ) = m ⎜ g + ⎟ = (1.0 kg) ⎢9.80 m s 2 +
⎥ = 13 N .
⎜
⎟
0.80 m ⎦⎥
r ⎠
⎝
⎣⎢
y
r
T
mg
(b) Strategy and Solution If the bob were at rest, the tension would be equal to the weight of the bob. Since the
bob is moving and its velocity is changing—its vertical component is increasing from zero—the tension must
be greater than the weight. Thus, the bob has an upward acceleration, so the net Fy must be upward and
greater than the weight of the bob.
40. Strategy The tension in the rope provides the radial acceleration that moves the child in a circular arc. The
tension must be such that it counteracts the force due to gravity on the child and provides enough radial
acceleration to give the child the given linear speed. Draw a diagram. Use Newton’s second law and the
relationship between radial acceleration and linear speed.
Solution Find the tension.
⎛
v2 ⎞
ΣFr = T − mg = mar , so T = m( g + ar ) = m ⎜ g + ⎟ .
⎜
r ⎟⎠
⎝
⎛
(4.20 m s)2 ⎞
Therefore, the tension must be T = (35.0 kg) ⎜ 9.80 m s 2 +
⎟ = 438 N .
⎜
6.50 m ⎟⎠
⎝
y
r
T
mg
41. Strategy Use Newton’s second law and the relationship between radial acceleration and linear speed.
Solution The only forces acting on the car are gravity and the normal force of the ground pushing on the car. The
radial acceleration is downward, or toward the center of the radius of curvature. Let up be positive.
⎛ v2 ⎞
∑ Fr = N − mg = mar = m ⎜ − ⎟
⎜ r ⎟
⎝
⎠
When the car is just in contact with the ground, the normal force must be zero. If the car goes any faster it will
lose contact with the road. Solve for the speed.
mv 2
−
= N − mg = 0 − mg , so v 2 = gr , or v = gr = (9.80 m s 2 )(55.0 m) = 23.2 m s .
r
311
Chapter 5: Circular Motion
Physics
42. Strategy Use Eq. (5-21) to find the angular acceleration.
Solution
ωf 2 − ωi 2 = 2α∆θ , so α =
ωf 2 − ωi 2 (0.50 rev s)2 − 0 ⎛ 2π rad ⎞
2
=
⎜
⎟ = 0.39 rad s .
2∆θ
2(2.0 rev)
rev
⎝
⎠
43. Strategy Use Eq. (5-20) to find the constant angular acceleration.
Solution Since the cyclist starts from rest, the initial angular velocity is zero.
(
)
2(8.0 rev) 2πrevrad
1
1
1
2∆θ
∆θ = ωi ∆t + α (∆t )2 = (0)∆t + α (∆t )2 = α (∆t )2 , so α =
=
= 4.0 rad s 2 .
2
2
2
( ∆t ) 2
(5.0 s)2
44. Strategy Use Eq. (5-18) to find the average angular acceleration.
Solution
∆ω = α∆t , so α =
(
2π rad
∆ω (7200 rpm) rev
=
∆t
4.0 s
)( 160mins ) =
190 rad s 2 .
45. Strategy Equations (5-18) and (5-19) are ∆ω = ωf − ωi = α∆t and ∆θ = 12 (ωf + ωi )∆t , respectively.
Solution Solve for ωf in Eq. (5-18), then substitute the result into Eq. (5-19) and simplify.
∆ω = ωf − ωi = α∆t , so ωf = ωi + α∆t. Substitute.
∆θ =
1
1
1
1
(ω + ωi )∆t = (ωi + α∆t + ωi )∆t = (2ωi + α∆t )∆t = ωi ∆t + α (∆t )2 = ∆θ , which is Eq. (5-20).
2 f
2
2
2
46. Strategy Equations (5-18) and (5-19) are ∆ω = ωf − ωi = α∆t and ∆θ = 12 (ωf + ωi )∆t , respectively.
Solution Solve for ∆t in Eq. (5-18), then substitute the result into Eq. (5-19) and simplify.
ω − ωi
. Substitute.
∆ω = ωf − ωi = α∆t , so ∆t = f
α
∆θ =
1
1
⎛ ω − ωi ⎞ ωf 2 − ωi 2
(ωf + ωi )∆t = (ωf + ωi ) ⎜ f
, so ωf 2 − ωi 2 = 2α∆θ , which is Eq. (5-21).
⎟=
2
2
2α
⎝ α ⎠
47. Strategy Draw a free-body diagram for the bob. Use Newton’s second law and the relationship between radial
acceleration and linear speed.
Solution Refer to the figure.
ΣFy = T − mg cos θ = mar and ΣFx = mg sin θ = mat , so
at = g sin θ = (9.80 m s 2 ) sin15.0° = 2.54 m s 2
and
y
θ = 15.0°
T
x
2
v 2 (1.40 m s )
=
= 2.45 m s2 . The tension is
0.800 m
r
T = m(ar + g cos θ ) = (1.00 kg) ⎡ 2.45 m s 2 + (9.80 m s 2 ) cos15.0° ⎤ = 11.9 N .
⎣
⎦
ar =
mg
48. Strategy and Solution Refer to the figure. If the +y-direction is radial and the +x-direction is tangential to the
left, according to Newton’s second law, ΣFx = mg sin θ = mat , or at = g sin θ .
312
Physics
Chapter 5: Circular Motion
49. (a) Strategy Use Eq. (5-18) to find the magnitude of the constant angular acceleration.
Solution
∆ω = α ∆t , so α =
∆ω
∆t
=
(
(33.3 rpm) 2πrevrad
2.0 s
)( 160mins ) =
1.7 rad s 2 .
(b) Strategy Use Eq. (5-20) to find the number of revolutions.
Solution
1
1
∆θ = ωi ∆t + α (∆t ) 2 = (0)∆t + α (∆t ) 2
2
2
1
2
∆θ = α (∆t )
2
1
⎛ 1 rev ⎞
2
2 ⎛ 1 rev ⎞
⎜
⎟ ∆θ = 1.744 rad s (2.0 s) ⎜
⎟ = 0.56 rev
2
rad
2
π
⎝
⎠
⎝ 2π rad ⎠
(
)
50. Strategy Use Eq. (5-20) to find the constant angular acceleration.
1
1
1
2∆θ
2(90.0°)
∆θ = ωi ∆t + α (∆t ) 2 = (0)∆t + α (∆t ) 2 = α (∆t ) 2 , so α =
=
= 180° s −2 .
2
2
2
2
(∆t )
(1.0 s) 2
Solution Use Eqs. (5-18) and (5-20) to find the angles.
1
(a) ωi = α∆t = (180° s −2 )(1.0 s) = 180° s −1 , so ∆θ = (180° s −1 )(1.0 s) + (180° s −2 )(1.0 s) 2 = 270° .
2
1
(b) ωi = α∆t = (180° s −2 )(2.0 s) = 360° s −1 , so ∆θ = (360° s −1 )(1.0 s) + (180° s −2 )(1.0 s)2 = 450° .
2
51. (a) Strategy Since the car moves with constant acceleration, use the relationship between speed, acceleration,
and diplacement and C = 2π r to find the speed of the car.
C 2π r π r
Solution The distance traveled by the car is ∆x = =
. Find the final speed.
=
4
4
2
⎛πr ⎞
vfx 2 − vix 2 = vfx 2 − 0 = vt 2 = 2at ∆x = 2at ⎜ ⎟ = at π r , so
⎝ 2 ⎠
vt = at π r = (2.00 m s 2 )π (50.0 m) = 17.7 m s .
(b) Strategy Use the relationship between radial acceleration and linear speed.
Solution
v 2 a πr
ar = t = t
= π at = π (2.00 m s 2 ) = 6.28 m s 2
r
r
313
Chapter 5: Circular Motion
Physics
(c) Strategy Draw a diagram. Use the Pythagorean Theorem to find the magnitude of the total acceleration.
Then use trigonometry to find its direction.
Solution Find the total acceleration.
at
17.7°
a = ar 2 + at 2 = π 2 at 2 + at 2 = at π 2 + 1 = (2.00 m s 2 ) π 2 + 1
= 6.59 m
s2
Let +y be north and +x be east. Then, a x = at and a y = − ar .
Initial
position
−π at
− ar
= tan −1
= tan −1 (−π )
ax
at
at
= −72.3° or 17.7° east of south
G
Thus, a = 6.59 m s 2 at an angle of 17.7° east of south .
θ = tan −1
ay
= tan −1
52. (a) Strategy Use Eq. (5-21) to find the magnitude of the constant angular acceleration.
Solution
ωf 2 − ωi 2 = 2α∆θ , so α =
ωf 2 − ωi 2 (7π rad s)2 − (2π rad s)2
=
= 7.1 rad s 2 .
2 ∆θ
2(10π rad)
(b) Strategy Use Eq. (5-19) to find the time it took for the disk to rotate through 10π radians.
Solution
1
2∆θ
2(10π rad)
∆θ = (ωf + ωi )∆t , so ∆t =
=
= 2.2 s .
ωf + ωi 7π rad s + 2π rad s
2
(c) Strategy The tangential acceleration is related to the angular acceleration by at = rα .
Solution
at = r α = (0.050 m)
(7π rad s) 2 − (2π rad s) 2
= 0.35 m s 2
2(10π rad)
53. (a) Strategy Use Eq. (5-18) to find the time it takes for the rotor to come to rest.
Solution The acceleration is opposite the rotation of the rotor, so it is negative.
ω − ωi 0 − 5.0 × 105 rad s
∆t = f
=
= 1.3 × 106 s
α
−0.40 rad s 2
(b) Strategy Use Eq. (5-21) to find the number of revolutions the rotor spun before it stopped.
Solution
ω 2 − ωi 2
∆θ = f
2α
0 − (5.0 × 105 rad s)2 ⎛ 1 rev ⎞
⎛ 1 rev ⎞
10
⎜
⎟ ∆θ =
⎜
⎟ = 5.0 × 10 rev
2
π
π
2
rad
2
rad
⎝
⎠
⎠
2(−0.40 rad s ) ⎝
314
ar
a
N
Physics
Chapter 5: Circular Motion
54. (a) Strategy Use the relationship between linear and angular speed.
Solution Compute the initial speed.
vi = rωi = (0.200 m)(5.0 × 105 rad s) = 1.0 × 105 m s
(b) Strategy Use the relationship between tangential and angular acceleration.
Solution Compute the tangential acceleration component.
at = rα = (0.200 m)(0.40 rad s 2 ) = 0.080 m s 2
(c) Strategy Use the relationship between radial acceleration and linear speed.
Solution The maximum radial acceleration component is the initial radial acceleration.
v 2 (1.0 × 105 m s) 2
ar =
=
= 5.0 × 1010 m s 2
r
0.200 m
55. Strategy The strength of the artificial gravity is equal to the radial acceleration.
Solution Compute the magnitude of the radial acceleration.
2
(
)
2
( 4.0 rev s ) 2πrevrad (0.25 m)
⎛ ω 2r ⎞
2
ar = ω r = ⎜
g=
g = 16 g
⎜ g ⎟⎟
9.80 m s 2
⎝
⎠
56. Strategy Use the relationship between radial acceleration and linear speed.
Solution The magnitude of the radial acceleration must be the same as the magnitude of the gravitational field
strength.
ar =
gr
v2
1
= g , so v = gr = ω r = (2π f )r. Thus, f =
=
r
2π r 2π
g
1
=
r 2π
9.80 m s 2
= 0.045 Hz .
120 m
57. Strategy Use ar = ω 2 r to find the angular speed required.
Solution The magnitude of the radial acceleration must be the same as the magnitude of the gravitational field
strength.
ar = ω 2 r = g , so ω =
9.80 m s 2
g
=
= 7.0 rad s .
0.20 m
r
58. Strategy Use ar = ω 2 r to find the angular speed required. The total acceleration is equal to the vector sum of the
gravitational field strength and the radial acceleration.
Solution Compute the angular speed.
a 2 = ar 2 + g 2 = ω 4 r 2 + g 2 , so ω 4 r 2 = a 2 − g 2 , or
14
⎛ a2 − g 2 ⎞
ω = ⎜⎜
⎟⎟
2
⎝ r
⎠
14
⎛ 4.0 g 2 − g 2 ⎞
=⎜
⎟⎟
⎜
r2
⎝
⎠
14
⎡ 3.0(9.80 m s 2 ) 2 ⎤
=⎢
⎥
2
⎣⎢ (0.125 m)
⎦⎥
315
= 12 rad s .
Chapter 5: Circular Motion
Physics
59. (a) Strategy Earth rotates once every 24 hours. Use ar = ω 2 r.
Solution The radius r is the radius of Earth.
2
2
⎛ 2π rad ⎞ ⎛ 1 h ⎞
6
2
ar = ω 2 r = ⎜
⎟ ⎜
⎟ (6.371× 10 m) = 0.034 m s
24
h
3600
s
⎝
⎠ ⎝
⎠
(b) Strategy and Solution Since the object is on the outside of the Earth, the rotation of the Earth seemingly
“pushes outward” on the object. (This is sometimes referred to as the fictitious “centrifugal force.”) So, since
g − a < g for a > 0 , the object’s apparent weight is less than its true weight.
(c) Strategy Compare the radial acceleration to the gravitational field strength.
Solution Divide the radial acceleration by g.
a
0.0337
(100%) r = (100%)
= 0.34%
g
9.80
The actual weight is reduced by this amount, so the apparent weight is 0.34% smaller than the actual weight.
(d) Strategy and Solution The rotation of the Earth at the poles has no effect on the reading of a bathroom
scale, so the actual weight is measured.
G
G
60. (a) Strategy At the top, g and a are both directed downward. Draw a free-body diagram. Use Newton’s second
law and the relationship between radial acceleration and angular speed.
Solution
y
N
ΣFr = N − mg = mar = m(−ω 2 R), so W ′ = N = m( g − ω 2 R) .
mg
G
G
(b) Strategy At the bottom, g is directed downward and a is directed upward. Draw a free-body diagram.
Solution
ΣFr = N − mg = mar =
y
m(ω 2 R),
so W ′ = N =
m( g + ω 2 R )
N
.
mg
G
G
61. (a) Strategy At the top, g and a are both directed downward. Draw a free-body diagram. Use Newton’s second
law.
Solution
ΣFr = N − mg = mar = m(− a y ), so W ′ = N = mg − ma y .
y
The apparent weight is less than the true weight by ma y . Thus, the lower weight,
N
mg
518.5 N, is measured at the top.
G
G
(b) Strategy At the bottom, g is directed downward and a is directed upward. Draw a free-body diagram.
Solution
ΣFr = N − mg = mar = m(a y ), so W ′ = N = mg + ma y .
The apparent weight is greater than the true weight by ma y . Thus, the higher weight,
521.5 N, is measured at the bottom.
316
y
N
mg
Physics
Chapter 5: Circular Motion
′ = W (1 − a y g ), where W = mg . Use
(c) Strategy The apparent weight at the top is given by Wtop
ar = ω 2 r = a y to find the radius.
Solution Solve for the radius r.
′
⎛ ay ⎞
⎛ ω 2r ⎞
g ⎛ Wtop
′ = W ⎜1 −
, so r =
1−
Wtop
= W ⎜1 −
⎟
⎜
⎟
⎜
⎜
g ⎟⎠
g ⎟⎠
W
ω 2 ⎜⎝
⎝
⎝
⎞
9.80 m s 2 ⎛ 518.5 N ⎞
1−
⎟⎟ =
⎟ = 45 m .
2⎜
⎠ ( 0.025 rad s ) ⎝ 520.0 N ⎠
62. Strategy The distance from the rotation axis is r = RE cos θ , where θ = 40.2°.
Solution Compute the magnitude of the radial acceleration.
2
⎛ 2π rad ⎞
6
2
ar = ω 2 r = ω 2 RE cos θ = ⎜
⎟ (6.371× 10 m) cos 40.2° = 0.0257 m s
⎝ 86, 400 s ⎠
63. Strategy Use Eq. (5-18).
Solution After one minute:
ωf − ωi = 0.80ω − ω = −0.20ω = α∆t
After three minutes:
ωf − ωi = ωf − ω = α (3∆t ) = 3α∆t = 3(−0.20ω ), so ωf = −0.60ω + ω = 0.40ω .
64. Strategy Earth rotates once per 24.0 hours.
Solution Compute the tangential speed of Mt. Kilimanjaro.
⎛ 2π rad ⎞⎛ 1 h ⎞
6
v = ωr = ⎜
⎟⎜
⎟ (6.378 × 10 m + 5895 m) = 464 m s
⎝ 24.0 h ⎠⎝ 3600 s ⎠
65. Strategy Use the relationship between linear speed and angular speed.
Solution Compute the linear speed of the tip of the nylon cord.
v = ω r = ( 660 rad s ) (0.23 m) = 150 m s
66. Strategy Use the definition of average angular velocity.
Solution Compute the number of degrees that the drill rotates.
∆θ
⎛ 360° ⎞
6
ωav =
, so ∆θ = ωav ∆t = (3.14 × 104 rad s) ⎜
⎟ (1.00 s) = 1.80 × 10 degrees .
∆t
⎝ 2π rad ⎠
1.00 rev
(376.8 s) = 2.00 revolutions, so his
188.4 s
2π r 2π (90.0 m)
=
= 3.00 m s. Therefore,
direction is due east. The jogger’s constant speed is
T
188.4 s
G
v = 3.00 m s east .
67. (a) Strategy and Solution At t = 376.8 s, the jogger has made
1.00 rev
(94.2 s) = 0.500 revolution, so his
188.4 s
direction is opposite his starting direction, or due west. The jogger’s speed is the same as before, therefore,
G
v = 3.00 m s west .
(b) Strategy and Solution At t = 94.2 s, the jogger has made
317
Chapter 5: Circular Motion
Physics
68. (a) Strategy Since ω = v r and the radius of gear B is shorter than that of gear A, gear B has a larger magnitude
angular velocity than gear A. Gear B rotates in the direction opposite to the rotation of gear A, so gear B
rotates clockwise.
Solution Find the magnitude of the angular rotation of gear B. Since the gears are in contact (without
slipping), vB = v A .
ωB vB rB rA
r
2r
=
= , so ω B = A ω A = B ω A = 2 ω A = 2(6.00 Hz) = 12.0 Hz.
rB
rB
ω A v A rA rB
So, the angular velocity of gear B is 12.0 Hz clockwise.
(b) Strategy Let v = vB = v A .
Solution Compute the linear speed of a point on the tip of either gear.
v = r ω = rA ω A = (0.100 m)2π (6.00 Hz) = 3.77 m s = v A = vB
69. Strategy Use Eq. (5-20) to find the number of rotations each gear goes through in 2.0 s. Refer to Problem 68.
Solution
1
∆θ A = ωi ∆t + α (∆t ) 2
2
1
⎛ 1 rotation ⎞
⎛ 1 rotation ⎞ ⎡
2
2⎤
⎜
⎟ ∆θ A = ⎜
⎟ ⎢ 2π (0.955 Hz)(2.0 s) + (3.0 rad s )(2.0 s) ⎥ = 2.9 rotations
2
⎝ 2π rad ⎠
⎝ 2π rad ⎠ ⎣
⎦
From Problem 68, we know that gear B has an angular speed that is twice that of gear A. Thus, gear B rotates
twice for each rotation of gear A. Therefore,
1
⎛ 1 rotation ⎞
⎛ 1 rotation ⎞ ⎡
2
2⎤
⎜
⎟ ∆θ B = 2 ⎜
⎟ ⎢ 2π (0.955 Hz)(2.0 s) + (3.0 rad s )(2.0 s) ⎥ = 5.7 rotations .
2
⎝ 2π rad ⎠
⎝ 2π rad ⎠ ⎣
⎦
70. (a) Strategy The apparent angular speed of the Sun is approximately the angular speed of the Earth.
Solution Compute the apparent angular speed of the Sun.
⎛ 2π rad ⎞⎛ 1 h ⎞
−5
ω =⎜
⎟⎜
⎟ = 7.3 × 10 rad s
24
h
3600
s
⎝
⎠⎝
⎠
(b) Strategy An estimate for an arm-length is 1 m, and for a finger-width an estimate is 2 cm.
Solution Use s = r ∆θ to estimate the angle subtended. Then s ≈ 2 cm, r ≈ 1 m, and
∆θ =
s 0.02 m
≈
= 0.02 rad .
r
1m
(c) Strategy Use the results of parts (a) and (b).
Solution
0.02 rad
⎛ 60 min ⎞
(24 h) ⎜
⎟ = 5 min
2π rad
⎝ h ⎠
318
Physics
Chapter 5: Circular Motion
71. (a) Strategy Use the relationship between linear and angular speed.
Solution
v = r ω = (1500 rad s)(0.025 m) = 38 m s
(b) Strategy and Solution To get the same relative speed between the tape and the heads, the tape would have
to move at 38 m s.
⎛ 3600 s ⎞
You would need (37.5 m s)(1 h) ⎜
⎟ = 135 km of tape to record one hour .
⎝ 1h ⎠
72. Strategy Use the relationship between linear and angular speed.
Solution
1y
⎞
⎛ 2π ⎞ 2π r 2π (2 × 1017 km) ⎛
v = r ω = r⎜
=
⎟=
⎜
⎟ = 200 km s
6
7
T
⎝ T ⎠
200 × 10 y ⎝ 3.156 × 10 s ⎠
73. (a) Strategy Use the relationship between radial acceleration and linear speed.
Solution
v 2 (2.0π m s)2
=
= 8.0π 2 m s 2 = 79 m s 2
ar =
0.50 m
r
(b) Strategy Use Newton’s second law.
Solution
ΣFr = T = mar , so T = (0.50 kg)(8.0π 2 m s 2 ) = 4.0π 2 N = 39 N .
74. Strategy Let the outer string be 2 and the inner string be 1. The outer string only supports the mass m2 . The
inner string supports both masses. Draw a diagram. Use the relationship between radial acceleration and angular
speed.
Solution According to Newton’s second law,
+
2
⎛ 2π rad ⎞
ΣF2 = T2 = m2 a2 = m2ω 2 r2 = (0.030 kg)(1.5 rev s )2 ⎜
⎟ (0.75 m)
⎝ 1 rev ⎠
= 2.0 N and
T1
m1
T2
T2
m2
ΣF1 = T1 − T2 = m1a1 = m1ω 2 r1 , so T1 = T2 + m1a1 = m2ω 2 r2 + m1ω 2 r1
2
⎛ 2π rad ⎞
= ω 2 (m2 r2 + m1r1 ) = (1.5 rev s)2 ⎜
⎟ [ (0.030 kg)(0.75 m) + (0.050 kg)(0.40 m) ] = 3.8 N .
⎝ 1 rev ⎠
75. Strategy Use the relationship between radial acceleration and linear speed. Recall that the circumference of a
circle is given by C = 2π r.
Solution The distance is C 2 = π r = v∆t and ar = v 2 r , so v = ar r . The time to complete the U-turn is given
by ∆t = π r v = π r
ar r = π r ar , so the larger the radius the greater the time to complete the U-turn.
Therefore, the smallest possible radius should be used to make the turn. Calculating the minimum time required to
complete the U-turn, we find that ∆tmin = π (5.0 m) (3.0 m s 2 ) = 4.1 s .
319
Chapter 5: Circular Motion
Physics
76. (a) Strategy Use the relationships between radial acceleration and angular speed and angular speed and period.
Solution
2
2
4π 2 r 4π 2 (2 × 1020 m) ⎛
1y
⎞
⎛ 2π ⎞
−10 m s 2
ar = ω 2 r = ⎜
⎟ r= 2 =
⎜
⎟ = 2 × 10
6
2
7
⎝ T ⎠
(200 × 10 y) ⎝ 3.156 × 10 s ⎠
T
(b) Strategy and Solution According to Newton’s second law,
ΣF = mar = (2 × 1030 kg)(2 × 10−10 m s2 ) = 4 × 1020 N .
77. Strategy For each revolution of the flagellum, the bacterium moves the distance of the pitch.
Solution Compute the speed of the bacterium.
v = (1.0 µm rev)(110 rev s) = 110 µm s
78. Strategy When the radial component of the net force is equal to the maximum force of static friction on the
penny, the penny will just begin to slide. Draw a free-body diagram for the penny. Use Eq. (5-18) and Newton’s
second law.
Solution
N
2
2
ΣFr = fs = µs N = mar = mω r and ΣFy = N − mg = 0, so µs N = µs mg = mω r.
µs g
Solving for ω , we have ω =
r
ω − ωi
ωf − ωi = α∆t , so ∆t = f
=
α
fs
mg
. Substitute this into Eq. (5-18) and solve for ∆t.
µs g
r
−0
α
=
1
α
µs g
r
=
1
2.00 rad s
2
y
0.350(9.80 m s 2 )
= 2.93 s .
0.100 m
79. Strategy When the radial component of the net force is equal to the maximum force of static friction on the coin,
the coin will just begin to slide. Draw a free-body diagram for the coin. Use the relationship between radial
acceleration and angular speed and Newton’s second law.
Solution
y
2
N
2
ΣFr = fs = µs N = mar = mω r and ΣFy = N − mg = 0, so µs N = µs mg = mω r.
Solve for r to find the distance from the center of the record where the coin can be placed
without it slipping off.
µ g
0.1(9.80 m s 2 )
= 8 cm
r= s =
2 2π rad 2
2
rev
ω2
1
min
33.3 min
rev
60 s
(
)(
)(
)
320
fs
mg
Physics
Chapter 5: Circular Motion
80. (a) Strategy Use the relationship between linear speed and angular speed.
Solution
⎛ 45.0 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞
v = r ω = (0.127 m) ⎜
⎟⎜
⎟⎜
⎟ = 0.60 m s
⎝ min ⎠⎝ rev ⎠⎝ 60 s ⎠
(b) Strategy When the radial component of the net force is greater than or equal to the maximum force of static
friction on the dolls, the dolls will slide. Use the relationship between radial acceleration and angular speed
and Newton’s second law.
Solution
y
N
2
ΣFr = fs = µs N ≥ mar = mω r and ΣFy = N − mg = 0, so
µs N = µs mg ≥ mω 2 r. Eliminate m and check the validity of the inequality.
µs g
fs
≥ ω 2r
2
2
mg
2
rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞
⎛
0.13(9.80 m s 2 ) ≥ ⎜ 45.0
⎟ ⎜
⎟ ⎜
⎟ (0.127 m)
min
⎝
⎠ ⎝ rev ⎠ ⎝ 60 s ⎠
1.3 m s 2 ≥ 2.8 m s 2 False
The inequality is false, so the dolls do not stay on the record.
81. Strategy Use the relationship between linear speed and angular speed.
Solution
⎛ 0.65 m ⎞
⎛ 1 km ⎞⎛ 3600 s ⎞
v=r ω =⎜
⎟ (101 rad s ) ⎜
⎟⎜
⎟ = 120 km h
⎝ 2 ⎠
⎝ 1000 m ⎠⎝ h ⎠
82. Strategy Draw a free-body diagram. Use the relationship between radial acceleration and angular speed and
Newton’s second law.
Solution
(4.25 m) sin 45.0°
mg
ΣFy = T cos θ − mg = 0, so T =
, and ΣFx = T sin θ = mar .
cos θ
Use these results to find the angular speed.
T sin θ = mar
mg
sin θ = mar
cos θ
g tan θ = ω 2 r
ω=
g tan θ
(9.80 m s 2 ) tan 45.0°
=
= 1.04 rad s
6.00 m + (4.25 m) sin 45.0°
r
321
6.00 m
45.0°
T
mg
45.0°
y
x
Chapter 5: Circular Motion
Physics
83. (a) Strategy Use the relationship between radial acceleration and angular speed.
Solution
2
2
2
rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞
g
⎛
= 90 g
ar = ω 2 r = ⎜1.0 × 103
⎟ ⎜
⎟ ⎜
⎟ (0.080 m)
min
rev
60
s
⎝
⎠ ⎝
⎠ ⎝
⎠
9.80 m s 2
(b) Strategy Use Newton’s second law.
Solution
ΣF = mar = m(89.5 g ) = (9.0 × 10−14 kg)(89.5)(9.80 m s 2 ) = 7.9 × 10−11 N
(c) Strategy Use Newton’s second law.
Solution
ΣF = mar = m(89.5 g ) = (5.0 × 10−21 kg)(89.5)(9.80 m s 2 ) = 4.4 × 10−18 N
(d) Strategy Use the relationship between radial acceleration and angular speed.
Solution
2
2
2
rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞
g
⎛
= 5.0 × 105 g
ar = ω 2 r = ⎜ 75, 000
⎟ ⎜
⎟ ⎜
⎟ (0.080 m)
2
min ⎠ ⎝ rev ⎠ ⎝ 60 s ⎠
⎝
9.80 m s
84. (a) Strategy Draw a free-body diagram. Use Newton’s second law.
Solution ΣFr = T sin 55° = mar and ΣFy = T cos 55° − mg = 0, so
T=
mg
T sin 55°
and ar =
=
m
cos 55°
= g tan 55° = (9.80 m
mg
sin 55°
cos 55°
s 2 ) tan 55° =
T
center
of
circle
m
14 m
s2
55°
.
mg
(b) Strategy and Solution Since the radial acceleration is equal to g tan 55°, the multiple is tan 55° = 1.4 .
(c) Strategy Use the relationship between radial acceleration and linear speed.
Solution Find the linear speed of the roller coaster.
v2
ar =
= g tan 55°, so v = gr tan 55° = (9.80 m s 2 )(80.0 m) tan 55° = 33 m s .
r
85. Strategy and Solution The cutting tool moves one inch in the time ∆t =
d
1 in
=
. The lathe chuck must
v 0.080 in s
complete 18 revolutions in the time ∆t. Thus, the rotational speed must be
322
18 rev
1 in
0.080 in s
= 1.4 rev s .
Physics
Chapter 5: Circular Motion
86. (a) Strategy and Solution For uniform circular motion, the acceleration and, therefore, the magnetic force must
be perpendicular to the particle’s velocity. So, the angle between the magnetic force and the particle’s
velocity must be 90°.
(b) Strategy F = kv where k is a constant of proportionality. Use Newton’s second law.
Solution Find an expression for the period.
2π
2π m
. Therefore, T =
.
k
T
Identical particles have the same mass and k is constant, so T is the same for each. (T does not depend on v.)
ΣFr = F = kv = mar , so kv = kω r = mar = mω 2 r. Thus, k = mω = m
(c) Strategy and Solution v =
2π r 2π r k
m
=
= r , so r = v .
2π m
k
T
m
k
Since k m is constant, r is proportional to v.
87. Strategy Use Newton’s second law and law of universal gravitation.
Solution The period is T = 86,400 s (24 h).
GmM E
C 2π r
v2
v= =
and ΣFr =
= mar = m . Solve for r.
T
T
r
r2
GM E
r2
r=3
=
2
T 2GM E
v 2 ⎛ 2π r ⎞ 1 4π 2 r
=⎜
=
= r 3 and
, so
⎟
2
2
r ⎝ T ⎠ r
T
4π
(86, 400 s)2 (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
4π 2
323
= 42, 200 km .
REVIEW AND SYNTHESIS: CHAPTERS 1–5
Review Exercises
1. Strategy Replace the quantities with their units.
Solution Find the units of the spring constant k.
F = kx, so k =
F
kg ⋅ m s 2
, and the units of k are N m =
= kg s 2 .
x
m
2. (a) Strategy Find Harrison’s total displacement. The return trip has the same magnitude but the opposite
direction. Draw a diagram.
Solution Compute the displacement using the component method.
∆x = x1 + x2 + x3 = −2.00 km + (5.00 km) cos 233° + (1.00 km) cos120°
∆y = y1 + y2 + y3 = 0 + (5.00 km) sin 233° + (1.00 km) sin120°
Compute the magnitude.
y
53.0°
60.0°
r3
r2
r1
x
N
∆r = [−2.00 km + (5.00 km) cos 233° + (1.00 km) cos120°]2 + [(5.00 km) sin 233° + (1.00 km) sin120°]2
= 6.33 km
Compute the direction.
∆y
(5.00 km) sin 233° + (1.00 km) sin120°
θ = tan −1
= tan −1
= 29.6° below the − x-axis (S of W)
∆x
−2.00 km + (5.00 km) cos 233° + (1.00 km) cos120°
Since Harrison must travel in the opposite direction, his return displacement should be
6.33 km at 29.6° north of east .
(b) Strategy and Solution Since Harrison will travel 6.33 km at a speed of 5.00 m s , his return trip will take
∆t =
∆r 6.33 × 103 m ⎛ 1 min ⎞
=
⎜
⎟ = 21.1 min .
v
5.00 m s ⎝ 60 s ⎠
3. (a) Strategy Divide the length of the roadway by the distance between reflectors. There are 1760 yards in one
mile.
Solution Find the distance between reflectors.
2.20 mi 1760 yd
×
= 220 markers
yd
1 mi
17.6
marker
(b) Strategy Divide the length of the roadway by the distance between reflectors. There are 1000 meters in one
kilometer.
Solution Find the distance between reflectors.
3.54 km 1000 m
×
= 221 markers
1 km
16.0 m
marker
324
Physics
Review and Synthesis: Chapters 1–5
4. Strategy Use the conversion between mL and tsp.
Solution Find the factor by which the baby was overmedicated.
mL
(3 4 tsp) 4.9
1 tsp
= 4.9
0.75 mL
The baby was overmedicated by a factor of 4.9.
5. (a) Strategy Find the total distance traveled and the time of travel. Then divide the distance by the time to obtain
the average speed.
Solution Find Mike’s average speed.
∆x = 50.0 m + 34.0 m = 84.0 m and ∆t =
vav =
50.0 m
34.0 m
+
= 48.2 s, so the average speed is
1.84 m s 1.62 m s
∆x 84.0 m
=
= 1.74 m s .
∆t
48.2 s
(b) Strategy Find Mike’s total displacement and divide it by the time found in part (a) to obtain his average
velocity. Let his initial direction be positive.
G
Solution Mike’s total displacement is ∆r = 50.0 m forward − 34.0 m back = 16.0 m forward. So, his average
G
∆r 16.0 m forward
G
=
= 0.332 m s forward, or
velocity is v av =
∆t
48.2 s
0.332 m s in his original direction of motion .
6. Strategy Assume constant acceleration. Use vfx 2 − vix 2 = v = 2a x ∆x and Newton’s second law.
Solution Find the acceleration of the jet.
v 2
vfx 2 − vix 2 = vfx 2 − 0 = 2a x ∆x, so a x = fx .
2 ∆x
Estimate the average force on the jet due to the catapult.
ΣFx = Feng + Fcata = ma x , so
Fcata = ma x − Feng =
mvfx 2
2∆x
2
− Feng =
(33, 000 kg)(160 mi h)2 ⎛ 0.4470 m s ⎞
⎛ 1N ⎞
⎜
⎟ − 2(27, 000 lb) ⎜
⎟
2(90 m)
⎝ 0.2248 lb ⎠
⎝ 1 mi h ⎠
= 700 kN .
7. Strategy Use the conversion between feet and meters.
Solution Find the difference between the incorrect and correct altitudes.
1m
1500 m − (1500 ft)
= 1000 m = 3300 ft
3.281 ft
The captain thought they were 3300 ft or 1000 m above the correct altitude.
325
Review and Synthesis: Chapters 1–5
Physics
8. Strategy To reach the other side of the river in as short a time as possible, Paula must swim in the direction
perpendicular to the river’s flow. Find the time it takes for Paula to cross. Then use this time and the speed of the
river to find how far downstream she travels while crossing.
Solution It takes Paula a time ∆t = ∆x vswim to cross the river. During this time, she travels
⎛ ∆x
∆y = vriver ∆t = vriver ⎜⎜
⎝ vswim
10.2 m
0.833 m/s
⎞ (1.43 m s)(10.2 m)
= 17.5 m downstream.
⎟⎟ =
0.833 m s
⎠
1.43 m/s
9. Strategy Let north be up. Using a ruler and a protractor, draw the force vectors to scale; then, find the sum of the
force vectors graphically.
Solution Draw the diagram and measure the length and angle of the sum of the force vectors.
F
N
12°
2.3F
1.5F
The net force has a magnitude of about 2.3F, where F is the magnitude of the force with which Sandy pulls. The
net force is at an angle of about 12° south of east. So, the cart will go off the road toward south.
10. Strategy Use Newton’s second law.
Solution
(a) Find the magnitude of the horizontal force exerted on the tire by the wind.
W
ΣFx = F − T sin θ = 0, so F = T sin θ ; ΣFy = T cos θ − W = 0, so T =
.
cos θ
W
Thus, F =
sin θ = W tan θ = W tan12° .
cos θ
T
y
12°
W
W
W
(b) From part (a), T =
=
.
cos θ
cos12°
326
x
F
Physics
Review and Synthesis: Chapters 1–5
11. (a) Strategy According to Newton’s third law, when the astronaut exerts a force of 250 N on the asteroid, the
asteroid exerts an equal and opposite force on the astronaut. Use Newton’s second law and the equations for
constant acceleration.
Solution Let the astronaut be 1 and the asteroid be 2. Find the accelerations of the astronaut and the asteroid.
F = ma, so a = F m . The acceleration of the astronaut during the 0.35-s time interval is
250 N
250 N
= 4.167 m s 2 and that of the asteroid is a2 =
= 6.25 m s 2 .
60.0 kg
40.0 kg
The distances traveled by the astronaut and the asteroid during the initial 0.35-s time interval are
1
1
d1i = a1 (∆ti )2 = (4.167 m s 2 )(0.35 s) 2 = 0.255 m and
2
2
1
1
d 2i = a2 (∆ti )2 = (6.25 m s 2 )(0.35 s)2 = 0.383 m. The speeds of the astronaut and the asteroid after the
2
2
a1 =
acceleration are v1 = a1∆ti = (4.167 m s 2 )(0.35 s) = 1.458 m s and
v2 = a2 ∆ti = (6.25 m s 2 )(0.35 s) = 2.188 m s. The distances traveled by the astronaut and the asteroid
during the final 5.00-s time interval are d1f = v1∆tf = (1.458 m s)(5.00 s) = 7.29 m and
d 2f = v2 ∆tf = (2.188 m s)(5.00 s) = 10.94 m. The sum of all the computed distances is the total distance
between the astronaut and the asteroid.
0.255 m + 0.383 m + 7.29 m + 10.94 m = 19 m (rounded to two significant figures)
(b) Strategy and Solution The relative speed between the astronaut and the asteroid is the sum of the two
speeds found in part (a): 1.458 m s + 2.188 m s = 3.6 m s .
12. (a) Strategy and Solution Answers will vary, but a reasonable magnitude of the force required to pull out a
single hair is 1 N.
(b) Strategy The total force exerted on all of the hairs is the weight of the prince, W = mg . Dividing his weight
by the number of hairs gives the average force pulling on each strand of hair.
Solution Will Rapunzel be bald?
W
(60 kg)(9.80 N kg)
Fper hair =
=
= 6 mN
100, 000
100,000
Since 6 mN << 1 N, Rapunzel will most certainly not be made bald by the prince climbing up her hair.
13. Strategy Use Newton’s second law to find the acceleration due to friction. Then use the acceleration and distance
the plate must travel to determine the necessary initial speed.
Solution According to Newton’s second law, ΣFy = N − mg = 0 and ΣFx = − f k = ma.
µ N
µ mg
f
= − µk g . Find the initial speed.
So, a = − k = − k = − k
m
m
m
vf 2 − vi 2 = 0 − vi 2 = 2a∆x = −2 µk g ∆x, so
vi = 2 µk g ∆x = 2(0.32)(9.80 m
s 2 )(0.44
m) = 1.7 m s .
327
y
N
fk
x
mg
Review and Synthesis: Chapters 1–5
Physics
14. Strategy Draw a free-body diagram for each crate. Use Newton’s second law.
Solution Let the 9.00-kg crate be (1) and the 14.0-kg crate
be (2). The crates have the same acceleration.
Crate 1:
ΣFx = T − f1 = T − µ N1 = m1a and
ΣFy = N1 − m1 g = 0, so T = µ m1 g + m1a (I).
y
f1
N1
N2
x
T
m 1g
F
T
20.0°
m 2g
f2
Crate 2:
ΣFx = F cos θ − T − f 2 = F cos θ − T − µ N 2 = m2 a and
ΣFy = N 2 + F sin θ − m2 g = 0, so F cos θ − T − µ (m2 g − F sin θ ) = m2 a (II).
Substitute I into II and solve for a.
F cos θ − µ m1 g − m1a − µ (m2 g − F sin θ ) = m2 a
F cos θ − µ m1 g − µ m2 g + µ F sin θ = (m1 + m2 )a
a = F (cos θ + µ sin θ ) (m1 + m2 ) − µ g
So, the magnitude of the acceleration is
(195 N)(cos 20.0° + 0.550sin 20.0°)
a=
− 0.550(9.80 m s 2 ) = 4.2 m s 2 .
9.00 kg + 14.0 kg
Compute the tension.
T = m1 ( µ g + a ) = (9.00 kg)[0.550(9.80 m s 2 ) + 4.172 m s 2 ] = 86 N
15. (a) Strategy Since the coefficient of static friction between the blocks is greater than that between the bottom
block and the floor, the two blocks will just begin to slide as a unit ( a ≈ 0). The boy must push with a
horizontal force equal to the maximum force of static friction on the bottom block due to the floor. Let the top
block be 1 and the bottom block be 2. Use Newton’s second law.
Solution
ΣFy = N − (m1 + m2 ) g = 0 and
y
N
ΣFx = Fboy − fs, max = Fboy − µs N = Fboy − µs (m1 + m2 ) g = 0, so
Fboy = µs (m1 + m2 ) g = 0.220(5.00 kg + 2.00 kg)(9.80 m s 2 ) = 15.1 N .
m1
fs
m2
x
Fboy
(m1 + m2)g
(b) Strategy The block is now sliding and the boy is still pushing. There are two forces acting on the system of
blocks, the push of the boy and the force of kinetic friction between the bottom block and the floor. There are
also two forces acting on each block alone; for the top block, the forces are the push of the boy and the force
of static friction due to the bottom block; for the bottom block, the force of static friction due to the top block
and the force of kinetic friction due to the floor. Use Newton’s second law.
Solution For the two-block system:
y
Fboy
N
x
ΣFx = Fboy − µk (m1 + m2 ) g = (m1 + m2 )a, so a =
− µk g .
m1 + m2
m1
Fboy
fs
For the top block alone:
m 1g
ΣFy = N − m1 g = 0 and ΣFx = Fboy − fs = Fboy − µs N = Fboy − µs m1 g = m1a.
Substitute the expression for the acceleration a found above and solve for the maximum force.
⎛ Fboy
⎞
Fboy − µs m1 g = m1a = m1 ⎜
− µk g ⎟ , so
⎜m +m
⎟
2
⎝ 1
⎠
( µs − µk )m1 g (m1 + m2 ) (0.400 − 0.200)(5.00 kg)(9.80 m s 2 )(5.00 kg + 2.00 kg)
=
= 34.3 N .
Fboy =
2.00 kg
m2
328
Physics
Review and Synthesis: Chapters 1–5
16. Strategy The point at which the gravitational field is zero is somewhere along the line between the centers of the
two stars. Use Newton’s law of universal gravitation.
Solution The distance between the stars is d = d1 + d 2 , where d1 is the distance to the F = 0 point from the star
with mass M1 and d 2 is the distance to the F = 0 point from the other star. The forces are equal in magnitude and
opposite in direction at the F = 0 point. Let m be a test mass at the F = 0 point.
GM1m G (4.0 M1 )m
=
, so d 2 = 2.0d1.
d12
d 22
d = d1 + d 2 = d1 + 2.0d1 = 3.0d1 , so d1 =
d
= 0.33d .
3.0
The gravitational field is zero approximately one third (0.33) of the distance between the stars as measured
from the star with mass M1.
17. Strategy and Solution Consider a cord attached to a wall at one end and pulled by one of the boys at the other
end. The cord does not accelerate when the boy pulls it; thus, the force on the cord from the wall must be equal in
magnitude to the pulling force. This situation is identical to the one in which the two boys pull from opposite ends
of the cord—the tension in the cord is the same as the case when only one boy is pulling. However, if both pull
from one end, the tension is doubled, so Stefan’s plan is superior and thus more likely to work.
18. Strategy Use the definition of average acceleration and Newton’s second law.
Solution
(a) Compute the trout’s average acceleration.
G
∆v 2 m s
aav =
=
= 40 m s 2 , so aav = 40 m s 2 in the direction of motion .
∆t 0.05 s
(b) Find the average net force on the trout.
G
W
40 m s 2
Fav = maav = aav = W
= 4W , so Fav = 4W in the direction of motion .
2
g
9.80 m s
(c) The trout pushes backward on the water, which then pushes the trout forward.
19. Strategy Use components of the displacement and the Pythagorean theorem to find the distance the trawler must
travel to catch up to the tuna. (This is the same as the distance the tuna is from the trawler’s initial position in 4.0
hours.) Let +y be north and +x be west.
Solution Find the distance.
y = 100.0 km + (5.00 km h)(4.0 h) cos 45° and x = (5.00 km h)(4.0 h) sin 45°, so
r = x 2 + y 2 = (5.00 km h)2 (4.0 h)2 sin 2 45° + [100.0 km + (5.00 km h)(4.0 h) cos 45°] = 115 km.
2
115 km
= 29 km h .
4.0 h
Find the direction.
y
100.0 km + (5.00 km h)(4.0 h) cos 45°
= 83° N of W
θ = tan −1 = tan −1
x
(5.00 km h)(4.0 h) sin 45°
G
Thus, v = 29 km h at 83° N of W .
So, vtrawler =
329
Review and Synthesis: Chapters 1–5
Physics
20. Strategy Find the time it takes for the newspaper to hit the ground; then use the time to find the horizontal
distance the newspaper travels before it hits the ground. Finally, use the speed of the car and the coefficient of
kinetic friction to find the distance the newspaper slides.
Solution Find the time it takes for the newspaper to hit the ground.
1
2∆y
2(1.00 m)
∆y = gt 2 , so t =
=
= 0.45 s.
2
g
9.8 m s 2
Find the horizontal distance traveled by the newspaper as it travels through the air.
∆x1 = vt = (15 m s)(0.45 s) = 6.8 m
Use Newton’s second law to find the acceleration due to friction that stops the newspaper.
ΣFx = − f k = − µk mg = ma, so a = − µk g .
Find the horizontal distance traveled by the newspaper after it hits the ground.
∆v = vf2 − vi2 = 0 − vi2 = 2a∆x2 = −2 µk g ∆x2 , so ∆x2 =
vi2
2µk g
=
(15 m s)2
2(0.40)(9.8 m s 2 )
= 29 m.
Julia should drop the paper a distance of 6.8 m + 29 m = 36 m.
21. (a) Strategy Use vfy 2 − viy 2 = vf 2 − vi 2 = 2 gh. Let the +y-direction be down and neglect air resistance.
Solution Let the initial speed for all three rocks be vi and the vertical distance from the cliff to the ground
be h. For the first rock (thrown straight down):
vfy 2 − viy 2 = vf 2 − vi 2 = 2 gh, so vf = vi 2 + 2 gh .
For the second rock (thrown straight up):
vfy 2 − viy 2 = vf 2 − vi 2 = 2 gh, so vf = vi 2 + 2 gh .
For the third rock (thrown horizontally):
vfy 2 − viy 2 = vfy 2 − 0 = vfy 2 = 2 gh and vfx = vix = vi , so vf = vfx 2 + vfy 2 = vi 2 + 2 gh .
Therefore, just before the rocks hit the ground at the bottom of the cliff, all three have the same final speed.
(b) Strategy Use the result obtained for the final speed in part (a).
Solution Compute the final speed.
vf = vi 2 + 2 gh = (10.0 m s)2 + 2(9.80 m s 2 )(15.00 m) = 19.8 m s
22. Strategy The question is: Will the ball be 10 ft above the ground when it has traveled 45 yd (135 ft) horizontally?
Use the equations of motion with a changing velocity.
Solution The ball will travel 135 ft in the time ∆t = ∆x vx = ∆x (vi cos θ ).
At this time, the ball will be
⎛ ∆x
1
∆y = vi sin θ∆t − g (∆t )2 = vi sin θ ⎜⎜
2
⎝ vi cos θ
⎞ 1 ⎛ ∆x
⎟⎟ − g ⎜⎜
⎠ 2 ⎝ vi cos θ
2
⎞
⎟⎟
⎠
vi sin35°
330
35°
vi cos 35°
2
⎤
1 ⎛ ∆x ⎞
1
135 ft
2 ⎡
= ∆x tan θ − g ⎜⎜
⎟⎟ = (135 ft) tan 35° − (32.2 ft s ) ⎢
⎥
2 ⎝ vi cos θ ⎠
2
⎣ (21 m s )( 3.281 ft m ) cos 35° ⎦
= 2.4 ft above the ground.
So, the ball will not clear the goal post and your favorite team will win.
vi = 21 m/s
2
Physics
Review and Synthesis: Chapters 1–5
23. Strategy Use Newton’s second law to find the maximum force of static friction. Then, use the equations for
circular motion.
Solution
ΣFr = fs = µs N = mar = m
v2
r
and ΣFy = N − mg = 0, so µs g =
v2
r
ar
N
, or v = µs gr .
mg
fs
Now, v = rω and ω = α∆t.
Substitute for v and solve for ∆t to find the time it takes for the coin to slide.
µs gr = v = rω = rα∆t , so ∆t =
1
α
µs g
r
=
1
1.20 rad s 2
0.110(9.80 m s 2 )
= 2.40 s
0.130 m
24. (a) Strategy Use Newton’s second law to find the acceleration down the slope.
Then find Carlos’s speed when Shannon starts from rest. Finally, find the time it
takes for Shannon to catch up to Carlos, and use this time to find the distance
traveled.
Solution Find the acceleration down the slope.
ΣFx = − f k + mg sin θ = − µk N + mg sin θ = ma x and ΣFy = N − mg cos θ = 0,
so N = mg cos θ . Thus, a x = − µk g cos θ + g sin θ = g (sin θ − µk cos θ ).
Find Carlos’s speed when Shannon starts from rest.
y
N
fk
θ
mg
θ
x
mg sinθ
∆v = vf2 − vi2 = vc2 − 0 = 2ac ∆x1 , so vc = 2ac ∆x1 , where ∆x1 = 5 m.
Find the time when Shannon and Carlos meet.
∆xs = ∆x1 + ∆xc
1 2
1
a t = ∆x1 + vc t + ac t 2
2 s
2
2
as t = 2∆x1 + 2t 2ac ∆x1 + ac t 2
0 = (as − ac )t 2 − t 8ac ∆x1 − 2∆x1
We can solve for t using the quadratic formula.
−b ± b 2 − 4ac
, where
2a
a = as − ac = g (sin θ − µs cos θ ) − g (sin θ − µc cos θ ) = g ( µc − µs ) cos θ
t=
= (9.8 m s 2 )(0.10 − 0.010) cos12° ≈ 0.8627 m s 2
b = − 8ac ∆x1 = − 8 g (sin θ − µc cos θ )∆x1 = − 8(9.8 m s 2 )(sin12° − 0.10 cos12°)(5 m) ≈ − 6.569 m s
c = −2∆x1 = −2(5 m) = −10 m
t ≈ 8.915 s (The other answer, − 1.3 s, is extraneous.)
The distance traveled is
∆xs = 12 as t 2 = 12 g (sin θ − µs cos θ )t 2 = 12 (9.8 m s 2 )(sin12° − 0.010 cos12°)(8.915 s)2 = 77 m .
(b) Strategy Shannon is moving faster than Carlos when the two meet. Shannon’s speed relative to Carlos is the
difference of the two speeds.
Solution Compute the relative speed.
vs − vc = as t − (vc + ac t ) = (as − ac )t − 2ac ∆x1 = (as − ac )t − 2 g (sin θ − µc cos θ )∆x1
= (0.8627 m s 2 )(8.915 s) − 2(9.8 m s 2 )(sin12° − 0.10 cos12°)(5 m) = 4.4 m s
331
Review and Synthesis: Chapters 1–5
Physics
25. (a) Strategy Use Kepler’s third law. Ignore the mass of the cable.
Solution Find the height H.
13
⎛ GmET 2
R=⎜
⎜ 4π 2
⎝
⎞
⎟
⎟
⎠
⎛ GmET 2
H =⎜
⎜ 4π 2
⎝
⎞
⎟
⎟
⎠
13
= RE + H , so
13
⎛ [6.674 × 10−11 m3 (kg ⋅ s 2 )](5.974 × 1024 kg)(86, 400 s)2 ⎞
− RE = ⎜
⎟⎟
⎜
4π 2
⎝
⎠
− 6.371× 106 m
≅ 3.6 × 107 m
(b) Strategy Use Newton’s second law and law of universal gravitation.
Solution Find the tension in the cable.
H
3.5874 × 107 m
= 6.371× 106 m +
= 2.4308 × 107 m
2
2
⎛ 4π 2 ⎞
GmE m
mv 2
ΣFy = Fg − T = ma =
= mω 2 R, so T =
− m⎜
R.
⎜ T 2 ⎟⎟
R
R2
⎝
⎠
R = RE +
T=
[6.674 × 10−11 m3 (kg ⋅ s 2 )](5.974 × 1024 kg)(100 kg)
(2.4308 × 107 m)2
−
4π 2 (100 kg)(2.4308 × 107 m)
(86, 400 s)2
= 55 N
26. Strategy Draw a free-body diagram for the package and use Newton’s second law.
Solution Find the maximum acceleration the truck can have without the
package falling of the back.
ΣFx = fs − mg sin10° = µs N − mg sin10° = ma x and
ΣFy = N − mg cos10° = 0, so N = mg cos10°.
The maximum acceleration allowed for the truck is
a x = g ( µs cos10° − sin10°)
= (9.80 m s 2 )(0.380 cos10° − sin10°) = 2.0 m s 2 .
332
N
y
fs
x
10°
mg cos 10°
mg sin 10°
10° mg
Physics
Review and Synthesis: Chapters 1–5
27. Strategy Draw a diagram. Use Newton’s second law and the relationship between radial acceleration and linear
speed. Refer to Example 5.7.
Solution From Example 5-7, the banking angle is given by
θ=
N
tan −1[vb 2
(rg )], where vb is the speed that a car can navigate the curve
without any friction. Find the relationship for the slowest speed the car can go
around the curve without sliding down the bank.
v2
ΣFy = N y − mg + f y = 0 and ΣFx = N x − f x = mar = m , so
r
v2
rN
m
= N x − f x = N sin θ − µs N cos θ , or v =
(sin θ − µs cos θ ).
r
m
Find N.
θ
y
fs
θ
x
mg
θ
mg
N y − mg + f y = N cos θ − mg + µs N sin θ = 0, so N (cos θ + µs sin θ ) = mg , or N =
.
cos θ + µs sin θ
Substitute for N in v and substitute for θ .
v=
=
r⎛
mg
⎜⎜
m ⎝ cos θ + µs sin θ
⎞
⎟⎟ (sin θ − µs cos θ ) =
⎠
gr[tan tan −1 vb 2 (rg ) − µs ]
1 + µs tan tan −1 vb 2 (rg )
=
gr (sin θ − µs cos θ )
cos θ + µs sin θ
gr[vb 2 (rg ) − µs ]
1 + µs [vb 2 (rg )]
Thus, the slowest allowed speed is v =
=
=
gr (tan θ − µs )
1 + µs tan θ
vb 2 − µs gr
1 + µs vb 2 (rg )
(15.0 m s)2 − 0.120(9.80 m s 2 )(75.0 m)
1 + 0.120 (15.0 m s) 2 [(9.80 m s 2 )(75.0 m)]
= 11.5 m s .
28. Strategy The rope has the same tension throughout its length. Use Newton’s second law.
Solution Since the box is to be lifted with constant velocity, the acceleration must be zero. For the pulley on the
left, we have ΣFy = 2T − mg = 0, so T = mg 2. Since the tension is the same along the length of the rope, the
minimum force required is equal to the tension. Thus, the pull force is
mg (98.0 kg)(9.80 m s 2 )
F =T =
=
= 480 N .
2
2
333
Review and Synthesis: Chapters 1–5
Physics
29. Strategy Use Newton’s second law to determine the acceleration of each block. Then, use the accelerations to
describe the motion of each block.
Solution Let the positive x-direction be to the right.
F
F
The accelerations of each block are a A = A and aB = B . In the time ∆t , the blocks are displaced by
mA
mB
⎞
2
⎟⎟ (∆t ) . At the end of ∆t , the blocks move
⎠
F
F
with constant velocities until they meet. These velocities are v A = a A ∆t = A ∆t and vB = aB ∆t = B ∆t. Let the
mA
mB
∆x A1 =
1
1⎛ F ⎞
1
1⎛ F
a A (∆t ) 2 = ⎜⎜ A ⎟⎟ (∆t ) 2 and ∆xB1 = aB (∆t ) 2 = ⎜⎜ B
2
2 ⎝ mA ⎠
2
2 ⎝ mB
time between ∆t and the moment when the two blocks meet be ∆t2 . Then, the displacements of each block
during ∆t2 are ∆x A2 = v A ∆t2 =
FA
F
∆t ∆t2 and ∆xB 2 = vB ∆t2 = B ∆t ∆t2 . Let d = 3.40 m. Then, we have the
mA
mB
F
F
1⎛ F ⎞
1⎛ F ⎞
equation d = ⎜⎜ A ⎟⎟ (∆t ) 2 + A ∆t ∆t2 − ⎜⎜ B ⎟⎟ (∆t ) 2 − B ∆t ∆t2 , where the displacements of block B are
2 ⎝ mA ⎠
2 ⎝ mB ⎠
mA
mB
subtracted because they are negative. Solving this equation for ∆t2 gives
( ∆t ) 2
∆t 2 =
d− 2
(
FA
mA
(
−
FA
mA
FB
mB
F
− mB
) ∆t
B
) = 3.40 m −
(
(
(0.100 s) 2 2.00 N
2
0.225 kg
2.00 N
0.225 kg
)
−5.00 N
− 0.600
kg
−5.00 N (0.100 s)
− 0.600
kg
) = 1.92 s. So, the elapsed time from t = 0 until
the blocks meet is ∆t + ∆t2 = 0.100 s + 1.92 s = 2.02 s . The distance from block B’s initial position to where
the two blocks meet is negative the displacement of B.
F
1⎛ F ⎞
1 ⎛ −5.00 N ⎞
2 −5.00 N (0.100 s)(1.924 s) = 1.65 m
− ⎜⎜ B ⎟⎟ (∆t ) 2 − B ∆t ∆t2 = − ⎜
⎟ (0.100 s) −
2 ⎝ mB ⎠
2 ⎝ 0.600 kg ⎠
0.600 kg
mB
The blocks meet at 1.65 m to the left of B’s initial position.
30. (a) Strategy Use the equations for circular motion.
Solution Find the angular acceleration.
∆ω = ωf − ωi = ωf − 0 = α∆t , so α =
ωf
∆t
.
Find the tangential acceleration.
ω
1.00 Hz ⎛ 2π rad ⎞
2
a = rα = r f = (0.100 m)
⎜
⎟ = 0.785 m s
0.800 s ⎝ cycle ⎠
∆t
(b) Strategy While the hamster is running, the forces on it are due to gravity and the normal force due to the
wheel. Use Newton’s second law.
Solution Find the normal force on the hamster.
v2
(rω )2
ΣFr = N − mg = mar = m
=m
= mrω 2 , so
r
r
N = m( g + rω 2 ) = (0.100 kg)[9.80 m s 2 + (0.100 m)(1.00 Hz) 2 (2π rad cycle) 2 ] = 1.37 N .
334
Physics
Review and Synthesis: Chapters 1–5
31. Strategy Use the equations of motion for changing velocity.
Solution ∆t = ∆t1 + ∆t2 = 0.10 s + 0.15 s = 0.25 s has elapsed since the first pellet was fired from the toy gun.
The components of its velocity are given by v1x = vi cos θ and v1 y = vi sin θ − g ∆t. The components of the second
pellet’s velocity are given by v2 x = vi cos θ and v2 y = vi sin θ − g ∆t2 . The horizontal components of the two
velocities are the same, but the vertical components differ. Compute this difference.
v1 y − v2 y = vi sin θ − g ∆t − (vi sin θ − g ∆t2 ) = − g (∆t − ∆t2 ) = −(9.80 m s 2 )(0.25 s − 0.15 s) = −0.98 m s
So, the velocity of the first pellet with respect to the second after the additional 0.15 s have passed is
0.98 m s directed downward .
32. Strategy Use Newton’s second law and the equations of motion for changing velocity.
Solution
(a) Find the acceleration of the crate.
ΣFx = mg sin 35.0° = ma x , so a x = g sin 35.0°.
Find the distance traveled by the crate.
1
1
∆x = a x (∆t )2 = ( g sin 35.0°)(∆t )2
2
2
1
2
= (9.80 m s ) sin 35.0°(2.50 s)2 = 17.6 m
2
x
35.0°
35.0°
mg
mg sin 35.0°
(b) The speed of the crate after 2.50 s of travel is
v = ax ∆t = g sin 35.0°∆t = (9.80 m s 2 ) sin 35.0°(2.50 s) = 14.1 m s .
33. Strategy Use the equations of motion for constant vertical acceleration.
Solution
(a) The vertical and horizontal components of the projectile are given by
1
∆y = (vi sin θ )∆t − g (∆t )2 and ∆x = (vi cos θ )∆t.
2
When a projectile returns to its original height, ∆y = 0.
2v sin θ
1
1
0 = (vi sin θ )∆t − g (∆t ) 2 = vi sin θ − g ∆t , so ∆t = i
.
2
2
g
Substitute this value for ∆t into R = ∆x = (vi cos θ )∆t to find the range.
⎛ 2v sin θ
R = (vi cos θ )∆t = vi cos θ ⎜ i
g
⎝
2vi 2 sin θ cos θ
⎞
=
⎟
g
⎠
(b) Using the equation for the range found in part (a), we find that the range of the projectile if it is not
2(50.0 m s)2 sin 30.0° cos 30.0°
= 221 m .
intercepted by the wall is R =
9.80 m s 2
335
Review and Synthesis: Chapters 1–5
Physics
(c) Find the time of flight in terms of vi , ∆x, and θ .
∆x
.
vi cos θ
Find the height at which the cannonball strikes.
∆x = (vi cos θ )∆t , so ∆t =
⎛ ∆x
1
y = yi + (vi sin θ )∆t − g (∆t )2 = yi + (vi sin θ ) ⎜⎜
2
⎝ vi cos θ
= 1.10 m + (215 m) tan 30.0° −
⎞ 1 ⎛ ∆x
⎟⎟ − g ⎜⎜
⎠ 2 ⎝ vi cos θ
(9.80 m s 2 )(215 m) 2
2(50.0 m s) 2 cos 2 30.0°
2
⎞
g ( ∆x ) 2
⎟⎟ = yi + ∆x tan θ − 2
2vi cos 2 θ
⎠
= 4m
34. (a) Strategy Let +y be down for m1 and up for m2 , since m1 >> m2 . Use Newton’s second law.
Solution Find the acceleration of each block.
For m1: ∑ Fy = m1 g − T = m1a y , so T = m1 g − m1a y .
y
For m2 : ∑ Fy = T − m2 g = m2 a y , so T = m2 g + m2 a y .
T and a y are identical in these two equations. Eliminate T.
m1 g − m1a y = m2 g + m2 a y
T
T
m 2g
y
m2 a y + m1a y = m1 g − m2 g
(m1 + m2 )a y = (m1 − m2 ) g
ay =
m1 − m2
m1 + m2
g
m 1g
Since m1 >> m2 , m1 − m2 ≈ m1 and m1 + m2 ≈ m1 , so a y ≈ g .
(b) Strategy Use the results from part (a) for the tension and the vertical component of the acceleration.
Solution Find the tension.
⎛ m − m2 ⎞
m1 + m2 + m1 − m2
2m1m2
g = m2 g ⎜⎜ 1 + 1
=
g
⎟⎟ = m2 g
m1 + m2
m1 + m2
m1 + m2
⎝ m1 + m2 ⎠
2m m
Since m1 >> m2 , m1 + m2 ≈ m1 , so T ≈ 1 2 g = 2m2 g .
m1
T = m2 g + m2 a y = m2 g + m2
m1 − m2
35. (a) Strategy The circumference of the track is given by C = 2π r. The distance traveled is three-fourths of this.
Solution Compute the distance traveled by the runner before the collision.
3
3
distance traveled = (2π r ) = π (60.0 m) = 283 m
4
2
(b) Strategy Let the center of the circle be the origin, and let the runner begin at
θ = 0° and collide at θ = 270°.
Solution Find the components of the runner’s displacement.
rix = 60.0 m, riy = 0, rfx = 0 and rfy = −60.0 m.
Find the magnitude of the displacement.
G
∆r = (rfx − rix )2 + (rfy − riy )2 = (−60.0 m) 2 + (−60.0 m) 2 = 84.9 m
336
y
60.0 m
x
∆r
Finish
Start
Physics
Review and Synthesis: Chapters 1–5
36. Strategy Let the +x-direction be along the dashed line away from the sun, and let the +y-direction be
perpendicular to the dashed line away from the Earth. Use Newton’s second law.
Solution
(a) Find the net force acting on the sailplane.
∑ Fx = (8.00 × 102 N) cos 30.0° − 173 N = ma x = Fx
∑ Fy = (8.00 × 102 N) sin 30.0° − 1.00 × 102 N = ma y = Fy
Compute the magnitude of the net force.
F = Fx 2 + Fy 2 = [(8.00 × 102 N) cos 30.0° − 173 N]2 + [(8.00 × 102 N) sin 30.0° − 1.00 × 102 N]2
= 6.00 × 102 N
Compute the direction of the net force.
Fy
(8.00 × 102 N) sin 30.0° − 1.00 × 102 N
= tan −1
= 30.0°
θ = tan −1
Fx
(8.00 × 102 N) cos 30.0° − 173 N
G
So, F = 6.00 × 102 N directed along the 8.00 × 102 -N vector .
(b) Find the acceleration of the sailplane.
G
F 6.00 × 102 N
a= =
= 0.0414 m s 2 , so a = 0.0414 m s 2 in the same direction as the force .
14,500 kg
m
37. (a) Strategy Use Newton’s law of universal gravitation. Assume the mass of the galaxy is concentrated at its
center.
Solution Estimate the mass of the galaxy.
GMm mv 2
=
, so
R
R2
v 2 R (2.75 × 105 m s)2 (40, 000 ly) ⎛ 9.461× 1015 m ⎞
=
⎜
⎟⎟
1 ly
G
6.674 × 10−11 m3 (kg ⋅ s 2 ) ⎜⎝
⎠
= 4.3 × 1041 kg or about 216 billion solar masses .
M =
(b) Strategy Compute the ratio of the visible mass to the estimated mass.
Solution
1011
2.16 × 1011
= 0.46
337
Review and Synthesis: Chapters 1–5
Physics
38. Strategy Draw and analyze the vector diagrams. The velocity of the water relative to the sailboat is opposite to
the velocity of the sailboat relative to the water.
Solution
(a) a = air, w = water, s = sailboat
Case (1):
Case (2):
Case (3):
vaw
vaw
vws
vws
vas
vas
vaw
vas
vws
(b) According to the vector diagrams, the apparent wind speed is greater than the true wind speed in cases
1 and 2.
(c) According to the vector diagrams, the apparent wind direction is forward of the true wind in all three cases.
MCAT Review
1. Strategy and Solution Gravity contributes an acceleration of − g . Air resistance is always opposite an object’s
direction of motion, so the vertical component of the acceleration contributed by air resistance is negative as well.
According to Newton’s second law, F = ma, so the magnitude of the acceleration due to air resistance is
aR = FR m = bv 2 m . Since we want the vertical component of acceleration, the correct answer is D ,
− g − (bvv y ) (0.5 kg).
2. Strategy Use the result for the range derived in Problem 4.48b.
Solution Assuming air resistance is negligible, the horizontal distance the projectile travels before returning to
the elevation from which it was launched is R =
vi 2 sin 2θ (30 m s) 2 sin[2(40°)]
=
= 90 m. Thus, the correct
g
9.80 m s 2
answer is C .
3. Strategy and Solution The magnitude of the horizontal component of air resistance is
FR cos θ = bv 2 cos θ = bv(v cos θ ) = bvvx . Thus, the correct answer is D .
338
Physics
Review and Synthesis: Chapters 1–5
4. Strategy Use Newton’s second law to analyze each case. For simplicity, consider only vertical motion.
Solution Let the positive y-direction be up.
On the way up:
bv 2
ΣFy = − mg − bv 2 = ma y , so a y = − g −
.
m
On the way down:
bv 2
ΣFy = −mg + bv 2 = ma y , so a y = − g +
.
m
The magnitude of the acceleration is greater on the way up than on the way down. On the way up, the magnitude
of the acceleration is never less than g. On the way down, it may be as small as zero. The projectile must travel
the same distance in each case. So, when a projectile is rising, it begins with an initial speed which is reduced to
zero relatively quickly due to the relatively large negative acceleration it experiences. When a projectile is falling,
it begins with zero speed and is accelerated toward the ground by a smaller acceleration relative to when it is
rising. Thus, it must take the projectile longer to reach the ground than to reach its maximum height; therefore, the
correct answer is C .
5. Strategy Find the time it takes to cross the river. Use this time and the speed of the river to find how far
downstream the raft travels while crossing. Then use the Pythagorean theorem to find the total distance traveled.
Solution Let x be the width of the river and y be the distance traveled down the river during the crossing.
The raft takes the time ∆t = x vraft to cross the river. During this time, the raft travels the distance
y = vriver ∆t = vriver ( x vraft ) down the river. Compute the distance traveled.
⎛v
x 2 + y 2 = x 2 + ⎜⎜ river
⎝ vraft
2
2
2
⎞
⎛v
⎞
⎛2 m s⎞
x ⎟⎟ = x 1 + ⎜⎜ river ⎟⎟ = (200 m) 1 + ⎜
⎟ = 283 m
v
⎝2 m s⎠
⎠
⎝ raft ⎠
The correct answer is C .
6. Strategy To row directly across the river, the component of the raft’s velocity that is antiparallel to the current of
the river must equal the speed of the current, 2 m s.
Solution Since the angle is relative to the shore, the antiparallel component of the raft’s velocity is (3 m s) cos θ .
Set this equal to the speed of the current and solve for θ .
2
2
(3 m s) cos θ = 2 m s , so cos θ = , or θ = cos −1 . The correct answer is D .
3
3
1
7. Strategy Use ∆y = viy ∆t + a y (∆t ) 2 .
2
Solution Find the time it takes the rock to reach the ground.
1
1
2 ∆y
2(0 − 100 m)
∆y = viy ∆t + a y (∆t ) 2 = (0)∆t − g (∆t ) 2 , so ∆t = −
= −
= 4.5 s.
2
2
g
10 m s 2
The correct answer is A .
339
Chapter 6
CONSERVATION OF ENERGY
Conceptual Questions
1. Assuming the object can be treated as a point particle, the total work done on it by external forces is equal to the
change in its kinetic energy. An object moving in a circle may be changing its speed as it goes around, so the total
work done on it is not necessarily zero.
2. The force exerted on the backpack by your back and shoulders is directed upward and is perpendicular to your
horizontal displacement. Hence it does not do any work on the backpack. (1) Now there is a component of your
displacement directed downward, anti-parallel to the force on the backpack, so the force does negative work on
the backpack. (2) In this case the backpack’s kinetic energy is increasing as you gain speed. The force exerted on
the backpack is no longer vertical, but has a horizontal component in the direction you are moving. Thus, it does
positive work on the backpack.
3. When the roads leading up a mountain wind back and forth, the angle of inclination of the road is less than that of
a road going straight up the mountain. This reduces the force necessary to drive the car up the road and the
required power output of the engine as well. The length of road is increased, however, so the total work that must
be done to reach the top of the mountain remains the same (or may be larger if frictional forces are taken into
account).
4. During the fall, the force of gravity on the mango is parallel to the mango’s displacement, so it does positive
work. The force of gravity on the Earth due to the mango is directed upward, toward the mango, and has the same
magnitude as the force of gravity on the mango (Newton’s third law). The Earth moves upward by a very small
(imperceptible) amount as the mango falls, so the mango’s gravitational field does positive work on the Earth.
Wm and WE are both positive, but WE is a very small number, close to zero, so Wm >> WE .
5. Yes, static friction can do work. As an example, imagine a book on a conveyor belt that carries it up an incline.
The force of static friction on the book is directed upward along the surface of the belt and has a component that is
parallel to the book’s displacement. Thus, the force of static friction does positive work on the book. (At the same
time, the work done on the book by gravity is negative and the total work done on the book is zero.)
6. Work is done on the roller coaster by a tow chain or some other mechanism designed to increase its height with
respect to the ground and thus to increase its gravitational potential energy. At the apex of the first hill, the kinetic
energy of the roller coaster is negligibly small so that its total initial energy is equal to its potential energy. Energy
is dissipated along the trip around the track as a result of frictional effects and air resistance. Thus, unless
additional energy is added to the system—via another tow chain for example—the energy available to the roller
coaster to climb subsequent hills is less than the original total energy—the hills must therefore be shorter.
7. When the ball reaches the ground, the gravitational potential energy it originally possessed will have been
converted into the kinetic energy of its motion. If the ball is a rigid point-like particle, its kinetic energy will be
conserved during the bounce and the ball’s velocity will be the same immediately before and after rebounding.
Most balls however are made of deformable materials like rubber that compress when bouncing. The deformation
process changes the state of the molecules that make up the ball—increasing the ball’s internal energy. The
energy required for this process must be obtained via a decrease in the kinetic energy of the ball after the
rebound—the maximum height attained by the ball will therefore be lower.
340
Physics
Chapter 6: Conservation of Energy
8. The total mechanical energy of the gymnast swinging in a vertical circle about a crossbar has the same constant
value for each point along the path (ignoring the relatively small amount of work done by the gymnast’s muscles
during the swing). The gravitational potential energy of the gymnast is lowest at the bottom of the path and
greatest at the top. Because the gymnast’s total energy is constant, this implies that the kinetic energy of the
gymnast must be lowest at the top of the circle and greatest at the bottom. Correspondingly, the gymnast’s
velocity is a minimum at the top of the loop and a maximum at the bottom.
9. The bicyclist requires a minimum amount of energy to climb the hill. This quantity is independent of the means
that the bicyclist employs to acquire the energy and is solely a function of the height of the hill (the energy
required is also affected by the work done by frictional and drag forces—the magnitude of this effect is
approximately equal for any method used by the bicyclist to climb the hill and therefore doesn’t affect our
reasoning). After beginning the ascent, a component of the gravitational force acts in the direction opposite to the
displacement thereby increasing the amount of negative work done on the rider with respect to the amount done
while riding on flat land. Therefore, the rate at which the rider must do work to acquire the necessary energy is
greater when pedaling uphill than when on flat land. It is thus advantageous to acquire as much energy as possible
before the ascent when the amount of kinetic energy gained per amount of work done by the rider is greatest.
10. When pushing the crate with a force parallel to the ground, the force of friction acting to impede its motion is
proportional to the normal force acting on the crate—in this situation, the normal force is equal to the crate’s
weight. When pulling the crate with a rope angled above the horizontal, the normal force on the crate is less than
its weight—the force of friction is therefore reduced. To keep the crate moving across the floor, the applied force
in the parallel direction must be greater than or equal to the force of friction—pulling on the rope therefore
requires a smaller parallel applied force. The work done in moving an object is equal to the product of the
displacement through which it has been moved and the force component parallel to the direction of motion. The
applied force component parallel to the ground is smaller when pulling the crate with the rope—thus, the work
done to move the crate with the rope must be less, regardless of the weight of the crate or the displacement.
11. Such animals have larger than average leg muscles located predominantly inside the body such that they don’t
have to move with the legs. As a result, the legs of these animals are thinner than the legs of slower animals of
similar size. The less massive legs require less work to accelerate and decelerate—more of the animal’s energy
can therefore go into increasing its kinetic energy and thus its speed.
12. Because an ideal spring has zero mass, Newton’s second law implies that the net force exerted on it must be zero
(provided it has a finite acceleration). The forces exerted by a spring on objects attached to its ends are equal and
opposite to the forces exerted by those objects on the spring, according to Newton’s third law. The forces exerted
by the spring must therefore be equal in magnitude and opposite in direction so that the spring experiences a zero
net force. The work done on the two attached objects is not necessarily the same, because the distances they move
can be different. Consider for example a spring attached to a heavy lead ball on one end and a Ping-Pong ball on
the other. If the spring were initially in a stretched position and then released, the end attached to the lead ball
would hardly move at all compared to the other end. The work done by the spring on the Ping-Pong ball would
therefore be greater than that done on the heavy ball.
13. Zorba is correct. You get to a top speed sooner on the first slide, so it takes less time to get to the bottom, but the
final speeds are the same from mgh = 12 mv 2 .
341
Chapter 6: Conservation of Energy
Physics
Problems
1. Strategy Use Eq. (6-1).
Solution Find the work done by Denise dragging her basket of laundry.
30.0 N
W = F ∆r cos θ = (30.0 N)(5.0 m) cos 60.0° = 75 J
60.0°
5.0 m
x
2. Strategy The distance is equal to the speed times the time interval. Use Eq. (6-1).
Solution Find the work done by the rope on the sled.
W = F ∆r cos θ = Tv∆t cos θ = (240 N)(1.5 m s)(10.0 s) cos 30.0° = 3.1 kJ
3. Strategy and Solution Since the book undergoes no displacement, no work is done on the book by Hilda.
4. Strategy The angle between the tension and the displacement is zero. Use Eq. (6-1).
Solution Find the work done by the towrope on the water-skier.
W = F ∆r cos θ = (240 N)(54 m) cos 0° = 13 kJ
5. Strategy Use Newton’s second law and Eq. (6-2).
Solution Find the net force on the barge.
ΣFy = T sin θ − T sin θ = 0 and ΣFx = T cos θ + T cos θ = Fx .
Find the work done on the barge.
W = Fx ∆x = (2T cos θ )∆x = 2(1.0 kN) cos 45°(150 m) = 210 kJ
y
1.0 kN
45°
45°
x
150 m
1.0 kN
6. (a) Strategy The force is equal to the weight of the pile driver. Use Eq. (6-1).
Solution Find the work done to raise the pile driver.
W = F ∆r cos θ = mg ∆r cos 0° = (402 kg)(9.80 N kg)(12 m) = 47 kJ
(b) Strategy and Solution The work done by gravity is negative the work done to raise the pile driver since the
force of gravity is opposite the driver’s motion, so Wgravity, up = −Wdriver = −47 kJ .
(c) Strategy and Solution The motion of the driver is in the same direction as the force of gravity, so the work
done by gravity is opposite that found in part (b); therefore, Wgravity, down = −Wgravity, up = 47 kJ .
7. (a) Strategy Consider the work done on the carton by Jennifer and the work done of the carton by gravity.
Solution Jennifer does positive work on the carton because the carton moves up in the direction of the force
applied by Jennifer. Gravity does negative work on the carton because the carton moves in the direction
opposite the force due to gravity. The absolute value of each amount of work done is the same, so the total
work done on the carton is zero.
(b) Strategy Use Eq. (6-1). Let the +y-axis point downward.
Solution The force of gravity is parallel to the displacement of the litter.
W = F ∆r cos θ = mg ∆r cos θ = (1.2 kg)(9.80 m s 2 )(0.75 m) cos 0° = 8.8 J
342
Physics
Chapter 6: Conservation of Energy
8. Strategy Use Eq. (6-2). Let the x-axis point in the direction of motion.
Solution The force of friction is opposite the motion of the box. Dirk’s horizontal
force is in the direction of motion.
66.0 N
4.80 N
x
W = Fx ∆x = ( F − f k )∆x = (66.0 N − 4.80 N)(2.50 m) = 153 J
9. Strategy Use Eq. (6-2). Let the x-axis point in the direction of motion.
Solution The force of friction is opposite the motion of the box, and according to Newton’s
second law, it is equal to f k = µk N = µk mg . Juana’s horizontal force is in the direction of
motion. Solve for the displacement.
W = Fx ∆x, so
W
W
W
74.4 J
∆x =
=
=
=
= 1.3 m .
Fx F − f k F − µk mg 124 N − 0.120(56.8 kg)(9.80 m s 2 )
N
x
F
fk
mg
10. Strategy Use Eq. (6-6).
Solution Compute the kinetic energy of the automobile.
1
1
K = mv 2 = (1600 kg)(30.0 m s)2 = 720 kJ
2
2
11. Strategy The work done on the briefcase by the executive is equal to the change in kinetic energy of the
briefcase. Use Eqs. (6-6) and (6-7).
Solution Find the work done by the executive on the briefcase.
1
1
W = ∆K = m(vf 2 − vi 2 ) = (5.00 kg) ⎡ (2.50 m s)2 − 0 ⎤ = 15.6 J
⎣
⎦
2
2
12. Strategy Use Eq. (6-6) for each case.
Solution Compute the kinetic energies.
1 2 1
mv = (70.5 kg)(27.8 m s) 2 = 27.2 kJ .
2
2
1 2 1
For Howard and his bike, the kinetic energy was K = mv = (70.5 kg)(68.04 m s) 2 = 163 kJ .
2
2
The kinetic energy of Murphy and his bike was K =
13. Strategy The kinetic energy of the sack is equal to the work done on it by Sam. Use Eqs. (6-2), (6-6),
and (6-7).
Solution
(a) Compute the kinetic energy of the sack.
1
1
∆K = m(vf 2 − vi 2 ) = mv 2 − 0 = K = W = Fx ∆x, so K = (2.0 N)(0.35 m) = 0.70 J .
2
2
(b) Solve for the speed of the sack.
1
2K
2(0.70 J)
K = mv 2 , so v =
=
= 0.37 m s .
2
m
10.0 kg
343
Chapter 6: Conservation of Energy
Physics
14. Strategy Use Eqs. (6-2), (6-6), and (6-7).
Solution Find the magnitude of the force.
1
1
mv 2 (12 kg)(0.40 m s) 2
W = Fx ∆x = ∆K = m(vf 2 − vi 2 ) = mv 2 − 0, so Fx =
=
= 0.12 N .
2
2
2∆x
2(8.0 m)
15. Strategy Use Eq. (6-6) for the initial and final kinetic energies.
Solution Compute the change in the kinetic energy of the ball.
1
1
∆K = m(vf 2 − vi 2 ) = (0.10 kg) ⎡(2.0 m s) 2 − (2.0 m s) 2 ⎤ = 0
⎣
⎦
2
2
Since the ball bounced back with the same speed, its kinetic energy did not change.
16. Strategy The sum of the work done on Jim and his skateboard by gravity and that of friction is equal to the
change in kinetic energy of Jim and his skateboard. Use Eqs. (6-6) and (6-7).
Solution Find the work done by friction on Jim and his skateboard. Let the y-axis point upward.
1
1
Wtotal = Wgravity + Wfriction = − mg ∆y + Wfriction = ∆K = K f − Ki = mv 2 − 0 = mv 2 , so
2
2
1 2
2
2
Wfriction = mv + mg ∆y = (65.0 kg)[(9.00 m s) 2 + (9.80 m s )(0 − 5.00 m)] = −550 J .
2
17. Strategy Use Eqs. (6-6) and (6-7).
Solution Compute the work done by the wall on the skater.
1
1
Wtotal = ∆K = K f − Ki = 0 − mv 2 = − (69.0 kg)(11.0 m s)2 = −4.17 kJ
2
2
18. (a) Strategy Use Eqs. (6-6) and (6-7). The weight is equal to W = mg.
Solution Calculate the work done on the plane by the cables.
1
mg
220 × 103 N
W = ∆K = m(vf 2 − vi 2 ) =
(0 − vi 2 ) = −
(67 m s) 2 = −50 MJ
2
2g
2(9.80 N kg)
The work done on the plane by the cables is −50 MJ .
(b) Strategy The force due to the cables is opposite the direction of motion. Use Eq. (6-1).
Solution Find the force exerted on the plane by the cables.
5.0 × 107 J
W
=−
= − 600 kN.
W = F ∆r cos180°, so F = −
∆r
84 m
The force exerted on the plane by the cables is 600 kN opposite the plane’s direction of motion.
344
Physics
Chapter 6: Conservation of Energy
19. Strategy Use Eq. (6-6) to compute the kinetic energies; then form a ratio to compare.
Solution Compute the kinetic energies of the car and the meteoroid.
1
1
K meteoroid = mv 2 = (0.0050 kg)(48 × 103 m s)2 = 5.8 MJ
2
2
1
K car = (1100 kg)(29 m s)2 = 0.46 MJ
2
Form the ratio.
K meteoroid
5.8 MJ
=
> 12
0.46 MJ
K car
The meteoroid has more than 12 times the kinetic energy of the car.
20. Strategy Since U = 0 at ground level, the potential energy of Sean and the parachute at the top of the tower is
equal to the negative of the work done by gravity as Sean climbed the tower.
Solution Find the potential energy of Sean and the parachute at the top of the tower.
U = mghtower = (68.0 kg)(9.80 m s 2 )(82.3 m) = 54.8 kJ .
21. (a) Strategy and Solution Since the floor is level, the motion of the desk is perpendicular to the force due to
gravity; therefore, the change in the desk’s gravitational potential energy is zero.
(b) Strategy The motion of the desk is in the direction of the applied constant force. Use Eq. (6-2).
Solution Compute the work done by Justin.
W = Fx ∆x = (340 N)(10.0 m) = 3.4 kJ
(c) Strategy and Solution Justin did work against friction, not gravity, so the energy has been dissipated as heat
by friction between the bottom of the desk and the floor.
22. (a) Strategy The energy saved is equal to the potential energy that the paint would have had had it been lifted by
the plane to cruising altitude. Use Eq. (6-9), where the ground level is zero and the height of the plane is h.
Solution Find the energy saved.
U = mgh = (100 kg)(9.80 m s 2 )(12, 000 m) = 12 MJ
(b) Strategy Use the work-kinetic energy theorem.
Solution Find the energy saved.
1
1
W = ∆K = mvf 2 − 0 = (100 kg)(250 m s)2 = 3.1 MJ
2
2
23. Strategy Use Eq. (6-9).
Solution
(a) Since the orange returns to its original position (∆y = 0) and air resistance is ignored, the change in its
potential energy is 0 .
(b) Let the y-axis point upward and the initial position be y = 0.
∆U grav = mg ∆y = (0.30 kg)(9.80 m s 2 )(−1.0 m − 0) = −2.9 J
345
Chapter 6: Conservation of Energy
Physics
24. (a) Strategy Find the change in potential energy from the change in height. Use this to find the change in kinetic
energy and, thus, the speed.
Solution Find the change in height.
∆h = A sin θ = (2.00 m) sin 30.0° = 1.00 m
Find the speed of the brick.
1
∆U = mg ∆h = ∆K = mv 2 , so v = 2 g ∆h = 2(9.80 m s 2 )(1.00 m) = 4.43 m s .
2
(b) Strategy Friction does negative work on the brick, slowing it.
Solution Find the work done by friction.
Wf = F ∆x = ( µ mg cos θ )A = µ mg A cos θ
Find the speed of the brick.
1
∆K = mv 2 = mg ∆h − µ mg A cos θ = mg A sin θ − µ mg A cos θ , so
2
v = 2 g A(sin θ − µ cos θ ) = 2(9.80 m s 2 )(2.00 m)(sin 30.0° − 0.10 cos 30.0°) = 4.03 m s .
25. (a) Strategy and Solution Since there are two pulleys, only half the force is required to move the mass (but
twice the length of rope must be pulled), so the pulley system multiplies the force exerted by a factor of 2 .
(b) Strategy Use ∆U = mg ∆h.
Solution Find the change in potential energy of the weight.
∆U = mg ∆h = (48.0 kg)(9.80 m s 2 )(4.00 m) = 1.88 kJ
(c) Strategy and Solution By conservation of energy, the work done to lift the mass is equal to its change in
potential energy, so W = ∆U = 1.88 kJ .
(d) Strategy and Solution Twice the length of rope must be pulled to do a given amount of work while applying
half the force, so the length of rope pulled is 8.00 m.
26. Strategy Use Newton’s second law and Eq. (6-10).
Solution The total work is given by Wtotal = Wvs. friction + Wvs. gravity .
Find the work done against friction.
ΣFy = N − mg cos φ = 0, so f = µ N = µ mg cos φ .
Wvs. friction = f ∆x = µ mg cos φ L = µ mg
2
L2 − h 2
⎛L⎞
L = µ mgh ⎜ ⎟ − 1 and the work done against gravity is
L
⎝h⎠
2
2
⎛
⎞
⎛
⎞
⎛L⎞
⎛ 4.0 m ⎞
⎟ = 2.5 kJ .
Wvs. gravity = mgh. So, Wtotal = mgh ⎜ µ ⎜ ⎟ − 1 + 1⎟ = (1400 N)(1.0 m) ⎜ 0.20 ⎜
1
1
−
+
⎟
⎜ ⎝h⎠
⎟
⎜
⎟
⎝ 1.0 m ⎠
⎝
⎠
⎝
⎠
346
Physics
Chapter 6: Conservation of Energy
27. (a) Strategy Use conservation of energy.
Solution Find the speed of the cart as it passes point 3.
1
1
1
1
E1 = mv12 if y1 = 0 and E3 = mv32 + mgy3 . Ef = E3 = mv32 + mgy3 = Ei = E1 = mv12 , so
2
2
2
2
v3 = v12 − 2 gy3 = (20.0 m s) 2 − 2(9.81 m s 2 )(10.0 m) = 14.3 m s .
(b) Strategy Use the result of part (a), replacing 3 with 4. If the result is real—the argument of the square root is
nonnegative—the cart will reach position 4.
Solution Compute the speed of the cart at position 4.
v4 = v12 − 2 gy4 = (20.0 m s) 2 − 2(9.81 m s 2 )(20.0 m) = 3 m s
The answer is yes; the cart will reach position 4.
28. Strategy Use conservation of energy.
Solution Find a general expression for the speed of the cart.
1
1
E4 = K 4 + U 4 = mv42 + mgy4 = Ei , so En = K n + U n = mvn 2 + mgyn , where n = 1, 2, or 3.
2
2
Solve for vn .
1
1
mv 2 + mgyn = Ei = mv42 + mgy4 , so vn = v42 + 2 g ( y4 − yn ).
2 n
2
Compute the speed at each position.
Ef =
v1 = (15 m s)2 + 2(9.80 m s 2 )(20.0 m − 0) = 25 m s
v2 = (15 m s)2 + 2(9.80 m s 2 )(20.0 m − 15.0 m) = 18 m s
v3 = (15 m s) 2 + 2(9.80 m s 2 )(20.0 m − 10.0 m) = 21 m s
29. Strategy The initial height of the rope is l cos θ where l is the length of the rope and θ is the angle it makes with
the vertical. Then ∆y = l cos θ − l = l (cos θ − 1). Use conservation of energy.
Solution Find Bruce’s speed at the bottom of the swing.
1
1
∆K = mv 2 − 0 = mv 2 = −∆U = − mg ∆y = mgl (1 − cos θ ), so
2
2
v = 2 gl (1 − cos θ ) = 2(9.80 m s 2 )(20.0 m)(1 − cos 35.0°) = 8.42 m s .
θ
l cos θ
l − l cos θ
30. Strategy Use conservation of energy.
Solution Find the maximum height achieved by the swinging child.
1
K f − Ki = mv 2 − 0 = U i − U f = mgytop − mgybottom , so
2
v2
(4.9 m s) 2
+ ybottom =
+ 0.70 m = 1.9 m .
ytop =
2g
2(9.80 m s 2 )
347
l
Chapter 6: Conservation of Energy
Physics
31. Strategy Use Eq. (6-10) to find the nonconservative work.
Solution Calculate the work done by friction and air resistance during the run.
1
Wtotal = Wc + Wnc = ∆K = mvf 2 , so
2
1
1
1
2
Wnc = mvf − Wc = mvf 2 − mgh = (75 kg)(12 m s)2 − (75 kg)(9.80 m s 2 )(78 m) = −52 kJ .
2
2
2
32. Strategy Assume frictional forces and air resistance are negligible. Use conservation of energy.
Solution Find h, the highest position the car reaches above the bottom of the hill.
∆U = mgh − mghi = −∆K =
1
mv 2 − 0, so h =
2 i
vi 2
2g
+ hi =
(20.0 m s)2
2(9.80 m s 2 )
+ 5.0 m = 25 m .
33. Strategy Use conservation of energy.
Solution
(a) Solve for the final speed of the ball.
1
1
∆K = mvf 2 − mv 2 = −∆U = mgh, so vf =
2
2
v 2 + 2 gh .
(b) By inspection of the equation found in part (a), we find that the final speed is independent of the angle.
34. Strategy Since energy is conserved and nonconservative forces do no work, ∆K = −∆U . Let the y-axis point
upward.
Solution The initial speeds are zero and the final speeds are the same (due to the rope). Since m1 < m2 , block 1
moves up the incline and block 2 falls. Let d be the distance block 1 moves along the incline, then ∆r1 = d sin θ
and ∆r2 = − d .
1
1
m v 2 + m2 v 2 = −∆U = −∆U1 − ∆U 2 = − m1 g ∆r1 − m2 g ∆r2 = − m1 gd sin θ − m2 g (− d ), so
2 1
2
2 gd (−m1 sin θ + m2 )
2(9.80 m s 2 )(1.4 m)[−(12.4 kg) sin 36.9° + 16.3 kg]
=
= 2.9 m s .
m1 + m2
12.4 kg + 16.3 kg
∆K = ∆K1 + ∆K 2 =
v=
35. Strategy Plot the force on the vertical axis and the spring length on the horizontal axis.
Solution The graph is shown at the right.
(b) Find the y-intercept.
F = 1.00 N = kx + b = (0.286 N cm)(14.5 cm) + b, so b = −3.15 N.
The force on the spring is zero when the spring is relaxed. Set F = 0.
b
−3.15 N
0 = kx0 + b, so x0 = − = −
= 11.0 cm .
k
0.286 N cm
348
5.00
Force, F (N)
(a) Determine the slope of the line to find k, since F = kx.
5.00 N − 1.00 N
4.00 N
k=
=
= 0.286 N cm
28.5 cm − 14.5 cm 14.0 cm
4.00
3.00
2.00
1.00
0.00
0.0
10.0 20.0 30.0
Spring length, x (cm)
Physics
Chapter 6: Conservation of Energy
36. (a) Strategy Since the gravitational field is uniform, the work done by gravity is Wgrav = Fy ∆y = − mg ∆y, where
the y-axis points up.
Solution Note that the slope is inclined at 15.0° to the horizontal.
Wgrav = − mg ∆y = −(75.0 kg)(9.80 m
s 2 )[0 − (32.0
m) sin15.0°]
32.0
15.0°
m
(32.0 m) sin 15.0°
= 6.09 kJ
The normal force is perpendicular to the motion of the skier, so the work done by the normal force is 0 J.
(b) Strategy Refer to part (a). Use conservation of energy and Newton’s second law.
Solution The work done by gravity is the same as found in part (a), 6.09 kJ. As before, the normal force is
0 J. The total work done on the skier is equal to the sum of the work done by gravity and the work done by
friction. We use Eqs. (6-6) and (6-7) to find the work done by friction.
1
1
Wtotal = Wgravity + Wfriction = − mg ∆y + Wfriction = ∆K = K f − Ki = mv 2 − 0 = mv 2 , so
2
2
1 2
Wfriction = mv + mg ∆y
2
1
= (75.0 kg)(10.0 m s) 2 + (75.0 kg)(9.80 m s 2 )[0 − (32.0 m) sin15.0°] = −2.34 kJ .
2
Now that we know the work done by friction, we use Eq. (6-2) to find the force of friction.
W
−2337 J
W = Fx ∆x = f k ∆x = Wfriction , so f k = friction =
= −73.0 N.
32.0 m
∆x
The force of friction is 73.0 N opposite the direction of motion. To find the coefficient of kinetic friction, we
draw a diagram and use Newton’s second law.
N
y
x
mg
15.0°
15.0°
mg cos 15.0°
ΣFy = N − mg cos15.0° = 0, so N = mg cos15.0°. Since f k = µk N , the coefficient of kinetic friction is
f
fk
73.0 N
µk = k =
=
= 0.103 .
N mg cos15.0° (75.0 kg)(9.80 m s 2 ) cos15.0°
37. Strategy Use the result for escape speed found in Example 6.8.
Solution Replace the values for Earth with those for the Moon.
v=
2GM Moon
RMoon
=
2(6.674 × 10−11 N ⋅ m 2 kg 2 )(7.35 × 1022 kg)
1.74 × 106 m
= 2.37 km s
38. Strategy Use the result for escape speed found in Example 6.8.
Solution The magnitude of the gravitational field is given by GM R 2 = 30.0 m s 2 . Find the escape speed.
vesc =
2GM
⎛ GM
= 2⎜
R
⎝ R2
⎞
2
7
⎟ R = 2(30.0 m s )(6.00 × 10 m) = 60.0 km s
⎠
349
Chapter 6: Conservation of Energy
Physics
39. Strategy Use conservation of energy and the result for escape speed found in Example 6.8.
Solution Replacing the values for Earth with those for the Zoroaster, we find that the escape speed for Zoroaster
is given by vesc = 2GM Z RZ . Find the speed of the meteor when it hits the surface of the planet.
∆K =
GM Z m
1
1
mvf 2 − mvi 2 = −∆U =
, so
2
2
RZ
vf = vi 2 +
2GM Z
RZ
= vi 2 + vesc 2 = (5.0 km s) 2 + (12.0 km s) 2 = 13.0 km s .
40. Strategy In the equation for the escape speed found in Example 6.8, replace the values for Earth with appropriate
values for the fictional planet. Use proportional reasoning and the relationship between the volume of a sphere
and its radius to relate the mass and radius of the planet with those of Earth.
Solution Find the escape speed.
Earth:
2GM E
ME
vesc =
and ρ E = density =
.
4πR 3
RE
E
3
Planet:
vesc =
=
2GM
=
R
2G
ρ V =
R E
2G
2 RE
⎛ M
⎞
E ⎟ ⎡ 4 π (2 R )3 ⎤ = 8GM E
⎜
E ⎥
⎜ 4 π RE3 ⎟ ⎢⎣ 3
RE
⎦
⎝3
⎠
8(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
6.37 × 106 m
= 22.4 km s
41. Strategy Use Eq. (6-14) and form a ratio.
Solution Find the ratio of the potential energies at perigee and apogee.
U perigee
U apogee
GmM
=
− 2R E
E
GmM
− 4R E
E
=
4
= 2
2
42. Strategy The initial kinetic and potential energies are zero. Neglect the drag force on the meteor due to the
atmosphere. Use conservation of energy and Eq. (6-14).
Solution Let M be the mass of Earth, m be the mass of the meteor, R be the radius of Earth, and h be the height of
the stratosphere. Find the minimum speed of the meteor when it reaches the stratosphere.
1
GMm
Ei = 0 = Ef = mvf 2 −
, so
2
R+h
2GM
2(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
=
= 11.2 km s .
vf =
R+h
6.371× 106 m + 40 × 103 m
350
Physics
Chapter 6: Conservation of Energy
43. Strategy Ignore air resistance. Use conservation of energy.
Solution Find the required initial speed of the projectile.
GM E m
GM E m
1
Ei = mvi 2 −
, so
= Ef = 0 −
RE
2
5 RE
vi =
8GM E
5 RE
=
8(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
5(6.37 × 106 m)
= 10.0 km s .
44. Strategy Use conservation of energy.
Solution Find the comet’s speed at perihelion.
1
GmM
1
GmM
= K a + U a = mva 2 −
K p + U p = mvp 2 −
, so
2
rp
2
ra
⎛1 1⎞
vp = va 2 + 2GM ⎜ − ⎟
⎜ rp ra ⎟
⎝
⎠
1
1
⎛
⎞
= (10.0 × 103 m s) 2 + 2(6.674 × 10−11 N ⋅ m 2 kg 2 )(1.987 × 1030 kg) ⎜
−
⎟
10
12
⎝ 8.9 × 10 m 5.3 × 10 m ⎠
= 55 km s
45. Strategy Use Newton’s second law and law of universal gravitation.
Solution Calculate the orbital speed.
v 2
GM E m
∑ Fr =
= mar = m orb , so vorb =
4.0 RE
(4.0 RE )2
GM E
4.0 RE
.
Calculate the escape speed.
GM E m
1
Ki + U i = mvesc 2 −
= K f + U f = 0 + 0, so vesc =
2
4.0 RE
GM E
2.0 RE
.
Find the change in speed.
GM E
GM E ⎛ 1
1 ⎞ GM E
∆v = vesc − vorb =
−
=⎜
−
⎟
2.0 RE
4.0 RE ⎝ 2.0 2.0 ⎠ RE
1 ⎞ (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
⎛ 1
=⎜
−
= 1.6 km s
⎟
6.371× 106 m
⎝ 2.0 2.0 ⎠
351
Chapter 6: Conservation of Energy
Physics
46. Strategy The small velocity toward Earth is just enough to upset the equilibrium of the rock and send it toward
Earth, but we can neglect it in the calculation of the rock’s final speed and thus neglect the initial kinetic energy of
the rock as well. Use conservation of energy.
Solution Let the distances from the equilibrium point to the centers’ of Earth and the Moon be d E and d M ,
respectively; and let the distance between the centers of the Earth and the Moon be R. Let h = 700, 000 m, and let
mr , m, and M be the masses of the rock, Moon, and Earth, respectively. Find the speed of the rock when it
encounters Earth’s atmosphere.
Gmr M Gmr m
Gmr M Gmr m
1
Ki + U i = 0 + U i = −
+
= K f + U f = mr v 2 −
+
, so
dE
dM
2
h
R−h
⎛ M
m M
m ⎞
v = 2G ⎜⎜ −
+
+
−
⎟⎟ .
⎝ dE dM h R − h ⎠
We need to find d E and d M . At the equilibrium point, the forces due to Earth and the Moon on the rock are
equal; so according to Newton’s law of universal gravitation, −
Gmr M
dE
2
=−
Gmr m
dM
2
, or
dE
=
dM
M
.
m
M
m
R
R
+ dM = dE + dE
, so d M =
and d E =
.
m
M
1+ M m
1+ m M
Substitute these into the equation for v.
Now, R = d E + d M = d M
⎡ M (1 + m M ) m(1 + M m ) M
m ⎤
m ⎞
⎛m−M M
v = 2G ⎢ −
+
+
−
+
−
⎥ = 2G ⎜
⎟
R
R
h R − h ⎥⎦
h R−h⎠
⎝ R
⎢⎣
G = 6.674 × 10−11 N ⋅ m 2 kg 2 , m = 7.349 × 1022 kg, M = 5.974 × 1024 kg, R = 3.845 × 108 m, and
h = 700, 000 m + 6.371× 106 m. Substituting these values into the equation for v gives a speed for the rock of
10,500 m s .
47. Strategy Use proportional reasoning.
Solution The force is linear with respect to the displacement of the string. If the string is pulled back half as far
(20.0 cm) as in Example 6.9 (40.0 cm), the average force is only half that as in Example 6.9. Therefore, the work
⎛ 1 ⎞⎛ 1 ⎞
done is ⎜ ⎟ ⎜ ⎟ (32 J) = 8 J .
⎝ 2 ⎠⎝ 2 ⎠
48. Strategy The work done on the spring is negative the work done by the spring. Use the relationship between
work and the extension or compression of a spring.
Solution Find the work done to stretch the spring.
1
1
W = kx 2 = (20.0 N m)(0.40 m)2 = 1.6 J
2
2
49. Strategy The work done by the hammer on the object is represented by the area between the curve and the x-axis.
Solution Compute the work done in driving the nail.
W = (50 N)(0.012 m) + (120 N)(0.050 m − 0.012 m) = 5.2 J
352
Physics
Chapter 6: Conservation of Energy
50. (a) Strategy The increase in the force is F and the length that the tendon increases is x if the tendon is modeled
as a spring and Hooke’s law is used.
Solution Compute the spring constant.
F 4800 N − 3200 N 1600 N
k= =
=
= 3200 N cm
x
0.50 cm
0.50 cm
(b) Strategy The triangular area under a force of the muscle vs. the stretch of the tendon graph is equal to the
work done by the muscle.
Solution Compute the work done by the muscle in stretching the tendon.
1
1
W = Fx = (4800 N − 3200 N)(0.0050 m) = 4.0 J
2
2
51. (a) Strategy Use Hooke’s law and form a proportion.
Solution
F
k= 1=
x1
Form the proportion.
F2
x2
Solve for x2 to find the amount that the spring stretches.
x2 =
F2
7.0 N
x =
(3.5 cm) = 4.9 cm
F1 1 5.0 N
(b) Strategy Substitute known values for F1 and x1 to find k.
Solution Compute the spring constant.
F
5.0 N
k= 1=
= 1.4 N cm
x1 3.5 cm
(c) Strategy The triangular area under a forces on the spring vs. the stretch of the spring graph is equal to the
work done by the forces.
Solution Compute the work done by the forces on the spring.
1
1
W = Fx = (5.0 N)(0.035 m) = 88 mJ
2
2
52. (a) Strategy Solve for k in Hooke’s law.
Solution Compute the spring constant.
F 120 N ⎛ 109 nm ⎞
k= =
⎜
⎟ = 6.0 × 1010 N m
x 2.0 nm ⎜⎝ 1 m ⎟⎠
(b) Strategy Solve for x in Hooke’s law and use the value for k found in part (a).
Solution Compute the compression of the block.
F
480 N
x= =
= 8.0 nm
k 6.0 × 1010 N m
353
Chapter 6: Conservation of Energy
Physics
(c) Strategy Since the forces due to the block are opposite to the directions of compression, the block does
negative work during the compression. The work done by the applied forces is positive and equal to the
negative of the work done by the block.
Solution Compute the work done.
1
1
W = −Wblock = kx 2 = (6.0 × 1010 N m)(8.0 × 10−9 m) 2 = 1.9 µJ
2
2
53. (a) Strategy Set the weight of the mass equal to the force in Hooke’s law.
Solution Compute the spring constant.
mg (1.4 kg) ( 9.80 N kg )
W = mg = F = kx, so k =
=
= 1.9 N cm .
x
7.2 cm
(b) Strategy Use Eq. (6-24).
Solution Compute the elastic potential energy stored in the spring.
1
1 ⎡ (1.4 kg) ( 9.80 N kg ) ⎤
2
U elastic = kx 2 = ⎢
⎥ (0.072 m) = 0.49 J
2
2⎣
0.072 m
⎦
(c) Strategy Solve for m in the equation for k found in part (a).
Solution Compute the second mass.
⎛m g⎞x
kx
x
12.2 cm
m2 = 2 = ⎜⎜ 1 ⎟⎟ 2 = 2 m1 =
(1.4 kg) = 2.4 kg
g
x
g
x
7.2 cm
1
⎝ 1 ⎠
54. Strategy W = Fx ∆x and the work is represented by the area under the curve, A = (1 2)bh.
Solution Find the work done in each situation.
(a) W =
1
(0.20 m)(15 N) = 1.5 J
2
(b) W =
1
1
(0.20 m)(15 N) − (0.10 m)(7.5 N) = 1.1 J
2
2
55. Strategy and Solution Since the displacement of the model airplane is zero, zero work has been done on it by
the string.
56. Strategy The work done by the force on the object is represented by the area between the curve and the x-axis.
The area under the axis represents negative work done.
Solution Compute the work done by the force.
1
1
W = (2.0 N)(1.0 m) + (1.0 N)(1.0 m) + (−1.0 N)(1.0 m) = 0.5 J
2
2
57. Strategy and Solution Let E be the elastic energy stored in the legs. This energy is converted into gravitational
potential energy, mgh, when the kangaroo jumps. Since only one leg is used, and since h ∝ E , the kangaroo can
only jump half as high, or (0.70 m) 2 = 0.35 m .
354
Physics
Chapter 6: Conservation of Energy
58. Strategy The mechanical energy is constant, so we set the elastic potential energy of the spring on the toy gun
equal to the gravitational potential energy of the rubber ball. Assume x h.
Solution mgh = (1 2)kx 2 , so h ∝ x 2 . Thus, the height reached by the ball is proportional to the square of the
compression of the spring. Form a proportion.
h2
h1
=
2
x 2
⎛ 2x ⎞
, so h2 = 2 h1 = ⎜ ⎟ h = 4h .
⎝ x ⎠
x12
x12
x22
59. Strategy The elastic potential energy of the catapult, (1 2)kx 2 , is converted into gravitational potential energy of
the pebble, mgh.
Solution Find the maximum height achieved by the pebble.
1
kx 2
(320 N m)(0.20 m) 2
mgh = kx 2 , so h =
=
= 13 m .
2
2mg 2(0.051 kg)(9.80 m s 2 )
60. Strategy Use conservation of energy.
Solution
(a) Compute the speed of the block as it passes through the equilibrium point.
1
1
k
Ki + U i = 0 + kd 2 = K f + U f = mv 2 + 0, so v = d
.
2
2
m
(b) Find the maximum distance below the equilibrium point that the block will reach.
1
1
m
k m
=d
= d .
Ki + U i = mv 2 + 0 = K f + U f = 0 + kx 2 , so x = v
2
2
k
m k
61. Strategy Take the surface of the unstretched trampoline to be y = 0. Use conservation of energy and Newton’s
second law.
Solution Find the spring constant from the gravitational potential energy.
1
2mgh
mgh = kymin 2 , so k =
.
2
ymin 2
Use Newton’s second law for the situation where the gymnast is at rest.
∑ F = ky − mg = 0, so
⎛y 2⎞ y 2
mg
(−0.75 m)2
y=
= mg ⎜ min ⎟ = min =
= 8.7 cm .
⎜ 2mgh ⎟
k
2h
2(2.5 m + 0.75 m)
⎝
⎠
62. Strategy The stretched length of the bungee cord (and the distance George falls) must be no more than
55.0 m − 2.00 m = 53.0 m. The gravitational potential energy decrease of George as he falls must equal the elastic
potential energy increase of the bungee cord.
Solution To find the spring constant, set Eqs. (6-24) and (6-13) equal and solve for k.
2mgyfall 2(75.0 kg)(9.80 m s 2 )(53.0 m)
1
kystretch 2 = mgyfall , so k =
=
= 115 N m .
2
ystretch 2
(53.0 m − 27.0 m)2
355
Chapter 6: Conservation of Energy
Physics
63. (a) Strategy The increase in kinetic energy of the block is equal to the decrease in its potential energy. Let the
potential energy be zero at y = 0.25 m.
Solution To find the speed of the block, set Eqs. (6-6) and (6-13) equal and solve for v.
1 2
mv = mgy, so v = 2 gy = 2(9.80 m s 2 )(0.25 m) = 2.2 m s .
2
(b) Strategy The elastic potential energy increase of the spring is equal to the decrease in gravitational potential
energy of the block. Let the potential energy be zero at y = 0 m.
Solution To find the compression of the spring, set Eqs. (6-24) and (6-13) equal and solve for x.
1 2
kx = mgy, so x =
2
2mgy
=
k
2(2.0 kg)(9.80 m s 2 )(0.50 m)
= 0.21 m .
450 N m
(c) Strategy and Solution Since the surface is frictionless, no nonconservative forces do work on the block. So,
the block will return to its previous height, or 0.50 m.
64. Strategy As Lars climbs the stairs, he increases his gravitational potential energy. The rate of potential energy
increase must be equal to the rate he does work.
Solution To find the time for Lars to climb the stairs, use Eqs. (6-26) and (6-9) and solve for ∆t.
Pav =
∆E ∆U mg ∆y
mg ∆y (82.4 kg)(9.80 m s 2 )(12.0 m − 0)
=
=
, so ∆t =
=
= 13.0 s .
∆t
∆t
∆t
Pav
746 W
65. Strategy Watts are joules per second and there are 3600 seconds in 1 hour.
Solution Show that 1 kW ⋅ h = 3.6 MJ.
J
3600 s
1 kW ⋅ h = 103 ⋅ h ⋅
= 3.6 × 106 J = 3.6 MJ
s
1h
66. Strategy Use the definition of average power and the potential energy in a uniform gravitational field.
Solution Find the minimum time required for the man to lift the boxes.
∆E
Pav =
and ∆E = ∆U = mtotal gh, so
∆t
∆E 50mgh 50(10.0 kg)(9.80 m s 2 )(2.00 m) ⎛ 1 min ⎞
∆t =
=
=
⎜
⎟ = 4.08 min .
Pav
Pav
40.0 W
⎝ 60 s ⎠
67. Strategy As Rosie lifts the trunk, she does work on it and increases its gravitational potential energy.
Solution To find the average rate of work Rosie does on the trunk (that is, the power she supplies), use Eqs.
(6-26) and (6-13).
∆E ∆W mgh (220 N)(4.0 m)
=
=
=
= 22 W
Pav =
∆t
∆t
∆t
40 s
356
Physics
Chapter 6: Conservation of Energy
68. Strategy Assume that friction is negligible. Use Eq. (6-27).
Solution The rate at which gravity does work on the bicycle and rider is
G
G
P = mgv cos θ , where θ is the angle between v and g, or θ = 90° + φ .
Find φ .
v
φ
θ
φ
g
5.0 m
5.0 m
100 m
, so φ = tan −1 0.050.
100 m
The power output of the rider is equal to the rate of change of potential energy, which equals −P. Therefore,
tan φ =
Prider = − P = −(75 kg)(9.80 m s 2 )(4.0 m s) cos(90° + tan −1 0.050) = 150 W .
69. Strategy Use Eq. (6-27).
Solution
(a) Find the force exerted on the cyclist by the air.
G
∆U
G
Pa + Pc = 0 since
= 0. v is antiparallel to Fa .
∆t
Pa
−120 W
Pa = Fa v cos θ , so Fa =
=
= 20 N .
v cos θ (6.0 m s) cos180°
(b) Find the speed of the cyclist.
Pc
120 W
v=
=
= 6.7 m s
Fc cos θ (−18 N) cos180°
70. (a) Strategy Use the definition of average power. The change in energy is equal to the change in kinetic energy.
Solution Find the average mechanical power output.
2
2
2
1
⎡
⎤
∆E ∆K 2 m(vf − vi ) (1000.0 kg) ⎣ (40.0 m s) − 0 ⎦
Pav =
=
=
=
= 80.0 kW
2(10.0 s)
∆t
∆t
∆t
(b) Strategy Relate the mechanical energy required to the chemical energy provided per liter of gasoline.
Solution Find the volume of gasoline consumed.
mechanical energy required
K
=
=
chemical energy provided
(efficiency)(46 MJ L)
per liter
1 (1000.0
2
kg)(40.0 m s)2
(46 × 106 J L)(0.22)
= 0.079 L
71. Strategy Use the definition of average power. The change in energy is equal to the change in kinetic energy.
Solution Determine the average mechanical power the engine must supply.
2
2⎤
2
2
1
⎡
∆E 2 m(vf − vi ) (1200 kg) ⎣(30.0 m s) − (20.0 m s) ⎦
=
=
= 60 kW
Pav =
∆t
∆t
2(5.0 s)
357
Chapter 6: Conservation of Energy
Physics
72. (a) Strategy Use the definition of average power. The change in energy is equal to the change in gravitational
potential energy.
Solution Find the woman’s average power output.
∆E ∆U mgh (62 kg)(9.80 m s 2 )(5.0 m)
Pav =
=
=
=
= 510 W
6.0 s
∆t
∆t
∆t
(b) Strategy and Solution The body would have to be 100% efficient for the answer to part (a) to be equal to
the average power input. So, the answer is no.
73. Strategy Relate the change in gravitational potential energy of the person to the energy provided by the
carbohydrate.
Solution Find the mass of carbohydrate required for the person to climb the stairs.
energy
mgh
(74 kg)(9.80 m s 2 )(15 m)
=
=
= 6.2 g
energy available per gram 0.100(energy per gram)
0.100(17.6 × 103 J g)
The other 90% of the energy is dissipated as heat.
74. Strategy The instantaneous power is given by Eq. (6-27), where θ = 0°. Obtain the necessary values of the force
from the graph.
Solution
(a) Compute the instantaneous power.
Pav = Fv cos θ = Fx vx = (800 N)(11 m s) = 8.8 kW
(b) As in part (a), we have Pav = Fv cos θ = Fx vx = (400 N)(16 m s) = 6.4 kW .
75. Strategy Use the definition of average power. The change in energy is equal to the change in kinetic energy.
Solution Find the engine’s average power output.
2
2
2
1
⎡
⎤
∆E ∆K 2 m(vf − vi ) (500.0 kg) ⎣(125 m s) − 0 ⎦
=
=
=
= 930 kW
Pav =
∆t
∆t
∆t
2(4.2 s)
76. (a) Strategy Use Eq. (6-9).
Solution Calculate the change in gravitational potential energy of the water.
∆U = mg ( yf − yi ) = (1 kg)(9.80 N kg)(0 − 50 m) = −500 J
(b) Strategy The mass flow rate times the potential energy change per unit mass gives the rate at which
gravitational potential energy is lost by the river.
Solution Compute the rate.
P = (5.5 × 106 kg s)(500 J kg) = 3 GW
(c) Strategy The electrical power output divided by the power per household gives the total number of
households supplied.
Solution Compute the number of households.
0.10(3 × 109 W)
= 300, 000 households
1× 103 W household
358
Physics
Chapter 6: Conservation of Energy
77. Strategy Use conservation of energy.
Solution Compute the required speed of the high jumper.
1
1
Ki + U i = mv 2 + 0 = mv 2 = K f + U f = 0 + mgh = mgh, so
2
2
v = 2 gh = 2(9.80 m s 2 )(1.2 m) = 4.8 m s .
78. Strategy Use conservation of energy.
Solution Find the maximum height of the pole-vaulter’s center of gravity.
1
1
Ki + U i = mv 2 + mghi = K f + U f = 0 + mghf , so mv 2 = mg (hf − hi ), or
2
2
v2
(10.0 m s) 2
hf =
+h =
+ 1.0 m = 6.1 m .
2 g i 2(9.80 m s 2 )
79. Strategy Use conservation of energy. Neglect drag.
Solution Find the speed of the hang glider.
1
1
∆K = mvf 2 − mvi 2 = −∆U = mgh, so vf = 2 gh + vi 2 = 2(9.80 m s 2 )(8.2 m) + (9.5 m s) 2 = 16 m s .
2
2
80. Strategy Use the work-kinetic energy theorem. The work done by friction is equal to the force of friction,
f k = µk N = µk mg , times the distance the car skidded d.
Solution Relate the speed to the distance.
1
1
∆K = mvf 2 − 0 = mvf 2 = W = f k d = µk mgd
2
2
Let the first case be represented by the subscript 1, and the second by 2:
1
1
mv 2 = µ mgd1 and mv22 = µ mgd 2 . Form a proportion to find the distance of the skid.
2 1
2
1 mv 2
1
2
1 mv 2
2
2
2
=
2
⎛v ⎞
µ mgd1
⎛ 60 mi h ⎞
, so d 2 = ⎜⎜ 2 ⎟⎟ d1 = ⎜
⎟ (50 ft) = 200 ft .
v
µ mgd 2
⎝ 30 mi h ⎠
⎝ 1⎠
81. Strategy Use Newton’s second law and law of universal gravitation.
Solution Prove that U = −2 K for any gravitational circular orbit.
∑ Fr =
m
v2
r
mv 2
1 2
mv
2
K
U
=
GMm
r2
GMm
= mar =
mv 2
, so
r
r2
GMm
=
r
1 ⎛ GMm ⎞
= ⎜
⎟
2⎝ r ⎠
1
=− U
2
= −2 K
359
Chapter 6: Conservation of Energy
Physics
82. Strategy Use conservation of energy.
Solution v y can be found from the kinetic energy gained due to gravity.
1
mv 2 = mgh, so v y 2 = 2 gh.
2 y
vx can be found from the kinetic energy gained from the spring’s elastic potential energy.
1
1
k ( xi 2 − xf 2 )
mvx 2 = k ( xi 2 − xf 2 ), so vx 2 =
.
2
2
m
Compute the speed of the ball when it hits the gound.
v = vx 2 + v y 2 =
k ( xi 2 − xf 2 )
(28 N m)[(0.18 m)2 − (0.12 m)2 ]
+ 2 gh =
+ 2(9.80 m s 2 )(1.4 m)
m
0.056 kg
= 6.0 m s .
83. Strategy Use Hooke’s law and Newton’s laws.
Solution
(a) The mass connected to the lower spring exerts a force on the lower spring equal to its weight, W. The spring
stretches an amount x = F k = W k . The lower spring exerts a force on the upper spring equal to F = W , and
causes it to stretch by x = F k = W k . Thinking of the two springs as a single spring:
2x =
F F 2F
k
+ =
= x′, so F = x′ = k ′x′. Therefore,
2
k k
k
k
= k ′, the effective spring constant.
2
(b) Sum the forces on the mass.
F + F − W = kx + kx − W = 2kx − W = 0, so W = 2kx = k ′x.
Therefore, 2k = k ′, the effective spring constant.
84. (a) Strategy Use conservation of energy.
Solution Find the speed of the car at the top of the loop.
1
1
1
∆K = mvf 2 − mvi 2 = mvf 2 − 0 = −∆U = − mg (hf − hi ) = mg (hi − hf ), so
2
2
2
vf = 2 g (hi − hf ) = 2(9.80 m s 2 )(40.0 m − 20.0 m) = 19.8 m s .
(b) Strategy Use Newton’s second law.
Solution Find the normal force on the car exerted by the track.
⎛ v2
⎞
ΣFr = mar = N + mg , so N = m(ar − g ) = m ⎜ − g ⎟ .
⎜ r
⎟
⎝
⎠
⎡ 2 g (hi − hf )
⎤
⎛h −h 1⎞
⎛ 40.0 m − 20.0 m 1 ⎞
− g ⎥ = 2mg ⎜ i f − ⎟ = 2(988 kg)(9.80 m s 2 ) ⎜
− ⎟ = 29.0 kN
N = m⎢
2⎠
10.0 m
2⎠
r
⎝
⎣
⎦
⎝ r
(c) Strategy Use Newton’s second law.
Solution Set the normal force on the roller coaster car equal to zero at the top to find the minimum height.
10.0 m
v2
r
N = 0 = m − mg , so gr = v 2 = 2 g (hi − hf ). Thus, hi = + hf =
+ 20.0 m = 25.0 m .
2
2
r
360
Physics
Chapter 6: Conservation of Energy
85. Strategy Use Eqs. (6-10) and (6-11) for the work and energy solution. Use Newton’s second law and Eq. (2-13)
for the force solution. Let d = 8.0 m, d1 = 5.0 m, and d 2 = 8.0 m − 5.0 m = 3.0 m.
Solution First method: Find the constant tension using work and energy.
Wtotal = Wc + Wnc = ∆K = 0, since the speeds at the top and bottom of the incline
.0 m
d=8
15°
d sin 15°
are zero. Also, Wc = −∆U and Wnc = Td 2 .
Wnc = Td 2 = −Wc = ∆U , so
∆U mg ∆y mg (0 − d sin θ )
(4.0 kg)(9.80 m s 2 )(8.0 m) sin15°
mgd sin θ
=
=
=−
=−
= −27 N.
3.0 m
d2
d2
d2
d2
The sign is negative because the work done by the tension is opposite the box’s motion. So, the magnitude of the
tension is 27 N.
T=
Second method: Find the speed of the block just before the person grasps the cord
using Newton’s second law.
ΣFy = N − mg cos θ = 0 and ΣFx = mg sin θ = ma.
vf 2 − vi 2 = vf 2 − 0 = 2 g sin θ d1
N
T
y
x
mg
Let vf = v.
Find the tension.
15°
mg cos 15°
mg sin 15°
T
ΣFx = −T + mg sin θ = ma x , so a x = − + g sin θ .
m
⎛ T
⎞
vf 2 − vi 2 = 0 − v 2 = −2 g sin θ d1 = 2a x ∆x = 2 ⎜ − + g sin θ ⎟ d 2 , so
⎝ m
⎠
⎛ d1 + d 2 ⎞
d
⎛ 8.0 m ⎞
2
T = mg sin θ ⎜⎜
⎟⎟ = mg sin θ d = (4.0 kg)(9.80 m s ) sin15° ⎜⎝ 3.0 m ⎟⎠ = 27 N.
d
⎝
⎠
2
2
86. (a) Strategy The spring does work against gravity. Use Eq. (6-20).
Solution
1
2mg ∆y
2(780 N)(68 m − 182 m)
Wspring = − kx 2 = −Wg = mg ∆y, so k = −
=−
= 25 N m .
2
2
x
(182 m − 68 m − 30.0 m)2
(b) Strategy The kinetic energy gained is the sum of the positive work done by gravity (which increases the
kinetic energy during the fall) and the negative work done by the cord (which decreases the kinetic energy).
Solution Find the speed of the jumper.
1 2
1
mv = − mg ∆y − kx 2 , so
2
2
k 2
25.2 N m
v = −2 g ∆y − x = −2(9.80 m s 2 )(92 m − 182 m) −
(182 m − 92 m − 30.0 m)2 = 25 m s .
780 N
m
2
9.80 m s
87. Strategy Use conservation of energy.
Solution The elastic potential energy of the spring is converted to gravitational potential energy, so
1 2
kx = mgh = mg (l sin θ ) where l is the distance the object travels up the incline.
2
kx 2
(40.0 N m)(0.20 m) 2
Thus, l =
=
= 0.33 m .
2mg sin θ 2(0.50 kg)(9.80 N kg) sin 30.0°
361
Chapter 6: Conservation of Energy
Physics
88. (a) Strategy According to the work-kinetic energy theorem, the total work done on the stunt woman is equal to
her change in kinetic energy.
Solution
Wtotal = ∆K = K f − Ki =
1
1
mv 2 − 0 = (62.5 kg)(10.5 m s)2 = 3.45 kJ
2 f
2
(b) Strategy The work done by gravity is negative the change in gravitational potential energy of the stunt
woman.
Solution
Wgrav = −∆U = U i − U f = − mg ∆y = −(62.5 kg)(9.80 m s 2 )(−8.10 m) = 4.96 kJ
(c) Strategy Use Eq. (6-10) to find the nonconservative work done by air resistance.
Solution
Wtotal = Wcons + Wnc = Wgrav + Wair = ∆K , so
Wair = ∆K − Wgrav =
1
⎡1
⎤
mv 2 + mg ∆y = (62.5 kg) ⎢ (10.5 m s) 2 + (9.80 m s 2 )(−8.10 m) ⎥ = −1.52 kJ .
2 f
2
⎣
⎦
(d) Strategy Use Eq. (6-2) to find the magnitude of the average constant force of air resistance during the fall.
Solution Let the +y-axis point upward.
W
−1516 J
= 187 N
Wair = Fair ∆y, so Fair = air =
∆y
−8.10 m
Thus, the magnitude of the average force of air resistance is 187 N.
89. Strategy Use conservation of energy and Newton’s second law.
Solution Find k.
mg
ΣFy = kx1 − mg = 0, so k =
. Find vmax .
x1
1
1
Ki + K f = 0 + mvmax 2 = U i + U f = kxmax 2 + 0, so
2
2
k
g
9.80 m s 2
= xmax
= (0.100 m − 0.050 m)
= 1.6 m s .
vmax = xmax
m
x1
0.060 m − 0.050 m
F = kx1
0.20 kg
mg
90. (a) Strategy Use the definition of average power and the fact that the change in energy is equal to the person’s
increase in potential energy.
Solution
∆E ∆U mg ∆y (70 kg)(9.80 m s 2 )(740 m) ⎛ 1 h ⎞
Pav =
=
=
=
⎜
⎟ = 94 W
1.5 h
∆t
∆t
∆t
⎝ 3600 s ⎠
(b) Strategy Since the human body is only 25% efficient, it takes 4.0 units of chemical energy for every single
unit of potential energy gained.
Solution Find the amount of chemical energy used in the hike.
Echem = 4.0∆U grav = 4.0mg ∆y = 4.0(70 kg)(9.80 m s 2 )(740 m) = 2.0 MJ
362
Physics
Chapter 6: Conservation of Energy
(c) Strategy and Solution Using the given conversion factor, the number of Calories of food energy used for
⎛ 1 Calorie ⎞
the hike is 4(70 kg)(9.80 m s 2 )(740 m) ⎜
⎟ = 490 Calories .
⎝ 4.186 × 103 J ⎠
91. (a) Strategy Let +x be up the incline. Use Newton’s second law and Eq. (6-27).
Solution Compute the power the engine must deliver.
ΣFx = Fair − mg sin φ = 0 at terminal speed.
The rate at which air resistance dissipates energy is Pair = Fair v cos180° = − Fair v = − mg sin φ v.
(We use cos 180° since the force of air resistance is opposite the car’s velocity.)
The power the engine must deliver to drive the car on level ground is
Pengine = − Pair = mg sin φ v = (1500 kg)(9.80 m s 2 )(20.0 m s) sin 2.0° = 10 kW .
(b) Strategy The power available to climb the hill is the power delivered by the engine minus the dissipating
power of air resistance.
Solution From part (a), for a slope of φ :
P = mgv sin φ , so φ = sin −1
P
40.0 × 103 W − 10.26 × 103 W
= sin −1
= 5.8° .
mgv
(1500 kg)(9.80 m s 2 )(20.0 m s)
92. Strategy Use Hooke’s law and Eq. (6-24).
Solution Since F = kx, compressing a spring a certain distance x0 requires a force F0. After cutting the spring in
half, compressing the spring x0 is equivalent to compressing the original spring 2x0, which would require a force
of 2F0. This is equivalent to the short spring having a spring constant twice as large as the original spring.
Find Ushort.
1 2
1
kx and U short = (2k ) x 2 . Form a ratio.
2
2
1 (2k ) x 2
= 2
= 2, so U short = 2U long = 2(10.0 J) = 20.0 J .
1 kx 2
2
U long =
U short
U long
93. (a) Strategy Use the definition of average power. The change in energy is equal to the gravitational potential
energy.
Solution Find the average power the motor must deliver.
∆E mgh (1202 kg − 801 kg)(9.80 m s 2 )(40.0 m)
=
=
= 2.62 kW
Pav =
∆t
∆t
60.0 s
(b) Strategy Without the counterweight, the motor must deliver more power.
Solution Find the average power the motor must deliver.
(1202 kg)(9.80 m s 2 )(40.0 m)
Pav =
= 7.85 kW
60.0 s
The answer is significantly larger.
363
Chapter 6: Conservation of Energy
Physics
94. (a) Strategy Use the work-kinetic energy theorem.
Solution Compute the work done by the pitcher.
1
1
2
W = ∆K = mvf 2 − 0 = (0.153 kg) ( 40.2 m s ) = 124 J
2
2
(b) Strategy Divide the energy available to do work by the energy required to throw a fastball.
Solution Compute the number of fastballs required to “burn off” the meal.
energy available to do work 0.200(1520 × 103 cal) ⎛ 4.186 J ⎞
=
⎜
⎟ = 10,300 fastballs
J
energy per fastball thrown
123.6
⎝ cal ⎠
fastball
95. Strategy The basal metabolic rate is equal to the number of kilocalories per day required by a person resting
under standard conditions.
Solution
(a) Compute Jermaine’s basal metabolic rate.
⎛ 1 kcal ⎞⎛ 0.015 mol ⎞ ⎛ 1440 min ⎞
BMR = ⎜
⎟ = 2200 kcal day
⎟⎜
⎟⎜
⎝ 0.010 mol ⎠⎝ min ⎠ ⎝ day ⎠
(b) Find the mass of fat lost.
2160 kcal day ⎛ 2.2 lb ⎞
⎜
⎟ = 0.51 lb day
9.3 kcal g ⎜⎝ 103 g ⎟⎠
Since Jermaine is not resting the entire time, he loses more than 0.51 lb.
96. Strategy Tarzan’s kinetic energy must be great enough that his gravitational potential energy can increase by
mgh, where h = 1.7 m.
Solution If Tarzan just makes it across the gully, his final kinetic energy is zero. Let his initial potential energy
be zero. Set his initial kinetic energy equal to his final potential energy and solve for his initial speed.
1
mv 2 = mgh, so vi = 2 gh = 2(9.80 m s 2 )(1.7 m) = 5.8 m s .
2 i
97. (a) Strategy Draw a diagram and use trigonometry.
Solution Referring to the diagram, we see that when Jane is at the
lowest point of her swing, L = h + L cos 20°. Solving for h, we find
that h = L − L cos 20°.
L
20°
L cos 20°
h
(b) Strategy We assume that no nonconservative forces act (significantly) on Jane. Thus, ∆E = 0.
Solution Use conservation of energy to find Jane’s speed at the lowest point of her swing.
1
1
1
1
1
1
∆E = 0 = ∆K + ∆U = mvf 2 − mvi 2 + mg ∆y = mvf 2 − mvi 2 + mg (0 − h), so vf 2 = vi 2 + gh, or
2
2
2
2
2
2
vf = vi 2 + 2 gh = vi 2 + 2 gL(1 − cos 20°) = (4.0 m s)2 + 2(9.80 m s 2 )(7.0 m)(1 − cos 20°) = 4.9 m s .
364
Physics
Chapter 6: Conservation of Energy
(c) Strategy When Jane’s entire initial kinetic energy is converted into gravitational potential energy, she will
have reached her maximum height.
Solution Use conservation of energy to find how high Jane can swing (with respect to her lowest point).
1
1
∆E = 0 = ∆K + ∆U = 0 − mvi 2 + mg ∆y = − mvi 2 + mg (hmax − hmin ), so
2
2
2
vi 2
(4.0 m s)
hmax =
+ L(1 − cos 20°) =
+ (7.0 m)(1 − cos 20°) = 1.24 m .
2g
2(9.80 m s 2 )
98. Strategy Ignore drag due to the atmosphere and gravitational forces due to the Moon and the other planets. Use
conservation of energy and Eq. (6-14) for the potential energy.
Solution Find the minimum speed.
RE-S = Earth-Sun distance
Ei =
GM E m GM S m
1
mvi 2 −
−
= Ef = 0 + 0, so
RE
RE-S
2
⎛ 5.974 × 1024 kg 1.987 × 1030 kg ⎞
⎛M
M ⎞
vi = 2G ⎜⎜ E + S ⎟⎟ = 2(6.674 × 10−11 N ⋅ m 2 kg 2 ) ⎜
+
⎟ = 43.5 km s .
6
⎜
1.50 × 1011 m ⎟⎠
⎝ RE RE-S ⎠
⎝ 6.371× 10 m
99. Strategy Draw a diagram. Then, use Newton’s second law and conservation of energy. Let the positive direction
be away from the slope.
N
Solution According to Newton’s second law, ΣFr = N − mg cos θ = mar = − mv 2 R .
When the normal force becomes zero, we have mg cos θ = mv 2 R , or mgR cos θ = mv 2 .
When this condition is true, the skier leaves the surface of the ice. Note from the figure that
h = R cos θ , thus, the condition becomes mgh = mv 2 . Now, mgh is the final gravitational
potential energy (U i = mgR) and mv 2 is twice the final kinetic energy ( Ki = 0).
mg cos θ
mg
h
θ
So, the condition becomes U f = 2 K f , or K f = U f 2. Use conservation of energy to find h in terms of R.
1
3
3
3
0 = ∆K + ∆U = K f − Ki + U f − U i = U f − 0 + U f − U i = U f − U i , so mgh − mgR = h − R = 0, or
2
2
2
2
2
h=
R .
3
365
θ
R
Chapter 6: Conservation of Energy
Physics
100. (a) Strategy Use Hooke’s law and Newton’s laws of motion.
Solution According to Hooke’s law, F1 = k1 x1 and F2 = k2 x2 .
Imagine that one spring (1) is suspended from a ceiling and the other (2), attached to the bottom of the first,
has a mass m attached to its bottom end. Assume that the masses of the springs are negligible, and that the
system is in equilibrium. The mass connected to the lower spring exerts a force on the lower spring equal to
its weight, W. The spring stretches an amount x2 = F2 k2 = W k2 . The lower spring exerts a force on the
upper spring equal to F2 = W , and causes it to stretch by x1 = F1 k1 = F2 k1 = W k1. So, F1 = F2 , thus
k1 x1 = k2 x2 . Let F1 = F2 = F and x = x1 + x2 , and imagine the two springs in series as only one spring which
stretches an amount x in response to a force F. Find the effective spring constant, k.
x = x1 + x2 =
F1
k1
+
F2
k2
=
⎛1 1
F F
+
= F ⎜⎜ +
k1 k2
⎝ k1 k2
The effective spring constant is k =
k1k2
k1 + k2
⎞
⎛1 1
⎟⎟ , so F = ⎜⎜ +
⎠
⎝ k1 k2
⎞
⎟⎟
⎠
−1
×x=
k1k2
k1 + k2
x = kx.
.
(b) Strategy Use the result from part (a) and Eq. (6-24).
Solution Compute the potential energy stored in the spring.
1
1 k1k2 2
(500 N m)(300 N m)
U = kx 2 =
x =
(0.040 m) 2 = 0.15 J
2
2 k1 + k2
2(500 N m + 300 N m)
101. (a) Strategy Use Hooke’s law and Newton’s laws of motion.
Solution According to Hooke’s law, F1 = k1 x1 and F2 = k2 x2 .
Imagine that the springs are suspended from a ceiling such that the bottom of each is at the same height. Then
a mass m is attached to the bottom of both, the springs stretch, and the system comes to equilibrium. Assume
that the masses of the springs are negligible. Sum the vertical forces.
F1 + F2 − W = 0, so W = F1 + F2 = k1 x1 + k2 x2 .
Assuming the springs are attached to the same point on the top of the mass, x1 = x2 = x.
W = k1 x1 + k2 x2 = k1 x + k2 x = (k1 + k2 ) x = kx = W
So, in response to a force that stretches the springs (W, in this case), the springs act like one spring with a
spring constant k = k1 + k2 .
(b) Strategy Use the result from part (a) and Eq. (6-24).
Solution Compute the potential energy stored in the spring.
1
1
1
U = kx 2 = (k1 + k2 ) x 2 = (500 N m + 300 N m)(0.020 m)2 = 0.16 J
2
2
2
366
Physics
Chapter 6: Conservation of Energy
102. (a) Strategy Use conservation of energy and the relationship between radial acceleration and tangential speed.
Solution The total kinetic energy required for the bob to travel the full circle is equal to the gravitational
potential energy difference between the bottom and the top of the circle, mgh = mg[2( L − d )] = 2mg ( L − d ),
plus the kinetic energy required for the bob to have enough speed at the top of the circle to complete it. The
radial acceleration must be equal to that due to gravity at the top of the circle for the bob to just complete the
circle. So, ar = vtop 2 / r = g , or vtop = gr = g ( L − d ).
Find the total kinetic energy of the bob at the bottom of the circle.
1
1
1
K total = mv 2 = 2mg ( L − d ) + mvtop 2 = 2mg ( L − d ) + mg ( L − d ), so v =
2
2
2
5g (L − d ) .
(b) Strategy h = L(1 − cos θ ) since θ = 0° gives h = L(1 − 1) = 0 and θ = 90° gives h = L(1 − 0) = L. Use
conservation of energy.
Solution Find the minimum angle.
U i + Ki = U f + K f
1
mgh + 0 = 0 + mv 2
2
1
mgL(1 − cos θ ) = m[5 g ( L − d )]
2
5⎛ L−d ⎞ 5⎛ d ⎞
1 − cos θ = ⎜
⎟ = ⎜1 − ⎟
2⎝ L ⎠ 2⎝ L⎠
5 5d
5d 3
− cos θ = −
−1 = −
+
2 2L
2L 2
⎛ 5d 3 ⎞
θ = cos −1 ⎜ − ⎟
⎝ 2L 2 ⎠
103. Strategy Use Newton’s second law, Hooke’s law, and Eq. (6-24).
Solution Find the distance that the tendon stretches.
T
4.7 kN
∑ F = T − kx = 0, so x = =
= 1.3 cm .
k 350 kN m
Find the stored elastic energy.
1
1 T2 T2
(4.7 × 103 N) 2
U = kx 2 = k
=
=
= 32 J
2
2 k 2 2k 2(350 × 103 N m)
104. (a) Strategy The work is represented by the area under the curve. Estimate the work done during stretching and
contraction; then, find the total work done.
Solution Estimate the work.
1
1
Wstretch ≈ (0.34 m)(18 N) = 3.1 J and Wcontract ≈ (0.34 m − 0.02 m)(16 N) = 2.6 J.
2
2
Therefore, the total work = 3.1 J − 2.6 J = 0.5 J
(b) Strategy and Solution Hooke’s Law is a conservative force, so the total work to stretch and contract the
rubber band would be zero.
(c) Strategy and Solution The work done on the rubber band does not all go into increasing its elastic potential
energy; some of the energy is dissipated as heat.
367
Chapter 6: Conservation of Energy
Physics
105. (a) Strategy Use Newton’s second law and Eq. (6-10).
Solution Find the speed at the bottom of the incline.
1
2 fd
∆K = mv 2 − 0 = Wc + Wnc = mgh − fd so v = 2 gh −
.
2
m
Use Newton’s second law with +y perpendicular to the incline and +x down the
incline.
ΣFy = N − mg cos θ = 0, so N = mg cos θ .
N
f
mg cos θ
θ
mg sinθ
mg
Now, d = 0.85 m, h = d sin θ , and f = µ N = µ mg cos θ .
Substitute.
2 µ mgd cos θ
= 2 gd (sin θ − µ cos θ )
v = 2 gd sin θ −
m
Find the maximum compression.
1
1
K f + U f = 0 + kx 2 = Ki + U i = mv 2 + 0, so
2
2
x=v
m
=
k
2mgd
(sin θ − µ cos θ ) =
k
2(0.50 kg)(9.80 m s 2 )(0.85 m)
(sin 30.0° − 0.25cos 30.0°)
35 N m
= 26 cm .
(b) Strategy When the block is accelerated by the spring, it attains its previous kinetic energy and speed.
Solution Find the distance along the incline, d ′.
1
∆K = 0 − mv 2 = Wc + Wnc = − mgh − fd ′ = − mgd ′ sin θ − µ mgd ′ cos θ = − d ′[ mg (sin θ + µ cos θ )], so
2
v2
2 gd (sin θ − µ cos θ )
sin 30.0° − 0.25cos 30.0°
d′ =
=
= (85 cm)
= 34 cm .
2 g (sin θ + µ cos θ ) 2 g (sin θ + µ cos θ )
sin 30.0° + 0.25cos 30.0°
368
Physics
Chapter 6: Conservation of Energy
106. (a) Strategy The blades sweep out a circle of radius L. The air moves though the circular area at a speed v;
therefore, the distance the air moves in a time ∆t is d = v∆t. The volume of the air is equal to the area of the
circle swept out by the blades times the distance d.
Solution Find the volume of air.
V = Ad = π L2 v∆t = π (4.0 m)2 (10 m s)(1.0 s) = 500 m3
(b) Strategy The mass of air is equal to its volume times its density.
Solution Find the mass of the air.
m = V ρ = (500 m3 )(1.2 kg m3 ) = 600 kg
(c) Strategy Use Eq. (6-6).
Solution Find the translational kinetic energy of the air.
1
1
K = mv 2 = (600 kg)(10 m s)2 = 30 kJ
2
2
(d) Strategy Use the definition of average power. The power output is 40% of the kinetic energy per unit time.
Solution Find the electrical power output.
∆E 0.40(30 kJ)
P=
=
= 12 kW
∆t
1.0 s
(e) Strategy Use the results of parts (a) through (d).
Solution Form a proportion.
3
2
1
1 ρVv 2
1
3
P5 ⎛ 2 v ⎞
∆E 2 mv
1
1
ρπ L2 v∆tv 2 ρπ L2 3
2
3
⎟ = ⎛⎜ ⎞⎟ = .
P=
v ∝ v , so
=
=
=
=
=⎜
P10 ⎜ v ⎟
2 ∆t
2
8
∆t
∆t
∆t
⎝2⎠
⎝
⎠
So, the power output would decrease to 1 8 of its previous value.
The power production of wind turbines is inconsistant, since modest changes in wind speed
produce large changes in power output.
107. Strategy and Solution The kinetic energy of a volume of wind passing through the circular area swept out by the
rotor blades in time ∆t is 12 mv 2 , where m = ρV = ρ Ad = ρ (π L2 )(v∆t ) and v is the speed of the wind; therefore,
∆E ε ∆K
=
, where ε is
the kinetic energy is given by K = 12 ρπ L2 ∆tv3 . The average power generated is Pav =
∆t
∆t
the efficiency of the energy conversion from kinetic energy to electrical energy.
ε ∆K ερπ L2 3
Therefore, Pav =
=
v ∝ v3 .
∆t
2
369
Chapter 6: Conservation of Energy
Physics
108. Strategy Use the method outlined in the problem statement, the relationship between mass, density, and volume,
and Eq. (6-9).
Solution Find how the speed with which animals of similar shape can run up a hill depends upon the size of the
animals.
∆U mg ∆y
∆U
Pmax ∝ L2 and
=
= mgv ∝ mv. If ρ is the mass density, then m = ρV ≈ ρ L3 , so
∝ L3v.
∆t
∆t
∆t
∆U
, then L3v ∝ L2 , or v ∝ 1 L .
Thus, if Pmax =
∆t
109. Strategy Use conservation of energy.
Solution According to the graph, the potential energy in the region under consideration is 300 J. So, U f = 300 J.
Initially, the kinetic energy is 200 J and the potential energy is 0, so Ki = 200 J and U i = 0. Compute the final
kinetic energy; that is, the kinetic energy in the region under consideration.
E = ∆K + ∆U = K f − Ki + U f − U i = K f − 200 J + 300 J − 0 = K f + 100 J = 0, so K f = −100 J, which is
impossible, since kinetic energy cannot be negative. Therefore, the answer is no, the particle cannot enter the
region 3 cm < x < 8 cm. Since the particle cannot enter the region, it must remain in the region x < 3 cm.
110. Strategy Use conservation of energy.
Solution According to the graph, the potential energy in the region under consideration is 300 J. So, U f = 300 J.
Initially, the kinetic energy is 400 J and the potential energy is 0, so Ki = 400 J and U i = 0. Compute the final
kinetic energy; that is, the kinetic energy in the region under consideration.
E = ∆K + ∆U = K f − Ki + U f − U i = K f − 400 J + 300 J − 0 = K f − 100 J = 0, so K f = 100 J .
Since the final kinetic energy is positive, the answer is yes, the particle can enter the region 3 cm < x < 8 cm.
370
Chapter 7
LINEAR MOMENTUM
Conceptual Questions
1. The likelihood of injury resulting from jumping from a second floor window is primarily determined by the
average force acting to decelerate the body.
(a) The deceleration time interval for a person landing stiff legged on pavement is very short. The impulsemomentum theorem tells us that the average force acting on the person’s feet must therefore be very large—
such a person is likely to incur injuries.
(b) Jumping into a privet hedge increases the time interval over which the body decelerates. This decreases the
average force on the person’s limbs and therefore decreases the likelihood of injury.
(c) Jumping into a firefighter’s net is the best option of the three. The net stretches downward, gradually bringing
the person to rest. Additionally, the firefighters lower the net with their hands as the person lands to further
lengthen the time interval during which the person is brought to rest.
2. (a) A body’s momentum change is equal to the impulse that has acted on it. Impulse is defined as the product of
the average force acting on a body and the time interval over which it acts—the bodies therefore experience
the same impulse and so have equal momentum changes.
(b) The change in a body’s velocity is defined as the ratio of the change in its momentum to its mass—the less
massive body therefore incurs a larger velocity change.
(c) The acceleration of a body is defined as the ratio of the force acting on it to its mass—the less massive body
therefore has the larger acceleration.
3. The muzzle speed is determined by the change in the bullet’s momentum. The impulse-momentum theorem tells
us that this momentum change is determined by the impulse acting on the bullet. The force acting on the bullet
due to the expanding hot gases is roughly constant throughout the muzzle. A shorter muzzle produces a shorter
time interval over which the bullet is accelerated by the firing force. This results in a smaller impulse and
therefore a smaller momentum change—thus producing lower bullet velocities.
4. After the explosion, each piece of the firecracker has a momentum vector associated with it that points in the
direction of its motion. The law of conservation of linear momentum tells us that the vector sum of the momentum
of all the pieces of the firecracker must equal the initial momentum of the whole firecracker—in this case, both
the initial and the final net momentum vectors equal zero.
5. The law of the conservation of linear momentum states that in the absence of external interactions, the linear
momentum of a closed system is constant. Floating in free space, the astronaut and the wrench form a closed
system free from interactions with other bodies. If the astronaut throws the wrench in the direction opposite the
ship, conservation of momentum dictates that he must in turn move toward the ship.
6. The horizontal component of the golf ball’s momentum is conserved since no external force acts on the ball in the
horizontal direction. The vertical component of the ball’s momentum is not conserved however because the
Moon’s gravitational force interacts with it and changes its momentum.
371
Chapter 7: Linear Momentum
Physics
7. In an elastic collision between the hammer and nail, the kinetic energy of the system is conserved while in a
perfectly inelastic collision, the greatest percentage of the kinetic energy is lost. The energy lost by the system in a
perfectly inelastic collision is used to do the work required to bring the hammer and nail together. In an elastic
collision, this work is available to drive the nail into the wood—the total work available to drive the nail is
therefore greater for an elastic collision. Thus, for equal applied forces, the hammer will drive the nail further into
the wood if the collision is elastic.
8. The momentum of the squid (including the water inside its cavity) comprises a system for which momentum is
conserved. The means the momentum of the squid (plus water) must be the same before and after some of the
water has been ejected. When the squid expels some water, the water gains momentum in the direction it is being
expelled. To conserve momentum, the squid must gain an equal amount of momentum in the opposite direction,
propelling it forward. Similarly, a rocket engine expels exhaust from burning fuel to propel itself forward.
9. First law: The momentum of an object is constant unless acted upon by an external force. Second law: The net
force acting on an object is equal to the rate of change of the object’s momentum. Third law: When two objects
interact, the changes in momentum that each imparts to the other are equal in magnitude and opposite in direction.
10. Noting that the (translational) kinetic energy can be written as p 2 /(2m), and that both objects have the same
kinetic energy, it is evident that the object with the greater mass has the larger magnitude of momentum.
11. The woman’s center of mass is not necessarily 0.80 m above the floor, because her mass is not necessarily
distributed uniformly with height. Normally, the upper body of a person is more massive than the lower body and
thus we would expect the woman’s center of mass to be slightly higher than 0.80 m.
12. The frictional force of the road on the tires supplies the external force to change the bicycle’s momentum.
Changes in the bicycle’s kinetic energy do not require an external force. For example, the rider could throw her
helmet away hard, increasing both her and the helmet’s speed. The kinetic energy of the system (bicycle, rider,
and helmet) would increase, while the momentum would remain the same. Note that the work-energy theorem
(total work done equals change in kinetic energy) cannot be used here, because the internal structure of the system
cannot be ignored.
13. An impulse must be supplied to the egg to change its momentum and bring it to rest. A good strategy is to make
the time interval over which the stopping force is applied as large as possible. This will reduce the magnitude of
the force required to stop the egg. One should therefore attempt to catch the egg with a swinging motion, moving
the hand backwards as it is being caught, to bring it to rest as slowly and gently as possible.
14. The collisions of the balls in the “executive toy” are nearly perfectly elastic. The kinetic energy of the system just
before and after a collision must therefore be the same. This is the reason we never see three balls moving away
after a collision in which two balls were initially pulled back and released—such an event would not conserve
kinetic energy.
15. According to the impulse-momentum theorem, the change in momentum of the baseball is equal to the impulse it
receives from the bat. Impulse is equal to the average force times the time interval over which the force is applied.
To give the ball the greatest possible momentum, one should attempt to maximize the amount of time during
which the force is being applied.
16. Jeremy has it right. By momentum conservation, Micah needs to throw the balls forward if he wants to propel
himself backward, but the balls need not strike any surface. You can also consider Newton’s third law and see that
it is the force by the balls on Micah’s hand that pushes Micah backward.
17. Daryl has done his homework. If he falls when rock climbing, his rope will stretch and stop him more gradually
than the rope Mary wants to buy. In a fall, the climber’s momentum must go from some initial value to zero. If the
time over which the momentum is decreased to zero is longer, the average force delivered by the rope is smaller.
372
Physics
Chapter 7: Linear Momentum
Problems
1. Strategy Use the definition of linear momentum.
Solution Find the magnitude of the total momentum of the system.
G
G G
G
G
G G
G
G
p total = p1 + p 2 = mv1 + mv 2 = m( v1 + v 2 ) = m[ v1 + (− v1 )] = 0, so the magnitude is 0 .
2. Strategy Use the definition of linear momentum.
Solution Find the momentum of the automobile.
G
9800 N
G W G
p = mv = v =
(35 m s south) = 3.5 × 104 kg ⋅ m s south
g
9.80 m s 2
3. Strategy and Solution Impulse = F ∆t , so the SI unit is N ⋅ s = kg ⋅ m s 2 ⋅ s = kg ⋅ m s. p = mv, so the SI unit is
kg ⋅ m s. Therefore, the SI unit of impulse is the same as the SI unit of momentum.
4. Strategy Use the impulse-momentum theorem.
Solution Find the final speed of the cue ball.
F ∆t (24 N)(0.028 s)
∆p = pf − pi = mvf − m(0) = Fav ∆t , so vf = av =
= 4.2 m s .
m
0.16 kg
5. Strategy Add the momenta of the three particles.
Solution Find the total momentum of the system.
G
G G
G
G
G
G
p tot = p1 + p 2 + p3 = m1v1 + m2 v 2 + m3 v3 = m1v1 north + m2 v2 south + m3v3 north
= (m1v1 − m2 v2 + m3v3 ) north = [ (3.0 kg)(3.0 m s) − (4.0 kg)(5.0 m s) + (7.0 kg)(2.0 m s) ] north
= 3 kg ⋅ m s north
6. (a) Strategy Form a ratio of the magnitudes of the final and initial momenta.
Solution Compute the ratio.
pf mvf vf 60.0 mi h
=
=
=
= 3.00
pi mvi vi 20.0 mi h
(b) Strategy Form a ratio of the final and initial kinetic energies.
Solution Compute the ratio.
Kf
=
Ki
1 mv 2
f
2
1 mv 2
i
2
⎛v
= ⎜⎜ f
⎝ vi
2
⎞
2
⎟⎟ = 3.00 = 9.00
⎠
7. Strategy The initial momentum is toward the wall and the final momentum is away from the wall.
Solution Find the change in momentum.
∆p = pf − pi = mvf − mvi = m(vf − vi ) = (5.0 kg)(−2.0 m s − 2.0 m s) = −20 kg ⋅ m s , so
G
∆p = 20 kg ⋅ m s in the −x-direction .
373
Chapter 7: Linear Momentum
Physics
8. Strategy The final and initial velocities are the same, since air resistance is ignored. Use the definition of the
1
linear momentum. Use Eq. ∆y = viy ∆t − g (∆t )2 to find the initial speed. Let up be the positive direction.
2
Solution Find the initial speed.
1
1
∆y = 0 = viy ∆t − g (∆t )2 , so viy = g ∆t.
2
2
Find ∆p.
⎛1
⎞
∆p = pfy − piy = m(vfy − viy ) = m(−viy − viy ) = −2m ⎜ g ∆t ⎟ = − mg ∆t = −(3.0 kg)(9.80 m s 2 )(3.4 s)
⎝2
⎠
2
= −1.0 × 10 kg ⋅ m s
G
So, ∆p = 1.0 × 102 kg ⋅ m s downward .
9. Strategy Use the definition of linear momentum. Let up be the positive direction.
Solution vf = vfy = viy − g ∆t = − g ∆t , since the object starts from rest.
Find ∆p.
∆p = pf − pi = m(vf − vi ) = m(− g ∆t − 0) = − mg ∆t = −(3.0 kg)(9.80 m s 2 )(3.4 s) = −1.0 × 102 kg ⋅ m s, so
G
∆p = 1.0 × 102 kg ⋅ m s downward .
10. Strategy Use the impulse-momentum theorem.
Solution Find the average force.
∆p m∆v (50.0 kg)(3.0 m s − 0)
Fav =
=
=
= 7.5 N
∆t
∆t
20.0 s
The force necessary is 7.5 N in the direction of the sled’s velocity.
11. Strategy Use the impulse-momentum theorem. Let the forward direction be positive.
Solution Find the time interval for which the engine must be fired.
∆p m∆v (3800 kg)(1.1× 104 m s − 2.6 × 104 m s)
∆t =
=
=
= 320 s
Fav
Fav
−1.8 × 105 N
374
Physics
Chapter 7: Linear Momentum
12. (a) Strategy Use the component method of subtracting vectors.
Solution Compute the magnitude of the change in momentum.
∆p x = m∆vx and ∆p y = m∆v y .
y
G
∆p = m (∆vx )2 + (∆v y )2 = (0.15 kg) [0 − (−20 m s)]2 + (15 m s − 0) 2 = 3.8 kg ⋅ m s
x
pf
pi
Find the angle of the change in momentum.
∆p y
∆v y
15
= tan −1
= tan −1
= 37°
θ = tan −1
20
∆p x
∆vx
The change in momentum of the baseball is
G
3.8 kg ⋅ m s at 37° above the horizontal direction opposite vi .
(b) Strategy Use the impulse-momentum theorem.
Solution Find the average force of the bat on the ball.
G
G
∆p 3.75 kg ⋅ m s
Fav =
=
= 75 N, so F = 75 N in the same direction as ∆p .
∆t
0.050 s
13. Strategy Use the impulse-momentum theorem. Let the positive direction be in the direction of motion.
Solution Find the average horizontal force exerted on the automobile during breaking.
∆p m(vf − vi ) (1.0 × 103 kg)(0 − 30.0 m s)
Fav =
=
=
= −6.0 × 103 N
5.0 s
∆t
∆t
G
So, Fav = 6.0 × 103 N opposite the car’s direction of motion .
14. Strategy Use the impulse-momentum theorem.
Solution
(a) Compute the changes in momenta for each direction.
∆pnorth = 0 and ∆peast = Fav ∆t = m∆veast = mveast .
G
Find the magnitude and direction of v f .
15 m/s
N
15 N
2
2
⎡ (15 N)(4.0 s) ⎤
⎛ F ∆t ⎞
vf = vnorth 2 + veast 2 = vnorth 2 + ⎜ av ⎟ = (15 m s) 2 + ⎢
⎥ = 25 m s and
m
⎝
⎠
⎣ 3.0 kg ⎦
v
15 m s
G
θ = tan −1 north = tan −1
= 37° north of east, so v f = 25 m s at 37° north of east .
veast
20 m s
(b) Let +y be north and +x be east. Compute the change in momentum.
∆p = Fav ∆t = (15 N)(4.0 s) = 60 kg ⋅ m s.
G
The entire change in momentum is due to the force, so ∆p = 60 kg ⋅ m s east .
375
∆p
pi
pf
y
x
Chapter 7: Linear Momentum
Physics
15. (a) Strategy Use the definition of linear momentum. Use vf2y − viy 2 = 2a y ∆y to find the speed after the fall.
Solution Find the initial speed, which is the final speed after the fall.
vf2y − viy 2 = vf2y − 0 = −2 g ∆y = 2 gh, so vfy = 2 gh .
G
G
If up is positive, v y = 2 gh down = − 2 gh up = vi .
(
)
∆p = m(vf − vi ) = m ⎡ 0 − − 2 gh ⎤ = m 2 gh = (60.0 kg) 2(9.80 m s 2 )(8.0 m) = 750 kg ⋅ m s , so
⎣
⎦
G
∆p = 750 kg ⋅ m s upward .
(b) Strategy The impulse on the net is equal to the boy’s weight times ∆t plus the change in momentum of the
G
boy due to the net, −∆p.
Solution Find the impulse on the net.
G
mg ∆t downward − ∆p = (60.0 kg)(9.80 N kg)(0.40 s) downward + 750 kg ⋅ m/s downward
= 990 N ⋅ s downward
(c) Strategy Use Eq. (7-3).
Solution Find the average on the net due to the boy.
G
G
∆p 990 kg ⋅ m/s downward
Fav =
=
= 2500 N downward
∆t
0.40 s
16. Strategy The impulse is equal to the area under the graph. Use the impulse-momentum theorem. Let the positive
direction be to the right.
Solution Each rectangle of the grid is equal to (100 N)(0.0010 s) = 0.10 kg ⋅ m s. The area can be divided easily
into three right triangles and one rectangle. Thus, there are
1 (7)(4) + 1 (6)(2) + 1 (8)(6) + (6)(4)
2
2
2
= 68 rectangles
under the graph and the magnitude of the impulse is ∆p = 68(0.10 kg ⋅ m s) = 6.8 kg ⋅ m s = m∆v. The impulse is
opposite the direction of motion of the initial velocity. Compute the final speed.
∆p
∆p
6.8 kg ⋅ m s
∆v =
= vf − vi , so vf = vi +
= −30 m s +
= 29 m s .
m
m
0.115 kg
17. (a) Strategy Use conservation of energy.
Solution Find the speed with which the pole-vaulter lands.
1
∆K = mv 2 = −∆U = mgh, so v = 2 gh = 2(9.80 m s 2 )(6.0 m) = 11 m s .
2
(b) Strategy Use the impulse-momentum theorem.
Solution Find the average force on the pole-vaulter’s body.
∆p = Fav ∆t , so
∆p m(vf − vi ) m[0 − (−v)] mv m
60.0 kg
Fav =
=
=
=
=
2 gh =
2(9.80 m s 2 )(6.0 m) = 1300 N .
∆t
∆t
∆t
∆t ∆ t
0.50 s
376
Physics
Chapter 7: Linear Momentum
18. Strategy Use conservation of momentum.
Solution Find the recoil speed of the rifle.
m
G
G
G
G
0.0100 kg
G
G
G
G
p rf + p bf = mr v rf + mb v bf = p ri + p bi = 0 + 0, so v rf = b − v bf =
(820 m s) = 1.8 m s .
mr
4.5 kg
19. (a) Strategy Right after the collision, the bullet and baseball combination must have the same momentum as the
bullet had just before it stuck the baseball.
Solution Before the collision, the momentum of the bullet is pi = mbullet vbullet . After the collision, the
momentum of the bullet and baseball combination is pf = (mbullet + mbaseball )vf = pi .
Thus, the speed of the bullet and baseball combination right after the collision was
mbullet vi
pi
(0.030 kg)(200 m s)
vf =
=
=
= 33 m s .
mbullet + mbaseball mbullet + mbaseball
0.030 kg + 0.15 kg
(b) Strategy Use conservation of energy to determine the work done by air resistance on the bullet and baseball
combination.
Solution Determine the work done by air resistance. Let up be the positive direction.
Wtotal = Wc + Wnc = −∆U + Wair = ∆K , so
1
⎡ 1
⎤
Wair = ∆K + ∆U = 0 − mv 2 + mg ∆y = (0.18 kg) ⎢ − (33.333 m s)2 + (9.80 m s 2 )(37 m) ⎥ = −34.73 J.
2
⎣ 2
⎦
W
−34.73 J
Since Wair = Fair, av ∆y, Fair, av = air =
= −0.94 N. Therefore, the average force of air resistance
∆y
37 m
was 0.94 N down .
20. Strategy Use conservation of momentum.
Solution Find the recoil speed of the submarine.
m G
G
G
G
G
250 kg
G
G
G
(100.0 m s) = 0.010 m s .
psf + p tf = ms vsf + mt v tf = psi + p ti = 0 + 0, so vsf = t − v tf =
ms
2.5 × 106 kg
21. Strategy Use conservation of momentum.
Solution Find the recoil speed of the thorium nucleus.
G
G
pi = 0 = −pf , so if n = nucleus and p = particle,
mp G
G
G
4.0 u ⎡
G
G
G
p n + p p = mn v n + mp v p = 0, so v n =
− vp =
0.050(2.998 × 108 m s) ⎤ = 2.6 × 105 m s .
⎦
mn
234 u ⎣
22. Strategy Use the law of conservation of linear momentum to determine the speed Dash must throw the balls.
Solution According to the law of conservation of linear momentum, Dash and his skateboard will move
backward with linear momentum equal in magnitude to the magnitude of the combined momentum of the balls.
Find the speed of the balls.
m v
(60 kg)(0.50 m s)
pb = 3mb vb = mD vD = pD , so vb = D D =
= 100 m s (224 mph) .
3mb
3(0.10 kg)
Since 224 mph is faster than any human can throw a ball, Dash will not succeed .
377
Chapter 7: Linear Momentum
Physics
23. Strategy Use the law of conservation of linear momentum to determine the astronaut’s speed.
Solution According to the law of conservation of linear momentum, the astronaut will move toward the ship with
linear momentum equal in magnitude to the magnitude of the combined momentum of the objects thrown. Find
the speed of the astronaut after he throws the mallet.
Σpobjects = pw + ps + pm = mw vw + ms vs + mm vm = pA = mA vA , so
vA =
mw vw + ms vs + mm vm
mA
(0.72 kg)(5.0 m s) + (0.80 kg)(8.0 m s) + (1.2 kg)(6.0 m s)
=
= 0.30 m s .
58 kg
24. Strategy Use conservation of energy to determine the skier’s speed and momentum just before he grabs the
backpack. Then, use conservation of linear momentum to find his new speed after he grabs the backpack. Finally,
from Chapter 4, use the equations for motion with a changing velocity.
Solution Use conservation of energy to find the speed of the skier just before he grabs the backpack.
1
1
E = ∆K + ∆U = K f − Ki + U f − U i = ms vs 2 − 0 + 0 − ms gh = ms vs 2 − ms gh = 0, so vs = 2 gh .
2
2
Use conservation of linear momentum to find his new speed after he grabs the backpack.
ms vs
m 2 gh
pi = ms vs = pf = (ms + mb )vx , so vx =
= s
.
ms + mb ms + mb
Now, find the time it takes for an object to fall 2.0 m from rest.
1
1
1
2∆y
∆y = viy ∆t − g (∆t ) 2 = (0)∆t − g (∆t ) 2 = − g (∆t ) 2 , so ∆t = −
.
2
2
2
g
The skier will travel a horizontal distance of
m 2 gh
2∆y 2ms −∆yh 2(65 kg) −(−2.0 m)(5.0 m)
∆ x = v x ∆t = s
−
=
=
= 4.8 m .
ms + mb
g
ms + mb
65 kg + 20 kg
25. Strategy Use conservation of momentum.
Solution Find the recoil speed of the railroad car.
G
G
pi = 0 = −p f , and since we are only concerned with the horizontal direction, we have:
mc vcx = ms vsx , so vcx =
ms
mc
vsx =
98 kg
5.0 × 104 kg
(105 m s) cos 60.0° = 0.10 m s .
26. Strategy Use conservation of momentum.
Solution Find the mass of the man and the car.
G
G
pi = 0 = −pf , and since we are only concerned with the horizontal direction, we have:
mmc vmc = mb vb , so mmc =
vb
vmc
mb =
(173 m s) cos 30.0°
1.0 × 10−3 m s
378
(0.010 kg) = 1500 kg .
Physics
Chapter 7: Linear Momentum
27. Strategy Use the component form of the definition of center of mass.
Solution Find the location of particle B.
Find xCM .
y (cm)
Β?
CM
(2.0, 5.0)
5
mA xA + mB xB 0 + mB xB
=
, so
mA + mB
mA + mB
m + mB
30.0 g + 10.0 g
xB = A
xCM =
(2.0 cm) = 8.0 cm.
mB
10.0 g
Similarly,
30.0 g + 10.0 g
yB =
(5.0 cm) = 20 cm.
10.0 g
The coordinates of particle B are ( xB , yB ) = (8.0 cm, 20 cm) .
xCM =
0
Α
0
5
x (cm)
28. Strategy Use the component form of the definition of center of mass.
Solution Find the location of the center of mass.
Find xCM and yCM .
xCM
yCM
y (cm)
25
m x + mB xB (5.0 g)(0) + (1.0 g)(25 cm)
= A A
=
= 4.2 cm
mA + mB
5.0 g + 1.0 g
m y + mB yB (5.0 g)(0) + (1.0 g)(0)
= A A
=
=0
mA + mB
5.0 g + 1.0 g
0
CM?
Α
0
Β
25
x (cm)
The location of the center of mass is ( xCM , yCM ) = (4.2 cm, 0) .
29. Strategy Since no y-components of the positions have changed, the center of mass moves only in the x-direction.
Use the component form of the definition of center of mass.
Solution Find the displacement of the center of mass of the three bodies.
mx + mx2i + mx3i x1i + x2i + x3i 1 m + 2 m + 3 m
xi = 1i
=
=
=2m
m+m+m
3
3
x + x2f + x3f x1i + x2i + x3i + 0.12 m 6 m + 0.12 m
xf = 1f
=
=
3
3
3
6 m + 0.12 m
0.12 m
∆x = xf − xi =
−2 m =
= 4.0 cm
3
3
The center of mass moves 4.0 cm in the positive x-direction.
30. Strategy Use the component form of the definition of center of mass.
Solution Find the location of the center of mass.
m x + m2 x2 + m3 x3 (4.0 kg)(4.0 m) + (6.0 kg)(2.0 m) + (3.0 kg)(−1.0 m)
xCM = 1 1
=
m1 + m2 + m3
4.0 kg + 6.0 kg + 3.0 kg
= 1.9 m
m y + m2 y2 + m3 y3 (4.0 kg)(0) + (6.0 kg)(4.0 m) + (3.0 kg)(−2.0 m)
yCM = 1 1
=
= 1.4 m
4.0 kg + 6.0 kg + 3.0 kg
m1 + m2 + m3
The location of the center of mass is ( xCM , yCM ) = (1.9 m, 1.4 m) .
379
y (m)
2
(2.0, 4.0)
CM?
1
3
( 1.0, 2.0)
(4.0, 0) x (m)
Chapter 7: Linear Momentum
Physics
31. Strategy Use symmetry and the component form of the definition of center of mass to determine the center of
mass of each object with respect to the origin at the top left corner of the sculpture.
Solution The centers of mass are as follows:
rectangle: (1.0 m, − 0.25 m); circle: (0 m, − 2.5 m); square: (1.4 m, − 1.9 m); octagon: (2.0 m, − 3.0 m)
Find the components of the center of mass of the entire sculpture.
m x + mc xc + ms xs + mo xo
xCM = r r
mr + mc + ms + mo
(2.0 kg)(1.0 m) + (5.0 kg)(0 m) + (2.0 kg)(1.4 m) + (3.0 kg)(2.0 m)
=
= 0.900 m
2.0 kg + 5.0 kg + 2.0 kg + 3.0 kg
m y + mc yc + ms ys + mo yo
yCM = r r
mr + mc + ms + mo
(2.0 kg)(−0.25 m) + (5.0 kg)(−2.5 m) + (2.0 kg)(−1.9 m) + (3.0 kg)(−3.0 m)
= −2.15 m
=
2.0 kg + 5.0 kg + 2.0 kg + 3.0 kg
The center of mass of the sculpture is (0.900 m, − 2.15 m) .
32. Strategy Use the component form of the definition of center of mass.
Solution The center of mass of the upper leg is at (17.5 cm, 0); the center of mass of the lower leg is at
[35 cm + (20 cm) sin 30.0°, − (20 cm) cos 30.0°].
Find the components of the center of mass of Jane’s leg.
Mxupper + mxlower (20 kg)(17.5 cm) + (10 kg)[35 cm + (20 cm) sin 30.0°]
=
= 27 cm
xCM =
M +m
20 kg + 10 kg
Myupper + mylower (20 kg)(0) + (10 kg)[−(20 cm) cos 30.0°]
=
= −5.8 cm
yCM =
M +m
20 kg + 10 kg
The center of mass of the leg is (27 cm, − 5.8 cm) .
33. Strategy The x-coordinate of each three-dimensional shape is midway along its horizontal dimension.
Solution Find the x-component of the center of mass of the composite object.
m x + mc xc + mr xr (200 g)(5.0 cm) + (450 g)(10 cm + 17 2 cm) + (325 g)(10 cm + 17 cm + 16 2 cm)
xCM = s s
=
ms + mc + mr
200 g + 450 g + 325 g
= 21 cm
34. Strategy The two masses fall at the same rate, so their center of mass, which lies on the (horizontal) line
between the two bodies, falls at the same rate as the two bodies. Use ∆v = a y ∆t to find the speed. Let the positive
y-direction be up.
Solution Find the velocity of the center of mass.
G
vfy = viy − g ∆t = 0 − g ∆t = −(9.80 m s 2 )(10.0 s) = −98.0 m s , so v CM = 98.0 m s downward .
380
Physics
Chapter 7: Linear Momentum
35. Strategy The total momentum of the system is equal to the total mass of the system times the velocity of the
center of mass.
Solution Find the total momentum.
G
G
m v + mB v B
G
G G
G
G
G
G
G
p = Mv CM = mA v A + mB v B since p = p A + p B . Thus, v CM = A A
.
mA + mB
G
Find the components of v CM .
x
7 m/s
A
14 m/s
(3 kg)(14 m s) + (4 kg)(0)
=6 m s
3 kg + 4 kg
mA + mB
(3 kg)(0) + (4 kg)(−7 m s)
=
= −4 m s
3 kg + 4 kg
vCMx =
vCMy
mA vAx + mB vBx
y
B
=
So, the components are (vCMx , vCMy ) = (6 m s , − 4 m s) .
36. Strategy The total momentum of the system is equal to the total mass of the system times the velocity of the
center of mass. Let east be in the positive direction.
Solution Find the total momentum.
G
G
m v + mB v B
G
G G
G
G
G
G
G
p = Mv CM = mA v A + mB v B since p = p A + p B . Thus, v CM = A A
.
mA + mB
Find the velocity of the center of mass.
(5.0 kg)(10 m s) + (15 kg)(−10 m s)
G
vCM =
= −5 m s , so v CM = 5 m s west .
5.0 kg + 15 kg
10 m/s
5.0 kg A
15 kg
B
N
10 m/s
37. (a) Strategy Draw a diagram and use conservation of linear momentum.
Solution
G
M G M G
5M G
G
p = Mv CM =
v1 +
v2 +
v = 0 Use components.
4
3
12 3
M
M
5M
v1x + v2 x +
v = 3v1x + 4v2 x + 5v3 x = 0, so
4
3
12 3 x
3v + 4v2 x
3(5.0 m s) cos 37° + 4(4.0 m s) cos135°
=−
= −0.13 m s.
v3 x = − 1x
5
5
M
M
5M
v1 y + v2 y +
v = 3v1 y + 4v2 y + 5v3 y = 0, so
4
3
12 3 y
3v1 y + 4v2 y
3(5.0 m s) sin 37° + 4(4.0 m s) sin135°
=−
= −4.1 m s.
v3 y = −
5
5
4.0 m/s
45°
y
5.0 m/s
37°
x
The velocity components are (−0.13 m s , − 4.1 m s) .
(b) Strategy and Solution Due to the law of conservation of linear momentum,
the center of mass of the system remains at the origin after the explosion .
G
G
38. Strategy Prove that ∑ Fext = MaCM .
Solution
G
G
G
G
G
G
G
G
Mv CMf − Mv CMi
∆v CM
∆v CM
p − pi
G
G
∆p
∑ Fext = lim
= lim f
= lim
= M lim
= MaCM since aCM = lim
.
∆t
∆t →0 ∆t
∆t →0 ∆t
∆t →0
∆ t → 0 ∆t
∆t →0 ∆t
381
Chapter 7: Linear Momentum
Physics
39. Strategy Use conservation of momentum. Let the positive direction be to the right.
Solution Find the final velocity of the helium atom.
mHe vHef + mO vOf = mHe vHei + mO vOi , so
vHef =
=
mO (vOi − vOf ) + mHe vHei
mO
mHe
mHe
(vOi − vOf ) + vHei =
32.0 u
( 412 m s − 456 m s ) + 618 m s = 270 m s.
4.00 u
Thus, the velocity of the helium atom after the collision is 270 m s to the right .
40. Strategy Linear momentum is conserved, so pf = pi .
Solution Find the change in speed of the car.
pf = (mcar + mclay )vf = pi = mcar vi , so vf =
∆v = vf − vi =
mcar
mcar + mclay
mcar
mcar + mclay
vi , and
⎛
⎞
mcar
⎛
⎞
120 g
− 1⎟ vi = ⎜
− 1⎟ (0.75 m s) = −0.15 m s .
vi − vi = ⎜
⎜ mcar + mclay
⎟
⎝ 120 g + 30.0 g ⎠
⎝
⎠
41. Strategy Use conservation of momentum.
Solution
(a) The collision is perfectly inelastic, so v1f = v2f = vf . Find the speed of the two cars after the collision.
v
1.0 m s
m1v1i + m2 v2i = mv1i + 4.0m(0) = m1v1f + m2 v2f = mvf + 4.0mvf , so vf = 1i =
= 0.20 m s .
5.0
5.0
(b) The cars are at rest after the collision, so v1f = v2f = 0.
mv1i + 4.0mv2i = 0, so v2i = −
v1i
4.0
=−
1.0 m s
= −0.25 m s. The initial speed was 0.25 m s .
4.0
42. Strategy Use conservation of momentum. The block is initially at rest, so v2i = 0. Let east be in the +x-direction.
Solution Find the final velocity of the block.
m1v1f + m2 v2f = m1v1i + m2 v2i = m1v1i + m2 (0), so
m (v − v ) 0.020 kg
v2f = 1 1i 1f =
[ 200.0 m s − (−100.0 m s)] = 3.0 m s.
m2
2.0 kg
G
Thus, v block = 3.0 m s east .
43. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f = v2f = vf . Also, the block
is initially at rest, so v2i = 0.
Solution Find the speed of the block of wood and the bullet just after the collision.
m1v1f + m2 v2f = (m1 + m2 )vf = m1v1i + m2 v2i = m1v1i + m2 (0), so
m1
0.050 kg
vf =
v =
(100.0 m s) = 5.0 m s .
m1 + m2 1i 0.050 kg + 0.95 kg
382
Physics
Chapter 7: Linear Momentum
44. Strategy The collision is perfectly inelastic since the bullet embeds in the wood. Friction does negative work on
the block and bullet combination. Use Newton’s second law and conservation of momentum.
Solution Let the +x-direction be in the direction of motion. Find the acceleration of the
block and bullet due to friction.
ΣFy = N − (mbullet + mblock ) g = 0, so N = (mbullet + mblock ) g .
ΣFx = − f k = − µk N = − µk (mbullet + mblock ) g = (mbullet + mblock )a x , so a x = − µk g .
Find the initial speed of the block and bullet (just after the collision).
N
x
fk
(mbullet + mblock)g
vfx 2 − vix 2 = 0 − vix 2 = 2a x ∆x = −2 µk g ∆x, so vix = 2 µk g ∆x = v.
Use conservation of momentum to find the speed of the bullet just before its collision with the block.
mbullet vbullet = (mbullet + mblock )v = (mbullet + mblock ) 2 µk g ∆x , so
vbullet =
(mbullet + mblock ) 2 µk g ∆x (2.02 kg) 2(0.400)(9.80 m s 2 )(1.50 m)
=
= 350 m s .
mbullet
0.020 kg
45. Strategy Use conservation of momentum.
Solution Find the total momentum of the two blocks after the collision.
∆p2 = −∆p1
p2f − p2i = p1i − p1f
p1f + p2f = p1i + p2i
(m1 + m2 )vf = m1v1i + m2 v2i
pf = (2.0 kg) (1.0 m s ) + (1.0 kg)(0) = 2.0 kg ⋅ m s = p1i
Since p1i was directed to the right, and pf = p1i , the total momentum of the two blocks after the collision is
2.0 kg ⋅ m s to the right .
46. Strategy The collision is perfectly inelastic, so v1f = v2f = vf . Use conservation of momentum. Let the positive
direction be the initial direction of motion.
Solution Find the speed of the man (1) just after he catches the ball (2).
m1v1f + m2 v2f = m1vf + m2 vf = m1v1i + m2 v2i = m1 (0) + m2 v2i , so
m2
0.20 kg
(25 m s) = 0.066 m s .
vf =
v =
m1 + m2 2i 75 kg + 0.20 kg
47. Strategy Use conservation of momentum. Let the positive direction be in the initial direction of motion.
Solution Find the speed of the Volkswagen after the collision.
∆pV = mV vVf − mV vVi = −∆pB = − mB ∆vB , so
vVf =
mV vVi − mB ∆vB
mV
=
(1.0 × 103 kg) ( 25 m s ) − (2.0 × 103 kg)(33 m s − 42 m s)
1.0 × 103 kg
383
= 43 m s .
Chapter 7: Linear Momentum
Physics
48. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic energy is
conserved.
Solution The 100-g ball is (1) and the 300-g ball is (2). Note that m2 = 3m1.
m1v1i + m2 v2i = m1v1i + m2 (0) = m1v1i = m1v1f + m2 v2f , so v1i = v1f +
m2
v = v1f + 3v2f .
m1 2f
1
1
1
1
1
1
1
m1v1i 2 + m2 (0)2 = m1v1i 2 = m1v1f 2 + m2 v2f 2 = m1v1f 2 + (3m1 )v2f 2 , so v1i 2 = v1f 2 + 3v2f 2 .
2
2
2
2
2
2
2
Substitute for v1i .
( v1f + 3v2f )2 = v1f 2 + 6v1f v2f + 9v2f 2 = v1f 2 + 3v2f 2 , so 6v1f v2f
= −6v2f 2 , or v1f = −v2f .
Find the final velocities of each ball.
v1i = v1f + 3v2f = −v2f + 3v2f = 2v2f , so v2f =
1
1
v1i = (5.00 m s) = 2.50 m s.
2
2
Since v1f = −v2f , v1f = −2.50 m s. So, the 300-g ball moves at 2.50 m s in the +x-direction
and the 100-g
ball moves at 2.50 m s in the − x-direction .
49. Strategy Use conservation of momentum. Let the positive direction be the initial direction of motion.
Solution Find the speed of the 5.0-kg body after the collision.
m1v1f + m2 v2f = m1v1i + m2 v2i , so
m (v − v ) + m2 v2i (1.0 kg) [10.0 m s − (−5.0 m s) ] + (5.0 kg)(0)
v2f = 1 1i 1f
=
= 3.0 m s .
m2
5.0 kg
50. Strategy The collision is perfectly inelastic, so v1f = v2f = vf . Use conservation of momentum. Let the positive
direction be the initial direction of motion.
Solution Find the speed of the combination.
m (0) + m2 v2i
3.0 kg
m1vf + m2 vf = m1v1i + m2 v2i , so vf = 1
=
(8.0 m s) = 4.8 m s .
m1 + m2
2.0 kg + 3.0 kg
51. Strategy The spring imparts the same (in magnitude) impulse to each block. (The same magnitude force is
exerted on each block by the ends of the spring for the same amount of time.) So, each block has the same final
magnitude of momentum. (The initial momentum is zero.)
Solution Find the mass of block B.
mB vB = mA vA , so
v
d ∆t
d
1.0 m
mB = A mA = A
m = Am =
(0.60 kg) = 0.20 kg .
vB
d B ∆t A d B A 3.0 m
−p
p
A
B
52. (a) Strategy Use conservation of momentum. Let the +x-direction be to the right.
Solution Find the final velocity of the other glider.
mv1f + mv2f = mvi + mvi , so v2f = vi + vi − v1f = 0.50 m s + 0.50 m s − 1.30 m s = −0.30 m s.
So, the velocity of the other glider is 0.30 m s to the left .
384
Physics
Chapter 7: Linear Momentum
(b) Strategy Form a ratio of the final to the initial kinetic energies.
Solution Compute the ratio.
1 mv 2 + 1 mv 2
Kf
v 2 + v2f 2 (1.30 m s) 2 + (0.30 m s) 2
1f
2f
2
= 2
= 1f
=
= 3.6
Ki
2vi 2
2(0.50 m s)2
2 1 mvi 2
(2
)
The final kinetic energy is greater than the initial kinetic energy. The extra kinetic energy
comes from the elastic potential energy stored in the spring.
53. Strategy The collision is perfectly inelastic, so v1f = v2f = v. The block is initially at rest, so v2i = 0 and
v1i = vi . Use conservation of momentum.
Solution Find the speed of the bullet and block system.
mbul
(mbul + mblk )v = mbul vi + mblk (0), so v =
v.
mbul + mblk i
Determine the time it takes the system to hit the floor.
1
1
2h
∆y = −h = viy ∆t − g (∆t )2 = 0 − g (∆t )2 , so ∆t =
.
2
2
g
h = 1.2 m
∆x
Find the horizontal distance traveled.
∆x = vix ∆t = v∆t =
mbul
2h
0.010 kg
2(1.2 m)
vi
=
= 0.49 m
( 400.0 m s )
mbul + mblk
g
0.010 kg + 4.0 kg
9.80 m s 2
54. Strategy Use conservation of momentum. Ki = Kf , since the collision is elastic.
Solution Show that the final speed of each object is the same as the initial speed.
m1v1f + m2 v2f = m1v1i + m2 v2i and p1i = − p2i. So, m1v1i = − m2 v2i and m1v1f + m2 v2f = 0, or m1v1f = − m2 v2f .
1
1
1
1
m1v1i 2 + m2 v2i 2 = m1v1f 2 + m2 v2f 2 , so m1v1i 2 + m2 v2i 2 = m1v1f 2 + m2 v2f 2 .
2
2
2
2
Eliminate v1i and v1f .
2
2
⎛ m
⎞
⎛ m
⎞
m1 ⎜⎜ − 2 v2i ⎟⎟ + m2 v2i 2 = m1 ⎜⎜ − 2 v2f ⎟⎟ + m2 v2f 2
⎝ m1
⎠
⎝ m1
⎠
2
⎛ m2
⎞ 2 ⎛ m22
⎞ 2
+ m2 ⎟ v2i = ⎜
+ m2 ⎟ v2f
⎜
⎜ m
⎟
⎜ m
⎟
⎝ 1
⎠
⎝ 1
⎠
v2i 2 = v2f 2
Therefore, the initial and final speeds of object 2 are the same. Eliminate v2i and v2f .
2
2
⎛ m
⎞
⎛ m
⎞
m1v1i 2 + m2 ⎜⎜ − 1 v1i ⎟⎟ = m1v1f 2 + m2 ⎜⎜ − 1 v1f ⎟⎟
⎝ m2
⎠
⎝ m2
⎠
2
2
⎛
⎞
⎛
⎞
m
m
⎜ m1 + 1 ⎟ v1i 2 = ⎜ m1 + 1 ⎟ v1f 2
⎜
⎟
⎜
m
m2 ⎟⎠
2 ⎠
⎝
⎝
v1i 2 = v1f 2
Therefore, the initial and final speeds of object 1 are the same.
385
Chapter 7: Linear Momentum
Physics
55. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic energy is
conserved as well.
Solution Let the +x-direction be in the original direction of motion of the 2.0-kg object. The 2.0-kg object is (1)
and the 6.0-kg object is (2). Note that m2 = 3m1.
m1v1i + m2 v2i = m1v1i + m2 (0) = m1v1i = m1v1f + m2 v2f , so v1i = v1f +
m2
m1
v2f = v1f + 3v2f .
1
1
1
1
1
1
1
m1v1i 2 + m2 (0)2 = m1v1i 2 = m1v1f 2 + m2 v2f 2 = m1v1f 2 + (3m1 )v2f 2 , so v1i 2 = v1f 2 + 3v2f 2 .
2
2
2
2
2
2
2
Substitute for v1i .
( v1f + 3v2f )2 = v1f 2 + 6v1f v2f + 9v2f 2 = v1f 2 + 3v2f 2 , so 6v1f v2f
= −6v2f 2 , or v1f = −v2f .
Find the final speed of the 6.0-kg object.
v1i = v1f + 3v2f = −v2f + 3v2f = 2v2f , so v2f =
1
1
v = (10 m s) = 5.0 m s .
2 1i 2
56. Strategy Look at the collision in its center of mass frame. Assume a one-dimensional collision. The initial
velocities are v1ix and v2ix . The masses are m1 and m2.
Solution Transform the initial velocities to the CM frame by subtracting vCMx from each.
v1ix′ = v1ix − vCMx
v2ix′ = v2ix − vCMx
According to the result from Problem 54, the final speeds of the objects must be the same as the initial speeds, but
the final and initial velocities are oppositely directed since the objects rebound after colliding. Therefore,
v1fx′ = −v1ix′ = −v1ix + vCMx
v2fx′ = −v2ix′ = −v2ix + vCMx
Transform back to the original frame of reference.
v1fx = v1fx′ + vCMx = −v1ix + 2vCMx
v2fx = v2fx′ + vCMx = −v2ix + 2vCMx
The relative speed after the collision is v1fx − v2fx = −v1ix + 2vCMx + v2ix − 2vCMx = −v1ix + v2ix , which is the
relative speed before the collision.
57. Strategy Use conservation of momentum. Let each of the first two pieces be 45° from the positive x-axis (one
CW, one CCW).
Solution Find the speed of the third piece.
Find v3 x .
p1x + p2 x + p3 x = mv1x + mv2 x + mv3 x = 0, so v3 x = −v1x − v2 x = −v cos 45° − v cos(−45°) = −
v
2
−
v
2
= −v 2.
Similarly,
v3 y = −v1 y − v2 y = −v sin 45° − v sin(−45°) = −
v
2
+
v
2
386
= 0, so v3 = v3 x = v 2 = (120 m s ) 2 = 170 m s .
Physics
Chapter 7: Linear Momentum
58. Strategy Use conservation of momentum.
Solution Find vBfx .
pix = MvAix = pfx = MvAfx + MvBfx , so vBfx = vAix − vAfx .
Find vBfy .
piy = 0 = pfy = MvAfy + MvBfy , so vBfy = −vAfy .
Calculate vBf .
vBf = (vAix − vAfx ) 2 + (−vAfy )2 =
( 6.0
2
2
m s − 1.0 m s ) + ( −2.0 m s ) = 5.4 m s
59. Strategy Use conservation of momentum. Refer to Practice Problem 7.11.
Solution
(a) Find the momentum change of the ball of mass m1.
⎡1
⎤
∆p1x = −∆p2 x = m2 v2ix − m2 v2fx = m2 (0 − v2fx ) = − m2 v2fx = −5m1 ⎢ vi cos(−36.9°) ⎥ = −1.00m1vi
4
⎣
⎦
⎡1
⎤
∆p1 y = −∆p2 y = m2 v2iy − m2 v2fy = 5m1 (0 − v2fy ) = −5m1v2fy = −5m1 ⎢ vi sin(−36.9°) ⎥ = 0.751m1vi
⎣4
⎦
(b) Find the momentum change of the ball of mass m2 .
∆p2 x = −∆p1x = m1 (v1ix − v1fx ) = m1 (vi − 0) = m1vi
∆p2 y = −∆p1 y = m1 (v1iy − v1fy ) = m1 (0 − v1 ) = − m1v1 = − m1 (0.751vi ) = −0.751m1vi
The momentum changes for each mass are equal and opposite.
60. Strategy Use conservation of momentum. Let right be +x and +y be in the initial direction of the puck.
y
Solution Find v2fx .
mv1fx + mv2fx = mv1ix + mv2ix = 0 + 0, so v2fx = −v1fx .
v1i
v1f
θ1 = 37°
x
Find v2fy .
mv1fy + mv2fy = mv1iy + mv2iy = v1iy + 0, so v2fy = v1iy − v1fy .
Calculate v2f .
v2f = v2fx 2 + v2fy 2 = (−v1fx )2 + (v1iy − v1fy )2 = (−v1f sin θ1 )2 + (v1i − v1f cos θ1 )2
2
2
= ⎡⎣ − ( 0.36 m s ) sin 37°⎤⎦ + ⎡⎣0.45 m s − ( 0.36 m s ) cos 37° ⎤⎦ = 0.27 m s
Calculate the direction of the second puck.
− ( 0.36 m s ) sin 37°
v
−v1fx
θ = tan −1 2fx = tan −1
= tan −1
= 53° to the left
v2fy
v1iy − v1fy
0.45 m s − ( 0.36 m s ) cos 37°
G
Thus, v 2f = 0.27 m s at 53° to the left .
387
Chapter 7: Linear Momentum
Physics
61. Strategy Use conservation of momentum.
Solution Find v2f in terms of v1f .
mv1fy + mv2fy = mv1f sin θ1 + mv2f sin θ 2 = mv1iy + mv2iy = 0 + 0, so
v2f =
− sin θ1
− sin 60.0°
v =
v = 1.73v1f .
sin θ 2 1f sin(−30.0°) 1f
y
v1i
v1f
60.0°
30.0° x
v2f
62. Strategy The collision is perfectly inelastic, so the final velocities of the blocks are identical. Use conservation of
momentum.
Solution Find the initial speed of block B.
pix = mA vAix + mB vBix = 0 + mB vBix = mB vBix = pfx = (mA + mB )vfx ,
(m + mB )vfx (mA + mB )vf cos θ
so vBix = A
=
= vBi . Compute vBi .
mB
mB
vBi =
y
vf
vBi
42.5°
(220 g + 300 g)(3.13 m s) cos(180° − 42.5°)
= −4.0 m s
300 g
N
x
vAi
Thus the initial speed of block B was 4.0 m s .
63. Strategy Use conservation of momentum. Let +x be along the initial direction of the projectile.
Solution Find the magnitude of the momentum of the target body after the collision.
Find p2fx .
∆p2 x = p2fx − p2ix = p2fx − 0 = −∆p1x = p1ix − p1fx = mvi − mvfx = m(vi − vf cos θ ), so p2fx = m(vi − vf cos θ ).
Find p2fy .
∆p2 y = p2fy − p2iy = p2fy − 0 = −∆p1 y = p1iy − p1fy = 0 − mvf sin θ , so p2fy = − mvf sin θ .
Calculate p2f .
p2f =
p2fx 2 + p2fy 2 = m (vi − vf cos θ )2 + vf 2 sin 2 θ
2
2
= (2.0 kg) ⎡⎣5.0 m s − ( 3.0 m s ) cos 60.0°⎤⎦ + ( 3.0 m s ) sin 2 60.0° = 8.7 kg ⋅ m s
388
Physics
Chapter 7: Linear Momentum
64. (a) Strategy The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation
of momentum.
Solution Let the 1500-kg car be (1) and the 1800-kg car be (2).
y
m1
pix = m1v1ix + m2 v2ix = m1v1ix + 0 = pfx = (m1 + m2 )vfx , so vfx =
v .
m1 + m2 1ix
m2
piy = m1v1iy + m2 v2iy = 0 + m2 v2iy = pfy = (m1 + m2 )vfy , so vfy =
v .
m1 + m2 2iy
N
v2i
v1i
x
Compute the final speed.
2
2
⎛ mv
⎞ ⎛ m2 v2iy ⎞
[(1500 kg)(17 m s)]2 + [(1800 kg)( −15 m s)]2
vf = vfx 2 + vfy 2 = ⎜⎜ 1 1ix ⎟⎟ + ⎜
=
⎟
⎜
⎟
1500 kg + 1800 kg
⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠
= 11 m s
Compute the direction.
vfy
(1800 kg)(−15 m s)
θ = tan −1
= tan −1
= −47°
vfx
(1500 kg)(17 m s)
Thus, the final velocity of the cars is 11 m s at 47° S of E .
(b) Strategy Find the change in kinetic energy.
Solution
1
1
1
(m + m2 )vf 2 − m1v1i 2 − m2 v2i 2
2 1
2
2
1
1
1
2
= (1500 kg + 1800 kg)(11.254 m s) − (1500 kg)(17 m s)2 − (1800 kg)(15 m s)2 = −210 kJ
2
2
2
∆K = K f − K i =
Thus, 210 kJ of the initial kinetic energy was converted to another form of energy during the collision.
65. Strategy The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation of
momentum.
Solution Let the 1700-kg car be (1) and the 1300-kg car be (2).
pix = m1v1ix + m2 v2ix = m1v1ix
m1
+ 0 = pfx = (m1 + m2 )vfx , so vfx =
v .
m1 + m2 1ix
m1v1iy + m2 v2iy
piy = m1v1iy + m2 v2iy = pfy = (m1 + m2 )vfy , so vfy =
.
m1 + m2
Compute the final speed and the direction.
⎛ mv
vf = vfx + vfy = ⎜⎜ 1 1ix
⎝ m1 + m2
2
2
2
⎞ ⎛ m1v1iy + m2 v2iy
⎟⎟ + ⎜⎜
m1 + m2
⎠ ⎝
⎞
⎟⎟
⎠
y
N
v1i
2
[(1700 kg)(14 m s) cos 45°]2 + [(1700 kg)(14 m s) sin 45° + (1300 kg)(−18 m s)]2
= 6.0 m s
1700 kg + 1300 kg
vfy
(1700 kg)(14 m s) sin 45° + (1300 kg)(−18 m s)
θ = tan −1
= tan −1
= −21°
vfx
(1700 kg)(14 m s) cos 45°
=
Thus, the final velocity of the cars is 6.0 m s at 21° S of E .
389
v2i
x
Chapter 7: Linear Momentum
Physics
66. Strategy Use conservation of momentum.
Solution Find the components of the deuteron’s velocity after the collision.
Find vdfx .
y
vi / 3
mn
vi
mn vnix + md vdix = mn vi + 0 = mn vnfx + md vdfx = 0 + md vdfx , so vdfx =
v.
md i
x
Find vdfy .
v
m
1
mn vniy + md vdiy = 0 + 0 = mn vnfy + md vdfy = mn i + md vdfy , so vdfy = −
vi n .
3
3 md
Find the components, vdfx and vdfy .
⎛m
m ⎞ ⎛ m
m
1
1
vi n ⎟⎟ = ⎜⎜ n vi , −
vi n
(vdfx , vdfy ) = ⎜⎜ n vi , −
3 md ⎠ ⎝ 2mn
3 2mn
⎝ md
⎞
vi ⎞
⎛ vi
⎟⎟ = ⎜ 2 , − 2 3 ⎟
⎝
⎠
⎠
67. Strategy Use conservation of linear momentum.
y
Solution Find vBfx .
pix = mvAix = pfx = mvAfx + mvBfx , so vBfx = vAix − vAfx .
Find vBfy .
piy = mvAiy = 0 = pfy = mvAfy + mvBfy , so vBfy = −vAfy .
vAf
vAi
60°
x
Compute the final speed of puck B.
vBf = (vBfx ) 2 + (vBfy ) 2 = (vAix − vAfx ) 2 + (−vAfy )2
= [2.0 m s − (1.0 m s) cos 60°]2 + [−(1.0 m s) sin 60°]2 = 1.7 m s
Compute the direction of puck B.
vBfy
−(1.0 m s) sin 60°
θ = tan −1
= tan −1
= −30°
2.0 m s − (1.0 m s) cos 60°
vBfx
Thus, the speed and direction of puck B after the collision is 1.7 m s at 30° below the x -axis .
390
Physics
Chapter 7: Linear Momentum
68. Strategy The collision is perfectly inelastic, so the final velocities of the acrobats are identical. Use conservation
of momentum.
Solution Let the first acrobat be (1) and the second acrobat be (2).
pix = m1v1ix + m2 v2ix = pfx = (m1 + m2 )vfx , so vfx = (m1v1ix + m2 v2ix ) (m1 + m2 ).
piy = m1v1iy + m2 v2iy = pfy = (m1 + m2 )vfy , so vfy = (m1v1iy + m2 v2iy ) (m1 + m2 ).
y
2.0 m/s
3.0 m/s
10°
20°
x
Compute the final speed.
(m1v1ix + m2 v2ix )2 + (m1v1iy + m2 v2iy ) 2
vf = vfx + vfy =
m1 + m2
2
2
[(60)(3.0 m s) cos10° + (80)(2.0 m s) cos160°]2 + [(60)(3.0 m s) sin10° + (80)(2.0 m s) sin160°]2
60 + 80
= 0.64 m s
Compute the direction.
vfy
(60)(3.0 m s) sin10° + (80)(2.0 m s) sin160°
= tan −1
= 73°
θ = tan −1
(60)(3.0 m s) cos10° + (80)(2.0 m s) cos160°
vfx
=
Thus, the final velocity of the acrobats is 0.64 m s at 73° above the +x-axis .
69. Strategy Use conservation of momentum.
Solution Let swallow 1 and its coconut be (1) and swallow 2 and its coconut be (2)
(before the collision). After the collision, let swallow 1’s coconut be (3), swallow 2’s
coconut be (4), and the tangled-up swallows be (5).
m v + m4 v4 x
pix = m1v1x + m2 v2 x = 0 + 0 = pfx = m3v3 x + m4 v4 x + m5 v5 x , so v5 x = − 3 3 x
.
m5
piy = m1v1 y + m2 v2 y = m1v1 + m2 v2 = pfy = m3v3 y + m4 v4 y + m5v5 y , so
m1v1 + m2 v2 − m3v3 y − m4 v4 y
v5 y =
.
m5
y
v2 30°
v4
N
x
10°
v3
v1
Compute the final speed of the tangled swallows, v5 .
v5 = v5x 2 + v5y 2 =
1
[− (m3v3 x + m4 v4 x )]2 + (m1v1 + m2 v2 − m3v3 y − m4 v4 y ) 2
m5
[(0.80 kg)(13 m s) cos 260° + (0.70 kg)(14 m s) cos 60°]2
+[(1.07 kg)(20 m s) + (0.92 kg)(−15 m s) − (0.80 kg)(13 m s) sin 260° − (0.70 kg)(14 m s) sin 60°]2
=
0.270 kg + 0.220 kg
= 20 m s
Compute the direction.
v5 y
θ = tan −1
v5 x
(1.07 kg)(20 m s) + (0.92 kg)(−15 m s) − (0.80 kg)(13 m s) sin 260° − (0.70 kg)(14 m s) sin 60°
= tan −1
−(0.80 kg)(13 m s) cos 260° − (0.70 kg)(14 m s) cos 60°
= −72°
Since v5 x < 0 and v5 y > 0, the velocity vector is located in the second quadrant, so the angle is
180° − 72° = 108° from the positive x-axis or 18° west of north. Thus, the velocity of the birds immediately
after the collision is 20 m s at 18° W of N .
391
Chapter 7: Linear Momentum
Physics
70. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f = v2f = vf .
Solution Find the speed of the sled once the book is on it.
m1v1f + m2 v2f = (m1 + m2 )vf = m1v1i + m2 v2i = m1vi + 0, so
m1
5.0 kg
vf =
v =
(1.0 m s) = 0.83 m s .
m1 + m2 i 5.0 kg + 1.0 kg
71. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f = v2f = vf .
Solution Find the speed of the cars just after the collision.
m1v1f + m2 v2f = (m1 + m2 )vf = m1v1i + m2 v2i = m1vi + 0, so
m1 g
13.6 kN
vf =
vi =
(17.0 m s ) = 10.2 m s .
(m1 + m2 ) g
13.6 kN + 9.0 kN
72. Strategy Use Eqs. (7-10) and (7-11).
Solution Find the velocity of the center of mass of the system.
G
G
G
G
G
p CM = Mv CM = m1v1 + m2 v 2 + m3 v3 , so
(3.0 kg)(290 m s) + (5.0 kg)(−120 m s) + (2.0 kg)(52 m s)
= 37 m s
3.0 kg + 5.0 kg + 2.0 kg
G
= 0, so v CM = 37 m s in the +x-direction .
vCMx =
vCMy
73. Strategy Use the definition of linear momentum.
Solution Find the magnitude of the total momentum of the ship and the crew.
ptot = mtot v = (2.0 × 103 kg + 4.8 × 104 kg)(1.0 × 105 m s) = 5.0 × 109 kg ⋅ m s
74. Strategy Use the definition of linear momentum and the impulse-momentum theorem.
Solution
(a) Compute the magnitude of the change in momentum of the ball.
∆p = pf − pi = mvf − mvi = m(vf − vi ) = (0.145 kg) ⎡⎣37 m s − ( −41 m s ) ⎤⎦ = 11 kg ⋅ m s
(b) Compute the impulse delivered to the ball by the bat.
Impulse = ∆p = 11 kg ⋅ m s
(c) Compute the magnitude of the average force exerted on the ball by the bat.
∆p 11.31 kg ⋅ m s
=
= 3.8 kN
Fav =
∆t
3.0 × 10−3 s
392
Physics
Chapter 7: Linear Momentum
75. Strategy Use the impulse-momentum theorem.
y
Solution Find the average force exerted by the ground on the ball.
∆p
x
Fav =
∆t
54 m/s
53 m/s
m (∆vx )2 + (∆v y ) 2
18°
22°
=
∆t
0.060 kg
2
2
=
⎡( 53 m s ) cos18° − ( 54 m s ) cos(−22°) ⎤⎦ + ⎡⎣( 53 m s ) sin18° − ( 54 m s ) sin(−22°) ⎤⎦ = 34 N
0.065 s ⎣
76. Strategy The center of each length is its center of mass. Use the component form of the definition of center of
mass.
Solution Find the location of the center of mass of the rod.
mx +mx +mx
1
1 3 2
3 3
xCM = 3
= (0 + 5.0 cm + 10.0 cm) = 5.00 cm
3
m
my +my +my
1
1 3 2 3 3
yCM = 3
= (5.0 cm + 10.0 cm + 5.0 cm) = 6.67 cm
3
m
Thus, ( xCM , yCM ) = (5.00 cm, 6.67 cm) .
77. Strategy The center of mass of each block is its center. Add up the individual center of mass components to find
the components of the center of mass of the block structure.
Solution
mx + 2mx2 + 5mx3 + 4mx4 x1 + 2 x2 + 5 x3 + 4 x4 0 + 2(1.0 in) + 5(2.0 in) + 4(3.0 in)
xCM = 1
=
=
= 2.0 in
12m
12
12
6my1 + 4my2 + my3 + my4 6 y1 + 4 y2 + y3 + y4 6(0) + 4(1.0 in) + 2.0 in + 3.0 in
yCM =
=
=
= 0.75 in
12m
12
12
9mz1 + 3mz2 9 z1 + 3 z2 9(0) + 3(1.0 in)
zCM =
=
=
= 0.25 in
12m
12
12
The center of mass of the block structure is located at (2.0 in, 0.75 in, 0.25 in).
78. Strategy Use the impulse-momentum theorem.
Solution Compute the force exerted by the stream on a person in the crowd.
∆p m∆v m
=
= ∆v = (24 kg s)(17 m s) = 410 N
Fav =
∆t
∆t
∆t
79. Strategy Use the impulse-momentum theorem.
Solution Compute the average forces imparted to the two gloved hands during the catches.
(
3
(130 km h ) 10kmm
∆p m∆v
=
= (0.14 kg)
Inexperienced: Fav =
∆t
∆t
10−3 s
Experienced: Fav = (0.14 kg)
(130
(
3
km h ) 10kmm
10 × 10−3
)(
1h
3600 s
s
393
)
)(
1h
3600 s
= 500 N
)
= 5000 N
Chapter 7: Linear Momentum
Physics
80. Strategy The fly splatters on the windshield, so the collision is perfectly inelastic (vfly, final = vcar, final = vf ).
Use conservation of momentum. Let the positive direction be along the velocity of the automobile.
Solution
(a) Compute the change in momentum.
∆pcar = −∆pfly = − mfly (vfly, f − vfly, i ) ≈ − mfly (vcar, i − 0) = −(0.1× 10−3 kg)(100 km h) = −0.01 kg ⋅ km h
So, the change in the car’s momentum due to the fly is 0.01 kg ⋅ km/h opposite the car’s motion.
(b) Compute the change in momentum.
∆pfly = −∆pcar = 0.01 kg ⋅ km h , or 0.01 kg ⋅ km h along the car’s velocity.
(c) Compute the number of flies N required to slow the car.
(1000 kg) ( −1 km h )
m ∆v
N ∆pfly = − mcar ∆vcar , so N = − car car = −
= 105 flies .
∆pfly
0.01 kg ⋅ km h
81. (a) Strategy The initial momentum of the baseball is pi = mvi . The final momentum is zero.
Solution Compute the change in momentum.
∆p = pf − pi = 0 − mvi = − mvi = −(0.15 kg)(35 m s) = −5.3 kg ⋅ m s
Thus, the change in momentum was 5.3 kg ⋅ m s opposite the ball’s direction of motion .
(b) Strategy and Solution According to the impulse-momentum theorem, the impulse applied to the ball is
equal to the change in the momentum of the ball, or 5.3 kg ⋅ m s opposite the ball’s direction of motion .
(c) Strategy Use the impulse momentum theorem.
Solution Since the acceleration is assumed constant, the time it takes for the ball to come to a complete stop
is ∆t = ∆x vav . Compute the average force applied to the ball by the catcher’s glove.
G
G
G
∆p
∆p ⎛ 35 m s ⎞ 5.25 kg ⋅ m s opposite the direction of motion
= vav
=⎜
Fav =
⎟
0.050 m
∆t
∆x ⎝ 2 ⎠
= 1.8 kN opposite the ball’s direction of motion
394
Physics
Chapter 7: Linear Momentum
82. Strategy Use conservation of momentum. The collision is perfectly elastic, so Ki = K f . Also, v1ix = v1i and
v1fy = v1f .
Solution Find the speed of the target body after the collision.
x-direction:
m1v1fx + m2 v2fx = 0 + m2 v2fx = m1v1ix + m2 v2ix = m1v1ix + 0
y-direction:
m1v1fy + m2 v2fy = m1v1iy + m2 v2iy = 0 + 0, so m2 v2fy = − m1v1fy .
y
v1f = 6.0 m/s
v1i = 8.0 m/s
x
Square the results and add.
m2
m22 v2fx 2 + m22 v2fy 2 = m22 v2f 2 = m12 v1ix 2 + m12 v1fy 2 = m12 v1i 2 + m12 v1f 2 , so m2 v2f 2 = 1 (v1i 2 + v1f 2 ).
m2
Calculate the kinetic energies.
1
1
1
1
1
m1v1f 2 + m2 v2f 2 = m1v1i 2 + m2 v2i 2 = m1v1i 2 + 0, so m1v1f 2 + m2 v2f 2 = m1v1i 2 .
2
2
2
2
2
2
v 2 + v1f 2
m
Thus, m2 v2f 2 = m1 (v1i 2 − v1f 2 ) = 1 (v1i 2 + v1f 2 ) and m2 = m1 1i
.
m2
v1i 2 − v1f 2
From the kinetic energies,
v2f =
m1
(v 2 − v1f 2 ) =
m2 1i
m1 (v1i 2 − v1f 2 )
⎛ v +v ⎞
m1 ⎜ 1i 2 1f 2 ⎟
⎝ v1i −v1f ⎠
2
2
=
(v1i 2 − v1f 2 )2
2
v1i + v1f
2
=
[(8.0 m s)2 − (6.0 m s)2 ]2
(8.0 m s)2 + (6.0 m s)2
= 2.8 m s .
83. Strategy Use conservation of momentum. Let e = electron, ν = neutrino, and n = nucleus.
Solution
(a) Find the direction of motion of the recoiling daughter nucleus.
G
G
G
pe + pν + p n = 0, so
pnx = − pex − pν x = − pex − 0 = − pe and pny = − pey − pν y = 0 − pν y = − pν ( pν < 0).
Find the angle with respect to the electron’s direction.
−p
5.00 × 10−19
= −31.4° + 180° = 148.6° CCW from the electron’s direction
θ = tan −1 ν = tan −1
− pe
−8.20 × 10−19
(b) Find the momentum of the recoiling daughter nucleus.
pn =
pnx 2 + pny 2 = (− pe )2 + (− pν )2 = (−8.20 × 10−19 kg ⋅ m s)2 + (5.00 × 10−19 kg ⋅ m s)2
= 9.60 × 10−19 kg ⋅ m s
G
Thus, p n = 9.60 × 10−19 kg ⋅ m/s in the direction found in (a) .
395
Chapter 7: Linear Momentum
Physics
84. Strategy Use conservation of momentum and the definition of center of mass. Let the pier be to the left of the
raft and woman at x = 0.
Solution
G
(a) Since ∆p CM = 0, as the woman walks toward the pier, the raft moves away from the pier, and the center of
mass does not change. So, xCM =
mw xwi + mr xri
=
mw xwf + mr xrf
.
mw + mr
mw + mr
Initially, xCM is to the right of xri . When the woman has walked to the other end of the raft, xCM is to the
left of xrf . By symmetry, the distance xCM − xri equals the distance xrf − xCM , thus
xrf − xCM = xCM − xri , so xrf = 2 xCM − xri .
The final distance of the raft from the dock, df , is equal to the difference between xrf and half its length,
3.0 m.
d f = xrf − 3.0 m = 2 xCM − xri − 3.0 m = 2 xCM − (3.0 m + 0.50 m) − 3.0 m = 2 xCM − 6.5 m
Calculate xCM .
xCM =
(60.0 kg)(6.5 m) + (120 kg)(3.5 m)
= 4.5 m, so d f = 2(4.5 m) − 6.5 m = 2.5 m .
60.0 kg + 120 kg
(b) Find the distance the woman walked relative to the pier.
∆xw = xwf − xwi = df − xwi = 2.5 m − 6.5 m = 4.0 m
85. Strategy We must determine the initial speeds of the two cars. The collision is perfectly inelastic, so the final
velocities of the cars are identical. Use conservation of momentum and the work-kinetic energy theorem.
y
Solution Let the 1100-kg car be (1) and the 1300-kg car be (2). Use the work-kinetic
energy theorem to determine the kinetic energy and, thus, the initial speed of the wrecked
vf
N
cars, which is the final speed of the collision.
v1i
30°
1
x
W = F ∆r = f k ∆r = − µk mg ∆r = ∆K = 0 − mvi 2 , so vi = 2 µk g ∆r .
v2i
2
Thus, the final speed of the collision is vf = 2 µk g ∆r .
Find the initial speeds.
pix = m1v1ix + m2 v2ix = m1v1i + 0 = pfx = (m1 + m2 )vfx , so
m + m2
m + m2
⎛ 1 km h ⎞
2400 kg
v1i = 1
vfx = 1
2 µk g ∆r cos150° =
2(0.80)(9.80 m s 2 )(17 m) cos150° ⎜
⎟
m1
m1
1100 kg
⎝ 0.2778 m s ⎠
= −110 km h .
piy = m1v1iy + m2 v2iy = 0 + m2 v2i = pfy = (m1 + m2 )vfy , so
m + m2
m + m2
⎛ 1 km h ⎞
2400 kg
v2i = 1
vfy = 1
2 µk g ∆r sin150° =
2(0.80)(9.80 m s 2 )(17 m) sin150° ⎜
⎟
m2
m2
1300 kg
⎝ 0.2778 m s ⎠
= 54 km h .
Since 110 > 70, the lighter car was speeding .
396
Physics
Chapter 7: Linear Momentum
86. Strategy Use the impulse-momentum theorem.
Solution Find the speed of the expelled gas relative to the ground.
Fav
∆p (∆m)vgas ∆m
6.0 × 104 N
Fav =
=
=
vgas , so vgas =
=
= 740 m s .
∆t
∆t
∆t
∆m ∆ t
81 kg s
87. Strategy Use the definition of linear momentum and the impulse-momentum theorem.
Solution Find the force the kinesin molecule needs to deliver in order to accelerate the organelle.
∆p m∆v (0.01× 10−15 kg)(1× 10−6 m s − 0)
Fav =
=
=
= 10−18 N
∆t
∆t
10 × 10−6 s
88. Strategy Use conservation of momentum. The collision is perfectly inelastic, so vAf = vBf = vf .
Solution Find the final speed in terms of the initial speed.
1
1
⎛1
⎞
mA vAf + mB vBf = (mA + mB )vf = ⎜ m + m ⎟ vf = mA vAi + mB vBi = mA vi + 0 = mvi , so vf = vi .
2
3
⎝2
⎠
Calculate the ratio of the final kinetic energy to the initial kinetic energy.
Kf
=
Ki
1 ( m + m )v 2
A
B f
2
1m v2
2 A i
( 1 m + m ) ( 13 vi )
= 2
1 mv 2
i
2
2
=
(
3m 1v2
2
9 i
1 mv 2
i
2
)=
1
3
89. Strategy Use conservation of energy and momentum. Let 2m = mB = 2mA .
Solution Find the maximum kinetic energy of A alone and, thus, its speed just before it strikes B.
1
∆K = mv12 − 0 = −∆U = mgh − 0, so v1 = 2 gh .
2
Use conservation of momentum to find the speed of the combined bobs just after impact. The collision is perfectly
inelastic, so vAf = vBf = v2 .
1
mA vAf + mB vBf = (m + 2m)v2 = mA vAi + mB vBi = mv1 + 0, so v2 = v1.
3
Find the maximum height.
2
1
1 ⎛1
1
⎞
2 gh ⎟ = −∆U = 0 − mgh2 , so h2 =
∆K = 0 − mv22 = − m ⎜
h .
2
2 ⎝3
9
⎠
90. Strategy The center of mass of the disk prior to drilling is ( xCM , yCM ) = (0, 0). Let S stand for the small circle
removed and L stand for the large circle that remains.
Solution Find the center of mass of the metal disk after the hole has been drilled.
m x + mS xS
xCM = L L
= 0, so
mL + mS
xL = −
mS
mL
xS = −
AS
AL
xS = −
π rS2
π rL 2 − π rS2
xS = −
(1.5 cm)2
(3.0 cm)2 − (1.5 cm)2
By symmetry, yL = yS = yCM = 0, so ( xL , yL ) = (0.50 cm, 0) .
397
(−1.5 cm) = 0.50 cm.
Chapter 7: Linear Momentum
Physics
91. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved.
Solution Find the speed of bob B immediately after the collision.
Momentum conservation:
mvAf + mvBf = mvAi + mvBi = mvAi + 0, so vBf = vAi − vAf .
Perfectly elastic collision ( Ki = K f ):
1
1
1
1
1
mv 2 + mvBf 2 = mvAi 2 + mvBi 2 = mvAi 2 + 0, so vAf 2 + vBf 2 = vAi 2 .
2 Af
2
2
2
2
Energy conservation:
1
mv 2 = mgh, so vAi = 2 gh .
2 Ai
Find vAf in terms of vAi .
vAi 2 = vAf 2 + vBf 2 = vAf 2 + (vAi − vAf )2 = vAf 2 + vAi 2 − 2vAi vAf + vAf 2 , so vAf (vAf − vAi ) = 0.
Thus, vAf = 0 or vAi . The only way vAf could equal vAi is if bob B didn’t exist, so vAf = 0.
Calculate vBf .
vBf = vAi − vAf = 2 gh − 0 = 2(9.80 m s 2 )(5.1 m) = 10 m s
92. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved. Let
the positive direction be to the right.
Solution Find the velocities of the gliders after the collision.
Momentum conservation:
mv1f + mv2f = mv1i + mv2i = mv1i + 0, so v2f = v1i − v1f .
Perfectly elastic collision due to bumpers ( Ki = K f ):
1
1
1
1
1
mv 2 + mv2f 2 = mv1i 2 + mv2i 2 = mv1i 2 + 0, so v1f 2 + v2f 2 = v1i 2 .
2 1f
2
2
2
2
Find v1f in terms of v1i .
v1f 2 + (v1i − v1f )2 = v1f 2 + v1i 2 − 2v1i v1f + v1f 2 = v1i 2 , so v1f (v1f − v1i ) = 0.
So, v1f = 0 or v1i . The only way v1f could equal v1i is if glider 2 didn’t exist, so v1f = 0.
Calculate v2f .
v2f = v1i − v1f = 0.20 m s − 0 = 0.20 m s
After the collision, glider 1 is stationary and glider 2 has a velocity of 0.20 m s in the direction of
glider 1’s initial velocity.
398
Physics
Chapter 7: Linear Momentum
93. Strategy Use conservation of momentum and Eq. (6-6) for the kinetic energies. Since the radium nucleus is at
G
G
rest, pi = p Ra = 0.
Solution
(a) Find the ratio of the speed of the alpha particle to the speed of the radon nucleus.
pf = mRn vRn + mα vα = pi = 0, so mα vα = − mRn vRn . Therefore,
vα
vRn
=
mRn 222 u 222
111
, where the negative was dropped because speed is nonnegative.
=
=
=
mα
4u
4
2
G
G
(b) Since the initial momentum is zero, p Rn = −pα ; therefore,
(c) Find the ratio of the kinetic energies.
Kα
K Rn
=
1m v 2
2 α α
1m v 2
2 Rn Rn
2
=
2
mα ⎛ vα ⎞
4 u ⎛ 111 ⎞
111
=
⎜
⎟
⎜
⎟ =
mRn ⎜⎝ vRn ⎟⎠
222 u ⎝ 2 ⎠
2
399
G
pα
p
= α = 1 .
G
pRn
p Rn
Chapter 8
TORQUE AND ANGULAR MOMENTUM
Conceptual Questions
1. To maximize the torque, locate it as far as possible from the rotation axis: along the lower edge.
2. The ease of driving a screw into a piece of wood is determined by the magnitude of the torque required to produce
the necessary downward force on the screw. The torque produced is equal to the product of the radius of the
screwdriver handle and the magnitude of the tangential force applied by the operator’s hand. Thus, the larger
diameter handle reduces the applied force required to create the necessary torque. The same amount of work is
done in driving the screw, but the task is made easier.
3. When you push near the edge, you have a larger moment arm. When you push in the middle, the moment arm is
half as much so you need to push with twice the force.
4. Of the three axes, the book has the smallest moment of inertia about the axis along the binding of the book
(axis 1). The moments of inertia about the other two axes are larger because the mass of the book is, on average,
farther from those axes.
5. For a body to be in equilibrium, both the net force and the net torque acting on it must equal zero. To satisfy the
first requirement, the two forces must be equal in magnitude and opposite in direction. To satisfy the second
requirement, the two forces must act along the same line—a net torque would otherwise act to rotate the object.
6. When the angular momentum of the main propeller changes, the body of the helicopter would suffer an opposite
change in angular momentum if no external torque about a vertical axis acts on the helicopter. The small propeller
attached to the tail of the helicopter is used to produce this external torque to keep the helicopter body from
rotating. Attaching the propeller at the tail produces the longest lever arm and therefore the greatest torque about
the vertical axis through its center of mass.
7. The total kinetic energy of a car is found by summing the translational and rotational kinetic energies of each of
the four wheels and the translational kinetic energy of the car’s body. The fraction of the car’s total kinetic energy
due to the rotation of the wheels depends on the ratio of the mass of the car’s body to the mass of the car’s wheels.
Thus, if two cars differ only in the mass of the body (while having wheels of the same mass), the more massive
car converts a greater fraction of its gravitational potential energy into translational kinetic energy—the heavier
car wins the race.
8. The force due to static friction acting on the barrel produces the torque that makes the barrel roll. If there were no
friction acting on the barrel due to the floor, the applied force would make the barrel slide along the floor without
rotating.
9. An object’s moment of inertia depends on how its mass is distributed with respect to the axis of rotation. The
farther the mass is from the axis, the greater the object’s moment of inertia. When animals have leg muscles that
are concentrated close to the hip joint, their legs have relatively small moments of inertia. This makes it easier for
them to rotate their legs, allowing them to run faster.
10. When the triceps muscle connects to the forearm as in Fig. 8.46a, the lever arm for the muscular force remains
relatively constant as the angle θ is varied. For an angle of about 90°, the lever arm in Fig. 8.46b is approximately
the same as the lever arm in the previous figure. If this angle is increased or decreased however, the lever arm
decreases significantly, resulting in a smaller torque about the elbow joint. For this reason, the situation depicted
in Fig. 8.46a is much more effective.
400
Physics
Chapter 8: Torque and Angular Momentum
11. With the forearm horizontal, the lever arms for the muscular forces on the forearm in Fig. 8.47a and Fig. 8.47b are
about the same, so the two arrangements would be about equally effective. However, for large angles (with the
arm nearly straight), the arrangement of Fig. 8.47b would have a very small lever arm and thus provide little
torque. The primary advantage of the arrangement shown in Fig. 8.47a is that the muscle is concentrated closer to
the shoulder, thereby reducing the moment of inertia of the forearm and the arm as a whole. This makes the arm
easier to move around.
G
G
12. The two forces, F12 and F21 , are not only equal and opposite; they also have the same line of action. Hence they
have the same lever arm as well. The torques they produce are therefore equal in magnitude and opposite in
direction.
13. The vertical component of the angular momentum of the system (merry-go-round and child) is conserved
throughout this process, since there are no external torques about the vertical axis of the merry-go-round. When
the child moves out to the rim, the rotational inertia of the system increases, because the child is located farther
from the axis. To conserve angular momentum, the angular velocity must therefore decrease. Noting that the
rotational kinetic energy can be written as L2 /(2 I ) and that L remains constant while I increases, we see that the
rotational kinetic energy of the system decreases.
14. The center of mass of the toy lies below the wire on which is it balancing. If the toy is pushed slightly off center,
the force of gravity acting at the center of mass produces a torque that tends to rotate the toy back toward the
center. If the center of mass were above the wire, this situation would be reversed and the toy would be unstable.
15. To knock a person over, their center of gravity must be moved until it is beyond the horizontal extent of their
support base. The force of gravity will then produce a net torque about the edge of their support base, and they
will topple over. The posture taken by defensive linemen makes them more difficult to push over because they
have a larger support base and a lower center of gravity. They must therefore be pushed (rotated) by a greater
amount to move their center of gravity beyond the edge of their support base. Four legged animals similarly have
a relatively large support base and low center of gravity compared to humans, making them naturally more stable.
Consequently, their neurological systems for maintaining balance do not need to be as complex as a human’s.
16. The location of the CG below the hips in birds makes them naturally more stable than humans. If the upper body
were displaced a little to the side, the torque produced by gravity about an axis through the hips would tend to
rotate the upper body back toward the center in birds and farther away from the center in humans.
17. The astronaut and satellite constitute an isolated system. The initial angular momentum of the system is zero.
When the astronaut tries to remove the bolt, both he and the satellite will rotate. They will rotate in opposite
directions so that the total angular momentum of the system remains zero. To put it another way, when the
astronaut applies a torque to a part of the satellite, the satellite applies an equal and opposite torque to him. The
astronaut must anchor the satellite and himself somehow before trying to remove the bolt.
18. The best place is as far from the hinge as possible so as to have the greatest possible moment arm for the torque
exerted by the stopper on the door. This way, the force required by the stopper to hold the door open will be as
small as possible, making the stopper less likely to slip on the floor.
19. Low gears are used for going uphill and high gears are used for downhill. The bicycle gears act like levers. The
energy remains the same so that the force you exert on the pedals times the distance the pedals move will equal
the force exerted on the rim of the wheel times the distance it moves. A low gear converts the force you exert at
the pedals into a lesser force that acts over a slightly longer distance, while a high gear converts the force you
exert on the pedals into a much smaller force exerted over a much longer distance. The gear ratio tells how many
times the rear bicycle wheel goes around for each time the pedals go around once. In a low gear ratio the bicycle
will go a shorter distance for each rotation of the pedals while in a high gear the bicycle goes a long distance for
each rotation of the pedals.
401
Chapter 8: Torque and Angular Momentum
Physics
20. The motion of the suspended irregular object is influenced by two external forces—the normal force from the nail
acting at its point of contact and the force due to the weight of the object acting at its center of mass. No torque is
produced by the normal force because it acts along the rotation axis. The net torque on the object is therefore
solely a result of the force from its weight. The object will rotate back and forth as determined by the direction of
the torque until frictional forces have brought its center of mass to rest directly beneath the rotation axis. At this
point, the applied force is parallel to the lever arm and no torque is produced. For each orientation of the object,
the line drawn will pass through its center of mass—the intersection of several such lines must therefore occur at
the center of mass location.
21. The melting of Earth’s polar ice caps would distribute some of its mass from locations near its rotation axis to
locations that are on average farther from its rotation axis. The rotational inertia of a sphere is greater if its mass is
distributed farther from its axis of rotation—the Earth’s moment of inertia would therefore increase. Angular
momentum conservation requires that the product of the Earth’s rotational inertia and its angular velocity be
constant. A larger moment of inertia must be accompanied by a smaller angular velocity—the melting of the caps
would therefore increase the length of the day.
Problems
1. Strategy and Solution I has units kg ⋅ m 2. ω 2 has units (rad s)2. So,
1
2
I ω 2 has units
kg ⋅ m 2 ⋅ rad 2 s 2 = kg ⋅ m 2 s 2 = J, which is a unit of energy.
2. Strategy The rotational inertia of a solid disk is I =
1
2
MR 2.
Solution Find the rotational inertial of the solid iron disk.
1
1
I = MR 2 = (49 kg)(0.200 m)2 = 0.98 kg ⋅ m 2
2
2
3. Strategy I =
2
5
MR 2 for a solid sphere and mass density is ρ = M V .
Solution
(a) M = ρV = ρ 43 π R3 for a solid sphere. Form a proportion.
3
3
M child ⎛ Rchild ⎞ ⎛ 1 ⎞
1
=⎜
=
= , so the mass is reduced by a factor of 8 .
⎟
M adult ⎝ Radult ⎠ ⎜⎝ 2 ⎟⎠
8
(b) Form a proportion.
2
2
I child 1 ⎛ Rchild ⎞
1⎛1⎞
1
= ⎜
⎟ = ⎜ ⎟ =
I adult 8 ⎝ Radult ⎠
8⎝ 2⎠
32
The rotational inertia is reduced by a factor of 32 .
402
Physics
Chapter 8: Torque and Angular Momentum
4. Strategy Use Eq. (8-2) to find the rotational inertia. Use Eq. (7-9) for the center of mass.
Solution
C
(a) I x = ∑ mi ri 2 = (200 g)(5.0 cm)2 + (300 g)(0 cm)2 + (500 g)(4.0 cm)2 = 13, 000 g ⋅ cm 2
i= A
C
(b) I y = ∑ mi ri 2 = (200 g)(3.0 cm)2 + (300 g)(6.0 cm) 2 + (500 g)(5.0 cm)2 = 25, 000 g ⋅ cm 2
i= A
C
(c) I z = ∑ mi ri 2 = (200 g)[(3.0 cm)2 + (5.0 cm)2 ] + (300 g)(6.0 cm)2 + (500 g)[(5.0 cm)2 + (4.0 cm)2 ]
i= A
= 38, 000 g ⋅ cm 2
(d) xCM =
yCM =
(200 g)( −3.0 cm) + (300 g)(6.0 cm) + (500 g)(−5.0 cm)
= −1.3 cm
200 g + 300 g + 500 g
(200 g)(5.0 cm) + (300 g)(0 cm) + (500 g)(−4.0 cm)
= −1.0 cm
200 g + 300 g + 500 g
5. Strategy Find the rotational inertia in each case by using Eq. (8-2).
Solution
(a) I = m(r 2 + 02 + 02 + r 2 ) = 2mr 2 = 2(3.0 kg)(0.50 m) 2 = 1.5 kg ⋅ m 2
(b) I = m(02 + r 2 + 02 + r 2 ) = 2mr 2 = 2(3.0 kg)(0.50 m
2)2 = 0.75 kg ⋅ m 2
(c) I = m(r 2 + r 2 + r 2 + r 2 ) = 4mr 2 = 4(3.0 kg)(0.50 m
2)2 = 1.5 kg ⋅ m 2
6. Strategy The rotational inertia of a solid disk is I =
1
2
MR 2. Use the work-kinetic energy theorem.
Solution Find the work done to spin the CD.
2
2
2
1
1⎛1
⎞ ⎡⎛ v ⎞ ⎛ v ⎞ ⎤ 1 ⎛ R ⎞
W = ∆K = I (ωf 2 − ωi 2 ) = ⎜ MR 2 ⎟ ⎢⎜ f ⎟ − ⎜ i ⎟ ⎥ = M ⎜ ⎟ (vf 2 − vi 2 )
2
2⎝2
⎠ ⎢⎣⎝ r ⎠ ⎝ r ⎠ ⎥⎦ 4 ⎝ r ⎠
2
=
1
⎡ (0.120 m) 2 ⎤
(0.0158 kg) ⎢
[(1.20 m s) 2 − 0] = 0.0512 J
⎥
4
⎣ 0.0200 m ⎦
7. Strategy I =
2
5
MR 2 for a solid sphere and I = MR 2 for the Earth about the Sun.
Solution Form a proportion.
2
2
I axis 5 MRE
2 RE 2
=
=
ISun
5Ro 2
MRo 2
So,
I axis
2 RE 2
=
, where RE is the Earth’s radius and Ro is Earth’s orbital radius about the Sun.
ISun
5 Ro 2
403
Chapter 8: Torque and Angular Momentum
Physics
8. Strategy Use Eq. (8-1) and form a proportion.
Solution Find the fraction of the total kinetic energy that is rotational.
(1 2)
( )
2 2 Iω
K rot
2
2
2
2
=
=
=
=
=
= 0.019
2
(79
kg)(0.32
m) 2
MR
K total 2 1 I ω 2 + 1 Mv 2 2 + Mv 2 2 + Mv 2
+
2
+
2
2
2
I
Iω 2
I (v 2 R 2 )
0.080 kg⋅m 2
9. (a) Strategy and Solution Since a significant fraction of the wheel’s kinetic energy is rotational, to model it as
if it were sliding without friction would be unjustified. So, the answer is no.
(b) Strategy Use Eq. (8-1) and form a proportion.
Solution Find the fraction of the total kinetic energy that is rotational.
(
)
4 12 I ω 2
K rot
1
=
=
=
Mv 2 + 1 1 +
K total 1 Mv 2 + 4 1 I ω 2
2
2
2
(
)
4Iω
1
Mv 2
4 I (v 2 R 2 )
=
1
1+
MR 2
4I
=
1
1+
(1300 kg)(0.35 m)2
4(0.705 kg⋅m 2 )
= 0.017
10. Strategy The total energy required to bring the centrifuge from rest to 420 rad/s is equal to the kinetic energy
when it rotates at ω = 420 rad/s. Use Eq. (8-1).
Solution Find the energy required to spin the centrifuge.
1
1
2
K = I ω 2 = (6.5 × 10−3 kg ⋅ m 2 ) ( 420 rad s ) = 570 J
2
2
11. Strategy Use Eq. (8-3).
Solution Find the magnitude of the torque applied to the wrench.
τ = rF⊥ = (0.16 m)(25 N) = 4.0 N ⋅ m
12. Strategy Use Eq. (8-3).
Solution Find the magnitude of the torque applied to the drum.
τ = rF⊥ = (0.0600 m)(75 N) = 4.5 N ⋅ m
13. Strategy Use Eq. (8-3).
Solution Find the magnitude of the torque.
τ = F⊥r = mgr = (40.0 kg)(9.80 N kg)(2.0 m) = 780 N ⋅ m
14. Strategy Use Eq. (8-3).
Solution Find the magnitude of the torque.
τ = F⊥r = mgr = (0.124 kg)(9.80 N kg)(0.25 m) = 0.30 N ⋅ m
15. Strategy The point of application of the force of gravity is at the geometrical center of the door, so
r⊥ = (1.0 m) 2. The force is equal to the weight of the door. Use Eq. (8-4).
Solution Find the magnitude of the torque.
τ = r⊥ F = r⊥ mg = [(1.0 m) 2](50.0 N) = 25 N ⋅ m
404
Physics
Chapter 8: Torque and Angular Momentum
16. Strategy Use Eqs. (8-3) and (8-4).
Solution
(a) The force is parallel to the lever arm at noon.
τ = Fr⊥ = F (0) = 0
(b) The torque is CCW (positive). The center of mass is (2.7 m) 2 from the axis.
τ = F⊥ r = mgr = (60.0 kg)(9.80 N kg)[(2.7 m) 2] = 790 N ⋅ m
17. Strategy Use Eq. (8-4).
Solution Find the net torque in each case.
(a) Στ = F (r2⊥ − r1⊥ ) = Fx2 − Fx1 = F ( x2 − x1) = Fd , since d = x2 − x1.
(b) Στ = F (r2⊥ − r1⊥ ) = F (r2 sin θ 2 ) − F (r1 sin θ1) = Fx2 − Fx1 = F ( x2 − x1) = Fd
18. Strategy Use Eq. (8-3) to compute the torque in each case.
Solution
(a) The force is applied perpendicularly to the door, so τ = rF = (1.26 m)(46.4 N) = 58.5 N ⋅ m .
(b) The force is applied at 43.0° from the door’s surface, so
τ = rF⊥ = rF sin θ = (1.26 m)(46.4 N) sin 43.0° = 39.9 N ⋅ m .
(c) Since the force is applied such that its line of action passes through the axis of the door hinges—the axis of
rotation—there is no perpendicular component of the force and the torque is 0 .
19. Strategy Use Eq. (8-3) to find the torque.
Solution Let the axis of rotation be a the hinge of the trap door. Since the door
is in equilibrium, the magnitude of the torque exerted on the door by the rope is
the same as that exerted by gravity. Compute the torque due to the rope.
τ = rF⊥
L
= mg cos 65.0°
2
1.65 m
(16.8 kg)(9.80 m s 2 ) cos 65.0°
=
2
= 57.4 N ⋅ m
L
mg
65.0°
20. Strategy The center of gravity is located at the center of mass.
Solution Find the center of gravity.
m x + m2 x2 + m3 x3 (5.0 kg)(0.0) + (15.0 kg)(5.0 m) + (10.0 kg)(10.0 m)
xCG = xCM = 1 1
=
= 5.83 m
M
5.0 kg + 15.0 kg + 10.0 kg
405
65.0°
Chapter 8: Torque and Angular Momentum
Physics
21. Strategy The center of gravity is located at the center of mass. Let the origin be at the center of the door.
Solution Due to symmetry, yCM = 0.
m x + m2 x2 m1(0) + m2 ( x) W2 x (5.0 N)(− 0.75 m)
xCM = 1 1
=
=
=
= − 0.012 m
M
M
W
5.0 N + 300.0 N
The center of gravity is located 1.2 cm toward the doorknob as measured from the center of the door.
22. Strategy The center of gravity is at the center of mass of the plate. Imagine that the plate consists of a rectangular
plate (on the left) and a square (on the right). The mass is proportional to the area for a uniform distribution.
Solution Find the center of gravity.
xCM =
yCM =
2
A1x1 + A2 x2 0.50s
=
A1 + A2
( 0.502 s ) + 0.502 s2 ( 0.50s + 0.502 s ) = 0.42s
(
0.50s 2 + 0.502 s 2
s
0.50s 2 (0.50s) + 0.502 s 2 0.50s + 0.50
2
0.50s 2 + 0.502 s 2
) = 0.58s
So, the center of gravity is located at (0.42s, 0.58s) .
23. Strategy Use Eqs. (8-6) and (8-4).
Solution Compute the work done on the stone.
20.0 N
W = τ ∆θ = r⊥ F ∆θ = (0.100 m)(20.0 N)(12 rev)(2π rad rev) = 150 J
10.0 cm
ω
24. (a) Strategy and Solution One revolution is equal to the circumference of the wheel, so
the rope unwinds
C = 2π r = 2π (0.500 m) = 3.14 m .
5.00 N 0.500 m
ω
(b) Strategy The work done by the rope on the wheel is equal to the force times the
distance.
Solution
W = Fd = (5.00 N)(3.14 m) = 15.7 J
(c) Strategy Use Eq. (8-4).
Solution Find the torque on the wheel due to the rope.
τ = r⊥ F = (0.500 m)(5.00 N) = 2.50 N ⋅ m
(d) Strategy and Solution There are 2π rad per revolution, so the angular displacement is
∆θ = (1.00 rev)(2π rad rev) = 6.28 rad .
(e) Strategy and Solution τ ∆θ = (2.50 N ⋅ m)(6.28 rad) = 15.7 J = W
406
Physics
Chapter 8: Torque and Angular Momentum
25. (a) Strategy Use the work-kinetic energy theorem.
Solution Find the work done spinning up the wheel.
1
1
W = ∆K = I ωf2 = (MR 2 )ωf2
2
2
1
= (182 kg)(0.62 m) 2[(120 rev min)(1 60 min s)(2π rad rev)]2 = 5.5 kJ
2
ω f = 120 rpm
0.62 m
(b) Strategy Use the equations for rotational motion with constant acceleration and the relationship between
work, torque, and angular displacement.
Solution Find the torque.
W = τ∆θ = τ (ωav ∆t ), so τ =
W
5.5 × 103 J
=
= 29 N ⋅ m .
ωav ∆t (120 rev min)(1 60 min s)(2π rad rev)(30.0 s) 2
26. (a) Strategy The rotational inertia of a hoop is MR 2. Use the work-kinetic energy theorem and Eq. (8-1).
Solution Find the work.
1
1
W = ∆K = I (ωf 2 − ωi 2 ) = (MR 2 )(ωf 2 − 0)
2
2
1
6
= (1.90 × 10 kg)(67.5 m) 2 (3.50 × 10−3 rad s)2 = 53.0 kJ
2
ω
67.5 m
(b) Strategy Constant torque implies constant angular acceleration, so ∆θ = ωav ∆t. Use Eq. (8-6).
Solution Find the torque.
⎛ ω + ωi ⎞
⎛ω +0⎞
∆t = τ ⎜ f
W = τ ∆θ = τωav ∆t = τ ⎜ f
⎟
⎟ ∆t , so
⎝ 2 ⎠
⎝ 2 ⎠
τ=
2W
2(53.0 × 103 J)
=
= 1.51 MN ⋅ m .
ωf ∆t (3.50 × 10−3 rad s)(20.0 s)
27. Strategy Choose the axis of rotation at the fulcrum. Use Eqs. (8-8).
Solution Find the force required to lift the load.
Στ = 0 = − FA cos θ (2.4 m) + Fload cos θ (1.2 m), so
1.2 m
FA =
Fload = 0.50mg = 0.50(20.0 kg)(9.80 N kg) = 98 N .
2.4 m
28. Strategy Choose the axis of rotation at the fulcrum. Use Eqs. (8-8).
Solution Find F.
Στ = 0 = − F (3.0 m) + (1200 N)(0.50 m), so F =
(1200 N)(0.50 m)
= 200 N .
3.0 m
407
Chapter 8: Torque and Angular Momentum
Physics
29. Strategy Choose the rotation axis at the edge of the base of the sculpture that is in contact with the floor as it is
tipped. The angle that the base makes with the floor is the same angle that the force due to gravity makes with the
vertical axis of the sculpture.
Solution Set the net torque equal to zero at the equilibrium point to find the maximum angle.
Στ = 0 = −mgb sin θ + mga cos θ , where b = 1.80 m and a = (1.10 m) / 2 = 0.550 m.
Solve for the angle.
a
0.550 m
= 17.0° .
b sin θ = a cos θ , so θ = tan −1 = tan −1
b
1.80 m
θ
mg
b
θ
a
30. (a) Strategy Choose the axis of rotation at the point at which the right-hand cable connects to the platform. Let
m1 = 75 kg and m2 = 20.0 kg. Let l = 5.0 m. The system is in equilibrium.
Solution Find the force exerted by the left-hand cable.
⎛l⎞
Στ = 0 = − FLl + m1g (l − d ) + m2 g ⎜ ⎟ , so
⎝2⎠
⎡ ⎛ d⎞ m ⎤
FL = g ⎢ m1 ⎜ 1 − ⎟ + 2 ⎥
l ⎠ 2 ⎦
⎣ ⎝
⎡
⎛ 2.0 m ⎞ 20.0 kg ⎤
= (9.80 N kg) ⎢(75 kg) ⎜1 −
⎥ = 540 N .
⎟+
2 ⎦
⎝ 5.0 m ⎠
⎣
l = 5.0 m
FL d = 2.0 m
FR
m 2g
m 1g
(b) Strategy Use Newton’s second law.
Solution Find the force exerted by the right-hand cable.
ΣF = 0 = −m1g − m2 g + FL + FR , so
FR = (m1 + m2 ) g − FL = (75 kg + 20.0 kg)(9.80 N kg) − 539 N = 390 N .
31. Strategy A system balances if its center of mass is above its base of support. Use Eq. (7-9) to find the center of
mass of the metersticks.
Solution Let the left end of the lowest meterstick be the origin.
mx + mx2 + mx3 + mx4 x1 + x2 + x3 + x4
xCM = 1
=
4m
4
0.5000 + (0.5000 + 0.3333) + (0.5000 + 0.3333 + 0.1667) + (0.5000 + 0.3333 + 0.1667 + 0.0833)
=
m
4
= 0.8542 m
Since the center of mass = 0.8542 m < 0.8600 m, so the system balances .
408
Physics
Chapter 8: Torque and Angular Momentum
32. Strategy Use Eqs. (8-8).
Solution Find the forces acting on the board.
Left support: Choose the axis of rotation at the top of the right support.
Στ = 0 = FL (1.2 m) − mb g (3.4 m − 2.5 m) − md g (3.4 m), so
(9.80 N kg)[(3.4 m)(55 kg + 65 kg) − (2.5 m)(55 kg)]
FL =
= 2.2 kN.
1.2 m
Since F > 0, the force is downward (CCW rotation for torque). Thus,
G
F = 2.2 kN downward .
FR
5.0 m
mbg
FL
1.2 m
mdg
3.4 m
Right support: Choose the axis of rotation at the top of the left support.
Στ = 0 = FR (1.2 m) − mb g (1.2 m + 3.4 m − 2.5 m) − md g (4.6 m), so
(9.80 N kg)[(55 kg)(2.1 m) + (65 kg)(4.6 m)]
FR =
= 3.4 kN.
1.2 m
G
Since F > 0, the force is upward (CCW rotation for torque). Thus, F = 3.4 kN upward .
33. Strategy Use Eqs. (8-8).
Solution Choose the axis of rotation at the point of contact between the driveway and the ladder.
ΣFx = 0 = f − N w , so f = N w .
cosθ ⎡
⎛ 15 m ⎞ ⎤
⎛ 3.0 m ⎞
(5.0 m) cos θ , so N w =
Στ = 0 = N w (4.7 m) − Wl (2.5 m) cos θ − Wp ⎜
⎢Wl (2.5 m) + Wp ⎜
⎟⎥ .
⎟
4.7 m ⎣
⎝ 4.7 ⎠ ⎦
⎝ 4.7 m ⎠
Find θ .
4.7
4.7 m = (5.0 m) sin θ , so θ = sin −1
.
5.0
Calculate f.
4.7
cos sin −1 5.0
⎡
⎛ 15 m ⎞ ⎤
f = Nw =
⎢(120 N)(2.5 m) + (680 N) ⎜
⎟ ⎥ = 180 N
4.7 m
⎝ 4.7 ⎠ ⎦
⎣
So, the force of friction is 180 N toward the wall .
34. Strategy Use Eqs. (8-8).
Solution
(a) Choose the axis of rotation at the point of contact between the vertical wall and the climber’s feet.
(0.91 m)Wc
(0.91 m)(770 N)
Στ = T cos θ (1.06 m) − Wc (0.91 m) = 0, so T =
=
= 730 N .
(1.06 m)cosθ (1.06 m) cos 25°
(b) ΣFx = 0 = Fx − T sin θ and ΣFy = 0 = Fy + T cos θ − Wc , so Fx = T sin θ and Fy = Wc − T cos θ .
Find the magnitude of the force.
F = Fx 2 + Fy 2 = T 2 sin 2 θ + Wc 2 + T 2 cos 2 θ − 2WcT cos θ = T 2 + Wc 2 − 2WcT cos θ
= (730 N)2 + (770 N)2 − 2(770 N)(730 N) cos 25° = 330 N
Find the direction.
Fy
W − T cos θ
770 N − (730 N) cos 25°
θ = tan −1
= tan −1 c
= tan −1
= 19°
Fx
T sin θ
(730 N) sin 25°
G
Thus, F = 330 N at 19° above the horizontal .
409
Chapter 8: Torque and Angular Momentum
Physics
35. Strategy Use Eqs. (8-8).
Solution Choose the axis of rotation at the hinge.
Στ = 0 = T (2.38 m) sin 35° − (80.0 N)(1.50 m) − (120.0 N)(3.00 m), so
(80.0 N)(1.50 m) + (120.0 N)(3.00 m)
T=
= 350 N .
(2.38 m) sin 35°
Find Fx and Fy .
ΣFx = 0 = −T cos 35° + Fx and ΣFy = 0 = Fy + T sin 35° − 80.0 N − 120.0 N, so
Fx = T cos 35° = (350 N) cos 35° = 290 N and Fy = −(351.6 N) sin 35° + 80.0 N + 120.0 N = −2 N .
The magnitude of Fy is small compared to that of Fx and T.
36. Strategy Use Eqs. (8-8).
Solution Choose the axis of rotation at the hinge.
l
mg 2 + W
.
Στ = 0 = Wl cos θ − Tl sin θ + mg cos θ , so T =
2
tan θ
For θ = 0, T → ∞, and for θ = 90°, T → 0.
37. Strategy Use Eqs. (8-8). Choose the axis of rotation at the point where the beam meets the store.
Solution The tension in the cable cannot exceed 417 N. Sum the torques.
Στ = 0 = T sin θ (1.50 m) − (50.0 N)(0.75 m) − (200.0 N)(1.00 m)
Solve for θ and substitute 417 N (the breaking strength) for T.
(50.0 N)(0.75 m) + (200.0 N)(1.00 m)
θ = sin −1
= 22.3°
(417 N)(1.50 m)
1.50 m θ
0.75 m
T
50.0 N
1.00 m
200.0 N
The minimum angle is 22.3° .
38. Strategy Use the results from Problem 37.
Solution We must add one term (for the cat) and substitute for the angle in the summation of the torques. Let d
be the distance between the store and the center of mass of the cat.
Στ = 0 = (417 N) sin 33.8°(1.50 m) − (50.0 N)(0.75 m) − (200.0 N)(1.00 m) − (8.7 kg)(9.80 m s2 )d , so
(417 N) sin 33.8°(1.50 m) − (50.0 N)(0.75 m) − (200.0 N)(1.00 m)
d=
= 1.3 m .
(8.7 kg)(9.80 m s 2 )
39. Strategy Use Eqs. (8-8). Choose the axis of rotation at the point of contact between the floor and the man’s feet.
Solution Find the forces exerted by the floor.
Palms:
mg (1.00 m) (68 kg)(9.80 N kg)(1.00 m)
Στ = 0 = − F (1.70 m) + mg (1.00 m), so F =
=
= 390 N .
1.70 m
1.70 m
Feet:
ΣFy = 0 = Fp + Ff − mg , so Ff = mg − Fp = (68 kg)(9.80 N kg) − 392 N = 270 N .
410
Physics
Chapter 8: Torque and Angular Momentum
40. Strategy Assuming Fb is (nearly) straight down, Fs is simply equal to the magnitude of the sum of the forces
due to gravity on your friend and the package.
Solution Find Fs .
Fs = Mg + mg = ( M + m) g = (55 kg + 10 kg)(9.80 N kg) = 640 N .
41. Strategy Use Eqs. (8-8). Choose the axis of rotation at the point of contact of the normal force.
Solution Find the tension in the Achilles tendon, FA .
Στ = 0 = − FA (4.60 cm + 12.8 cm) + FT (12.8 cm) and ΣFy = 0 = N + FA − FT , or FT = N + FA , so
(12.8 cm)(750 N)
= 2100 N.
4.6 cm
Find the force that the tibia exerts on the ankle joint, FT .
12.8
FT = N + FA = 750 N +
(750 N) = 2800 N
4.6
− FA (17.4 cm) + ( N + FA )(12.8 cm) = 0, or FA =
The forces are: tendon, 2100 N upward and tibia, 2800 N downward .
42. Strategy Use Eqs. (8-8). Choose the axis of rotation at the shoulder joint. One arm supports half of the person’s
weight, so Fp = 12 (700 N) = 350 N.
Solution Find the force each muscle exerts.
Στ = 0 = Fm (12 cm) sin15° − Fg (27.5 cm) − Fp (60 cm), so
Fm =
Fg (27.5 cm) + Fp (60 cm)
(12 cm) sin15°
=
(30.0 N)(27.5 cm) + (350 N)(60 cm)
= 7.0 kN .
(12 cm) sin15°
43. Strategy Use Eqs. (8-8). Choose the axis of rotation at the elbow.
Solution Find the force exerted by the biceps muscle.
Στ = 0 = −Wm (35.0 cm) − Wa (16.5 cm) + Fb (5.00 cm) sin θ , so
Fb =
Wm (35.0 cm) + Wa (16.5 cm) (9.9 N)(35.0 cm) + (18.0 N)(16.5 cm)
=
= 130 N .
30.0 cm
(5.00 cm) sin θ
(5.00 cm)
2
2
(30.0 cm) + (5.00 cm)
44. Strategy Use Eqs. (8-8). Choose the axis of rotation at the knee.
Solution Find the forces exerted by the patellar tendon.
(a) Στ = 0 = Fp (10.0 cm) sin 20.0° − Fw (41 cm) sin 30.0° − FL (22 cm) sin 30.0°, so
g sin 30.0°[mw (41 cm) + mL (22 cm)] (9.80 N kg) sin 30.0°[(3.0 kg)(41 cm) + (5.0 kg)(22 cm)]
=
(10.0 cm) sin 20.0°
(10.0 cm) sin 20.0°
= 330 N .
Fp =
(b) Στ = 0 = Fq (10.0 cm) sin 20.0° − Fw (41 cm) sin 90.0° − FL (22 cm) sin 90.0°, so
Fq =
g[mw (41 cm) + mL (22 cm)] (9.80 N kg)[(3.0 kg)(41 cm) + (5.0 kg)(22 cm)]
=
= 670 N .
(10.0 cm) sin 20.0°
(10.0 cm) sin 20.0°
411
Chapter 8: Torque and Angular Momentum
Physics
45. Strategy Refer to Figure 8.32. First find the magnitude of the force exerted by the back Fb by analyzing the
torques about an axis at the sacrum; then, find the horizontal component of the extreme force on the sacrum Fs .
Use Eqs. (8-8).
Solution Sum the torques to find Fb .
Στ = 0 = Fb (44 cm) sin12° − (10 kg)(9.80 m s 2 )(76 cm) − (55 kg)(9.80 m s 2 )(38 cm), so
(10 kg)(9.80 m s 2 )(76 cm) + (55 kg)(9.80 m s 2 )(38 cm)
= 3053 N.
(44 cm) sin12°
The only forces with components in the horizontal direction are those due to the back and the sacrum. Find the
horizontal component of the extreme force, Fsx .
Fb =
ΣFx = 0 = Fsx − Fb cos12°, so Fsx = Fb cos12° = (3053 N) cos12° = 3.0 kN .
(3053 N) cos12°
= 5.5, so the force is about 5.5 times larger than that from his torso alone!
540 N
46. Strategy Use Eqs. (8-8).
Solution
(a) The torque exerted by the erector spinae muscles must be equal in magnitude and opposite in direction to the
torque due to the mass of the upper body and the 60.0-kg mass.
Στ = xCMWub + mgx = (0.38 m)(455 N) + (60.0 kg)(9.80 N kg)(0.76 m) = 620 N ⋅ m
(b) τ = ( Fb sin θ )d where d = 0.44 m, Fb is the magnitude of the force due to the erector spinae muscles,
θ = 12°, and τ = 620 N ⋅ m.
τ
620 N ⋅ m
=
= 6800 N
Fb =
d sin θ (0.44 m) sin12°
The force exerted by the erector spinae muscles is 6800 N at 12° above the horizontal.
(c) The component of the force that compresses the spinal column is Fb cos θ = (6780 N) cos12° = 6600 N .
47. Strategy and Solution Torque has units N ⋅ m = kg ⋅ m ⋅ s−2 ⋅ m = kg ⋅ m 2 ⋅ s−2. Inertia times angular acceleration
has units kg ⋅ m 2 ⋅ s −2 = N ⋅ m. Thus, the units are consistent.
48. Strategy Use the rotational form of Newton’s second law.
Solution Find the frictional torque.
∆ω
0 − 20.0 rad s
Στ = I α = I
= (400.0 kg ⋅ m 2 )
= −26.7 N ⋅ m
∆t
300.0 s
The torque is 26.7 N ⋅ m opposite the flywheel’s rotation.
412
Physics
Chapter 8: Torque and Angular Momentum
49. Strategy Use the rotational form of Newton’s second law and Eq. (5-21).
Solution Find the torque that the motor must deliver.
I=
1
2
MR 2 for a uniform disk, so
(
)
2
0.305 m
2⎞
⎛ 2
( 3.49 rad s )
1
MR 2ωf 2 (0.22 kg)
2
2 ωf − ωi
Στ = I α = MR ⎜
=
= 0.0012 N ⋅ m .
⎟=
⎜ 2∆θ ⎟
2
4∆θ
4(2.0 rev)(2π rad rev)
⎝
⎠
2
50. Strategy The rotational inertia of the gear is I = 9.20 × 10−2 kg ⋅ m 2 and α = ∆ω ∆t . Let r = 15.0 cm, the length
of each spout, and let F be the force per spout. Use the rotational form of Newton’s second law.
Solution Find F.
Στ = 3Fr = I α = I
∆ω
I ∆ω (9.20 × 10−2 kg ⋅ m 2 )(2.2 rev s) ⎛ 2π rad ⎞
=
, so F =
⎜
⎟ = 0.88 N .
3r ∆t
3(0.150 m)(3.20 s)
∆t
⎝ 1 rev ⎠
51. Strategy The rotational inertia of the gear is I = 12 MR 2 and α = ∆ω ∆t . Use the rotational form of Newton’s
second law.
72.5 N
Solution Find the total frictional torque.
0.650 m
1.35 rev/s
1
∆ω
2
Στ = TR − τ frictional = I α = MR
, so
2
∆t
1
∆ω
τ frictional = TR − MR 2
2
∆t
1
1.35 rev s ⎛ 2π rad ⎞
= (72.5 N)(0.650 m) − (40.6 kg)(0.650 m)2
⎜
⎟ = 4.3 N ⋅ m .
2
1.70 s ⎝ 1 rev ⎠
52. Strategy Use the rotational form of Newton’s second law and the definition of rotational inertia.
Solution Find the torque required to cause the angular acceleration.
D
I = ∑ mi ri 2 = (m A + mB + mC + mD )r 2 , since all four masses are (0.75 m)/2 from the axis.
i= A
Στ = I α = (4.0 kg + 3.0 kg + 5.0 kg + 2.0 kg)[(0.75 m) 2]2 (0.75 rad s2 ) = 1.5 N ⋅ m
53. Strategy The rotational inertia of the wheel is I = MR 2. Use the rotational form of Newton’s second law.
Solution Find the magnitude of the average torque.
∆ω
⎛ 4.00 rev s ⎞ ⎛ 2π rad ⎞
Στ av = I α = MR 2
= (2 kg)(0.30 m)2 ⎜
⎟ ⎜ rev ⎟ = 0.09 N ⋅ m
∆t
⎝ 50 s
⎠⎝
⎠
413
4.00 rev/s
0.30 m
Chapter 8: Torque and Angular Momentum
Physics
54. (a) Strategy The rotational inertia of the merry-go-round is I =
1
2
MR 2 and that of the children is I = 2 MR 2 .
Use the rotational form of Newton’s second law.
Solution Find the torque on the merry-go-round.
⎛1
⎞ ∆ω
Στ = I α = ⎜ MR 2 + 2mR 2 ⎟
⎝2
⎠ ∆t
⎡1
⎤ ⎛ 25 rpm ⎞⎛ 2π rad ⎞ ⎛ 1 min ⎞
2
= ⎢ (350.0 kg)(1.25 m) + 2(30.0 kg)(1.25 m)2 ⎥ ⎜
⎟⎜
⎟⎜
⎟ = 48 N ⋅ m
2
⎣
⎦ ⎝ 20.0 s ⎠⎝ rev ⎠ ⎝ 60 s ⎠
(b) Strategy Let F be the magnitude of the tangential force with which each child
must push the rim.
Solution Find F.
FR + FR = Στ , so F =
F
1.25 m
ω
Στ
48 N ⋅ m
=
= 19 N .
2 R 2(1.25 m)
F
55. Strategy The rotational inertia is I = 12 MR 2. Use the rotational form of Newton’s second law and Eq. (5-18).
Solution
(a) α =
Στ FR + FR 4 F
4(10.0 N)
=
=
=
= 0.11 rad s 2
2
1
I
MR (180 kg)(2.0 m)
MR
2
(b) ωf = ωi + α ∆t = 0 + (0.11 rad s 2 )(4.0 s) = 0.44 rad s
10.0 N 2.0 m
ω
10.0 N
56. (a) Strategy and Solution This is just the relation between tangential acceleration and angular acceleration,
a = Rα .
(b) Strategy Use Eqs. (8-8).
Solution Find the net torque on the pulley about its axis of rotation.
Στ = T1R − T2 R = (T1 − T2 ) R
R
T1
T2
The motion is CCW, so Στ = (T1 − T2 ) R CCW .
(c) Strategy and Solution If m1 = m2 , T1 = T2 , so Στ = 0.
If m1 ≠ m2 , the blocks accelerate, so the pulley has an angular acceleration. Since a nonzero net torque is
required for the pulley to accelerate, T1 − T2 ≠ 0, thus T1 ≠ T2 .
414
Physics
Chapter 8: Torque and Angular Momentum
(d) Strategy The rotational inertia of a pulley is I = 12 MR 2 . Use Eqs. (8-8) and (8-9).
R
Solution Find the magnitudes of the tensions.
m1a = m1g − T1 ⇒ T1 = m1( g − a) and m2a = T2 − m2 g ⇒ T2 = m2 ( g + a) .
so a =
T2
T1
Find a.
Στ = (T1 − T2 ) R = (m1g − m1a − m2 g − m2a) R = I α =
T2
T1
a
1
a 1
MR 2 = MRa,
2
R 2
m1
m2
a
m 2g
m1g
(m1 − m2 ) g
.
M +m +m
1
2
2
(e) Strategy Use the result for the speed from Example 8.2.
Solution Check the answer for a.
2(m1 − m2 ) gh
. Find a.
From Example 8.2, v =
m1 + m2 + I R 2
2a y ∆y = 2ah = vfy 2 − viy 2 = v 2 − 0 =
2(m1 − m2 ) gh
2
, so a =
(m1 − m2 ) g
m1 + m2 + I R
m1 + m2 + I R 2
1
(m1 − m2 ) g
(m1 − m2 ) g
Now, I = MR 2 , so a =
=
.
M
2
2
1
2
m
m1 + m2 + 2 MR R
1 + m2 + 2
.
The expression for a is the same as that found in part (d).
57. Strategy Follow the steps to derive the rotational from of Newton’s second law.
Solution
(a) According to Newton’s second law, Fi = mi ai , so ai = Fi mi .
(b) The torque is the product of the perpendicular component of the force and the shortest distance between the
rotation axis and the point of application of the force, so τ i = Fi ri = mi ai ri .
(c) The tangential acceleration is related to the angular acceleration by ai = riα , so τ i = mi (riα )ri = mi ri 2α .
(d) Summing the torques and using the definition of rotational inertia, we have
N
N
⎛N
⎞
∑ τ i = ∑ mi ri 2α = ⎜ ∑ mi ri 2 ⎟ α = I α .
i =1
i =1
⎝ i =1
⎠
58. Strategy The rotational inertia of a uniform solid sphere is 52 MR 2 . Use the expression for the acceleration found
in Example 8.13.
Solution Find the acceleration of the solid sphere.
g sin θ
g sin θ
g sin θ (9.80 m s 2 ) sin 35°
aCM =
=
=
=
1+ 2 5
1 + I ( MR 2 ) 1 + 2 MR 2 ( MR 2 ) 1 + 2 5
5
35°
Mg
35°
= 4.0 m s 2
415
M g sin 35°
Chapter 8: Torque and Angular Momentum
Physics
59. Strategy Use conservation of energy. The rotational inertia of a uniform solid sphere is
2 MR 2 .
5
Solution Find the speed of the sphere.
1
1
1
1⎛2
⎞⎛ v ⎞
Mv 2 + I ω 2 − 0 = Mv 2 + ⎜ MR 2 ⎟⎜ ⎟
2
2
2
2⎝5
⎠⎝ R ⎠
1
1
7
= Mv 2 + Mv 2 = Mv 2 = −∆U = Mgh, so
2
5
10
2
∆K =
v=
60 cm
30°
10
10
gh =
(9.80 m s 2 )(0.60 m) = 2.9 m s .
7
7
60. Strategy Use Eqs. (6-6) and (8-1).
Solution Find the total kinetic energies of each object.
Solid sphere:
2
1 2 1 2 1 2 1 ⎛ 2 2 ⎞⎛ v ⎞
1
1
7
mv + I ω = mv + ⎜ mr ⎟ ⎜ ⎟ = mv 2 + mv 2 = mv 2
2
2
2
2⎝5
2
5
10
⎠⎝ r ⎠
Solid cylinder:
K tr + K rot =
2
1 2 1 2 1 2 1 ⎛ 1 2 ⎞⎛ v ⎞
1
1
3
mv + I ω = mv + ⎜ mr ⎟ ⎜ ⎟ = mv 2 + mv 2 = mv 2
2
2
2
2⎝2
2
4
4
r
⎠⎝ ⎠
Hollow cylinder:
K tr + K rot =
2
1 2 1 2 1 2 1 2⎛v ⎞
1
1
mv + I ω = mv + mr ⎜ ⎟ = mv 2 + mv 2 = mv 2
2
2
2
2
2
2
⎝r⎠
In order from smallest to largest, the total kinetic energies are
7
3
solid sphere: K = mv 2 ; solid cylinder: K = mv 2 ; hollow cylinder: K = mv 2 .
10
4
K tr + K rot =
61. Strategy The sphere is rolling on a horizontal surface, so its total energy is equal to its total kinetic energy. Use
conservation of energy.
Solution Compute the total energy.
2
Etotal = K tr + K rot =
=
1 2 1 2 1 2 1 ⎛ 2 2 ⎞⎛ v ⎞
1
1
7
mv + I ω = mv + ⎜ mr ⎟⎜ ⎟ = mv 2 + mv 2 = mv 2
2
2
2
2⎝5
2
5
10
⎠⎝ r ⎠
7
(0.600 kg)(5.00 m s)2 = 10.5 J
10
Find the height achieved by the sphere.
K
10.5 J
∆U = mgh = −∆K = K , so h =
=
= 1.79 m .
mg (0.600 kg)(9.80 N kg)
5.00 m/s
h
30°
416
Physics
Chapter 8: Torque and Angular Momentum
62. (a) Strategy Use conservation of energy. The rotational inertia of a uniform solid cylinder is
Solution Let h = 0.80 m, m be the mass of the bucket, and M be the mass of the cylinder.
The tangential speed of the cylinder is the same as the linear speed of the bucket, since
they are attached by a rope.
2
∆K =
1 2 1 2 1 2 1⎛1
1
1
⎞⎛ v ⎞
mv + I ω = mv + ⎜ MR 2 ⎟⎜ ⎟ = mv 2 + Mv 2 = −∆U = mgh, so
2
2
2
2⎝2
2
4
⎠⎝ R ⎠
v=
4mgh
=
2m + M
1 MR 2 .
2
R
M
T
4(2.0 kg)(9.80 m s 2 )(0.80 m)
= 3.0 m s .
2(2.0 kg) + 3.0 kg
mg
(b) Strategy Use the work-kinetic energy theorem.
Solution Find the tension T in the rope as the bucket falls a distance h.
1
Wtotal = ∆K = mv 2 = Wrope + Wgrav = −Th + mgh, so
2
⎛
⎡
8.96 m 2 s 2 ⎤
v2 ⎞
T = m ⎜ g − ⎟ = (2.0 kg) ⎢9.80 m s 2 −
⎥ = 8.4 N .
⎜
2h ⎟⎠
2(0.80 m) ⎦⎥
⎝
⎣⎢
(c) Strategy Use Newton’s second law.
Solution Find the acceleration of the bucket as it falls.
T
8.4 N
ΣFy = T − mg = ma y , so a y = − g + = −9.80 m s 2 +
= −5.6 m s 2 , or 5.6 m s 2 down .
m
2.0 kg
63. Strategy Let h = 17.0 m, m be the mass of the bucket, and M be the mass of the cylinder. The tangential speed of
the cylinder is the same as the linear speed of the bucket, since they are attached by a rope. Use conservation of
energy. The rotational inertia of a uniform solid cylinder is 12 MR 2 .
Solution Find the speed of the bucket when it reaches the bottom of the well.
2
1 2 1 2 1 2 1⎛1
1
1
4mgh
⎞⎛ v ⎞
mv + I ω = mv + ⎜ MR 2 ⎟ ⎜ ⎟ = mv 2 + Mv 2 = −∆U = mgh, so v =
.
2
2
2
2⎝2
R
2
4
2
m+M
⎠⎝ ⎠
Compute how long it will take for the bucket to fall to the bottom of the well.
1
∆y = h = (vfy + viy )∆t , so
2
h(2m + M )
2h
2h
(17.0 m)[2(1.10 kg) + 2.60 kg]
∆t =
=
=
=
= 2.75 s .
4 mgh
vfy + viy
mg
(1.10 kg)(9.80 m s 2 )
+0
∆K =
2m + M
417
Chapter 8: Torque and Angular Momentum
Physics
64. (a) Strategy and Solution The drilled cylinder takes more time because it converts a larger fraction of its
potential energy to rotational kinetic energy and a smaller fraction to translational kinetic energy than the
solid cylinder; the drilled cylinder takes more time because its rotational inertia is larger .
(b) Strategy Use conservation of energy and the result for the acceleration from Example 8.13.
Solution Find the speeds of the solid and drilled cylinders.
Solid cylinder:
Let m = the mass of the solid cylinder; its rotational inertia is Is = mR 2 2, where R is the radius. Let h be the
vertical height of the incline and vs be the speed of the solid cylinder at the bottom. From conservation of
energy,
2
1
1
1
1⎛1
1
3
gh
⎞⎛ v ⎞ 1
mgh = K tr + K rot = mvs2 + I sω 2 = mvs2 + ⎜ mR 2 ⎟ ⎜ s2 ⎟ = mvs2 + mvs2 = mvs2 , so vs = 2
.
⎜
⎟
2
2
2
2⎝2
4
4
3
⎠⎝ R ⎠ 2
Drilled cylinder:
Let m′ be the mass of the drilled cylinder. It has inner radius b = R 2 and outer radius a = R. From Table 8.1,
its rotational inertia is
2
1
1 ⎡
⎛R⎞ ⎤ 5
I d = m′(a 2 + b 2 ) = m′ ⎢ R 2 + ⎜ ⎟ ⎥ = m′R 2
2
2 ⎢
⎝ 2 ⎠ ⎥⎦ 8
⎣
Let vd be the speed of the drilled cylinder at the bottom. From conservation of energy,
m′gh =
2
1
1⎛5
5
13
gh
⎞⎛ v ⎞ 1
m′vd2 + ⎜ m′R 2 ⎟ ⎜ d2 ⎟ = m′vd2 + m′vd2 = m′vd2 , so vd = 4
.
⎜
⎟
2
2⎝8
2
16
16
13
⎠⎝ R ⎠
The ratio of the times to move down the incline is the inverse ratio of the final speeds. Why? Both move with
constant acceleration starting from rest, so their average velocities are one half of their final velocities. They
move the same distance—call it d—along the incline, so
1
1
d = vd ∆td = vs ∆ts
2
2
Then,
2 gh 3
∆td vs
1 13
=
=
=
≈ 1.0408
∆ts vd 4 gh 13 2 3
The time for the drilled cylinder to roll down the incline is 4.08% longer than that for the solid cylinder.
418
Physics
Chapter 8: Torque and Angular Momentum
65. (a) Strategy Use conservation of energy and the relationship between speed and radial acceleration.
Solution At the top of the loop, the sphere’s speed must be at least the speed that results in a radial
acceleration of g.
v2
= g , so v 2 = gr.
r
1
1
The sphere’s kinetic energy is mv 2 = mgr, and it must equal the potential energy difference
2
2
1
5
mgh − mg (2r ). Thus, r = h − 2r or h = r .
2
2
(b) Strategy The rotational inertia of a uniform solid sphere is
2 mr 2 .
5
Use conservation of energy.
Solution Find the kinetic energy of the sphere.
1
1⎛2
7
⎞ ⎛ v2 ⎞ 7
K = mv 2 + ⎜ mr 2 ⎟ ⎜ 2 ⎟ = mv 2 = mgr
⎜
⎟
2
2⎝5
10
10
⎠⎝ r ⎠
Find h.
7
27
∆K = mgr = −∆U = mgh − mg (2r ), so h =
r .
10
10
66. Strategy Use conservation of energy and the relationship between speed and radial acceleration.
Solution At the top of the loop, the cylinder’s speed must be at least the speed that results in a radial
acceleration of g.
v2
= g , so v 2 = gr.
r
2
1 2 1 2 1 2 1 2⎛v⎞
mv + I ω = mv + mr ⎜ ⎟ = mv 2 = mgr , and it must equal the
2
2
2
2
⎝r⎠
potential energy difference mgh − mg (2r ).
The cylinder’s kinetic energy is
Thus, mgr = mgh − 2mgr = mg (h − 2r ), so h = 3r .
67. Strategy Consider the rotational inertia of each object. Use conservation of energy and the relationship between
speed and radial acceleration.
Solution Since Isphere = 52 mr 2 < mr 2 = I hollow cylinder ,
h will decrease. The smaller the rotational inertia, the less gravitational energy will go into rotational energy,
and the more will go into translational energy.
Redo the calculation with the solid sphere. At the top of the loop, the sphere’s speed must be at least the speed
that results in a radial acceleration of g.
2
v2
1
1
1
1
7
7
⎛v⎞
= g , so v 2 = gr. Thus, its kinetic energy is mv 2 + I ω 2 = mv 2 + mr 2 ⎜ ⎟ = mv 2 = mgr , and the
r
2
2
2
5
10
10
⎝r⎠
kinetic energy must equal the potential energy difference mgh − mg (2r ). Find h.
7
7
mgr = mgh − 2mgr , so
r = h − 2r , or h = 2.7 r.
10
10
Problem 67 had a minimum of h = 3r. With a solid sphere, the minimum is h = 2.7r , which is a little less
than 3r.
419
Chapter 8: Torque and Angular Momentum
Physics
68. (a) Strategy Let r1 = 0.00500 m and r2 = 0.0200 m. The tangential speed of the axle and the speed of the yo-yo
are the same. Use conservation of energy.
Solution Find the speed of the yo-yo.
⎛ 2 ⎞ 1 2 1 ⎛ r2 ⎞2 2
1 2 1 2 1 2 1⎛1
2⎞ v
∆K = mv + I ω = mv + ⎜ mr2 ⎟ ⎜ 2 ⎟ = mv + m ⎜ ⎟ v = −∆U = mgh, so
2
2
2
2⎝2
4 ⎝ r1 ⎠
⎠ ⎝⎜ r1 ⎠⎟ 2
v=
4 gh
2 + (r2 r1)
2
=
4(9.80 m s 2 )(1.00 m)
2+
(
0.0200 m
0.00500 m
)
2
= 1.5 m s .
(b) Strategy Assume constant acceleration.
Solution Find the time is takes the yo-yo to fall.
1
v
2∆y 2(1.00 m)
∆y = vav ∆t = (vfy + viy )∆t = ∆t , so ∆t =
=
= 1.36 s .
2
2
v
1.476 m s
69. Strategy The rotational inertia of a uniform disk is I = 12 MR 2 . Use Eq. (8-14).
Solution Find the magnitude of the angular momentum of the turntable.
1
1
L = I ω = MR 2ω = (5.00 kg)(0.100 m) 2 (0.550 rev s)(2π rad rev) = 0.0864 kg ⋅ m 2 s
2
2
70. Strategy The rotational inertia of a uniform solid sphere is I = 52 MR 2 . Use Eq. (8-14).
Solution Find the magnitude of the angular momentum of the Earth.
2
2
⎛ 1 rev ⎞⎛ 2π rad ⎞ ⎛ 1 h ⎞
33
2
L = I ω = MR 2ω = (5.97 × 1024 kg)(6.37 × 106 m)2 ⎜
⎟⎜
⎟⎜
⎟ = 7.0 × 10 kg ⋅ m s
5
5
⎝ 24 h ⎠⎝ rev ⎠ ⎝ 3600 s ⎠
71. Strategy The rotational inertia of a hoop is I = MR 2 . Use Eq. (8-14).
Solution Find the magnitude of the angular momentum of the flywheel.
⎛ 2π rad ⎞⎛ 1 min ⎞
7
2
L = I ω = MR 2ω = (5.6 × 104 kg)(2.6 m)2 (350 rpm) ⎜
⎟⎜
⎟ = 1.4 × 10 kg ⋅ m s
⎝ rev ⎠⎝ 60 s ⎠
72. Strategy Since the torque is constant, it is equal to the change in angular momentum divided by the time interval.
Solution Find the applied torque.
∆L 115 kg ⋅ m 2 s − 240 kg ⋅ m 2 s
τ=
=
= −50 N ⋅ m.
∆t
2.5 s
The torque applied is 50 N ⋅ m opposite the rotation of the wheel .
73. Strategy Since the torque is constant, it is equal to the change in angular momentum divided by the time interval.
Solution Find the time to stop the spinning wheel
∆L
∆L −6.40 kg ⋅ m 2 s
τ=
, so ∆t =
=
= 1.60 s .
∆t
−4.00 N ⋅ m
τ
420
Physics
Chapter 8: Torque and Angular Momentum
74. Strategy Use conservation of angular momentum and Eq. (8-14).
Solution Find the skater’s new rate of rotation.
I
1
Li = I iωi = Lf = I f ωf , so ωf = i ωi =
(1.0 rev s) = 1.5 rev s .
If
0.67
75. Strategy Use conservation of angular momentum and Eq. (8-14).
Solution Find the skater’s final angular velocity.
I
2.50
Li = I iωi = Lf = I f ωf , so ωf = i ωi =
(10.0 rad s) = 15.6 rad s .
If
1.60
76. Strategy The initial rotational inertia is Ii = 12 MR 2 , and the final rotational inertia is I f = 12 MR 2 + mr 2 , where
M is the mass of the disk, m is the mass of the clod of clay, R is the radius of the disk, and r is the distance from
the center of the disk (axis of rotation) to the center of the clod. Use conservation of angular momentum.
Solution Solve for the final angular speed.
1
⎛1
⎞
Li = I iωi = MR 2ωi = Lf = I f ωf = ⎜ MR 2 + mr 2 ⎟ ωf , so
2
⎝2
⎠
ωf =
1 MR 2
2
1 MR 2 + mr 2
2
⎛ 2mr 2
ωi = ⎜⎜1 +
MR 2
⎝
−1
⎞
⎡ 2(0.12 g)(0.0800 m)2 ⎤
⎟⎟ ωi = ⎢1 +
⎥
(0.80 kg)(0.170 m)2 ⎦⎥
⎠
⎣⎢
ωf
−1
(18.0 Hz)
17.0 cm
8.00 cm
= 16.9 Hz .
77. Strategy The rotational inertias of the wheel and guinea pig are I w = MR 2 and I g = mR 2 , respectively, where M
is the mass of the wheel, m is the mass of the guinea pig, and R is the radius of the wheel. Use conservation of
angular momentum and v = rω.
Solution Find the angular velocity of the wheel.
Lw = Lg , so I w ωw = MR 2ωw = I gωg = mR 2ωg = mRvg .
Thus, ωw =
mvg
MR
=
(0.500 kg)(0.200 m s)
= 0.125 rad s .
(2.00 kg)(0.400 m)
78. Strategy Use conservation of angular momentum and Eq. (8-14).
Solution Find the diver’s initial angular velocity.
I
1 ⎛ 2.00 rev ⎞ ⎛ 2π rad ⎞
= 3.15 rad s .
Li = I iωi = Lf = I f ωf , so ωi = f ωf =
Ii
3.00 ⎜⎝ 1.33 s ⎟⎠ ⎜⎝ rev ⎟⎠
421
Chapter 8: Torque and Angular Momentum
Physics
79. Strategy Use Eqs. (5-2), and (8-14).
Solution
(a) Find the time elapsed during the dive in the tuck position.
1
1
2h
∆y = −h = viy ∆t − g (∆t )2 = 0 − g (∆t )2 , so ∆t =
.
2
2
g
Find the number of turns (revolutions).
L 2h
⎛ 1 rev ⎞
⎛ 1 rev ⎞ 106 kg ⋅ m 2 s 2(10.0 m)
∆θ = ω ∆ t =
∆
θ
=
= 3.0 .
, so ⎜
⎟
⎜ 2π rad ⎟
I g
9.80 m s 2
⎝ 2π rad ⎠
⎝
⎠ 8.0 kg ⋅ m 2
(b) Find the number of turns during the dive in the pike position.
⎛ 1 rev ⎞ 106 kg ⋅ m 2 s 2(10.0 m)
= 1.6
⎜ 2π rad ⎟
⎝
⎠ 15.5 kg ⋅ m 2 9.80 m s 2
80. Strategy The initial rotational inertia is Ii = 12 MR 2 , and the final rotational inertia is I f = 12 MR 2 + ( Fg g ) R 2 ,
where M is the mass of the merry-go-round, Fg is the weight of the child, R is the radius of the merry-go-round
and the distance from the center of the merry-go-round (axis of rotation) to the child. Use conservation of angular
momentum.
Solution Solve for the final angular speed.
Fg 2 ⎞
⎛1
1
Li = I iωi = MR 2ωi = Lf = I f ωf = ⎜ MR 2 +
R ⎟ ωf , so
⎜2
⎟
2
g
⎝
⎠
−1
−1
1 MR 2
⎛ 2 Fg ⎞
⎡
⎤
2(180 N)
2
=
+
=
+
1
1
ωf =
ω
ω
⎜
⎟
⎥ (0.75 rev s) = 0.61 rev s .
i ⎜
i ⎢
2
gM ⎟⎠
1 MR 2 + Fg R 2
⎣⎢ (9.80 m s )(160 kg) ⎦⎥
⎝
2
g
Compute the change in rotational kinetic energy.
Fg 2 ⎞ 2 1 ⎛ 1
1
1
1⎛1
⎞
∆K r = I f ωf 2 − Iiωi 2 = ⎜ MR 2 +
R ⎟ ωf − ⎜ MR 2 ⎟ ωi 2
⎟
2
2
2 ⎜⎝ 2
2
2
g
⎝
⎠
⎠
2
2
2
2
⎧
1 ⎪⎡ 1
(180 N)(2.0 m) ⎤ ⎛
rev ⎞ ⎡ 1
rev ⎞ ⎪⎫ ⎛ 2π rad ⎞
⎤⎛
0.61
= ⎨ ⎢ (160 kg)(2.0 m) 2 +
− ⎢ (160 kg)(2.0 m) 2 ⎥ ⎜ 0.75
⎥
⎬
⎜
⎟
⎟
⎜
⎟
2 ⎪ ⎣⎢ 2
s ⎠ ⎣2
s ⎠ ⎪ ⎝ rev ⎠
⎦⎝
9.80 m s 2 ⎥⎦ ⎝
⎩
⎭
= −660 J
422
Physics
Chapter 8: Torque and Angular Momentum
81. Strategy The average torque is equal to the magnitude of the change in angular momentum divided by the time
interval.
G
G
Solution Let Li = L in the +y-direction. Then ∆L has components ∆Lx = L sin θ and
∆Ly = L cos θ − L = L(cos θ − 1). So,
G
∆L = ( L sin θ )2 + [ L(cos θ − 1)]2 = L sin 2 60.0° + (cos 60.0° −1)2 = 1.00L.
Compute the magnitude of the required torque.
G
2
1
∆L 1.00 L 1.00I ω 2 mr ω
τ=
=
=
=
∆t
∆t
∆t
∆t
=
60.0°
y
Li
∆L
60.0°
x
60.0°
60.0°
60.0°
Lf
(1.00 × 105 kg)(2.00 m)2 (300.0 rpm) ⎛ 2π rad ⎞⎛ 1 min ⎞
6
⎜ rev ⎟⎜ 60 s ⎟ = 2.10 × 10 N ⋅ m
2(3.00 s)
⎝
⎠⎝
⎠
82. Strategy Consider how the angular momentum of the rotating disk affects the motion of the ship.
Solution
The disk should rotate in a horizontal plane so that the angular momentum vector is vertical. This does not
make it difficult to steer; the ship can change direction without affecting the direction of the angular
momentum.
83. Strategy The rotational inertia of the Moon is I = mr 2 . Use conservation of angular momentum and v = ω r.
Solution Find the ratio of the Moon’s orbital speed at perigee to that at apogee.
vp
vp ra 4.07 × 105 km
v
I aωa = mra 2 a = mra va = I pωp = mrp 2
= mrpvp , so
= =
= 1.14 .
ra
rp
va rp 3.56 × 105 km
84. Strategy The rotational inertia of each blade (uniform rod) is I = 13 ML2 , where L is the length of each blade.
Find the angular acceleration of the fan using the definition; and use Eq. (8-9) to find the torque applied to the fan
by the motor.
Solution The angular acceleration is α = ∆ω ∆t . Find the torque.
⎛1
⎞ ∆ω 4 ML2 ∆ω 4(0.35 kg)(0.60 m) 2 (1.8 rev s) ⎛ 2π rad ⎞
Στ = I α = 4 ⎜ ML2 ⎟
=
=
⎜
⎟ = 0.44 N ⋅ m
3∆t
3(4.35 s)
⎝3
⎠ ∆t
⎝ rev ⎠
85. Strategy Use Eq. (8-3). The force due to the weight is mg.
Solution Find the torque.
τ = F⊥ r = mgr = (10.0 kg)(9.80 N kg)(1.0 m) = 98 N ⋅ m
86. Strategy The rotational inertia of the rod is I = 13 mL2 . Use conservation of energy.
Solution Find the speed of the lower end of the uniform rod when moving at its lowest point.
2
∆K = K rot =
1 2 1 ⎛ 1 2 ⎞⎛ v ⎞
1
L
I ω = ⎜ mL ⎟ ⎜ ⎟ = mv 2 = −∆U = mgh = mg , so v =
2
2⎝3
6
2
⎠⎝ L ⎠
423
3gL .
Chapter 8: Torque and Angular Momentum
87. Strategy The rotational inertia of the gymnast is
Physics
1
m(2r )2 , where r = 1.0 m. Use conservation of energy.
3
Solution Find the angular speed at the bottom of the swing.
1
1 ⎡1
2
⎤
∆K = I ω 2 = ⎢ m(2r )2 ⎥ ω 2 = mr 2ω 2 = −∆U = mg (2r ) = 2rmg , so
2
2 ⎣3
3
⎦
ω=
3g
=
r
3(9.80 m s 2 )
= 5.4 rad s .
1.0 m
88. Strategy Use Eqs. (8-8). Choose the axis of rotation at the hinge attaching the crane to the cab (the pivot).
Solution Find T1.
Στ = 0 = T2 (12.2 m) sin10.0° + T1(12.2 m) sin 5.0° − (18 kN)(6.1 m) sin 40.0° − (67 kN)(12.2 m) sin 40.0° and
∑ Fy = T1 − 67 kN = 0, so T1 = 67 kN .
Find T2 .
[(18 kN)(6.1 m) + (67 kN)(12.2 m)]sin 40.0° − (67 kN)(12.2 m) sin 5.0°
, so T2 = 250 kN .
(12.2 m) sin10.0°
At the pivot:
∑ Fy = Fpy − 18 kN − 67 kN − T1 cos 45.0° − T2 cos 50.0° = 0, so
Fpy = 18 kN + 67 kN + (247.7 kN) cos 50.0° + (67 kN) cos 45.0° = 291.6 kN.
∑ Fx = Fpx − T1 sin 45.0° − T2 sin 50.0° = 0, so
Fpx = (247.7 kN) sin 50.0° + (67 kN) sin 45.0° = 237.1 kN.
T2 =
Find the magnitude.
Fp = (237.1 kN)2 + (291.6 kN) 2 = 380 kN
Find the direction.
291.6
θ = tan −1
= 51°
237.1
G
So, Fp = 380 kN at 51° with the horizontal .
424
Physics
Chapter 8: Torque and Angular Momentum
89. Strategy Use conservation of energy.
Solution Find the final speeds of each object.
Solid sphere:
2
∆K = K tr + K rot =
1 2 1 2 1 2 1 ⎛ 2 2 ⎞⎛ v ⎞
1
1
7
mv + I ω = mv + ⎜ mr ⎟ ⎜ ⎟ = mv 2 + mv 2 = mv 2 = −∆U = U i − U f = mgh,
2
2
2
2⎝5
r
2
5
10
⎠⎝ ⎠
10 gh
.
7
Hollow sphere:
so v =
2
6 gh
.
5
2
4 gh
.
3
1 2 1 ⎛ 2 2 ⎞⎛ v ⎞
1
1
5
mv + ⎜ mr ⎟⎜ ⎟ = mv 2 + mv 2 = mv 2 = mgh, so v =
2
2⎝3
r
2
3
6
⎠⎝ ⎠
Solid cylinder:
1 2 1 ⎛ 1 2 ⎞⎛ v ⎞
1
1
3
mv + ⎜ mr ⎟ ⎜ ⎟ = mv 2 + mv 2 = mv 2 = mgh, so v =
2
2⎝2
r
2
4
4
⎠⎝ ⎠
Hollow cylinder:
2
1 2 1
1
1
⎛v⎞
mv + (mr 2 ) ⎜ ⎟ = mv 2 + mv 2 = mv 2 = mgh, so v = gh .
2
2
r
2
2
⎝ ⎠
Cube:
1 2
mv = mgh, so v = 2 gh .
2
So, vcube > vsolid sphere > vsolid cylinder > vhollow sphere > vhollow cylinder .
The objects reach the bottom in the following order from first to last: cube, solid sphere, solid cylinder, hollow
sphere, and hollow cylinder.
90. (a) Strategy The rotational inertia of a uniform solid cylinder is I = 12 mr 2 . Use conservation of energy.
Solution Find the speed v of the cylinder after it has fallen a height h.
2
1 2 1 2
1
1⎛1
3
⎞⎛ v ⎞
mv + I ω − 0 = mv 2 + ⎜ mr 2 ⎟⎜ ⎟ = mv 2 = −∆U = mgh, so
2
2
2
2⎝2
4
⎠⎝ r ⎠
4
2
2
v 2 = gh = 2a y ∆y = 2a y (0 − h), or a y = − g = − (9.80 m s 2 ) = −6.53 m s 2 .
3
3
3
∆K =
The acceleration of the cylinder is 6.53 m s 2 down .
(b) Strategy Use Newton’s second law.
Solution The cords each pull upward on the cylinder with tension T.
1
1
ΣFy = 2T − mg = ma y , so T = m(a y + g ) = (2.6 kg)(−6.533 m s 2 + 9.80 m s 2 ) = 4.2 N .
2
2
425
Chapter 8: Torque and Angular Momentum
Physics
91. Strategy Compute the net torque on the piece of uniform metal.
Solution Let the length of the piece of metal be L.
L
mg
(53.0 kg)(9.80 m s 2 )
Στ = 0 = kx sin θ L − mg cos θ , so x =
=
= 0.792 m .
2
2k tan θ 2(275 N m) tan 50.0°
92. Strategy Let the subscripts be 1 for the painter, 2 for the can, and 3 for the plank.
Solution
(a) Choose the axis of rotation at the point of contact between the plank and the right sawhorse.
m d − m2d 2
Στ = 0 = m3 gd3 − m1gd1 − m2 gd 2 , so d1 = 3 3
.
m1
The distance from the right-hand edge is 1.40 m − d1 = d .
m d − m2d 2
(20.0 kg)(3.00 m − 1.40 m) − (4.0 kg)(1.40 m − 0.14 m)
d = 1.40 m − 3 3
= 1.40 m −
61 kg
m1
= 0.96 m from the RH edge
(b) Choose the axis of rotation at the point of contact between the plank and the left sawhorse.
m d + m3d3
.
Στ = 0 = m1gd1 − m2 gd 2 − m3 gd3, so d1 = 2 2
m1
The distance from the left-hand edge is 1.40 m − d1 = d .
m d + m3d3
(4.0 kg)(6.00 m − 1.40 m − 0.14 m) + (20.0 kg)(1.60 m)
d = 1.40 m − 2 2
= 1.40 m −
61 kg
m1
= 0.58 m from the LH edge
93. (a) Strategy The rotational inertia of a uniform solid disk is I = 12 MR 2 .
Solution Compute the rotational inertia.
1
1
I = MR 2 = (200.0 kg)(0.40 m) 2 = 16 kg ⋅ m 2
2
2
(b) Strategy Use Eq. (8-1).
Solution Compute the initial rotational kinetic energy.
1
1
K rot = I ω 2 = (16 kg ⋅ m 2 )(3160 rad s) 2 = 8.0 × 107 J
2
2
(c) Strategy and Solution The ratio of the rotational to the translational kinetic energies is
K rot
K
2(8.0 × 107 J)
= rot2 =
= 320 .
1 mv
K tr
(1000.0 kg)(22.4 m s)2
2
(d) Strategy Set the work done by air resistance equal to the stored energy in the flywheel.
Solution Find the distance d the car can travel.
K
8.0 × 107 J
Fd = K rot , so d = rot =
= 120 km .
670.0 N
F
426
Physics
Chapter 8: Torque and Angular Momentum
94. (a) Strategy The rotational inertia of a uniform solid sphere is I = 52 MR 2 . Use Eq. (8-1).
Solution Find the kinetic energy of the Earth.
2
K rot =
2
1 2 1⎛2
1
⎞
⎛ 2π rad ⎞ ⎛ 1 h ⎞
29
I ω = ⎜ MR 2 ⎟ ω 2 = (5.974 × 1024 kg)(6.371× 106 m) 2 ⎜
⎟ ⎜
⎟ = 2.6 × 10 J
2
2⎝5
5
⎠
⎝ 24 h ⎠ ⎝ 3600 s ⎠
(b) Strategy and Solution T =
2π
ω
and K rot ∝ ω 2 , so ω ∝ K rot and
Tf ωi
=
=
Ti ωf
Ki
. The change in the
Kf
⎛ Ki
⎞
⎛
⎞
1
⎛ 60 min ⎞
− 1⎟ Ti = ⎜
− 1 (24 h) ⎜
period is Tf − Ti = ⎜
⎟ = 7 min.
⎜ 0.990 ⎟⎟
⎜ K
⎟
⎝ 1h ⎠
f
⎝
⎠
⎝
⎠
The length of the day would increase by 7 minutes.
(c) Strategy Divide 1.0% of the Earth’s rotational kinetic energy by the world’s energy usage.
Solution One percent of the Earth’s rotational kinetic energy would supply the world’s energy needs (at
0.010(2.6 × 1029 J)
today’s usage) for
= 2.6 million years .
1.0 × 1021 J yr
95. Strategy Refer to the figure. Use Eqs. (7-9) and (8-4).
Solution
(a) τ i = Fi r⊥i = mi gri cos θ i = xi mi g
If ∑ ximi g > 0, then the system rotates CW ( τ < 0 ), and if ∑ ximi g < 0, then the system rotates CCW
( τ > 0 ). Therefore τ i = − ximi g.
⎛ Σx m ⎞
(b) Since τ i = − ximi g and the center of gravity is at ( xCG , yCG ), ∑ τ i = − ∑ xi mi g = − g ⎜ i i ⎟ M = − xCG Mg.
⎝ M ⎠
(c) − ∑ xi mi g = − xCG Mg
∑ xi mi g
= xCG g
M
xCM = xCG
427
Chapter 8: Torque and Angular Momentum
Physics
96. (a) Strategy The rotational inertia of a uniform disk is I = 12 MR 2 .
Solution Find the radius.
I=
1
MR 2 , so R =
2
2I
=
M
2(4.55 × 106 kg ⋅ m 2 )
7.27 × 105 kg
= 3.54 m .
(b) Strategy The rotational inertia of a hollow cylinder is I = MR 2 .
Solution Find the radius.
I = MR 2 , so R =
I
=
M
4.55 × 106 kg ⋅ m 2
7.27 × 105 kg
= 2.50 m .
(c) Strategy Use the definition of average power, the work-kinetic energy theorem, and Eq. (8-1).
Solution The rate at which the energy of the flywheel is decreased is
2 1
2
1
W ∆K 2 I ωf − 2 I ωi
I
(ωf 2 − ωi 2 ).
=
=
=
Pav =
2∆t
∆t
∆t
∆t
The average power supplied is − Pav .
2
2
I
4.55 × 106 kg ⋅ m 2
⎛ 2π rad ⎞ ⎛ 1 min ⎞
8
(ωi 2 − ωf 2 ) =
[(386 rpm)2 − (252 rpm)2 ] ⎜
⎟ ⎜ 60 s ⎟ = 4.3 × 10 W
2∆t
2(5.00 s)
rev
⎝
⎠ ⎝
⎠
97. Strategy The system is in equilibrium. Use Eqs. (8-8).
Solution Find h.
h
= tan 75°, so h = [(1.26 m) 2] tan 75° = (0.630 m) tan 75°.
(1.26 m) 2
At the top of the ladder, each leg exerts a horizontal force on the other. These forces are equal in magnitude and
opposite in direction, since the system is in equilibrium. Let the magnitude of this force be F. The tension T in the
rope is directed to the left at the connection point on the right leg, so for the right leg, we have
ΣFx = F − T = 0 or T = F .
Calculate the torque about the contact point of the right leg of the ladder and the ground.
(0.630 m)mg
(0.630 m)mg
mg
Στ = (0.630 m)mg − Fh = 0, so T = F =
=
=
.
h
(0.630 m) tan 75° tan 75°
The tension in the rope is the same along its length, so
mg
(42 kg)(9.80 N kg)
=
= 110 N .
Trope =
tan 75°
tan 75°
428
Physics
Chapter 8: Torque and Angular Momentum
98. Strategy The system is in equilibrium. Choose the axis of rotation at the point of contact between the ladder and
the floor.
Solution
(a) Find the vertical force.
Στ = 0 = F (6.0 m) cos 60.0° − mg (4.0 m) cos 60.0°, so
4.0
4.0
F=
mg =
(15 kg)(9.80 N kg) = 98 N .
6.0
6.0
(b)
F
2.0 m
8.0 m
This does not help the person trying to lift the ladder, since the torque
problem is not alleviated by exerting a force at the point of rotation.
mg
60.0°
99. (a) Strategy The rotational inertia of a uniform thin rod is I = 13 ML2.
Solution Compute the rotational inertia of the limb.
1
1
I = ML2 = (0.0280 kg)(0.0380 m) 2 = 1.35 × 10−5 kg ⋅ m 2
3
3
(b) Strategy Use Eq. (8-9).
Solution Compute the muscular force required to achieve the blow.
1
∆ω
(0.0280 kg)(0.0380 m) 2 (175 rad s)
Στ = Fr = I α = ML2
, so F =
= 524 N .
3
∆t
3(1.50 × 10−3 s)(3.00 × 10−3 m)
100. Strategy Use Eqs. (8-8) and (8-9). Let a x = a = − a y .
Solution For the two blocks, we have ∑ Fx = T1 = m1a x = m1a and ∑ Fy = 0;
∑ Fx = 0 and ∑ Fy = T2 − m2 g = m2a y = −m2a, so T2 = m2 g − m2a.
For the pulley, we have ∑ τ = −T1R + T2 R = I α = I
a
Ia
, so T1 − T2 = − 2 .
R
R
Find the acceleration of the blocks.
Ia
I ⎞
m2 g
⎛
.
T1 − T2 = m1a + m2a − m2 g = − 2 , so ⎜ m1 + m2 + 2 ⎟ a = m2 g or a =
R
R ⎠
m1 + m2 + I R 2
⎝
101. Strategy The rotational inertial of a uniform disk is I = 12 MR 2 . Use Eq. (8-14).
Solution Find the magnitude of the angular momentum of the disk.
1
1
L = I ω = MR 2ω = (2.0 kg)(0.100 m) 2 (3.0 rev s)(2π rad rev) = 0.19 kg ⋅ m 2 s
2
2
429
Chapter 8: Torque and Angular Momentum
Physics
102. (a) Strategy Since the hoop started at rest, the final angular velocity is twice the average angular velocity.
Solution Find the angular velocity of the hoop when it arrives at the bottom of the inclined plane.
v
4π (10.0 m)
⎛ 2π ⎞ ∆x
ωf = 2ωav = 2 av = 2 ⎜ ⎟
=
= 6.28 rad s
(2.00
m)(10.0 s)
∆
r
C
t
⎝
⎠
(b) Strategy The rotational inertia of a hoop is I = MR 2 . Use Eq. (8-14).
Solution Find the angular momentum of the hoop when it reaches the bottom of the incline.
2
2
⎛ C ⎞
⎛ 2.00 m ⎞
2
ω = (1.50 kg) ⎜
L = I ω = MR 2ω = M ⎜
⎟
⎟ (2π rad s) = 0.955 kg ⋅ m s
⎝ 2π ⎠
⎝ 2π ⎠
(c) Strategy Consider the forces acting on the hoop.
Solution The gravitational force acts on the hoop in the direction parallel to the line between the axis of the
hoop and the point of contact between the rim of the hoop and the inclined plane, so it supplies no torque.
The force of friction acts at the rim of the hoop, perpendicularly to the line between the axis of the hoop and
the point of contact between the rim of the hoop and the inclined plane; therefore, it is the force of friction
that supplied the net torque.
(d) Strategy The average torque on the hoop is equal to the change in angular momentum of the hoop divided by
the time interval of the change.
Solution Find the force of friction.
C
∆L
2π ∆L 2π (0.955 kg ⋅ m 2 s)
= τ av = fr = f
, so f =
=
= 0.300 N .
C ∆t
∆t
2π
(2.00 m)(10.0 s)
103. Strategy Since the mass is concentrated at the tip, I = MR 2 . Use Eq. (8-14).
Solution Compute the angular momenta of the second and hour hands of the clock.
⎛ 1 rev ⎞ ⎛ 2π rad ⎞
−4
2
(a) L = I ω = MR 2ω = (0.10 kg)(0.300 m)2 ⎜
⎟⎜
⎟ = 9.4 × 10 kg ⋅ m s
⎝ 60 s ⎠ ⎝ rev ⎠
⎛ 1 rev ⎞⎛ 1 h ⎞⎛ 2π rad ⎞
−6
2
(b) L = (0.20 kg)(0.200 m)2 ⎜
⎟⎜
⎟⎜
⎟ = 1.2 × 10 kg ⋅ m s
⎝ 12 h ⎠⎝ 3600 s ⎠⎝ rev ⎠
430
Physics
Chapter 8: Torque and Angular Momentum
104. (a) Strategy and Solution τ = F⊥ r = (0)r = 0, since the force due to gravity is parallel to the radial distance
between the planet and the Sun.
(b) Strategy The rotational inertial of a planet is I = mr 2 . Use Eq. (8-14).
Solution
L = I ω = mr 2ω
(c) Strategy and Solution If ∆t is small, the area swept out is approximately
A=
1
1
1
rv∆t = r (rω )∆t = r 2ω ∆t .
2
2
2
(Since ∆t is small, the area is approximately a triangle with height equal to r and base equal to v∆t.)
A 1 2
A
= r ω , and L = mr 2ω is constant, so r 2ω is constant. Thus,
, the area swept
∆t
∆t 2
out per unit time, is constant.
(d) Strategy and Solution
105. Strategy The system is in equilibrium. Choose the axis of rotation at the ankle.
Solution Find the force that each calf muscle needs to exert while the woman
is standing.
Στ = 0 = 2 F (4.4 cm) sin 81° − mg (3.0 cm), so
mg (3.0 cm)
(68 kg)(9.80 N kg)(3.0 cm)
F=
=
= 230 N .
2(4.4 cm) sin 81°
2(4.4 cm) sin 81°
3.0 cm
mg
4.4 cm
81°
Ankle
2F
106. (a) Strategy Use conservation of angular momentum and Eq. (8-14).
Solution Calculate the angular velocity after the child moves out to the rim of the merry-go-round.
Iiωi
Li = Iiωi = Lf = I f ωf = ( I i + I child )ωf = ( I i + mR 2 )ωf , so ωf =
.
Ii + mR 2
(b) Strategy Use Eqs. (8-1) and (8-14).
Solution Calculate the rotational kinetic energy and angular momentum before and after.
Before:
1
K rot = I iωi 2 and L = I iωi .
2
After:
2
K rot
⎛ I iωi ⎞
1
1
1 I i 2ωi 2
= I ω 2 = ( I i + mR 2 ) ⎜
=
and L = I iωi .
⎟
⎜ I + mR 2 ⎟
2
2
2 I i + mR 2
⎝ i
⎠
431
Chapter 8: Torque and Angular Momentum
Physics
107. (a) Strategy The weight is equal to the change in the combined readings of the scales.
Solution Compute the student’s weight.
W = 394.0 N + 541.0 N − 100.0 N − 100.0 N = 735.0 N
(b) Strategy The system is in equilibrium. Choose the axis of rotation at the point of contact between the plank
and scale B.
Solution Find x1.
( )
FA L − mp g L2
(2.2 m) ⎡⎣394.0 N − 12 (200.0 N) ⎤⎦
⎛L⎞
Στ = 0 = ms gx1 − FA L + mp g ⎜ ⎟ , so x1 =
=
= 0.88 m .
mg
735.0 N
⎝2⎠
(c) Strategy The height of the student is h = 1.60 m.
Solution Find the height y of the student’s center of gravity.
h
h
y = x1 = (0.88 m)
= 0.55h
1.60 m
h
108. (a) Strategy Use Eq. (8-9) and Newton’s second law. Let the +x-direction be down the plane.
Solution Find the tension in the thread.
Ia
Ia
Στ = Tr = I α = CM , so T = CM .
R
rR
Find the spool’s acceleration.
Ia
−T + mg sin θ = − CM + mg sin θ = maCM , so
rR
I ⎞
g sin θ
⎛
or aCM =
.
g sin θ = aCM ⎜1 +
⎟
I
1 + mrR
⎝ mrR ⎠
T
θ
x
mg
mg sin θ
θ
The spool spins and moves down the incline with aCM =
g sin θ
.
I
1 + mrR
(b) Strategy Use Eqs. (8-8).
Solution Find the magnitude and direction of the frictional force.
∑ Fx = mg sin θ − fs − T = 0 and ∑ Fy = N − mg cos θ = 0.
Choose the axis of rotation at the axis of the spool.
mg sin θ
Στ = 0 = fs R − Tr = fs R − (mg sin θ − fs )r , so fs =
.
1+ R r
The force of friction is
mg sin θ
up the incline .
1+ R r
T
N
y
fs
θ
mg
x
θ
f
mg sin θ
tan θ
(c) Strategy and Solution µs, min N = fs , so µs, min = s =
=
.
N mg cos θ (1 + R r )
1+ R r
432
mg sin θ
Physics
Chapter 8: Torque and Angular Momentum
109. Strategy Since the bike travels with constant velocity, the acceleration is zero and Στ = 0.
Solution Find the magnitude of the force with which the chain pulls.
r
Στ = 0 = fr2 − FCr1, so FC = 2 f = 6.0(3.8 N) = 23 N .
r1
110. Strategy Use conservation of energy.
Solution
(a) Find the speed with which the roustabout reaches the ground.
1
∆K = mv 2 = −∆U = mgL, so v = 2 gL .
2
(b) Find the speed with which the roustabout reaches the ground.
2
∆K =
(c) Since
1 2 1⎛1
1
L
⎞⎛ v ⎞
I ω = ⎜ ML2 ⎟⎜ ⎟ = Mv 2 = −∆U = Mg , so v =
2
2⎝3
6
2
⎠⎝ L ⎠
3gL .
2 gL < 3gL , the roustabout should jump.
111. Strategy Use conservation of angular momentum.
Solution Find the new rate of rotation.
Lf = I f ωf = Li = I iωi , so
ωf =
Ii
(2.40 kg ⋅ m 2 )(0.50 rev s)
ωi =
= 1.3 rev s .
If
2.40 kg ⋅ m 2 − 2 ⎡ 13 (3.00 kg)(0.65 m) 2 ⎤ − 2(1.00 kg)(0.65 m)2
⎣
⎦
2
1
⎡
⎤
+ 2 3 (3.00 kg)(0.22 m) + 2(1.00 kg)(0.22 m) 2
⎣
⎦
112. Strategy Choose the axis of rotation at the elbow. The scale pushes with an upward force of 96 N.
Solution Find the force exerted by the triceps muscle.
38
Στ = 0 = (96 N)(38 cm) − Ft (2.5 cm), so Ft =
(96 N) = 1.5 kN .
2.5
113. Strategy Choose the axis of rotation at the contact point between the horizontal surface and the tip of the left leg.
Solution Find the maximum wind speeds in which the blowfly and dog can stand.
(a) τ net = 0 = Fwind r sin θ − mgr cos θ , so
mg
mg
(0.070 × 10−3 kg)(9.80 m s 2 )
= Fwind = cAv 2 or v =
=
= 9.6 m s .
tan θ
cA tan θ
(1.3 N ⋅ s 2 m 4 )(0.10 × 10−4 m 2 ) tan 30.0°
(b) v =
(c) v =
(0.070 × 10−3 kg)(9.80 m s 2 )
(1.3 N ⋅ s 2 m 4 )(0.10 × 10−4 m 2 ) tan 80.0°
(10.0 kg)(9.80 m s 2 )
(1.3 N ⋅ s 2 m 4 )(0.030 m 2 ) tan 80.0°
= 3.1 m s
= 21 m s
433
Chapter 8: Torque and Angular Momentum
Physics
114. (a) Strategy Refer to Example 8.7. The system is in equilibrium until the ladder begins to slip.
Solution Use Newton’s second law.
ΣFx = N w − f = 0, so f = N w .
At the person’s highest point, the frictional force has its maximum possible magnitude, f = µs N f .
Thus, N w = µs N f .
ΣFy = N f − Mg − mg = 0, so N w = µs g ( M + m).
Choose the axis of rotation at the contact point between the ladder and the floor.
⎛1
⎞
Στ = 0 = − N w L sin θ + mg ⎜ L cos θ ⎟ + Mgd cos θ , so
⎝2
⎠
1
N w L sin θ − 2 mgL cos θ µs g ( M + m) L sin θ − 12 mgL cos θ ⎡⎣ µs ( M + m) sin θ − m2 cos θ ⎤⎦ L
d=
=
=
Mg cos θ
Mg cos θ
M cos θ
m
⎛ M +m
tan θ −
= ⎜ µs
M
2M
⎝
⎞
⎟L
⎠
(b) Strategy and Solution Since tan θ increases as θ increases on the interval 0 ≤ θ < 90°, and since d
increases if tan θ increases [which is evident from the equation found in part (a)], placing the ladder at a
larger angle θ allows a person to climb farther up the ladder without having it slip.
(c) Strategy Set d = L.
Solution Find the minimum angle that enables the person to climb all the way to the top of the ladder.
m ⎞
⎛ M +m
L = ⎜ µs
tan θ −
⎟L
M
2M ⎠
⎝
m
M +m
1+
tan θ
= µs
M
2M
2M + m
= µs ( M + m) tan θ
2
2M + m
= tan θ
2 µs ( M + m)
2M + m
2(60.0 kg) + 15.0 kg
θ = tan −1
= tan −1
= 63°
2 µs ( M + m )
2(0.45)(60.0 kg + 15.0 kg)
434
REVIEW AND SYNTHESIS: CHAPTERS 6–8
Review Exercises
1. (a) Strategy Multiply the extension per mass by the mass to find the maximum extension required.
Solution
⎛ 1.0 mm ⎞
⎛ 1000 g ⎞ ⎛ 1 m ⎞
⎜
⎟ (5.0 kg) ⎜
⎟⎜
⎟ = 0.20 m
⎝ 25 g ⎠
⎝ 1 kg ⎠ ⎝ 1000 mm ⎠
(b) Strategy Set the weight of the mass equal to the magnitude of the force due to the spring scale. Use
Hooke’s law.
Solution
Weight = mg = kx, so k =
mg (5.0 kg)(9.80 N kg)
=
= 250 N m .
x
0.20 m
2. Strategy Plot force on the y-axis and the spring length on the x-axis. Use the graph to answer the questions.
Solution Graph the data.
Force, F (N)
1.500
1.000
0.500
0.000
10.0
15.0
20.0
25.0
Spring length, x (cm)
(a) Determine the slope of the line to find k, since F = kx.
1.20 N − 0 N
1.20 N
=
= 0.15 N cm
k=
20.0 cm − 12.0 cm 8.0 cm
(b) The force on the spring is zero when the spring is relaxed, so from the figure, x0 = 12 cm .
3. Strategy Use conservation of energy and Newton’s second law.
Solution Relate the speed to the length of the cord.
1
1
L
∆K = mv 2 − 0 = mv 2 = −∆U = −mg ∆y = mg , so v 2 = gL.
2
2
2
Use Newton’s second law and solve for the tension.
v2
gL
∑ Fy = T − mg = mar = m
=m
= mg , so T = 2mg .
r
L
435
Review and Synthesis: Chapters 6–8
Physics
4. Strategy The work done by the muscles is 22% of the energy expended. The gravitational potential energy
gained by the person is equal to the work done by the muscles.
Solution 0.22 E = W = ∆U = mgh, so
mgh (80.0 kg)(9.80 N kg)(15 m)
E=
=
= 53 kJ .
0.22
0.22
5. (a) Strategy Use the conservation of energy.
Solution Find the work done by friction.
N
1
Wtotal = Wfriction + Wgrav = Wfriction + mgd sin θ = ∆K = 0 − mvi 2 , so
2
1
1
⎛
⎞
Wfriction = − mvi 2 − mgd sin θ = − m ⎜ vi 2 + gd sin θ ⎟
2
⎝2
⎠
⎡1
⎤
= −(100 kg) ⎢ (2.00 m s)2 + (9.80 m s 2 )(1.50 m) sin 30.0° ⎥
⎣2
⎦
= −940 J.
fk
d
mg
y
θ
mg cos θ
mg sin θ
x
θ
Thus, the energy dissipated by friction was 940 J .
(b) Strategy Use Newton’s second law.
Solution Find the normal force on the crate.
ΣFy = N − mg cos θ = 0, so N = mg cos θ .
Since vfx 2 − vix 2 = 0 − vi 2 = 2a x ∆x = 2a x d , the acceleration of the crate is − vi 2 (2d ).
Find the force of sliding friction.
v2
ΣFx = − f k + mg sin θ = − µk mg cos θ + mg sin θ = ma x = − m i , so
2d
vi 2
(2.00 m s)2
µk = tan θ +
= tan 30.0° +
= 0.734 .
2dg cos θ
2(1.50 m)(9.80 m s 2 ) cos 30.0°
6. Strategy Use conservation of energy.
Solution Find the speed of the packing carton at the bottom of the inclined
plane.
1
1
∆K = mvf 2 − mvi 2 = −∆U = −mg ∆y, so
2
2
2.0
m
30.0°
vf = vi 2 − 2 g ∆y = (4.0 m s)2 − 2(9.80 m s 2 )[0 − (2.0 m) sin 30.0°]
= 6.0 m s .
7. Strategy Use conservation of energy.
Solution Find the maximum height of the swing.
1
v2
(6.0 m s) 2
∆K = 0 − mv 2 = −∆U = mghi − mghmax , so hmax =
+ hi =
+ 0.50 m = 2.3 m .
2
2g
2(9.80 m s 2 )
436
(2.0 m) sin θ
Physics
Review and Synthesis: Chapters 6–8
8. Strategy Use conservation of energy and Newton’s second law.
Solution Find the normal force on the crate.
ΣFy = N − mg cos θ = 0, so N = mg cos θ .
fk
N
θ
Find the force of sliding friction.
ΣFx = − f k + mg sin θ = − µk mg cos θ + mg sin θ = ma x , so
y
a x = − µk g cos θ + g sin θ = −0.70(9.80 m s 2 ) cos 53° + (9.80 m s2 ) sin 53° = 3.7 m s2 .
x
Therefore, the acceleration of the block is 3.7 m s 2 down the ramp .
mg cos θ
mg sin θ
mg
θ
9. Strategy The collision is inelastic. Use conservation of momentum and energy.
Solution Write equations using conservation of momentum and energy.
momentum: mvi = (m + M )vf
1
energy: (m + M )vf2 = (m + M ) g ∆y
2
Find the initial speed of the putty.
2
1
⎛ mvi ⎞
(m + M ) ⎜
⎟ = ( m + M ) g ∆y
2
⎝m+M ⎠
2
⎛ m ⎞ 2
⎜
⎟ vi = 2 g ∆y
⎝m+M ⎠
2
2
⎛ 0.50 kg + 2.30 kg ⎞
⎛m+M ⎞
2
vi = 2 g ∆y ⎜
⎟ = 30 m s
⎟ = 2(9.8 m s )(1.50 m) ⎜
0.50 kg
m
⎝
⎠
⎝
⎠
10. Strategy Use conservation of energy. The rotational inertia of a hollow cylinder is I = mr 2 .
Solution Find d, the distance the cylinder travels up the incline.
2
1
1
1
1
⎛v ⎞
0 = ∆K + ∆U = − mvi 2 − I ω 2 + mgd sin θ = − mvi 2 − mr 2 ⎜ i ⎟ + mgd sin θ ,
2
2
2
2
⎝r⎠
2
2
v
(3.00 m s)
=
= 1.53 m .
so d = i
g sin θ (9.80 m s 2 ) sin 37.0°
vi
d
r
y
θ
x
11. Strategy The rotational inertia of a wheel about its central axis is I = 12 MR 2. Use the rotational form of
Newton’s second law.
Solution
(a) I =
1
1
MR 2 = (20.0 kg)(0.224 m)2 = 0.502 kg ⋅ m 2
2
2
(b) The torque required to overcome the friction must be added to that necessary to accelerate the wheel to 1200
rpm in 4.00 s in the absence of friction to get the net torque necessary to accelerate the wheel to 1200 rpm in
4.00 s. Find the torque.
⎡ ∆ω ⎛ ∆ω ⎞ ⎤
1
Στ = I α + I α f = I (α + α f ) = MR 2 ⎢
+⎜
⎟ ⎥
2
⎣ ∆t ⎝ ∆t ⎠f ⎦
=
1
⎛ 1200 rpm 1200 rpm ⎞ ⎛ 2π rad ⎞⎛ 1 min ⎞
+
(20.0 kg)(0.224 m)2 ⎜
⎟ = 17 N ⋅ m
2
60.0 s ⎟⎠ ⎜⎝ rev ⎟⎜
⎝ 4.00 s
⎠⎝ 60 s ⎠
437
Review and Synthesis: Chapters 6–8
Physics
12. Strategy Use the work-kinetic energy theorem. The rotational inertia of a thin hoop is I = mr 2 . The distance d
the bike travels while slowing is equal to the distance the friction force is applied to each wheel.
Solution The force of friction on one of the wheels due to one brake pad is f k = µk N = 0.90 N . Assuming
constant acceleration, the distance the bike travels in the time ∆t = 4.5 s is ∆x = d = (1 2)(vfx + vix )∆t = (1 2)vi ∆t.
Find the normal force on the wheel due to one brake pad.
1
⎛ v ∆t ⎞
⎛1
⎞
Wtotal = Wfriction = −4 f k d = −4(0.90) N ⎜ i ⎟ = −1.8 Nvi ∆t = ∆K = 0 − mb vi 2 − 2 ⎜ I w ω 2 ⎟
2
⎝2
⎠
⎝ 2 ⎠
2
1
1
⎛v ⎞
= − mb vi 2 − mw r 2 ⎜ i ⎟ = − mb vi 2 − mw vi 2 , so N =
2
2
⎝r⎠
1m v
2 b i
+ mw vi
1.8∆t
=
(7.5 m s)[ 12 (11 kg) + 1.3 kg]
1.8(4.5 s)
= 6.3 N .
13. Strategy Use conservation of energy. Let d = 2.05 m. Then, the ramp rises h = d sin 5.00°. The rotational inertia
of a uniform sphere is 52 mr 2 .
Solution Find the speed of the ball when it reaches the top of the ramp.
1
1
1
1
0 = ∆K + ∆U = mvf 2 + I ωf 2 − mvi 2 − I ωi 2 + mgh
2
2
2
2
2
2
v
1
1
2
1
1
⎛
⎞
⎛
⎞
⎛2
⎞⎛ v ⎞
= mvf 2 + ⎜ mr 2 ⎟ ⎜ f ⎟ − mvi 2 − ⎜ mr 2 ⎟ ⎜ i ⎟ + mgh
2
2⎝5
2
2⎝5
⎠⎝ r ⎠
⎠⎝ r ⎠
7
7
= mvf 2 − mvi 2 + mgh, so
10
10
vf = vi 2 −
d
h
5.00°
10
10
gh = (2.20 m s)2 − (9.80 m s 2 )(2.05 m) sin 5.00° = 1.53 m s .
7
7
14. Strategy Use conservation of angular momentum, Eq. (8-1), Eq. (8-14), and the relationship between period and
angular velocity.
Solution
(a) Since the angular momentum is conserved, the ratio is 1 .
(b) Since the rotational inertia is proportional to the square of the radius, ω =
Find the ratio of the angular velocities.
2
ωf ri 2 ⎛
1
⎞
8
= 2 =⎜
⎟ = 1.0 × 10
ωi rf
⎝ 1.0 × 10−4 ⎠
1 2 1⎛ L⎞ 2 1
I ω = ⎜ ⎟ ω = Lω.
2
2⎝ω ⎠
2
Find the ratio of the rotational kinetic energies.
K f ωf
=
= 1.0 × 108
Ki ωi
(c) The rotational kinetic energy is K rot =
(d) The period is related to the angular velocity by T =
2π
ω
.
Find the period of the star after collapse.
Tf ωi
ω
=
, so Tf = i Ti = (1.0 × 10−8 )(1.0 × 107 s) = 0.10 s .
Ti ωf
ωf
438
L
L
∝ 2.
I
r
Physics
Review and Synthesis: Chapters 6–8
15. Strategy Assume the collision time between the dart and the block is short so that the block’s motion during the
collision can be neglected. Let the dart be fired to the right and let the positive x-direction be to the right. Let the
origin be at the original position of the block. Use conservation of momentum during the collision and
conservation of energy after.
Solution Find the speed v of the dart and block just after the collision.
md
0.122 kg
pi = md vd = pf = (md + mb )v, so v =
vd =
(132 m s) = 3.144 m s.
0.122 kg + 5.00 kg
md + mb
Find the compression of the spring.
1
1
1
1
Wtotal = Wfriction + Wspring = − f k x − kx 2 = − µk Nx − kx 2 = − µk (md + mb ) gx − kx 2 = ∆K = 0 − (md + mb )v 2 ,
2
2
2
2
so 0 = kx 2 + 2µk (md + mb ) gx − (md + mb )v 2 . Solve for x.
x=
=
−2 µk (md + mb ) g ± [2 µk (md + mb ) g ]2 − 4k[−(md + mb )v 2 ]
2k
−(0.630)(5.122 kg)(9.80 m s 2 ) ± [(0.630)(5.122 kg)(9.80 m s 2 )]2 + (8.56 N m)(5.122 kg)(3.144 m s) 2
8.56 N m
= −3.69 m ± 4.42 m = 0.73 m or − 8.11 m
Since x > 0, the maximum compression is 0.73 m .
16. Strategy Use the conditions for equilibrium.
Solution Find the vertical components of the forces on each hinge.
mg (5.60 kg)(9.80 m s 2 )
ΣFy = 2 Fv − mg = 0, so Fv =
=
= 27.4 N .
2
2
Let the axis of rotation be a the midpoint of the left edge of the door. The only
horizontal forces are the horizontal components of the forces on the hinges,
therefore, these force are equal and opposite.
Στ = (0.735 m) Fh − (0.380 m)(5.60 kg)(9.80 m s 2 ) + (0.735 m) Fh = 0, so
(0.380 m)(5.60 kg)(9.80 m s 2 )
Fh =
= 14.2 N.
2(0.735 m)
The upper and lower horizontal forces on the hinges are 14.2 N
away from the door and 14.2 N toward the door, respectively.
439
0.760 m
Fv
0.735 m
Fh 0.380 m
2.030 m
0.735 m
mg
Fv
Fh
Review and Synthesis: Chapters 6–8
Physics
17. Strategy Use conservation of energy. The energy delivered to the fluid in the beaker plus the kinetic energies of
the pulley, spool, axle, paddles, and the block are equal to the work done by gravity on the block, which is
negative the change in the block’s gravitational potential energy. The rotational inertia of the pulley (uniform
solid disk) is 12 mp r 2 .
Solution Let the energy delivered to the fluid be E, the distance the block falls be h, and the rotational inertia of
the spool, axle, and paddles be Is = 0.00140 kg ⋅ m 2 . Since the radii of the pulley and the spool are the same (r),
their tangential speeds are the same, so let vp = vs = v.
2
2
1
1
1
1
1⎛1
1 ⎛v⎞
⎞⎛ v ⎞
mb vb 2 + I pωp 2 + I sωs 2 + E = mb vb 2 + ⎜ mp r 2 ⎟ ⎜ ⎟ + I s ⎜ ⎟ + E
2
2
2
2
2⎝2
2 ⎝r⎠
⎠⎝ r ⎠
The tangential speeds of the pulley and spool are equal to the speed of the block.
1
1
1 v2
1
1
1 v2
mb gh = mb vb 2 + mp v 2 + Is
+ E = mb v 2 + mp v 2 + Is
+ E , so
2
4
2 r2
2
4
2 r2
mb gh =
E = mb gh −
v 2 (2mb + mp + 2 I s r 2 )
4
= (0.870 kg)(9.80 m s 2 )(2.50 m) −
(3.00 m s)2 [2(0.870 kg) + 0.0600 kg + 2(0.00140 kg ⋅ m 2 ) (0.0300 m)2 ]
4
= 10.3 J .
18. Strategy Use conservation of linear momentum, the work-kinetic energy theorem, and Newton’s second law.
Solution According to Newton’s second law, the normal force of the ground on the players is N = (m1 + m2 ) g ,
where m1 = 85 kg and m2 = 95 kg. The force of friction is opposite the players direction of motion and has a
magnitude of f k = µk N = µk (m1 + m2 ) g . Find the initial speed v2 of the two-player combination.
pi = m1v1 = pf = (m1 + m2 )v2 , so v2 =
m1v1
.
m1 + m2
Find the distance d the players slide.
2
⎛ m1v1 ⎞
1
1
Wtotal = Wfriction = − f k d = − µk (m1 + m2 ) gd = ∆K = 0 − (m1 + m2 )v22 = − (m1 + m2 ) ⎜⎜
⎟⎟ , so
2
2
⎝ m1 + m2 ⎠
m12 v12
(85 kg) 2 (8.0 m s) 2
=
= 1.0 m .
d=
2
2 µk (m1 + m2 ) g 2(0.70)(85 kg + 95 kg)2 (9.80 m s 2 )
440
Physics
Review and Synthesis: Chapters 6–8
19. Strategy Since the collision is elastic, kinetic energy is conserved. Use conservation of linear momentum and
conservation of energy.
1
m v 2 = mA gh, so vAi = 2 gh , where h
2 A Ai
is the height fallen by bob A, 5.1 m. Since the mass of bob A is half that of bob B, let m = mA and 2m = mB .
Since kinetic energy is conserved, we have
1
1
1
1
mA gh = mA vAi 2 = mA vA 2 + mB vB2 = mvA 2 + mvB2 = mgh (1),
2
2
2
2
where vA and vB are the speeds of the bobs just after the collision.
Solution The kinetic energy of bob A just before is strikes bob B is
Use conservation of linear momentum to find vA in terms of vB .
pi = mvAi = pf = mvA + 2mvB , so vA = vAi − 2vB = 2 gh − 2vB . Substitute this into (1) and solve for vB .
1
1
2
m( 2 gh − 2vB )2 + mvB2 = m(2 gh − 4vB 2 gh + 4vB2 ) + mvB2 = mgh, so vB =
2 gh . Thus, we have,
2
2
3
vA = 2 gh − 2vB = 2 gh −
2 gh
4
2 gh = −
. Now, we use conservation of energy to find how high each bob
3
3
rises after the collision.
2
2 gh ⎞
1
1 ⎛
mgh
h 5.1 m
mghA = mvA 2 = m ⎜ −
, so hA = =
= 0.57 m .
⎟⎟ =
⎜
2
2 ⎝
3 ⎠
9
9
9
2
⎛ 2 2 gh ⎞
1
8mgh
4h 4(5.1 m)
2mghB = (2m)vB2 = m ⎜
, so hB =
=
= 2.3 m .
⎟⎟ =
⎜
2
9
9
9
⎝ 3 ⎠
441
Review and Synthesis: Chapters 6–8
Physics
20. Strategy Since the collision is elastic, kinetic energy is conserved. Use conservation of linear momentum. Let the
positive y-direction be along the shooter’s original velocity
Solution
(a) Let the mass of the marble be m, then the mass of the shooter is 3m. After the collision, let the speed of the
marble be v and the speed of the shooter be V.
1
1
1
(3m)Vi 2 = (3m)V 2 + mv 2 , so 3Vi 2 = 3Vx 2 + 3V y 2 + v 2 (1).
2
2
2
pix = 0 = pfx = 3mVx + mvx , so 0 = 3Vx − v sin 40° (2).
piy = 3mVi = pfy = 3mV y + mv y , so 3Vi = 3V y + v cos 40° (3).
3Vx
(4).
We have three equations and three unknowns (Vx , V y , and v). From (2), we have v =
sin 40°
Substituting this into (3) and solving for V y gives V y = Vi − Vx cot 40° (5). Substitute (4) and (5) into (1) and
solve for Vx .
⎛ 3Vx ⎞
3Vi 2 = 3Vx 2 + 3(Vi − Vx cot 40°)2 + ⎜
⎟
⎝ sin 40° ⎠
2
3Vi 2 = 3Vx 2 + 3(Vi 2 − 2ViVx cot 40° + Vx 2 cot 2 40°) +
0 = Vx 2 − 2ViVx cot 40° + Vx 2 cot 2 40° +
3Vx 2
9Vx 2
sin 2 40°
sin 2 40°
0 = Vx sin 40° − 2Vi cos 40° sin 40° + Vx cos 2 40° + 3Vx
2
0 = Vx (sin 2 40° + cos 2 40°) + 3Vx − 2Vi cos 40° sin 40°
0 = Vx (1) + 3Vx − 2Vi cos 40° sin 40°
4Vx = 2Vi cos 40° sin 40° = Vi sin 80°
V sin 80°
Vx = i
4
Use this result and (5) to find V.
2
⎤
⎛ V sin 80° ⎞ ⎡
⎛ Vi sin 80° ⎞
V = Vx 2 + V y 2 = Vx 2 + (Vi − Vx cot 40°)2 = ⎜ i
⎟ + ⎢Vi − ⎜
⎟ cot 40° ⎥
4
4
⎝
⎠ ⎣
⎝
⎠
⎦
2
= Vi
2
2
sin 2 80° ⎡ sin 80° cot 40° ⎤
sin 2 80° ⎡ sin 80° cot 40° ⎤
(3.2
m
s)
+ ⎢1 −
=
+ ⎢1 −
⎥
⎥ = 2.4 m s
16
4
16
4
⎣
⎦
⎣
⎦
(b) Substituting the result for Vx in (4) gives
3Vx
3V sin 80° 3(3.2 m s) sin 80°
v=
= i
=
= 3.7 m s .
sin 40°
4sin 40°
4sin 40°
(c) According to the way we set up the coordinate system, the tangent of θ is equal to Vx divided by V y instead
of the usual V y divided by Vx .
θ = tan −1
Vx
Vy
= tan −1
Vi sin 80° 4
sin 80°
= tan −1
= 19°
Vi − (Vi sin 80° 4) cot 40°
4 − sin 80° cot 40°
442
Physics
Review and Synthesis: Chapters 6–8
21. Strategy Use energy conservation to find the speed of Jones just before he grabs Smith. Then, use momentum
conservation to find the speed of both just after. Finally, again use energy conservation to find the final height.
Solution Find Jones’s speed, vJ .
1
mJ vJ 2 = mJ ghJ , so vJ = 2 ghJ .
2
Find the speed of both, v.
pi = mJ vJ = pf = (mJ + mS )v, so v =
m 2 ghJ
mJ vJ
= J
.
mJ + mS
mJ + mS
Find the final height, h.
2
⎛ m 2 ghJ ⎞
mJ 2 hJ
1
(78.0 kg) 2 (3.70 m)
⎟ , so h =
=
= 1.27 m .
(mJ + mS ) gh = (mJ + mS ) ⎜ J
⎜ m + mS ⎟
2
(mJ + mS )2 (78.0 kg + 55.0 kg)2
⎝ J
⎠
22. (a) Strategy Use the definition of angular acceleration.
Solution
∆ω 11 rad s − 0
α=
=
= 55 rad s 2 .
∆t
0.20 s
(b) Strategy Use Newton’s second law for rotation.
Solution Find the torque.
∆ω
11 rad s − 0
Στ = I α = I
= (1.5 kg ⋅ m 2 )
= 83 N ⋅ m .
∆t
0.20 s
(c) Strategy Use Eq. (5-21).
Solution Let ∆θ1 be the angle during spin-up and ∆θ 2 be the angle during spin-down.
ωf 2 − ωi 2 = ω 2 − 0 = 2α1∆θ1 , so ∆θ1 =
Find ∆θ1 + ∆θ 2 .
∆θ1 + ∆θ 2 =
ω2
−ω 2
. ωf 2 − ωi 2 = 0 − ω 2 = 2α 2 ∆θ 2 , so ∆θ 2 =
.
2α1
2α 2
⎞
ω 2 −ω 2 ω 2 ⎛ 1
1 ⎞ (11 rad s)2 ⎛
1
1
+
=
−
= 7.3 rad
⎜⎜
⎟
⎜⎜ −
⎟⎟ =
2
2α1 2α 2
2 ⎝ α1 α 2 ⎠
2
−9.8 rad s 2 ⎟⎠
⎝ 55 rad s
(d) Strategy Use Eq. (5-18) and the relationship between angular speed and linear speed.
Solution Find the speed of a point halfway along the radius of the disk 0.20 s after the accelerating torque is
removed.
v
ωf − ωi = − ωi = α∆t , so
r
0.115 m ⎡
v = r (α∆t + ωi ) =
(−9.8 rad s 2 )(0.20 s) + (11 rad s) ⎤ = 0.52 m s .
⎣
⎦
2
443
Review and Synthesis: Chapters 6–8
Physics
23. Strategy Use the work-kinetic energy theorem and Newton’s second law.
Solution ΣFy = N − mg cos θ = 0, so N = mg cos θ . Thus, f k = µk mg cos θ .
1
Wtotal = Wgrav + Wfriction = mgd sin θ − µk mg cos θ d = ∆K = mvf 2 , so
2
vf = 2 gd (sin θ − µk cos θ ) = 2(9.80 m
s 2 )(0.300
fk
N
y
θ
m)(sin 60.0° − 0.38cos 60.0°)
x d
= 2.0 m s .
mg cos θ
mg sin θ
mg
θ
24. Strategy The cylinder falls a vertical distance h = d sin θ = (0.300 m) sin 60.0° as it rolls down the incline. The
rotational inertia of a uniform solid cylinder is 12 mr 2 . Use conservation of energy.
Solution Find the cylinder’s final speed.
2
1
1
1
1⎛1
⎞⎛ v ⎞
mv 2 + I ωf 2 − 0 + 0 − mgh = mvf 2 + ⎜ mr 2 ⎟ ⎜ f ⎟ − mgd sin θ
2 f
2
2
2⎝2
⎠⎝ r ⎠
3
4
4
gd sin θ =
= mvf 2 − mgd sin θ , so vf =
(9.80 m s 2 )(0.300 m) sin 60.0° = 1.84 m s .
4
3
3
0 = ∆K + ∆U =
d
θ
25. Strategy Use conservation of linear momentum.
Solution
y
mb vb
.
pix = mb vb = pfx = (mb + mc )vfx , so vfx =
mb + mc
N
mc vc
.
piy = mc vc = pfy = (mb + mc )vfy , so vfy =
mb + mc
vc
vb
Compute the magnitude of the final velocity.
⎛ mb vb
v = vfx 2 + vfy 2 = ⎜⎜
⎝ mb + mc
2
⎞ ⎛ mc vc
⎟⎟ + ⎜⎜
⎠ ⎝ mb + mc
2
(mb vb )2 + (mc vc )2
⎞
=
⎟⎟
mb + mc
⎠
[(2.00 kg)(2.70 m s)]2 + [(1.50 kg)(−3.20 m s)]2
= 2.06 m s
2.00 kg + 1.50 kg
Compute the angle.
=
θ=
vfy
tan −1
vfx
=
mcvc
mb + mc
−
1
tan
mb vb
mb + mc
= tan −1
mc vc
mb vb
= tan −1
(1.50 kg)(−3.20 m s)
= − 41.6°
(2.00 kg)(2.70 m s)
The velocity of the block and the clay after the collision is 2.06 m s at 41.6° S of E .
444
x
vf
d sin θ
Physics
Review and Synthesis: Chapters 6–8
26. (a) Strategy Use conservation of angular momentum.
Solution Find the tangential speed of the skaters after they grab the rods.
v
v
Li = I1iω1i + I 2iω2i = m1r 2 1 + m2 r 2 2 = m1rv1 + m2 rv2 = Lf = I f ωf = m1rv + m2 rv, so
r
r
m1v1 + m2 v2 = (m1 + m2 )v.
Solve for the tangential speed.
m v + m2 v2 (60.0 kg)(6.0 m s) + (30.0 kg)(2.0 m s)
=
= 4.7 m s .
(m1 + m2 )v = m1v1 + m2 v2 , so v = 1 1
60.0 kg + 30.0 kg
m1 + m2
(b) Strategy and Solution According to the RHR, the angular momentum is upward, away from the ice before
and after the collision. Angular momentum is conserved in magnitude and direction.
27. (a) Strategy Use conservation of angular momentum, since no external torques act on the two-disk system.
Solution Find the final angular velocity.
ωi
ωi
Iω
Iiωi
=
=
.
Lf = I f ωf = Li = I iωi , so ωf = i i =
2
2
2
2
1
If
1 + mr [2(MR 2)] 1 + mr (MR 2 )
Ii + 2 mr
(b) Strategy and Solution
The total angular momentum does not change, since no external torques act on the system.
(c) Strategy Compute the initial and final total kinetic energies and compare their values.
Solution
1
1⎛1
1
⎞
Ki = Iiωi 2 = ⎜ MR 2 ⎟ ωi 2 = MR 2ωi 2
2
2⎝2
4
⎠
⎛
1
1⎛1
1
ωi
⎞
K f = I f ωf 2 = ⎜ MR 2 + mr 2 ⎟ ⎜⎜
2
2
2⎝2
2
⎠ ⎜ 1 + mr 2
MR
⎝
=
(
)
2
1 1 MR 2 ω 2
i
2 2
1 MR 2 + 1 mr 2
2
2
2
2
2
⎞
1 MR 2ω
⎛
⎞
i
⎟ = 1 ⎛ 1 MR 2 + 1 mr 2 ⎞ ⎜
2
⎟
⎟⎜ 1
⎟
2 ⎜⎝ 2
2
⎠ ⎝ 2 MR 2 + 12 mr 2 ⎟⎠
⎟
⎠
2
MR ωi
Ki
= 4
=
2
2
mr
1+ 2
1 + mr 2
1
MR
MR
So, K f ≠ Ki . Therefore, the answer is yes; the kinetic energy changes.
28. (a) Strategy The candy is release with a horizontal speed equal to the tangential speed of the pocket of the
rotating wheel. Use the relationship between angular and tangential speed and the equations of motion for a
changing velocity.
Solution Find the time it takes for the candy to land.
1
1
2 ∆y
∆y = a y (∆t )2 = (− g )(∆t )2 , so ∆t = −
.
2
2
g
Find the candy’s distance from its starting point.
⎛ 2π rad ⎞
2∆y
2(−0.240 m)
v = r ω , so ∆x = v∆t = r ω −
= (0.120 m)(1.60 Hz) ⎜
= 0.267 m .
⎟ −
g
9.80 m s 2
⎝ cycle ⎠
(b) Strategy Use the relationship between radial acceleration and angular speed.
Solution Find the radial acceleration of the candy.
ar = ω 2 r = (1.60 Hz)2 (2π rad cycle)2 (0.120 m) = 12.1 m s 2
445
Review and Synthesis: Chapters 6–8
Physics
29. (a) Strategy Consider the work-kinetic energy theorem and the impulse momentum theorem.
Solution Since the Romulan ship is twice as massive as the Vulcan ship, the Romulan ship will not travel as
far as the Vulcan ship for the same engine force, since ∆x = (1 2)a (∆t ) 2 = (1 2)( F m)(∆t ) 2 . Since
W = F ∆x = ∆K ,
the Vulcan ship will have the greater kinetic energy . Since ∆p = F ∆t ,
the ships will have the same momentum .
(b) Strategy Consider the work-kinetic energy theorem and the impulse momentum theorem.
Solution Since the distances and the forces are the same, and since W = F ∆x = ∆K ,
the ships will have the same kinetic energy . Since ∆x = (1 2)a (∆t ) 2 = (1 2)( F m)(∆t ) 2 , the more massive
Romulan ship will have to fire its engines longer than the Vulcan ship to travel the same distance. Since
∆p = F ∆t and the forces are the same, the Romulan ship will have the greater momentum .
(c) Strategy Refer to parts (a) and (b).
Solution For part (a), we have the following:
Vulcan:
⎡ F
⎤ (9.5 × 106 N) 2 (100 s) 2
∆K = W = F ∆ x = F ⎢
= 6.9 × 1012 J
( ∆t ) 2 ⎥ =
2(65, 000 kg)
⎣ 2m
⎦
∆p = F ∆t = (9.5 × 106 N)(100 s) = 9.5 × 108 kg ⋅ m s
Romulan:
⎡ F
⎤ (9.5 × 106 N) 2 (100 s) 2
∆K = W = F ∆ x = F ⎢
= 3.5 × 1012 J
( ∆t ) 2 ⎥ =
2
m
2(130,
000
kg)
⎣
⎦
∆p = F ∆t = (9.5 × 106 N)(100 s) = 9.5 × 108 kg ⋅ m s
In part (a), the momenta are the same, 9.5 × 108 kg ⋅ m s , but the kinetic energies differ:
Vulcan at 6.9 × 1012 J and Romulan at 3.5 × 1012 J.
For part (b), we have the following:
Vulcan:
∆K = W = F ∆x = (9.5 × 106 N)(100 m) = 9.5 × 108 J
Since K =
1 2 p2
mv =
, p = 2mK = 2(65, 000 kg)(9.5 × 108 J) = 1.1× 107 kg ⋅ m s.
2
2m
Romulan:
∆K = W = F ∆x = (9.5 × 106 N)(100 m) = 9.5 × 108 J
p = 2mK = 2(2 × 65, 000 kg)(9.5 × 108 J) = 1.6 × 107 kg ⋅ m s.
In part (b), the kinetic energies are the same, 9.5 × 108 J, but the momenta differ:
Vulcan at 1.1× 107 kg ⋅ m s and Romulan at 1.6 × 107 kg ⋅ m s.
446
Physics
Review and Synthesis: Chapters 6–8
30. (a) Strategy Use conservation of energy.
Solution Let d be the distance moved along the incline by m2 . Both masses move the same distance and
have the same speed, since they are connected by a rope.
1
1
0 = ∆K + ∆U = m1v 2 + m2 v 2 − 0 − 0 + m1 g (− d sin θ ) + m2 gd sin φ , so
2
2
v=
2 gd (m1 sin θ − m2 sin φ )
=
m1 + m2
2(9.80 m s 2 )(2.00 m)[(6.00 kg) sin 36.9° − (4.00 kg) sin 45.0°]
6.00 kg + 4.00 kg
= 1.7 m s .
(b) Strategy Use Newton’s second law.
Solution Let the positive direction be along the inclines from left to right.
For m1: ΣF = −T + m1 g sin θ = m1a, so T = m1 g sin θ − m1a.
For m2 : ΣF = T − m2 g sin φ = m2 a, so T = m2 g sin φ + m2 a.
Solve for the acceleration.
m2 g sin φ + m2 a = m1 g sin θ − m1a, so
g (m1 sin θ − m2 sin φ ) (9.80 m s 2 )[(6.00 kg) sin 36.9° − (4.00 kg) sin 45.0°]
=
= 0.76 m s 2 .
a=
6.00 kg + 4.00 kg
m1 + m2
Find the speed.
vf 2 − vi 2 = v 2 − 0 = 2ad , so v = 2ad = 2(0.76 m s 2 )(2.00 m) = 1.7 m s .
31. Strategy Refer to the figure and use conservation of energy.
Solution
(a) According to the graph, the particle’s potential energy is −550 J . Since E = K + U , the kinetic energy of
the particle is K = E − U = −100 J − (−550 J) = 450 J .
(b) The total energy is as given, −100 J . According to the graph, the potential energy is −100 J . The
kinetic energy is K = E − U = −100 J − (−100 J) = 0 .
(c) The kinetic energy is K = E − U = −100 J − (−300 J) = 200 J .
(d)
The particle has a kinetic energy of 450 J at t = 0, and we are told the motion is to the left. The particle
will continue moving left but the kinetic energy will decrease by 450/4.5 J for every cm of travel until it
reaches x = 1 cm. At this point K = 0, and the particle has stopped instantaneously. It will next move to
the right with an increasing K until it reaches x = 5.5 cm. At this point K = 450 J, and this kinetic energy
will be maintained as it continues moving right until it reaches x = 11 cm. At this point, its kinetic energy
will decrease by 450/2.5 J for every cm of travel until it reaches x = 13.5 cm. At this point K = 0, and
the particle has again stopped instantaneously. It will then turn around again.
447
Review and Synthesis: Chapters 6–8
Physics
32. (a) Strategy Use conservation of angular momentum at the moment of impact.
Solution Li = I ωi = Lf = I ωf + rp, where I is the rotational inertia of the blade, r is the distance from the
center of the blade to the location of impact, and p = mv is the momentum of the stone just after it is struck.
Find the speed of the stone.
I ωi = I ωf + rp = I ωf + rmvtan
I (ωi − ωf ) = rmvtan
2
2
1
1
I (ωi − ωf ) 12 ML (ωi − ωf ) 12 M (2r ) (ωi − ωf ) Mr (ωi − ωf )
=
=
=
rm
rm
rm
3m
(2.0 kg)(0.25 m)[2π (60 rev s − 55 rev s)]
=
= 52 m s
3(0.10 kg)
vtan =
(b) Strategy Find time it takes for the stone to reach the house. Then use this time to find the distance the stone
falls just before it reaches the window.
Solution Find the time.
∆x
10.0 m
∆x = v∆t , so ∆t =
=
= 0.191 s.
vtan 52.4 m s
Find the distance the stone falls.
1
1
∆y = − g (∆t )2 = − (9.8 m s 2 )(0.191 s)2 = − 0.18 m
2
2
Since 0.18 m is less than half of 1.00 m (0.50 m), the stone hits the window.
33. Strategy Use conservation of energy. m is the mass of one wheel. M is the total mass of the system. v is the speed
of the center of mass of the system (which is the same as the speed of a point on either wheel).
Solution
1 2 1 2
1
1
I ω = mv = K trans for one wheel. K rot,total = 2 ⋅ mv 2 = mv 2 and K trans,total = Mv 2.
2
2
2
2
K total = U i
1
mv 2 + Mv 2 = MgH
2
v 2 (2m + M ) = 2MgH
(a) K rot =
v=
2MgH
=
2m + M
2(80.0 kg)(9.8 m s 2 )(20.0 m)
= 19 m s
2(1.5 kg) + 80.0 kg
(b) Since the speed depends upon the combined total mass of the system, the speed at the bottom would not be
the same for a less massive rider. The answer is no .
448
Physics
Review and Synthesis: Chapters 6–8
34. Strategy Find the change is height from the initial height to the height at which the vine breaks; then use the
change in height to find Tarzan’s speed when the vine breaks. Use Newton’s second law to find the tension when
the vine breaks.
Solution
(a) Find L, the length of the vine.
opposite
5.00 m
5.00 m 5.00 m
sin θi =
, so L =
=
=
= 5.77 m.
hypotenuse
L
sin θi
sin 60°
Find the change in height.
5.00 m
(cos θ f − cos θi )
∆y = yf − yi = L cos θf − L cos θi =
sin θi
Find Tarzan’s speed when the vine breaks.
5.00 m
(cos θ f − cos θi )
vf2 − vi2 = v 2 − 0 = 2 g ∆y = 2 gL(cos θ f − cos θi ) = 2 g
sin θi
5.00 m
5.00 m
v = 2g
(cos θ f − cos θi ) = 2(9.80 m s 2 )
(cos 20.0° − cos 60.0°) = 7.05 m s
sin θi
sin 60.0°
Find the tension.
mv 2
ΣF = T − mg cos θf =
, so
r
⎛ v2
⎞
mv 2
mgv 2
T=
+ mg cos θf =
+ mg cos θ f = mg ⎜
+ cos θ f ⎟
⎜
⎟
r
gL
⎝ gL
⎠
⎡
⎤
(7.05 m s)2
= (900.0 N) ⎢
+ cos 20°⎥ = 1.64 kN
⎢⎣ (9.80 m s 2 )(5.77 m)
⎥⎦
(b) At the moment of the vine breaking, the distance to ground level is
8.00 m − L cos θ f = 8.00 m − (5.77 m) cos 20.0° = 2.58 m;
and the distance to the river’s edge is
L sin θf = (5.77 m) sin 20.0° = 1.97 m.
The time it takes for Tarzan to reach ground level is given by
1
1
∆y = −v sin θf ∆t − g (∆t ) 2 , or 0 = g (∆t )2 + v sin θf ∆t + ∆y.
2
2
Using the quadratic formula, we find ∆t = 0.52 s. The horizontal distance traveled in this time is
∆x = v cos θ f ∆t = (7.05 m s) cos 20.0°(0.52 s) = 3.4 m. Since 3.4 m > 1.97 m, Tarzan lands safely on the
other side. The answer is yes .
449
Review and Synthesis: Chapters 6–8
Physics
35. Strategy Use the relationship between energy and work to find the boy’s speed just before his friend lands on the
sled and the speed when the two reach the bottom. Use conservation of momentum to find the speed of the boys
just after the friend lands on the sled.
Solution Find the speed of the boy.
1
∆K = m1v12 − 0 = −∆U + Wfriction = m1gh1 − fd1 = m1gd1 sin θ − µ m1g cos θ d1, so
2
v1 = 2 gd1(sin θ − µ cos θ ) = 2(9.8 m s 2 )(20 m)(sin15° − 0.12 cos15°) = 7.48 m s.
Find the initial speed of the two boys.
(60 kg)(7.48 m s)
m1v1
pi = m1v1 = pf = (m1 + m2 )v2 , so v2 =
=
= 4.08 m s.
60 kg + 50 kg
m1 + m2
Find the final speed of the two boys.
∆K = −∆U + Wfriction
1
1
(m1 + m2 )v32 − (m1 + m2 )v22 = (m1 + m2 ) gh2 − fd 2 = (m1 + m2 ) gd 2 sin θ − µ (m1 + m2 ) g cos θ d 2
2
2
v32 = 2 gd 2 sin θ − 2µ g cos θ d 2 + v22
v3 = 2 gd 2 (sin θ − µ cos θ ) + v22
= 2(9.8 m s 2 )(50 m)(sin15° − 0.12 cos15°) + (4.08 m s)2 = 13 m s
36. Strategy Use conservation of momentum and the equations for motion with a constant acceleration.
Solution
(a) The banana will fall at the same rate as the monkey; therefore, you should throw the banana directly at the
monkey.
3.33 m + 1.67 m
5.00
tan θ =
, so θ = tan −1
= 59.0° above the horizontal .
3.00 m
3.00
(b) Since the banana will fall at the same rate as the monkey, regardless of the launch speed of the banana, the
launch angle is the same for all launch speeds. Relatively high launch speeds will reach the monkey relatively
sooner (and higher); relatively low launch speeds will reach the monkey relatively later (and lower).
(c) Find the time it takes the banana to reach the monkey.
1
2∆y
2(−1.67 m)
∆y = − g (∆t ) 2 , so ∆t = −
= −
= 0.5838 s = t.
2
g
9.80 m s 2
The banana reaches the monkey when it has traveled 3.00 m.
∆x
3.00 m
∆x = vx ∆t = v cos θ∆t , so v =
=
= 9.98 m s .
∆t cos θ (0.5838 s) cos 59.0°
450
Physics
Review and Synthesis: Chapters 6–8
(d) The speed of the monkey just before the collision is given by vmy = − gt and vmx = 0. The speed of the banana
at this time is given by vby = v sin θ − gt and vbx = v cos θ . Use conservation of momentum.
mvbx = (m + M )vfx , so
m
m
0.20 kg
vfx =
vbx =
v cos θ =
(9.98 m s) cos 59.0° = 0.321 m s.
m+M
m+M
0.20 kg + 3.00 kg
mvby + Mvmy = (m + M )vfy , so
vfy =
mvby + Mvmy
=
m(v sin θ − gt ) + M (− gt )
m
=
v sin θ − gt
m+M
m+M
m+M
0.20 kg
=
(9.98 m s) sin 59.0° − (9.80 m s 2 )(0.5835 s) = −5.18 m s.
0.20 kg + 3.00 kg
The time it takes for the monkey to hit the ground is given by
1
1
∆y = vfyt2 − gt22 , or 0 = gt22 − vfyt2 + ∆y = (4.90 m s 2 )t22 + (5.18 m s)t2 − 5.33 m.
2
2
Using the quadratic formula, we find t = 0.64 s. The horizontal distance is
d = vfxt = (0.321 m s)(0.64 s) = 0.21 m .
MCAT Review
1. Strategy Use conservation of momentum.
Solution
pi = mvi = pf = mvf + pwall , so pwall = m(vi − vf ) = (0.2 kg)[2.0 m s − (−1.0 m s)] = 0.6 kg ⋅ m s.
The correct answer is D .
2. Strategy Use Hooke’s law.
Solution Let up be the positive direction. The gravitational force on the mass is
F = mg = (0.10 kg)(−9.80 m s 2 ) = −0.98 N. Solving for the spring constant in Hooke’s law, we have
k =−
−0.98 N
F
=−
= 6.5 N m. Thus, the correct answer is D .
x
0.15 m
3. Strategy The net torque is zero.
Solution
m) − (1.0 × 10−7
0.60 m
s 2 )(0.40
Στ = 0 = F (0.60
kg)(9.80 m
m), so
7
2
−
(1.0 × 10 kg)(9.80 m s )(0.40 m)
F=
= 6.5 × 10−7 N.
0.60 m
The correct answer is B .
451
F
0.40 m
mg
Review and Synthesis: Chapters 6–8
Physics
4. Strategy Determine the speed of the first ball just before in collides with the second. The collision is completely
inelastic; that is, the balls stick together. Use conservation of momentum to find the speed of the balls after the
collision.
Solution Find the speed of the first ball just before the collision.
vfx − vix = v1 − 0 = a x ∆t , so v1 = (10 m s 2 )(2.0 s) = 20 m s.
Find the speed v of the balls just after the collision.
m1v1
(0.50 kg)(20 m s)
pi = m1v1 = pf = (m1 + m2 )v, so v =
=
= 6.7 m s.
m1 + m2
0.50 kg + 1.0 kg
The correct answer is B .
5. Strategy Use Newton’s second law and Eq. (6-27).
Solution The gravitational force working against the motion of the car as it climbs the hill is
mg sin10°, so the additional power required is
10°
mg
Pcar = − Pgrav = − Fv cos180° = (mg sin10°)v = (1000 kg)(10 m s 2 ) sin10°(15 m s)
= 1.5 × 105 × sin10°
mg sin 10°
W.
The correct answer is D .
6. Strategy Find the vertical distance the patient would have climbed had the treadmill been stationary (and very
long). Then, find the work done by the patient on the treadmill.
Solution The “distance” walked along the incline is (2 m s)(600 s) = 1200 m.
Thus, the vertical distance climbed is (1200 m) sin 30° = 600 m. The work done is
W = Fd = mgd = (90 kg)(10 m s 2 )(600 m) = 0.54 MJ.
0m
120
600 m
30°
The correct answer is C .
7. Strategy Find the angle between the force exerted by the patient and the patient’s velocity. Use Eq. (6-27).
Solution The force due to gravity is down, so the force exerted by the patient is up. The velocity is
directed at the angle of the incline, or 30° above the horizontal, so the angle between the force and
the velocity is 60°. Compute the mechanical power output of the patient.
P = Fv cos θ = mgv cos θ = (100 kg)(10 m
s 2 )(3
F
60°
v
m s) cos 60° = 1500 W
The correct answer is B .
8. Strategy and Solution The force pushing each friction pad is normal to the wheel; that is, it is the normal force
in f k = µk N . Solve for the normal force.
f
20 N
N= k =
= 50 N
0.4
µk
This is the total force. The force pushing each friction pad is half this, or 25 N. The correct answer is B .
452
Physics
Review and Synthesis: Chapters 6–8
9. Strategy Find the average tangential speed at the friction pads. Then, use the relationship between tangential
speed and radial acceleration.
Solution
The average tangential speed is v =
ar =
4800 m 1 min
×
= 4.0 m s. The radial acceleration is
20 min 60 s
v 2 (4.0 m s) 2
=
= 50 m s 2 . The correct answer is D .
r
0.3 m
10. Strategy Use the work-kinetic energy theorem.
Solution The work done by friction on the wheel is W = − f k d , where d is the linear distance the wheel passes
between the pads before it stops. Relate d to the kinetic energy of the wheel.
K
Wtotal = − f k d = ∆K = 0 − Ki , so d = i .
fk
Divide d by the circumference of a circle with radius 0.3 m to find the number of rotations.
Ki
d
30 J
=
=
= 0.8 rotations
2π r 2π rf k 2π (0.3 m)(20 N)
Since 0.8 < 1, the correct answer is A .
11. Strategy Compute the average mechanical power output of the cyclist and compare it to the power consumed by
the wheel at the friction pads.
Solution The metabolic power available for work is 535 W − 85 W = 450 W. Since the efficiency is 20%, the
average mechanical power output of the cyclist is 0.20 × 450 W = 90 W. The average tangential speed of the
4800 m 1 min
×
= 4.0 m s. Therefore, the power consumed by the friction pads is
wheel is v =
20 min 60 s
P = f k v = (20 N)(4.0 m s) = 80 W. Thus, the difference between the average mechanical power output of the
cyclist and the power consumed by the wheel at the friction pads is 90 W − 80 W = 10 W.
The correct answer is B .
12. Strategy and Solution Increasing the force on the friction pads would increase the power consumed by the wheel
at the friction pads (because P = Fv). So, if the cyclist is pedaling at the same rate and the power consumed by
the friction pads increases, the difference between the two decreases and the fraction of mechanical power output
of the cyclist consumed by the wheel at the friction pad increases. Thus, the correct answer is D .
13. Strategy Relate the cyclist’s average metabolic rate to the energy released per volume of oxygen consumed, the
time on the bike, and volume of oxygen consumed.
Solution The cyclist’s average metabolic rate while riding is 535 W. The total energy used during 20 minutes is
60 s
(535 W)(20 min)
= 642, 000 J. The total energy released by the consumption of oxygen is (20, 000 J L)V ,
1 min
where V is the volume of oxygen consumed. Equating these two expressions and solving for V gives the number
of liters of oxygen the cyclist consumes.
642, 000 J
(20, 000 J L)V = 642, 000 J, so V =
= 32 L ≈ 30 L. The correct answer is B .
20, 000 J L
453
Review and Synthesis: Chapters 6–8
Physics
14. Strategy and Solution Since the force has been reduced by 50% and the distance has been doubled, the cyclist
does the same amount of work [W = 0.50 F (2∆x) = F ∆x]. So, the energy transmitted in the second workout is
equal to the energy transmitted in the first. The correct answer is C .
15. Strategy The circumference of a circle is C = 2π r. A wheel moves a distance equal to its circumference during
each rotation. The wheel rotates twice during each rotation of the pedals.
Solution The circumference of a circle with a radius of 0.15 m is 2π (0.15 m). The circumference of a circle with
a radius of 0.3 m is 2π (0.3 m). During each rotation of the pedals, a point on the wheel at a radius of 0.3 m
moves a distance 2[2π (0.3 m)]. The ratio of the distance moved by a pedal to the distance moved by a point on
the wheel located at a radius of 0.3 m in the same amount of time is
2π (0.15 m)
= 0.25.
2[2π (0.3 m)]
The correct answer is A .
16. Strategy Use the definition of power.
Solution
∆E
∆E ⎛ 300 kcal ⎞ ⎛ 4186 J ⎞ ⎛ 1 min ⎞
P=
, so ∆t =
=⎜
⎟⎜
⎟⎜
⎟ = 41.9 min. The correct answer is D .
P ⎝ 500 W ⎠ ⎝ 1 kcal ⎠ ⎝ 60 s ⎠
∆t
17. Strategy Consider the distance a point on the wheel travels for each situation.
Solution The circumference of a circle with a radius of 0.3 m is 2π (0.3 m). The
circumference of a circle with a radius of 0.4 m is 2π (0.4 m). During each rotation, a point
on a wheel travels a distance equal to the circumference. The force on the wheel is the same
in each case, but the distance traveled by a point on the wheel is greater for a greater radius.
In this case, the distance is 0.4 m (0.3 m) = 1.33 times farther or 33%. Since work is equal
to the product of force times distance, the work done on the wheel per revolution is 33%
more. Thus, the correct answer is C .
454
0.3 m
0.4 m
Chapter 9
FLUIDS
Conceptual Questions
1. A manometer (with one side open) measures gauge pressure. A barometer measures absolute pressure. A tire
pressure gauge and a sphygmomanometer both measure gauge pressure.
2. (a) The water in the hose exerts a force on the water entering the narrower nozzle at the end, causing it to
accelerate and emerge at a high speed. The fast moving water being expelled exerts a force of equal
magnitude back on the hose and the firefighter who is holding the hose in place.
(b) If a constriction develops in the hose, the water speed in the constriction will be increased due to the smaller
area. If the speed is sufficiently large, the flow through the constriction will become turbulent, with
seemingly random variations in pressure and water velocity. This produces the pulsating and vibrating motion
of the hose.
3. While floating, the weight of the water displaced by an object is equal to the object’s weight. The “displacement”
listed in place of a boat’s weight is just the amount of water it displaces while floating.
4. Allowing the water to collect in tanks at atmospheric pressure every few floors limits the maximum pressure in
any of the pipes. Otherwise, in a tall building the pressure in the pipes near the bottom floors would be much too
large to safely contain the water. The water must also be supplied at each floor (in sinks, toilets, etc.) with a
relatively steady pressure. Without the exposure to atmospheric pressure every few floors, the water pressure at a
particular faucet would depend significantly on how many other faucets in the building were running at the time.
Another significant advantage to this arrangement is that the pumps don’t have to work continuously, turning on
and off every time someone flushes a toilet. They only need to run when the water level in a holding tank gets
low; then they work until the tank is refilled.
5. There is practically no atmosphere on the Moon and hence practically zero pressure. Drinking from a glass with a
straw would be impossible, since a pressure difference is required to push the liquid up the straw. With a sealed
juice box, the astronaut could supply the necessary pressure by squeezing the box.
6. To be accurate, one should say that wood floats because it is less dense than water, and a stone sinks because it is
more dense than water.
7. When measuring blood pressure, one is interested in the pressure of the blood in the heart to gauge how hard the
heart is working. Therefore, the blood pressure cuff is wrapped around the arm at the same level as the heart. At a
different level there would be a pressure change in the blood due to a difference in height.
8. (a) The pressure inside the hot air balloon is slightly higher than the pressure outside. The balloon is like a
bubble—it maintains its roughly spherical shape by stretching the balloon walls a bit. The walls in turn push
in on the air, keeping it at a slightly higher pressure.
(b) The density of the air inside the balloon is smaller. The average density of the entire balloon must be equal to
the density of the surrounding air for it to be floating in equilibrium.
9. The pressure of the atmosphere decreases with altitude. Therefore, the balloon gradually expands as it rises.
10. Although the fluid velocity may be different at different points in the liquid, the flow is steady if the velocity at
any particular point is constant in time.
455
Chapter 9: Fluids
Physics
11. The heat produced in the engine of a running car increases the temperature of the oil flowing through it. The
viscosity of a liquid such as oil decreases as the temperature increases. Thus, driving a car a few miles before
stopping for an oil change reduces the viscosity of the oil, making it easier to flush the oil from the car.
12. As the speed of the train increases, the pressure of the outside air moving by it decreases. Because the air inside
the train is in contact with the air outside, the inside pressure must also decrease. A passenger may experience a
popping sensation in their ears when the pressure within their ears adjusts to equilibrate with the decreased
pressure of the train.
13. The rate at which air moves up a chimney is determined by the size of the pressure change along the chimney’s
length. Bernoulli’s equation tells us that on a windless day, the pressure difference between the two ends of the
chimney is proportional to the height of the chimney. On a windy day, the velocity difference between the two
ends causes an additional pressure difference—resulting in an increased draft.
14. The excess pressure inside a bubble is inversely proportional to the radius of the bubble—a small bubble therefore
has a greater excess pressure than a large bubble. When the valve is opened, the higher pressure inside the smaller
bubble pushes air from its interior into the larger bubble—decreasing the size of the smaller bubble and increasing
the size of the larger one. If the surface tension of the small bubble is too large, the bubble may collapse.
15. (a) Given that points A and B are at equal heights, Bernoulli’s equation tells us that the pressure difference
between the two points is proportional to the difference in the square of the fluid velocities at the points.
Thus, the velocities at each point cannot both equal zero—the fluid cannot be static.
(b) Fluid flows from a region of higher pressure to a region of lower pressure. If the pressure at point C were to
increase while the pressure at point D remained unchanged, fluid would flow from point C to point D—thus
contradicting the assumption that the fluid is static.
16. Hydraulic systems have the advantage of providing a very large mechanical advantage with a very simple design.
Hydraulic systems also provide very sensitive controls with no mechanical lag because of the near
incompressibility of the fluids used in their construction.
17. To get optimal use from hydraulic systems, the fluid used in mediating the operation of the system must be very
nearly incompressible. Liquids meet this requirement, but gases like air are compressible. Thus, proper operation
of a hydraulic device requires that gases be “bled” from the system.
18. As the diver travels deeper beneath the sea, the pressure from the surrounding water increases. The snorkel
provides a direct connection from the inside of the diver’s body to the air at the surface—the pressure inside the
diver’s body is therefore equal to the pressure of the atmosphere. At some depth, the difference in the pressure on
the outside of the diver’s body due to the water and the pressure on the inside from the atmosphere will become
unsustainably large—the exterior pressure will ultimately cause the diver’s lungs to collapse.
19. The buoyant force is only slightly less than the weight of the bubble, so the bubble has a very small terminal
velocity. The bubble falls slowly and is easily lifted up by rising air currents as it drifts into their vicinity.
20. The weight of the bottle and water is the same before and after the squeezing with the strap. The pressure on the
bottom of the bottle increases because there is a higher column of water above the base.
456
Physics
Chapter 9: Fluids
Problems
1. Strategy Use the definition of average pressure.
Solution Compute the average pressure.
2
Pav =
F
500 N ⎛ 100 cm ⎞ ⎛
1 atm
⎞
=
= 49 atm
A 1.0 cm 2 ⎜⎝ 1 m ⎟⎠ ⎝⎜ 101.3 × 103 Pa ⎠⎟
2. Strategy The frictional force is equal to the net pressure times the cross-sectional area of the neck of the bottle.
Solution Compute the frictional force on the cork.
f = Pnet A = (4.5 atm)(1.013 × 105 Pa atm)π (0.010 m) 2 = 140 N
3. Strategy Use the definition of average pressure.
Solution Compute the average pressure on the soles of the feet.
F mg (90.0 kg)(9.80 N kg)
Pav = =
=
= 22 kPa
A
A
2(0.020 m 2 )
4. Strategy Use the definition of average pressure.
Solution
(a) Find the magnitude of the downward force.
F = Pav A = (1.0 × 105 Pa)(1.0 m 2 ) = 1.0 × 105 N
(b) Convert to pounds.
(1.0 × 105 N)(0.2248 lb N) = 2.2 × 104 lb
(c) The huge force doesn’t crush the desk because the pressure of the air under the desktop pushes upward
counteracting the downward force.
5. Strategy The average pressure is the force applied to the floor divided by the contact area.
Solution
mb g
4mb g
F
.
=
=
2
A 3 1πd 2
3
π
d
s
s
4
m
g
m g
F
a
= a .
The adult applies a pressure of Pa = =
A 4 1πd 2
π dc2
c
The baby applies a pressure of Pb =
(
)
(4
)
4mb g ⎛ ma g ⎞
=
The ratio of these two pressures is
⎜
⎟
Pa 3π ds 2 ⎜⎝ π dc 2 ⎟⎠
Pb
−1
=
The baby applies 2.0 times as much pressure as the adult.
457
4mb dc 2
3ma ds 2
=
4(10 kg)(0.060 m)2
3(60 kg)(0.020 m) 2
= 2.0.
Chapter 9: Fluids
Physics
6. (a) Strategy The force required to lift the lid is equal to the area of the box times the pressure difference
between atmospheric pressure and the pressure inside the box.
Solution Find the required force.
F = ∆PA = (1.013 × 105 Pa − 0.80 × 105 Pa)(0.15 m)(0.13 m) = 420 N
(b) Strategy Compare the atmospheric pressure at Denver to that in the box.
Solution
The atmospheric pressure at Denver is about 23 (1.013 × 105 Pa) = 0.6753 × 105 Pa. Since the atmospheric
pressure at Denver is less than that in the box, no force is needed
to remove the lid.
7. Strategy The gas pushes outward equally on each wall of the cube.
Solution Find the force on the south side of the container.
G
F = PA = (4.0 × 105 Pa)(0.10 m) 2 = 4.0 kN, so F = 4.0 kN southward .
8. Strategy The pressure increases by an amount equal to the force divided by the cross-sectional area of the piston.
Solution Compute the pressure increase.
F
4.40 N
∆P = =
= 88.0 kPa
A 5.00 × 10−5 m2
9. Strategy The work done by the small piston must equal that done on the car.
Solution Find the distance that the small piston must be pushed downward to raise the car 1.0 cm.
Fc /Fp = A/a since ∆P = Fc /A = Fp /a.
Wp = Fp d p = Wc = Fc dc , so d p =
Fc
A
dc = dc = 100.0(0.010 m) = 1.0 m .
Fp
a
10. (a) Strategy The pressure is the same throughout the hydraulic fluid, so P = Fs As = Fl Al .
Solution Find the force that must be applied to the small piston.
2
A
r2
⎛ 2.50 cm ⎞
(10.0 kN) = 625 N
Fa = Fs = s Fl = s Fl = ⎜
⎟
Al
⎝ 10.0 cm ⎠
rl2
(b) Strategy The work done at each piston must be equal.
Solution
F
625 N
(10.0 cm) = 6.25 mm .
Wl = Fl dl = Ws = Fs ds , so dl = s ds =
10.0 kN
Fl
(c) Strategy Compute the ratio of the weight to the applied force.
Solution Compute the mechanical advantage.
W
10.0 kN
=
= 16.0
Fa 0.625 kN
458
Physics
Chapter 9: Fluids
11. (a) Strategy The pressure in the fluid is equal to P = Fb Ab . The pressure is also equal to the normal force N
divided by the area of the brake pad piston A.
Solution Compute the normal force applied to each side of the rotor.
A
12.0 cm 2
N = PA =
Fb =
(7.5 N) = 30 N
Ab
3.0 cm 2
(b) Strategy The frictional force due to one pad is µ N , so the total force is 2µ N . Use Eq. (8-3).
Solution Find the torque on the rotor due to the two pads.
τ = F⊥ r = 2µ Nr = 1.6(30 N)(0.12 m) = 5.8 N ⋅ m
12. Strategy Use Eq. (9-4).
Solution Find the depth under water where the pressure is 4.0 atm.
P − Patm
4.0 atm − 1.0 atm
=
P = Patm + ρ gd , so d =
(1.0 × 105 Pa atm) = 31 m .
ρg
(1.0 × 103 kg m3 )(9.80 m s 2 )
13. Strategy Use Eq. (9-4).
Solution Compute the pressure on the fish.
1 atm
⎛
⎞
P = Patm + ρ gd = 1.0 atm + (1025 kg m3 )(9.80 m s 2 )(10 m) ⎜
⎟ = 2.0 atm
5
⎝ 1.013 × 10 Pa ⎠
14. Strategy Since d is depth, − d is height. Use Eq. (9-4).
Solution Find the height that water can be sucked up a straw.
P −P
10 kPa
P = Patm + ρ gd , so h = −d = atm
=
= 1.0 m .
ρg
(1.0 × 103 kg m3 )(9.80 m s 2 )
15. Strategy Use Eq. (9-2).
Solution Find the ratios of the volumes of the platinum and aluminum.
mPt
⎛ m ⎞⎛ ρ ⎞
VPt
⎛ 2702 ⎞
ρ
= mPt = ⎜ Pt ⎟ ⎜ Al ⎟ = 1.00 ⎜
⎟ = 0.126
Al
VAl
⎝ 21,500 ⎠
⎝ mAl ⎠ ⎝ ρ Pt ⎠
ρ
Al
16. Strategy Use the definition of average pressure and Eq. (9-4).
Solution Find the force that the Dutch boy must exert to save the town.
F
∆P =
and P − Patm = ∆P = ρ gd , so
A
F = Aρ gd = (1.0 × 10−4 m 2 )(1.0 × 103 kg m3 )(9.80 m s 2 )(3.0 m) = 2.9 N .
459
Chapter 9: Fluids
Physics
17. Strategy Use the definition of average pressure and Eq. (9-4).
Solution Find the magnitude of the force exerted by the water on the bottom of the container.
F
P = Patm + ρ gd = , so
A
F = ( Patm + ρ gd ) A = ⎡1.013 × 105 Pa + (1.0 × 103 kg m3 )(9.80 m s 2 )(11.0 m) ⎤ (5.00 m 2 ) = 1.0 MN .
⎣
⎦
18. Strategy Use Eq. (9-4).
Solution Find the maximum depth at which the organisms can live.
∆P
1000 atm
d=
=
(1.013 × 105 Pa atm) = 10 km
ρ g (1025 kg m3 )(9.80 m s 2 )
19. (a) Strategy Use Eq. (9-4).
Solution Compute the pressure increase.
∆P = ρ water gd = (1.00 × 103 kg m3 )(9.80 m s 2 )(35.0 m) = 343 kPa
(b) Strategy Use Eq. (9-3).
Solution Compute the pressure decrease.
∆P = − ρair gh = −(1.20 kg m3 )(9.80 m s 2 )(35 m) = −410 Pa
The pressure decreases by 410 Pa .
20. Strategy Use the appropriate conversion factors to convert the woman’s systolic blood pressure into the various
pressure units.
Solution
⎛ 101.3 kPa ⎞
(a) (160 mm Hg) ⎜
⎟ = 21 kPa
⎝ 760.0 mm Hg ⎠
⎛ 14.70 lb in 2 ⎞
(b) (160 mm Hg) ⎜
⎟ = 3.1 lb in 2
⎜ 760.0 mm Hg ⎟
⎝
⎠
⎛
⎞
1 atm
(c) (160 mm Hg) ⎜
⎟ = 0.21 atm
⎝ 760.0 mm Hg ⎠
⎛ 760.0 torr ⎞
(d) (160 mm Hg) ⎜
⎟ = 160 torr
⎝ 760.0 mm Hg ⎠
460
Physics
Chapter 9: Fluids
21. Strategy Use the appropriate conversion factors to convert the gauge pressure into the various pressure units.
Solution
⎛ 1.013 × 105 Pa ⎞
(a) (32 lb in 2 ) ⎜
⎟ = 2.2 × 105 Pa
⎜ 14.7 lb in 2 ⎟
⎝
⎠
⎛ 760.0 torr ⎞
(b) (32 lb in 2 ) ⎜
= 1700 torr
⎜ 14.7 lb in 2 ⎟⎟
⎝
⎠
⎛ 1 atm ⎞
(c) (32 lb in 2 ) ⎜
= 2.2 atm
⎜ 14.7 lb in 2 ⎟⎟
⎝
⎠
22. Strategy The gauge pressure of the IV fluid must be at least equal to that of the blood in the vein, which is equal
to a column of mercury 12 mm tall.
Solution Find the required height of the IV bag above the vein.
ρ
13, 600 kg m3
ρ1gh1 = ρ 2 gh2 , so h1 = 2 h2 =
(0.012 m) = 15 cm .
ρ1
1060 kg m3
23. Strategy Use Eq. (9-6).
Solution Determine the absolute pressure of the gas.
Pabs = Patm + Pgauge = 74.0 cm Hg + 40.0 cm Hg = 114.0 cm Hg
24. Strategy Use Eq. (9-5).
Solution Find the change in pressure of the gas.
∆P = ρ gd = (1.0 × 103 kg m3 )(9.80 m s 2 )(0.040 m) = 390 Pa
25. Strategy The amount the fluid rises is one-half the difference of levels, or ∆hoil 2. Use Eq. (9-5).
Solution
(a) Find the amount that the fluid level rises.
ρ Hg
∆hoil
13.6 g cm3
∆P = ρHg g ∆hHg = ρoil g ∆hoil , so
(0.74 cm Hg) = 5.6 cm .
=
∆hHg =
2
2 ρoil
2(0.90 g cm3 )
(b)
∆hoil ⎛ ρoil
⎜
2 ⎜⎝ ρ Hg
⎞
0.90 g cm3
⎟ = (5.6 cm)
= 0.37 cm
⎟
13.6 g cm3
⎠
26. Strategy Assume that the blood is not flowing. Use Eq. (9-3).
Solution Estimate the average blood pressure in a person’s foot.
⎛ 760.0 mm Hg ⎞
P = Paorta + ρ gd = 104 mm Hg + (1060 kg m3 )(9.80 m s 2 )(1.37 m) ⎜
⎟ = 211 mm Hg
⎝ 1.013 × 105 Pa ⎠
461
Chapter 9: Fluids
Physics
27. Strategy and Solution Since the goose has 25% of its volume submerged, its density is 25% of water’s, or about
250 kg m3 .
28. Strategy The weight of the barge must equal the weight of the displaced water.
Solution Find the depth of the barge below the waterline.
Weight water = ρ gV = ρ gdA = Weight barge = mg , so d =
m
3.0 × 105 kg
=
= 1.5 m .
ρ A (1.0 × 103 kg m3 )(20.0 m)(10.0 m)
29. (a) Strategy The relationship between the fraction of a floating object’s volume that is submerged to the ratio of
the object’s density to the fluid in which it floats is Vf Vo = ρo ρf . Since the water contains ice, use the
density of water at 0°C.
Solution Find the percent of the volume of ice that is submerged when it floats in water.
Vsubmerged
ρ
917 kg m3
= ice =
= 0.917, or 91.7%
ρ water 999.87 kg m3
Vice
(b) Strategy and Solution The specific gravity and the fraction of the object submerged in water are the same
for objects that float, so the specific gravity of ice is 0.917 .
30. Strategy The relationship between the fraction of a floating object’s volume that is submerged to the ratio of the
object’s density to the fluid in which it floats is Vf Vo = ρo ρf .
Solution
(a) Find the density of the object.
Vf ρo
V
=
, so ρo = f ρf = 0.14(999.87 kg m3 ) = 140 kg m3 .
Vo ρf
Vo
(b) Find the percentage of the object that is submerged if it is placed in ethanol by forming a proportion.
ρ
fraction submerged in ethanol ρo ρethanol
999.87
, so the fraction submerged in ethanol is
=
= water =
fraction submerged in water
790
ρo ρ water
ρethanol
999.87
(0.14) = 0.18, or 18% .
790
31. Strategy In (a), the buoyant force is equal to the weight of the ice. In (b), the buoyant force is equal to the weight
of the displaced water.
Solution Find the buoyant force in each situation.
G
(a) FB = mice g = (0.90 kg)(9.80 m s 2 ) = 8.8 N, so FB = 8.8 N upward .
(b) FB = mwater g =
G
ρ water
1.0 × 103 kg m3
(0.90 kg)(9.80 m s 2 ) = 9.6 N, so FB = 9.6 N upward .
mice g =
3
ρice
917 kg m
462
Physics
Chapter 9: Fluids
32. Strategy The ratio of the density of the wood to that of the oil is equal to the fraction of the volume of the wood
that is submerged.
Solution Find the density of the oil.
ρ w Vo
V
Vw
0.67 g cm3
=
= 0.74 g cm3 .
ρw =
, so ρo = w ρ w =
ρo Vw
Vo
0.900Vw
0.900
33. Strategy Compare the densities of a block of ebony and ethanol.
Solution The density of a block of ebony is between 1000 and 1300 kg m3 . The density of ethanol is
790 kg m3 . Since the density of ebony is more than that of ethanol, the block will sink; therefore, 100% of the
volume of the block of ebony is submerged.
34. Strategy Use Eqs. (9-2) and (9-9).
Solution
(a) Compute the specific gravity of the disk.
ρ
m
8.16 kg
S.G. =
=
=
= 0.910
3
ρ w ρ wV (1.00 × 10 kg m3 )(8.97 × 10−3 m3 )
(b) The ratio of the density of the disk to that of water is equal to the fraction of the volume of the disk that is
submerged.
ρdisk
V
dA
V
8.97 × 10−3 m3
= S.G. = water =
, so d = (S.G.) = 0.910
= 1.28 cm .
ρ water
Vdisk
V
A
0.640 m 2
(c) Find the distance of the top surface of the disk above the water level.
dA dA d
V
8.97 × 10−3 m3
(1 − 0.910) = 0.13 cm .
=
= = S.G., so h − d = h − (S.G.)h = (1 − S.G.) =
V
hA h
A
0.640 m3
35. Strategy The weight of the alcohol displaced is equal to the buoyant force. Use Eqs. (9-2) and (9-9).
Solution Find the specific gravity of the alcohol.
malcohol g = 1.03 N − 0.730 N = 0.30 N and
ρ
malcohol
malcohol g
0.30 N
S.G. = alcohol =
=
=
= 0.78 .
3
3
ρw
ρ wValcohol ρ wValcohol g (1.00 × 10 kg m )(3.90 × 10−5 m3 )(9.80 m s 2 )
36. Strategy Find an expression for the new density of the fish; then set this equal to the density of the water and
solve for the volume of the bladder.
Solution The new density of the fish is
m + ma mf + ρaVa
= m
= ρw.
ρ= f
f
Vf + Va
+ Va
ρf
Solve for Va.
1060 kg m3
ρ
1−
1 − ρw
ρw
1080 kg m3
f
= (0.0100 kg)
= 0.17 cm3 .
mf + ρaVa =
mf + ρ wVa , so Va = mf
ρf
ρ w − ρa
1060 kg m3 − 1.20 kg m3
463
Chapter 9: Fluids
Physics
37. Strategy Find the volume of the coin using the density of water and the mass of the displaced water. Then find its
density using its mass as measured in air and compare this to the density of gold.
Solution The mass of water displaced is 49.7 g − 47.1 g = 2.6 g. Since the density of water is about
1.00 g cm3 , the volume of the coin is 2.6 cm3 = 2.6 × 10−6 m3 . So, the density of the coin is
m 49.7 × 10−3 kg
=
= 19, 000 kg m3 . The density of gold is 19,300 kg m3 , so yes , you should get
V 2.6 × 10−6 m3
excited, since the coin may be genuine.
ρ=
38. Strategy The weight of the water displaced by the fish is mw g = 200.0 N − 15.0 N = 185.0 N. The volume of the
fish is equal to the volume of the displaced water. Use Eq. (9-2).
Solution Set the volumes of the fish and the displaced water equal.
m g
Vf = Vw = w
ρw g
Find the average density of the fish.
m
mf g g
m g
200.0 N
ρf = f =
= f ρw =
(1.00 × 103 kg m3 ) = 1080 kg m3
Vf
mw g ( ρ w g ) mw g
185.0 N
39. Strategy Let the +y-direction be upward. Use Newton’s second law and Eq. (9-7).
Solution
(a) ΣFy = FB − mg = ma, so
⎛ 1.00 g cm3
⎞
⎛ρ V
⎞
ρ gV
F
− g = g ⎜ w − 1⎟ = (9.80 m s 2 ) ⎜
− 1⎟ = 9.8 m s 2 .
a= B −g = w
3
⎜
⎟
m
m
⎝ ρV
⎠
⎝ 0.50 g cm
⎠
G
2
Thus, a = 9.8 m s upward .
⎛ 1.00 g cm3
⎞
G
− 1⎟ = 3.3 m s 2 , so a = 3.3 m s 2 upward .
(b) a = (9.80 m s 2 ) ⎜
3
⎜ 0.750 g cm
⎟
⎝
⎠
⎛ 1.00 g cm3
⎞
G
−
1
(c) a = (9.80 m s 2 ) ⎜
⎟ = 68.6 m s 2 , so a = 68.6 m s 2 upward .
⎜ 0.125 g cm3 ⎟
⎝
⎠
40. Strategy Let the +y-direction be downward. Use Newton’s second law and Eq. (9-7).
Solution Find the initial acceleration of the piece of metal.
⎛ ρ V
ρ gV
F
ΣFy = mg − FB = ma, so a = g − B = g − w
= g ⎜1 − w
ρV
m
m
⎝
G
Thus, a = 0.80 g downward .
⎞
⎟=
⎠
FB
1 ⎞
⎛
g ⎜1 −
⎟ = 0.80 g.
5.0
⎝
⎠
41. Strategy Use Eq. (9-13).
Solution Find the speed of the water as it passes through the nozzle.
2
πr2
A
⎛ 1.0 cm ⎞
A2 v2 = A1v1 , so v2 = 1 v1 = 1 v1 = ⎜
⎟ ( 2.0 m s ) = 50 m s .
A2
⎝ 0.20 cm ⎠
π r22
464
y
mg
Physics
Chapter 9: Fluids
42. Strategy Use Eq. (9-12).
Solution Find the average speed of the blood in the aorta.
1 ∆V 8.5 × 10−5 m3 s
v=
=
= 28 cm s
A ∆t
3.0 × 10−4 m 2
43. (a) Strategy Use Eq. (9-13).
Solution Find the speed of the water in the hose.
2
2
⎛r ⎞
A
πr2
⎛ 1.00 mm ⎞
v2 = 1 v1 = 1 2 v1 = ⎜ 1 ⎟ v1 = ⎜
⎟ (25.0 m s) = 39.1 cm s
A2
π r2
⎝ 8.00 mm ⎠
⎝ r2 ⎠
(b) Strategy Use Eq. (9-12).
Solution Compute the volume flow rate.
∆V
= A1v1 = π (1.00 × 10−3 m) 2 (25.0 m s) = 78.5 cm3 s
∆t
(c) Strategy Use Eq. (9-11).
Solution Compute the mass flow rate.
∆m
= ρ A1v1 = (1.00 g cm3 )(78.5 cm3 s) = 78.5 g s
∆t
44. Strategy Use Eq. (9-13).
Solution Find the speed of the water at the taper.
2
2
⎛d ⎞
A
πr2
⎛ 0.010 m ⎞
v2 = 1 v1 = 1 2 v1 = ⎜ 1 ⎟ v1 = ⎜
⎟ ( 0.20 m s ) = 3.2 m s
A2
π r2
⎝ 0.0025 m ⎠
⎝ d2 ⎠
45. Strategy Use Eqs. (9-13) and (9-14). Since the pipe is horizontal, y1 = y2 .
Solution Let the larger end be labeled 2. Find the speed of the water at the narrow end in terms of the speed at the
larger end.
A
A1v1 = A2 v2 , so v1 = 2 v2 .
A1
Find the pressure at the narrow end of the segment of pipe.
2⎤
⎡
1
1
1 ⎢ 2 ⎛ A2 ⎞ ⎥
2
2
P1 + ρ v1 = P2 + ρ v2 , so P1 = P2 + ρ v2 − ⎜⎜
v2 ⎟⎟
2
2
2 ⎢
⎝ A1 ⎠ ⎥⎦
⎣
2⎤
⎡ ⎛
1
50.0 cm 2 ⎞ ⎥
= 1.20 × 105 Pa + (1.00 × 103 kg m3 )(0.040 m s)2 ⎢1 − ⎜
⎟ = 1.12 × 105 Pa .
⎢ ⎜⎝ 0.500 cm 2 ⎟⎠ ⎥
2
⎣
⎦
465
Chapter 9: Fluids
Physics
46. Strategy Use Eq. (9-14) to find the pressure difference at the roof.
Solution Let the region above the roof be labeled 1. Assume the air under the roof is still. Since y1 ≈ y2 , assume
that the difference in height has negligible effect on the pressure. With these assumptions, the pressure difference
is ∆P = 12 ρ v12 . Therefore, the magnitude of the force on the roof is
1
ρ v 2A
2 air 1
2
2
2
2
1
⎛ 150 mi ⎞ ⎛ 1 h ⎞ ⎛ 1609 m ⎞
2 ⎛ 1m ⎞
5
= (1.20 kg m3 ) ⎜
⎟ ⎜
⎟ ⎜
⎟ (2000 ft ) ⎜
⎟ = 5.0 × 10 N .
2
1
h
3600
s
1
mi
3.281
ft
⎝
⎠ ⎝
⎠ ⎝
⎠
⎝
⎠
F = ∆PA =
47. Strategy Use Eq. (9-14).
Solution The potential energy difference is relatively small, so Bernoulli’s equation becomes
1
1
1
1
P1 + ρ v12 = P2 + ρ v22 , or P1 − P2 = ρ v22 − ρ v12 .
2
2
2
2
Estimate the force.
1
1
⎛1
⎞
F = ∆PA = ( P1 − P2 ) A = ⎜ ρ v22 − ρ v12 ⎟ A = Aρ (v22 − v12 )
2
2
⎝2
⎠
1
= (28 m 2 )(1.3 kg m3 )[(190 m s)2 − (160 m s)2 ] = 1.9 × 105 N
2
48. (a) Strategy According to Newton’s second law, the weight of the plane is equal to the force due to the pressure
difference.
Solution Estimate the weight of the plane.
Weight = F = ∆PA = (500 Pa)(200 m 2 ) = 1.0 × 105 N
(b) Strategy Use Eq. (9-14) with y1 = y2. Use the density of air from Problem 47.
Solution Estimate the air speed above the wings.
1
1
2∆P
2(500 Pa)
ρ v12 − ρ v22 = ∆P, so v1 =
+ v22 =
+ (80.5 m s)2 = 85 m s .
ρ
2
2
1.3 kg m3
49. Strategy Use Eqs. (9-13) and (9-14).
Solution y1 =
1
∆P = ρ v12 =
2
y2 and v1 >> v2 since A2 >> A1 (d 2 >> d1), so Bernoulli’s equation gives
1
(1.0 × 103 kg m3 )(25 m s)2 = 310 kPa .
2
466
Physics
Chapter 9: Fluids
50. Strategy Use Eq. (9-13) to relate the speeds at points A and B. Then, use Eq. (9-14) to find the speed of the air at
point B.
Solution Relate the air speeds at points A and B.
AB vB = AAv A , so v A =
AB
dB2
vB =
v B = 9v B .
AA
(d B 3)2
Find the speed of the air at point B. Note that y A = yB .
1
1
ρ vB 2 = PA + ρ v A2
2
2
1
81
2
ρ v = PA − PB + ρ vB 2
2 B
2
40 ρa vB 2 = ρ w gh
PB +
vB =
ρ w gh
(1.00 × 103 kg m3 )(9.80 m s 2 )(0.0175 m)
=
= 1.82 m s
40 ρa
40(1.29 kg m3 )
51. Strategy Use Eq. (9-12) to determine the speed of the water at the faucet. Then, use Eq. (9-14) to find the height
difference. Assume that the diameter of the tower is so large compared to that of the faucet that the water at the
top of the tower does not move. Also, the pressure at the tower and the faucet is the same.
Solution Let the tower be labeled as 1. Find the speed of the water at the faucet.
2
1
1 ∆V 4 π dc hc dc 2 hc
v2 =
=
=
1 π d 2 ∆t
A ∆t
df 2 ∆t
f
4
2
1
1 ⎛ d 2h ⎞
With the above assumptions, Bernoulli’s equation becomes ρ gy1 = ρ gy2 + ρ v22 = ρ gy2 + ρ ⎜ c c ⎟ , so
2
2 ⎜⎝ df 2 ∆t ⎟⎠
2
2
⎡ (44 cm)2 (0.52 m) ⎤
1 ⎛ dc 2 hc ⎞
1
the height difference is y1 − y2 =
⎜
⎟ =
⎢
⎥ = 8.6 m .
2 g ⎜⎝ df 2 ∆t ⎟⎠
2(9.80 m s 2 ) ⎢⎣ (2.54 cm)2 (12 s) ⎥⎦
52. (a) Strategy The rate at which the well does work on the water is equal to the potential energy change of the
water per unit time.
Solution Find the power output of the pump.
∆W mg ∆y
∆V
=
= ρ g ∆y
= (1.0 × 103 kg m3 )(9.80 m s 2 )(40.0 m)(2.0 × 10−4 m3 s) = 78 W
∆t
∆t
∆t
(b) Strategy Assume that the speed of the water at the top and bottom of the well is zero. Use Eq. (9-14).
Solution Find the pressure difference the pump must maintain.
∆P = ρ g ∆y = (1.00 × 103 kg m3 )(9.80 m s 2 )(40.0 m) = 392 kPa
(c) Strategy and Solution Since ∆P is greater than atmospheric pressure, the pump must be at the bottom so
that it can push the water up.
467
Chapter 9: Fluids
Physics
53. Strategy Use Eq. (9-15).
Solution Show that viscosity has SI units of pascal-seconds.
Solve for η.
∆P
∆V π L 4
π ∆P ∆ t 4
Pa s
⋅
⋅ m4 = Pa ⋅ s.
r , so η =
r . Thus, the units of η are
=
m m3
∆t
8 η
8 L ∆V
54. Strategy Use Eq. (9-15). Form a ratio of the volume flow rates.
Solution Find the total flow rate in the two pipes compared to the flow rate through the single pipe.
π∆P ⎡ D 4 + D 4 ⎤
∆V
1
1
8Lη ⎢ 4
4 ⎥
∆t total
⎣
⎦ = 256 + 256 = 16 = 1 the original flow rate
=
1
∆V
128
8
π∆P D 4
16
( )
( ∆t )original
( ) ( )
8 Lη ( 2 )
55. (a) Strategy Use Eq. (9-15).
Solution Find the pressure of the fluid in the syringe, Ps .
∆V π∆Pr 4
⎛ 8η L ⎞ ∆V
=
, so ⎜ 4 ⎟
= Ps − Pv and
∆t
8η L
⎝ π r ⎠ ∆t
3
⎛ 1.013 × 105 Pa ⎞ 8(2.00 × 10−3 Pa ⋅ s)(0.0300 m)
⎛ m ⎞
3
(0.250
cm
s)
Ps = (16.0 mm Hg) ⎜
⎟+
⎜ 100 cm ⎟ = 6850 Pa .
⎜ 760.0 mm Hg ⎟
π (0.000300 m)4
⎝
⎠
⎝
⎠
(b) Strategy Use the definition of average pressure.
Solution Find the force the must be applied to the plunger.
F = Pav A = (6850 Pa)(1.00 cm2 )(10−2 m cm)2 = 0.685 N
56. Strategy and Solution The volume flow rate for each of the pipes in system C is one quarter that of the pipe in
system A, since the total rates are the same and system C has four times as many pipes. So, since the flow speed
in each of the pipes in C is 3.0 m/s, the flow speed in A must be four times this, or 12 m/s.
57. Strategy Use Eq. (9-15). Form a ratio of the volume flow rates.
Solution Find the total flow rate in system C.
⎡ π∆P r 4 ⎤
4 ⎢ CL ⎥
∆V
5
∆t C
⎣ 8η ( 2 ) ⎦ = 4∆PC , so ∆V = 4∆PC ⎛ ∆V ⎞ = 4 ⎛⎜ 2.0 × 10 Pa ⎞⎟ (0.020 m3 s) = 0.040 m3 s .
=
∆V
⎜ 4.0 × 105 Pa ⎟
⎛ π∆P r 4 ⎞ ∆PB
∆t C ∆PB ⎜⎝ ∆t B ⎟⎠
⎝
⎠
∆t B
2 ⎜ 8ηBL ⎟
⎝
⎠
58. Strategy Use Eq. (9-15). Form a ratio of the volume flow rates.
Solution Find the pressure supplied by the pump in system A.
π∆PA r 4
8η L
⎡ π∆P r 4 ⎤
2 ⎢ BL ⎥
⎣ 8η ( 2 ) ⎦
=
P −P
∆PA
= A atm =
4∆PB 4( PB − Patm )
∆V
∆t A
∆V
∆t B
= 1, so PA = 4 PB − 3Patm = [4(5.0) − 3(1.0)] × 105 Pa = 17 × 105 Pa .
468
Physics
Chapter 9: Fluids
59. Strategy Use Eqs. (9-12) and (9-15). In (c), use the conversion factor for Pa and torr.
Solution
(a) Find the pressure difference required to make blood flow through an artery of the given size.
π∆Pr 4 ∆V
8η Lv 8(2.1× 10−3 Pa ⋅ s)(0.20 m)(0.060 m s)
⎛ 8η L ⎞
=
= Av, so ∆P = ⎜ 4 ⎟ π r 2v = 2 =
= 50 Pa .
∆t
8η L
r
(0.0020 m)2
⎝πr ⎠
(b) Find the pressure difference required to make blood flow through a capillary of the given size.
8(2.1× 10−3 Pa ⋅ s)(0.0010 m)(0.00060 m s)
∆P =
= 1100 Pa
(3.0 × 10−6 m)2
(c) Compare the answers in (a) and (b) to average blood pressure.
⎛ 101.3 kPa ⎞
(100 torr) ⎜
⎟ = 13 kPa
⎝ 760.0 torr ⎠
100 torr is approximately 13 kPa, which is much larger than both pressures found in parts (a) and (b).
60. Strategy Use Eq. (9-15).
Solution
(a) Show that Poiseuille’s law can be written in the form ∆P = IR.
π∆Pr 4 ∆V
∆V ⎛ 8η L ⎞
∆V
8η L
and R = 4 .
=
= IR. Therefore, ∆P = IR where I =
, so ∆P =
∆t
∆t
∆t ⎜⎝ π r 4 ⎟⎠
8η L
πr
(b) From part (a), R =
8η L
π r4
.
61. Strategy Use Eq. (9-16) and Newton’s second law.
Solution Find the viscosity of the second liquid.
4
ΣFy = FD + FB − ms g = 6πη rv + ml g − ms g = 6πη rv + (ml − ms ) g = 6πη rv − π r 3 ( ρs − ρl ) g = 0,
3
4 π r 3(ρ
s
3
− ρl ) g
FB
2r 2 ( ρ s − ρ l ) g
.
6π rv
9v
Find the viscosity of the second liquid by forming a proportion.
so η =
η2
=
η1
2r 2 ( ρs − ρ l ) g
9v2
2
2 r ( ρs − ρ l ) g
9v1
=
=
FD
y
m sg
v1
v
η
0.5 Pa ⋅ s
, so η2 = 1 η1 = 1 =
= 0.4 Pa ⋅ s .
v2
1.2v1
1.2
1.2
62. Strategy Use Eq. (9-16) and Newton’s second law.
Solution Find the viscosity of the liquid.
FB
4
⎛
⎞
ΣFy = FD + FB − ms g = 6πη rv + ml g − ms g = 6πη rv − (ms − ml ) g = 6πη rv − ⎜ ms − π r 3ρl ⎟ g = 0,
3
⎝
⎠
3
3
2
3
4
4
ms − 3 π r ρl g ⎡0.0120 kg − 3 π (0.010 m) (1200 kg m ) ⎤ (9.80 m s )
⎦
so η =
=⎣
= 2.4 Pa ⋅ s .
6π rv
6π (0.010 m)(0.15 m s)
(
)
469
FD
m sg
y
Chapter 9: Fluids
Physics
63. (a) Strategy Use Eq. (9-16).
Solution Find the drag force on the dinoflagellate in seawater.
1.0 × 10−3 m
FD = 6πη rv = 6π (0.0010 Pa ⋅ s)(35.0 × 10−6 m)
= 1.3 × 10−10 N
5.0 s
(b) Strategy Assuming the flagellate travels the 1.0 mm with constant speed, the force with which it pushes on
the water is equal in magnitude to the drag force. To find the power output, divide the work done by the
flagellate by the time of travel.
Solution Find the power output of the flagellate.
W FD d (1.3 × 10−10 N)(1.0 × 10−3 m)
P=
=
=
= 2.6 × 10−14 W
∆t
∆t
5.0 s
64. Strategy Use Eq. (9-16) and Newton’s second law.
Solution Find the terminal velocity of the air bubble.
ΣFy = FB − FD − ma g = mo g − 6πη rvt − ma g = 0, so
vt =
(mo − ma ) g
=
6πη r
4 π r 3(ρ
o
3
− ρa ) g
6πη r
=
FB
y
2
2r (0.85ρ w − ρa ) g
9η
2(0.0010 m) 2 (0.85 × 103 kg m3 − 1.20 kg m3 )(9.80 m s 2 )
=
= 1.5 cm s .
9(0.12 Pa ⋅ s)
FD
mag
65. Strategy For a viscous drag force, vt ∝ m. For a turbulent drag force, vt ∝ m . For the data in the table,
compute m vt and m vt 2 .
Solution The data, m vt , and m vt 2 are organized in the table.
m (g)
vt (cm s)
m
(g ⋅ s cm)
vt
m
vt 2
(g ⋅ s 2 cm 2 )
8
12
16
20
24
28
1.0
1.5
2.0
2.5
3.0
3.5
8
8.0
8.0
8.0
8.0
8.0
8
5.3
4.0
3.2
2.7
2.3
Since m vt is constant, the drag force is primarily viscous.
470
Physics
Chapter 9: Fluids
66. Strategy For a viscous drag force, vt ∝ m. For a turbulent drag force, vt ∝ m . For the data in the table,
compute m vt and m vt 2 .
Solution The data, m vt , and m vt 2 are organized in the table.
m (g)
5.0
11.3
20.0
31.3
45.0
80.0
vt (cm s)
1.0
1.5
2.0
2.5
3.0
4.0
m
(g ⋅ s cm)
vt
5.0
7.5
10
13
15
20
5.0
5.0
5.0
5.0
5.0
5.0
m
vt
2
(g ⋅ s 2 cm 2 )
Since m vt 2 is constant, the drag force is primarily turbulent.
67. Strategy Use Eq. (9-16) and Newton’s second law.
Solution Find the terminal velocity for the droplets.
ΣFy = FD + FB − mw g = 6πη rvt + ma g − mw g
4
= 6πη rvt − (mw − ma ) g = 6πη rvt − π r 3 ( ρ w − ρa ) g = 0, so
3
vt =
=
4 π r 3(ρ
w
3
− ρa ) g
6πη r
9(1.8 × 10−5 Pa ⋅ s)
y
m wg
2r 2 ( ρ w − ρ a ) g
=
9η
2(5.0 × 10−6 m) 2 (1.0 × 103 kg m3 − 1.20 kg m3 )(9.80 m s 2 )
FD
FB
= 3.0 mm s .
68. Strategy Use Eq. (9-16) and Newton’s second law.
Solution Find the terminal speed of the air bubble.
Aluminum sphere: ΣFy = F1D + F1B − m1g = 0
F1B
F1D
Air bubble: ΣFy = F2B − F2D − m2 g = 0
F2D
F2D 6πη rv2 F2B − m2 g mw g − m2 g
, so
=
=
=
F1D 6πη rv1 m1g − F1B
m1g − mw g
ρ
1.20
1− m2
1− ρ a
1 − 1001.8
mw − m2
w
v2 =
v1 = m
v1 = 2.7 ρ w v1 =
( 5.0 cm s ) = 2.9 cm s .
1
w
2.7 − 1
m1 − mw
1
−
1
−
m
ρ
w
y
y
Divide F2D by F1D.
m
F2B
w
69. Strategy Use Eq. (9-17).
Solution Form a proportion to find the pressure inside the air bubble.
2γ
∆P2
r
r
r
= 22γ , so ∆P2 = ∆P1 1 = (10 Pa) 1 = 5 Pa .
2r1
∆P1
r2
r1
471
m 1g
m2g
Chapter 9: Fluids
Physics
70. (a) Strategy Use Eq. (9-17) and the definition of average pressure.
Solution Find the maximum possible upward force on the foot due to surface tension of the water.
2γ
π r 2 = 2πγ r = 2π (0.070 N m)(0.02 × 10−3 m) = 9 × 10−6 N
F = ∆PA =
r
(b) Strategy Use Eq. (9-17) and Newton’s second law.
Solution Find the maximum mass of the water strider.
6 F 12πγ r 12π (0.070 N m)(0.02 × 10−3 m)
ΣF = 6 F − mg = 0, so m =
=
=
= 5 mg .
g
g
9.80 m s 2
71. Strategy Consider the relationships between work, force, energy, and displacement. Use Eq. (9-17).
Solution
(a) γ is the force per unit length with which the surface pulls on its edge, so W = F ∆s = γ L∆s .
(b) W = ∆E and ∆A = L∆s, so ∆E = γ ∆A , where ∆E is the increase in surface energy.
(c) γ ∆A = ∆E , so γ =
(d) F ∆s = γ L∆s, so
∆E
. Therefore, γ can be thought of as the surface energy (∆E ) per unit area (∆A).
∆A
F
∆E
= γ . Therefore, γ has units N/m. Since γ =
, γ has units J m2.
∆A
L
72. (a) Strategy and Solution The net force on the hemisphere is vertical and equal to the force on the flat surface
of the hemisphere. Each horizontal component of the force has an equal and opposite horizontal component at
the opposite side of the hemisphere that cancels its contribution to the net force. The flat part of the
hemisphere ultimately has the net vertical force exerted on it via the body of the hemisphere.
F = AP = π r 2 P
(b) Strategy Use Eq. (9-17) and the definition of average pressure.
Solution Show that the air pressure inside the bubble must exceed the water pressure outside by ∆P = 2γ r .
F 2π rγ 2γ
∆P = =
=
A π r2
r
472
Physics
Chapter 9: Fluids
73. (a) Strategy The mass of the water is equal to its density times its volume.
Solution The weight of the water in the straw is
mg = ( ρV ) g = ρ (π r 2 h) g = (1.00 × 103 kg m3 )π (0.00250 m)2 (8.00 m)(9.80 m s 2 ) = 1.54 N .
(b) Strategy Equate the pressures and solve for the force on the top of the barrel. Use the definition of pressure.
Solution Find the force with which the water in the barrel pushes up on the top of the barrel.
A
r2
(25.0 cm) 2
(1.54 N) = 1.54 × 104 N .
Fb = b Fs = b Fs =
2
As
rs
(0.250 cm)2
(c) Strategy Consider the nature of pressure in a column of fluid.
Solution For a given depth, the pressure is the same everywhere, so the very tall, narrow column of water is
as effective as having a whole barrel of water filled to the same height and pushing upward on the barrel top.
74. (a) Strategy Compute the volume of the block and use its density ( ρ = 2702 kg m3 ) to find its mass. Then,
compute its weight.
Solution The volume of the block is (0.0200 m)(0.0300 m)(0.0500 m) = 3.00 × 10−5 m3 . The weight of the
block is mg = ρVg = (2702 kg m3 )(3.00 × 10−5 m3 )(9.80 m s 2 ) = 0.794 N .
(b) Strategy Find the weight of the oil displaced by the block and subtract it from the weight of the block to find
the reading of the scale.
Solution The weight of the displaced oil is moil g = ρoilVblock g . Thus, the scale reading is
0.794 N − ρoilVblock g = 0.794 N − (850 kg m3 )(3.00 × 10−5 m3 )(9.80 m s 2 ) = 0.544 N .
75. (a) Strategy The canoe displaces 85.0 kg of water.
Solution
The volume of water displaced is V =
m
ρ
=
85.0 kg
1.00 × 103 kg m3
= 0.0850 m3 .
1
1
The volume of the canoe is V = π r 2 h = π (0.475 m) 2 (3.23 m) = 1.14 m3 .
2
2
0.0850
The percentage of the volume of the canoe below the waterline is
× 100% = 7.43% .
1.1447
(b) Strategy The canoe will begin to sink when it displaces a volume of water equal to its own volume.
Solution The total mass of a volume of water equal to the volume of the canoe is
1
m = ρV = (1.00 × 103 kg m3 ) π (0.475 m)2 (3.23 m) = 1140 kg. The additional mass that can be placed in
2
the canoe is 1145 kg − 85.0 kg = 1060 kg .
473
Chapter 9: Fluids
Physics
76. Strategy Consider how the densities of pine and steel compare to water.
Solution The density of pine is less than that of water, so the block of pine will float, displacing only an amount
of water that is equal to its weight. Thus, the scale reading for the pine doesn’t change. The density of steel is
greater than that of water, so the block of steel will be fully submerged, displacing an amount of water equal to its
own volume. Thus, the scale reading for the steel will increase by the difference in densities times the volume of
the block times g.
( ρs − ρ w )Vg = (7860 kg m3 − 1.00 × 103 kg m3 )(8.00 × 10−6 m3 )(9.80 m s 2 ) = 0.538 N
77. (a) Strategy Assume that the change in the height of the water level in the vat is negligible and that the
pressures at the top of the vat and at the outlet are the same. Use Bernoulli’s equation.
Solution Let the top of the vat be labeled 1. With the above assumptions, Bernoulli’s equation becomes
1
ρ gy1 = ρ gy2 + ρ v22 , so v2 = 2 g ( y1 − y2 ) = 2(9.80 m s 2 )(1.80 m) = 5.94 m s .
2
(b) Strategy and Solution The density “falls out” of Bernoulli’s equation in our calculation of the speed, so
as long as we can assume Bernoulli’s equation applies, it doesn’t matter what fluid is in the vat.
(c) Strategy and Solution Since the speed is directly proportional to the square root of the gravitational field
strength, 1.6 9.80 = 0.40, the speed would be reduced by a factor of 0.40.
78. Strategy The magnitude of the force on the spring is equal to the difference between the weight of the displaced
water and the weight of the cube. Use Hooke’s law to find the spring constant.
Solution
F ( ρ − ρc )Vg (1.00 × 103 kg m3 − 8.00 × 102 kg m3 )(0.0400 m)3 (9.80 m s 2 )
k= = w
=
= 12.5 N m
x
x
0.0100 m
79. Strategy The mass of the barrel is equal to the mass of the water it displaces. The extra mass that can be put in
the barrel is equal to the mass of a volume of water equal to the volume of the barrel between 33% of its length
from the bottom to 30 cm below its top.
Solution The volume of the barrel is Vb = 14 π d 2 h, where h is the height of the barrel. Thirty-three percent of the
barrel is submerged when empty and 30 cm of the barrel must remain above water, so
1.20 m − 0.33(1.20 m) − 0.30 m = 0.504 m more of the barrel can be submerged to accommodate the extra mass.
This volume is V = 14 π d ′2 h. The extra mass is equal to this volume times the density of water.
1
1
m = ρV = ρπ d ′2 h = (1.00 × 103 kg m3 )π (0.76 m)2 (0.504 m) = 230 kg
4
4
80. (a) Strategy Use Eq. (9-4).
Solution Find the water pressure.
P = Patm + ρ gd = 1.013 × 105 Pa + (1025 kg m3 )(9.80 m s2 )(10.915 × 103 m) = 1.10 × 108 Pa
(b) Strategy Use the definition of average pressure.
Solution Compute the force on the sub’s hull.
F = Pav A = (1.1× 108 Pa)(1.0 m 2 ) = 1.1× 108 N
474
Physics
Chapter 9: Fluids
81. Strategy Use Eq. (9-14) with v1 = v2 = 0.
Solution Find the height between the basement and the seventh floor.
P −P
4.10 × 105 Pa − 1.85 × 105 Pa
P1 + ρ gy1 = P2 + ρ gy2 , so y1 − y2 = 2 1 =
= 23.0 m .
ρg
(1.00 × 103 kg m3 )(9.80 m s 2 )
82. Strategy Divide the mass of fat by the total mass and multiply by 100%.
Solution Find the percentage of the person’s body weight that is composed of fat.
ρV
(890 kg m3 )(0.020 m3 )
mass of fat
× 100% =
× 100% = 20%
× 100% =
90.0 kg
m
total mass
83. Strategy Use Eq. (9-3).
Solution Find the height of a hill you must ascend for the barometer to read a pressure drop of 1.0 cm Hg.
⎛ 1.013 × 105 Pa ⎞
∆P
1.0 cm Hg
∆P = ρ gh, so h =
=
⎜
⎟ = 110 m .
ρ g (1.20 kg m3 )(9.80 m s 2 ) ⎜⎝ 76.0 cm Hg ⎟⎠
84. Strategy The weight in air is decreased by an amount equal to the weight of the displaced fluid.
Solution Find the scale reading for each situation.
(a) Wscale = W − ρ w gV = W −
(b)
ρs
2.50
gV = W −
1
1 ⎞
⎛
ms g = W ⎜ 1 −
⎟ = 0.600W
2.50
⎝ 2.50 ⎠
ρs
ρ
ρ
ρ
0.90
= 2.50 and o = 0.90, so s = o , or ρ o =
ρs .
2.50 0.90
2.50
ρw
ρw
Wscale = W − ρo gV = W −
0.90
0.90
⎛ 0.90 ⎞
ρs gV = W −
ms g = W ⎜1 −
⎟ = 0.64W
2.50
2.50
⎝ 2.50 ⎠
85. Strategy Use the continuity equation. The speed of an object that has fallen a distance h from rest is v = 2 gh .
Solution Find the diameter of the water flow after the water has fallen 30 cm.
v1
1
1
0.62 m s
= (2.2 cm)
= 1.1 cm .
A1v1 = π d12 v1 = A2 v2 = π d 22 v2 , so d 2 = d1
4
4
2 gh
2(9.80 m s 2 )(0.30 m)
86. Strategy Use Eq. (9-13).
Solution Find vc , the average speed of the blood in the capillaries.
2
vc =
2
π ra 2
Aa
1 ⎛ ra ⎞
1 ⎛ 0.01 m ⎞
=
=
va =
v
v
⎜
⎟
a
a
⎜
⎟ (0.3 m s) = 0.4 mm s
Ac
N ⎝ rc ⎠
N π rc 2
2 × 109 ⎝ 6 × 10−6 m ⎠
87. Strategy Use Eq. (9-15).
Solution Find the pressure drop between the ends of the aorta.
π∆Pr 4 ∆V
⎛ 8η L ⎞ ∆V 8(4.0 × 10−3 Pa ⋅ s)(0.400 m)(4.1× 10−3 m3 s)
=
=
= 27 kPa .
, so ∆P = ⎜
⎟
8η L
∆t
π (0.0050 m)4
⎝ π r 4 ⎠ ∆t
475
Chapter 9: Fluids
Physics
88. Strategy Use Eq. (9-14) with y1 = y2 and Eq. (9-12).
Solution Find how much on average the blood pressure inside the aneurysm is higher than in the unenlarged part
of the aorta.
1
1
P2 + ρ v22 = P1 + ρ v12
2
2
1
P2 − P1 = ρ (v12 − v22 )
2
2
1 ⎛ ∆V ⎞ ⎛ 1
1 ⎞
∆P = ρ ⎜
⎜ 2− 2⎟
⎟
2 ⎝ ∆t ⎠ ⎝⎜ A1
A2 ⎠⎟
2
2
⎤
1 ⎞ 1060 kg m3
1
1
ρ ⎛ ∆V ⎞ ⎛ 1
3 2⎛ 1m ⎞ ⎡
(120
cm
s)
=
−
=
−
⎜
⎟
⎢
⎥
⎜
⎟
⎜
⎟
2 ∆t
4
4⎟
2
4
4
⎜
r2 ⎠
2π ⎝
2π
(3.0 cm) ⎥⎦
⎠ ⎝ r1
⎝ 100 cm ⎠ ⎢⎣ (1.0 cm)
= 76 Pa
89. (a) Strategy Use Eqs. (9-7), (9-16), and Newton’s second law.
Solution Find the terminal velocity of the bubbles.
ΣFy = FB − FD − ma g = ρ w gV − 6πη rvt − ρa gV =
vt =
=
( ρ w − ρa ) 43 π r 3 g
=
= 0, so
2r 2 g ( ρ w − ρa )
6πη r
9η
2(1.0 × 10−3 m) 2 (9.80 m s 2 )(1.00 × 103 kg m3 − 1.20 kg m3 )
G
Thus, v t = 2.2 m s up .
9(1.0 × 10−3 Pa ⋅ s)
y
FB
−6πη rvt + ( ρ w − ρa ) 43 π r 3 g
FD
mag
= 2.2 m s.
(b) Strategy Divide the change in pressure by the change in time and use the result from part (a).
Solution
∆P = ρ g ∆y, so
∆P
∆y
= ρg
= ρ gvt = (1.00 × 103 kg m3 )(9.80 m s 2 ) ( 2.175 m s ) = 21 kPa s .
∆t
∆t
90. (a) Strategy Since a pump maintains a pressure difference between the outflow and intake pressures, the
maximum depth is given by Patm = ρ gd max .
Solution Find the maximum depth of a well for which a surface pump will work.
P
1.013 × 105 Pa
d max = atm =
= 10.3 m
ρ g (1.00 × 103 kg m3 )(9.80 m s 2 )
(b) Strategy Consider the differences between a pump at the surface and a pump at the bottom of a well.
Solution
A pump at the bottom of a well does not rely on a pressure difference to bring the water to the surface;
it pushes the water up from below.
476
Physics
Chapter 9: Fluids
91. (a) Strategy Use Eq. (9-2).
Solution Find the weight of the beach ball.
4
4
⎡
⎤
Weight = mb g + π r 3ρa g = (9.80 m s 2 ) ⎢0.10 kg + π (0.200 m)3 (1.3 kg m3 ) ⎥ = 1.4 N
3
3
⎣
⎦
(b) Strategy The buoyant force is equal to the mass of displaced air and is directed upward. Use Eq. (9-7).
Solution Find the buoyant force on the beach ball.
G
4
FB = ρ gV = (1.3 kg m3 )(9.80 m s 2 ) π (0.200 m)3 = 0.43 N, so FB = 0.43 N upward .
3
(c) Strategy Use Newton’s second law.
Solution Find the acceleration of the beach ball at the top of its trajectory.
ΣFy = FB − mg = ma, so
(
FB
)
⎡
⎤
ρa gV
F − mg FB
g
m −1
a= B
=
−g =
−g = m
− g = g ⎢ 1 + ρ Vb
− 1⎥
b
a
m
m
mb + ρaV
+1
⎣
⎦
ρV
y
v=0
mg
a
−1
⎧
⎫
⎤
⎪
0.10 kg
2 ⎪⎡
2
= (9.80 m s ) ⎨ ⎢1 +
−
1
⎬ = − 6.8 m s .
⎥
3 4
3
(1.3
kg
m
)
(0.200
m)
π
3
⎣
⎦
⎩⎪
⎭⎪
G
Thus, a = 6.8 m s 2 downward .
92. Strategy Since the net force is zero, the weight of the lead plus the weight of the wood is equal to the weight of
the displaced water. In addition, the volume of the lead is equal to the volume of the water displaced less the
volume of the wood.
Solution Find the volume of the displaced water in terms of the mass
of the lead and wood.
+ mwood
m
mlead g + mwood g = mw g = ρ w gVw , so Vw = lead
.
FB
ρw
Find the mass of the lead.
Vw − Vwood = Vlead
mlead + mwood
m
− Vwood = lead
ρw
mwoodg
mleadg
ρlead
mlead + mwood − ρ wVwood =
⎛
ρ
mlead ⎜⎜1 − w
⎝ ρlead
ρw
m
ρlead lead
⎞
⎟⎟ = ρ wVwood − mwood = ρ wVwood − ρ woodVwood
⎠
( ρ − ρ wood )Vwood (1.00 × 103 kg m3 − 0.78 × 103 kg m3 )(0.330 m)3
mlead = w
=
= 8.7 kg
ρ
1.00×103 kg m3
1− ρ w
1−
3
3
11.3×10 kg m
lead
477
Chapter 9: Fluids
Physics
93. Strategy Use Eq. (9-7) and Newton’s second law.
Solution Find an expression for d.
ΣFy = FB − mg = ρ gV − mg = ρ gAd − mg = ρ g (π r 2 )d − mg = 0, so d =
d is not a linear function of ρ : d =
m
πρ r 2
m
πρ r 2
.
.
94. (a) Strategy Let V be the volume of the liquid displaced and Vh be the volume of the hydrometer. Then
Vh − V = Ah where A is the cross-sectional area of the stem and h is the height above the liquid. Use
Newton’s second law.
Solution Find the distance from the top of the cylinder where the mark should be placed.
ΣFy = FB − mh g = ρ gV − mh g = (S.G.)ρ w gV − mh g = 1.00 ρ w g (Vh − Ah) − mh g = 0, so
h=
1⎛
mh ⎞
1
⎜ Vh −
⎟=
A⎝
1.00 ρ w ⎠ 0.400 cm 2
⎡
⎤
4.80 g
3
= 10.0 cm .
⎢8.80 cm −
3 ⎥
1.00(1.00 g cm ) ⎥⎦
⎢⎣
(b) Strategy Use the results of part (a),
Solution Find the specific gravity of the alcohol.
(S.G.)ρ w gV − mh g = (S.G.)ρ w g (Vh − Ah) − mh g = 0, so
mh
4.80 g
S.G. =
=
= 0.814 .
3
ρ w (Vh − Ah) (1.00 g cm )[8.80 cm3 − (0.400 cm 2 )(7.25 cm)]
(c) Strategy For the minimum S.G., the volume of the displaced liquid is equal to the volume of the hydrometer.
Solution Find the lowest specific gravity that can be measured with this hydrometer.
mh
4.80 g
(S.G.)ρ wVh = mh , so S.G.min =
=
= 0.545 .
ρ wVh (1.00 g cm3 )(8.80 cm3 )
95. Strategy Use Eqs. (9-13) and (9-14).
Solution
(a) Find the speed of the water as it exits the showerhead (v1).
P1 + ρ gy1 +
1
1
ρ v12 = P2 + ρ gy2 + ρ v22
2
2
1
ρ (v12 − v22 ) = P2 − P1 + ρ g ( y2 − y1)
2
2 Pgauge
A2
− 2 gh
v12 − 1 2 v12 =
ρ
A2
2 Pgauge
v1 =
ρ
− 2 gh
1 − ⎡ N π r12 (π r22 ) ⎤
⎣
⎦
2
=
2(410 ×103 Pa)
1.00 ×103 kg m3
− 2(9.80 m s 2 )(6.7 m)
(
)
4
mm
1 − 362 0.33
6.3 mm
(b) Find the speed of the water as it moves through the output pipe of the pump.
2
v2 =
A1
N π r12
⎛ 0.33 mm ⎞
v1 =
v1 = 36 ⎜
⎟ (26 m s) = 2.6 m s
2
A2
π r2
⎝ 6.3 mm ⎠
478
= 26 m s
Physics
Chapter 9: Fluids
96. Strategy Use Eq. (9-14) with y1 ≈ y2 and v2 = 0 (for the opening where the airspeed is zero). ∆P = ρ m g ∆ym ,
where m stands for mercury.
Solution Find the plane’s airspeed.
1
1
P1 + ρa v12 = P2 , so ρa v12 = ∆P and
2
2
v1 =
2∆P
ρa
=
2 ρ m g ∆ym
ρa
=
2(13, 600 kg m3 )(9.80 m s 2 )(0.25 m)
0.90 kg m3
= 270 m s .
97. Strategy Use the relationships between pressure, density, force, area, and height.
Solution Find the density of the liquid.
ρ gV
W
∆P = l = l l = ρ w g ∆yw , so
A
A
ρ w ∆yw A ρ w ∆ywπ r 2 ρ w ∆yw (1.0 g cm3 )[0.45 m − (0.50 m − 0.30 m)]
ρl =
=
=
=
= 0.83 g cm3 .
Vl
h
0.30 m
π r 2h
98. (a) Strategy Use Newton’s second law and the definition of average pressure.
Solution ΣFy = mair g = mair a y and ΣFx = 0. The force due to gravity on the atmosphere is equal to its
weight, mair g. The pressure that a vertical column of air exerts at the surface of the Earth is P = F A ,
where F is the weight of the column of air divided by its cross-sectional area.
(b) Strategy Assume that the column has uniform density. Use Eq. (9-3).
Solution Find the height of the column of air.
∆P
1.013 × 105 Pa
∆P = ρ gh, so h =
=
= 8.0 km .
ρ g (1.29 kg m3 )(9.80 m s 2 )
(c) Strategy and Solution The height found in (b) is the lower limit because a larger volume of air is required
to maintain the same weight if the density is decreasing.
99. (a) Strategy Use Eq. (9-3).
Solution Find the pressure difference between the top and bottom of the Sears Tower.
1 atm
⎛
⎞
∆P = ρ gh = (1.20 kg m3 )(9.80 m s 2 )(440 m) = (5.2 kPa) ⎜
⎟ = 0.051 atm
5
⎝ 1.013 × 10 Pa ⎠
So, the pressure is 5.2 kPa = 0.051 atm less at the top.
(b) Strategy Divide the pressure difference by the height of the tower.
Solution Find the number of pascals the pressure decreases for every meter of altitude.
∆P
= ρ g = (1.20 kg m3 )(9.80 m s 2 ) = 11.8 Pa m
∆y
479
Chapter 9: Fluids
Physics
(c) Strategy Set the pressure P2 in Eq. (9-3) equal to zero and let P1 = Patm . Since the pressure decreases, set
d = − h.
Solution Find the altitude at which the atmospheric pressure equals zero.
P
1.013 × 105 Pa
= 8.61 km .
0 = Patm − ρ gh, so h = atm =
ρ g (1.20 kg m3 )(9.80 m s 2 )
(d) Strategy and Solution Since the “zero-altitude” is inversely proportional to the air density, a decreasing air
density means that the atmosphere extends to a higher altitude.
100. Strategy Use Eq. (9-15).
Solution Find the absolute pressure at the bug’s end of the feeding tube.
π∆Pr 4 ∆V
=
8η L
∆t
⎛ 8η L ⎞ ∆V
Parm − Pbug = ⎜ 4 ⎟
⎝ π r ⎠ ∆t
⎛ 8η L ⎞ ∆V
Pbug = Parm − ⎜ 4 ⎟
⎝ π r ⎠ ∆t
3
8(0.0013 Pa ⋅ s)(0.20 × 10−3 m) ⎛ 0.30 cm3 ⎞ ⎛ 1 m ⎞ ⎛ 1 min ⎞
= −110 kPa
Pbug = 105 kPa −
⎜
⎟
⎜
⎟
⎜ 25 min ⎟ ⎝ 100 cm ⎠ ⎜⎝ 60 s ⎟⎠
π (5.0 × 10−6 m)4
⎝
⎠
101. (a) Strategy Use Eq. (9-15).
Solution Find the percentage that the blood pressure difference between the ends of the artery increased.
⎛ 8η L ⎞ ∆V ⎛ 8η L ⎞ ∆V
1 − 1
⎡⎛ r ⎞4 ⎤
⎜ π r 4 ⎟ ∆t − ⎜ π r 4 ⎟ ∆ t
∆Pa − ∆Pu
ra 4 ru 4
a ⎠
u ⎠
⎝
⎝
⎢ ⎜ u ⎟ − 1⎥ × 100%
× 100% =
× 100% =
×
=
100%
1
∆Pu
⎛ 8η L ⎞ ∆V
r
⎢
⎥
ru 4
⎣⎝ a ⎠
⎦
⎜ π r 4 ⎟ ∆t
⎝ u ⎠
⎡⎛ 1 ⎞4 ⎤
= ⎢⎜
⎟ − 1⎥ × 100% = 220%
⎢⎣ ⎝ 0.75 ⎠
⎥⎦
(b) Strategy Divide the absolute value of the change in flow rate by the original flow rate.
Solution Compute the factor of blood flow decrease.
π∆Pra 4
∆V − ∆V
∆V
⎛r
change in flow rate
8η L
∆t u
∆t a
∆t a
=
= 1 − ∆V = 1 −
= 1− ⎜ a
∆V
π∆Pru 4
original flow rate
⎝ ru
∆t u
∆t u
8η L
480
4
⎞
4
⎟ = 1 − (0.75) = 0.68
⎠
Chapter 10
ELASTICITY AND OSCILLATIONS
Conceptual Questions
1. Young’s modulus does not tell us which is stronger. Instead, it tells us which is more resistant to deformation for a
given stress. The ultimate strength would tell us which is stronger—i.e., which can withstand the greatest stress.
2. The pendulum should be lengthened to increase its period and slow down the clock.
3. When the block is struck, it initially begins to bend downward before actually breaking. Thus, the top of the block
is compressed, while the bottom is stretched. Since concrete has much less tensile strength than compressive
strength, it will break at the bottom first.
4. (a) Answer: 2F. Since stress is proportional to strain, the same force would produce the same strain on the bar of
half the length. However, strain is a relative change in length, defined as ∆L/L. Therefore, the change in
length would be half as much. To compress the bar by the same amount, then, would require a force twice as
great.
(b) Answer: F/4. To compress the bar by the same amount, with the same length as before, would require the
same stress. Stress is defined as F/A. With half the radius, the area is reduced to 1/4 of its initial value.
Therefore, to produce the same stress would require a force 1/4 as great as before.
5. The compressive force experienced by the columns is greater at the bottom than at the top, because the bottom
must support the weight of the column itself in addition to whatever the column is holding up. By increasing the
cross-sectional area of the bottom of the column, the stress it experiences is reduced. Tapering columns so that
they are thicker at the base prevents the stress at the bottom from being too large.
6. Although the distance traveled by the mass during each cycle is proportional to the amplitude of the oscillation,
the maximum velocity of the mass is as well. If the mass has farther to go, for example, it travels correspondingly
faster. This is how the period of the mass-spring system can be independent of amplitude.
7. Yes, the motion of the saw blade is SHM. The Scotch yoke effectively makes the horizontal displacement of the
saw blade equal to the x-component of the position of the knob, which is moving in a circle. When an object
moves in uniform circular motion, its x- (or y-) component exhibits SHM.
8. For the mass and spring system, the period will remain 1 s, because the period depends only on the mass and the
spring constant. For a pendulum, the period depends on the length and the gravitational field strength. With a
stronger gravitational field, the period of the pendulum would be less than 1 s.
9. The tension in the bungee cord at the lowest point would be greater than the person’s weight, because there is an
upward acceleration. In fact, the tension would have its maximum value at the bottom, because that is where the
upward acceleration is the greatest.
10. The breaking point of a rope is determined by the maximum strain it can withstand. The strain is the ratio of the
change in the length of the rope to the original length—the maximum strain is therefore independent of the rope’s
length. The strain is directly proportional to the stress—defined as the force per unit area. Thus, ropes of varying
length that are otherwise identical require the same force to reach the breaking point. The actual distance the rope
stretches before breaking is greater for a longer rope—more work, and thus more energy, is required to break a
longer rope.
481
Chapter 10: Elasticity and Oscillations
Physics
11. The velocity of the plane is lower at the top of its loop than at the bottom. Thus, the speed of the plane’s shadow
at the midpoint of its flight depends on the direction of travel—the shadow does not therefore exhibit simple
harmonic motion.
12. Concrete is much stronger under compressive stress than tensile stress. As a result of this, concrete is very strong
in vertical columns where most of the force is compressive. Concrete is weaker in horizontal columns because it
must withstand additional tensile stresses. Steel rods with a high tensile strength are therefore inserted into the
concrete to reinforce it against these tensile stresses.
13. To produce the same strain, the ratio of the force to the cross-sectional area must remain unchanged. The total
cross-sectional area of the two wires together is twice the original area. Thus, the force applied to the two wires
must be doubled as well. Modeling a thick wire as a bundle of thin wires, the preceding argument explains why
the force to produce a given strain must be proportional to the cross-sectional area—and thus why the strain
depends on the stress.
14. A given tensile stress produces a given stretch in the length of a single spring. Wires of different lengths can be
modeled as being comprised of varying numbers of springs in series. The elongation of any wire under an
identical tensile stress is therefore equal to the sum of the distances that each of the springs it is built from has
stretched. The elongation is thus proportional to the number of springs—but so too is the original length.
Therefore, a given tensile stress produces an elongation of the wire proportional to the wire’s initial length.
15. Using stress and strain to describe deformations provides the means to describe properties of materials in a way
that is independent of their dimensions. These concepts also allow physical laws to be expressed in terms of
general properties of materials—rather than properties particular to a given piece of a material.
16. The shock absorbing system in the car can be thought of as a set of springs that dampen out oscillations induced
by bumps in the road. These springs have a natural oscillation frequency just as a simple spring does. When the
frequency of encounters with bumps in the road matches the resonant frequency of the springs, the car’s
oscillations are amplified. The passengers in the car are experiencing resonance.
17. In the mass-spring system, the restoring force supplied by the spring is independent of the object’s mass. Thus, the
larger inertia of a more massive object produces a longer period. The restoring force for small amplitude
oscillations of the pendulum is the horizontal component of the tension in the string. In this case, the magnitude of
the tension is approximately equal to the weight of the bob. Although a more massive bob has more inertia, it also
has a proportionally larger restoring force. Thus, the period of oscillation of the pendulum is independent of the
mass.
18. The percent of kinetic energy (short dashes), potential energy (medium dashes) and total energy (long dashes) of a
mass connected to an ideal spring oscillating on a frictionless horizontal surface are plotted below as a function of
time.
482
Physics
Chapter 10: Elasticity and Oscillations
Problems
1. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find the vertical compression of the beam.
∆L F
FL
(5.8 × 104 N)(2.5 m)
Y
= , so ∆L =
=
= 0.097 mm .
L
A
YA (200 × 109 Pa)(7.5 × 10−3 m 2 )
2. Strategy The man is standing on two feet, so the force for one leg is equal to half his weight. The stress is
proportional to the strain. Use Eq. (10-4).
Solution Find the compression of the thighbone.
1 (91 kg)(9.80 N kg)(0.50 m)
FL
∆L =
= 2
= 29 µm
YA (1.1× 1010 N m 2 )(7.0 × 10−4 m 2 )
3. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find how much the wire stretches.
FL
(5.0 × 103 N)(2.0 m)
∆L =
=
= 2.2 cm
10
YA (9.2 × 10 Pa)(5.0 mm 2 )(10−3 m mm)2
4. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find Young’s modulus for the wire.
FL
(1.00 × 103 N)(5.00 m)
Y=
=
= 7.69 × 1010 Pa
A∆L (0.100 cm 2 )(10−2 m cm)2 (6.50 × 10−3 m)
5. Strategy Form a proportion of the elongations of the left and right wires. Use Eq. (10-4).
Solution Find how far the midpoint moves.
FL
∆LL YAL
A
π (2r )2
= FL = R =
= 4, so ∆LL = 4∆LR and ∆L = ∆LL + ∆LR is the total elongation, 1.0 mm.
∆LR YA
AL
π r2
R
∆L = ∆LL + ∆LR = ∆LL +
1
5
4
4
∆LL = ∆LL , so ∆LL = ∆L = (1.0 mm) = 0.80 mm .
4
4
5
5
6. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Compute the compression of the abductin ligament.
FL
(1.5 N)(1.0 × 10−3 m)
∆L =
=
= 0.48 mm
YA (4.0 × 106 N m 2 )(0.78 mm 2 ) 1 m 2
1000 mm
(
)
7. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find the required stretch of the string.
FL
(20 N)(0.50 m)
∆L =
=
= 5.0 mm
YA (2.0 × 109 N m 2 )(1.0 × 10−6 m 2 )
483
Chapter 10: Elasticity and Oscillations
Physics
8. (a) Strategy The average power required is equal to the kinetic energy change divided by the elapsed time it
takes for the flea to reach its peak velocity.
Solution Find the average power required.
2
2
1
∆K 2 m(vf − vi ) mvf2 (0.45 × 10−6 kg)(0.74 m s)2
Pav =
=
=
=
= 1.2 × 10−4 W
∆t
∆t
2∆t
2(1.0 × 10−3 s)
(b) Strategy and Solution Compute the power output of the flea.
(60 W kg)(0.45 × 10−6 kg)(0.20) = 5.4 × 10−6 W < 1.2 × 10−4 W
No, the flea’s muscle cannot provide the power needed.
(c) Strategy There are two pads, so the total energy stored is E = 2U = 2[ 12 k (∆L)2 ] = k (∆L)2. Use Eq. (10-5).
Solution Find the energy stored in the resilin pads.
A
L2 2
E = k (∆L)2 = Y (∆L)2 = Y
L = YL3 = (1.7 × 106 N m 2 )(6.0 × 10−5 m)3 = 3.7 × 10−7 J
L
L
(d) Strategy Use the definition of average power.
Solution Compute the power provided by the resilin pads.
∆E 3.7 × 10−7 J
Pav =
=
= 3.7 × 10−4 W > 1.2 × 10−4 W
∆t 1.0 × 10−3 s
Yes, enough power is provided for the jump.
9. Strategy Refer to Fig. 10.4c. The stress is proportional to the strain.
Solution Calculate Young’s moduli for tension and compression of bone.
Tension:
For tensile stress and strain, the graph is far from being linear, but for relatively small values of stress and strain, it
is approximately linear. So, for small values of tensile stress and strain, Young’s Modulus is
stress 5.0 × 107 N m 2
Y=
=
= 1.5 × 1010 N m 2 .
strain
0.0033
Compression:
−4.5 × 107 N m 2
= 9.0 × 109 N m 2 .
Similarly, for small values of compressive stress and strain, Y =
−0.0050
10. Strategy Set the stress equal to the elastic limit to find the minimum diameter.
Solution Find the minimum diameter of the wire required to support the acrobat.
4F
4(55 kg)(9.80 N kg)
F
F
=
= 1.7 mm .
elastic limit = =
, so d =
2
1
π (elastic limit)
A
πd
π (2.5 × 108 Pa)
4
11. Strategy Set the stress equal to the tensile strength of the hair to find the diameter of the hair.
Solution Find the diameter of the hair.
F
F
4F
4(1.2 N)
, so d =
=
= 8.7 × 10−5 m .
tensile strength = =
8
A 1 πd2
π (tensile strength)
(2.0
×
10
Pa)
π
4
484
Physics
Chapter 10: Elasticity and Oscillations
12. Strategy Compare the substances in each case using ratios.
Solution
(a) Tendon:
tensile strength 80.0 × 106 Pa
=
= 7.3 × 104 Pa ⋅ m3/kg
density
1100 kg m3
Steel:
0.50 × 109 Pa
= 6.5 × 104 Pa ⋅ m3/kg
7700 kg m3
Tendon is stronger than steel.
(b) Bone:
160 × 106 Pa
1600 kg m3
Concrete:
0.40 × 109 Pa
2700 kg m3
= 1.0 × 105 Pa ⋅ m3/kg
= 1.5 × 105 Pa ⋅ m3/kg
Concrete is stronger than bone.
13. Strategy The stress on the copper wire must be less than its elastic limit.
Solution Find the maximum load that can be suspended from the copper wire.
F
< elastic limit, so F < π r 2 (elastic limit) = π (0.0010 m)2 (2.0 × 108 Pa) = 630 N .
A
14. Strategy The stress on the copper wire must be less than its tensile strength.
Solution Find the maximum load that can be suspended from the copper wire.
F
< tensile strength, so F < π r 2 (tensile strength) = π (0.0010 m)2 (4.0 × 108 Pa) = 1300 N .
A
15. Strategy Set the stresses equal to the compressive strengths to determine the effective cross-sectional areas.
Solution Find the effective cross-sectional areas.
5 × 104 N
F
= 1.6 × 108 Pa, so A =
= 3 cm 2 .
Human:
8
A
1.6 × 10 Pa
Horse: A =
10 × 104 N
8
1.4 × 10 Pa
= 7.1 cm 2
16. Strategy Assume that the stress is proportional to the strain up to the breaking point. Use Eq. (10-4).
Solution Find the stress at the breaking point of the steel wire.
∆L
⎛ 0.20 ⎞
8
= (2.0 × 1011 N m 2 ) ⎜
stress at breaking point = Y
⎟ = 4.0 × 10 Pa
L
100
⎝
⎠
485
Chapter 10: Elasticity and Oscillations
Physics
17. Strategy Use Eqs. (10-1), (10-2), and (10-4).
Solution
(a) stress =
7.0 × 104 N
F
=
= 2.8 × 107 Pa
A 25 × 10−4 m 2
(b) strain =
∆L F
7.0 × 104 N
=
=
= 4.7 × 10−4
−
10
4
2
L YA (6.0 × 10 Pa)(25 × 10 m )
(c) ∆L =
FL
(7.0 × 104 N)(2.0 m)
=
= 9.3 × 10−4 m
YA (6.0 × 1010 Pa)(25 × 10−4 m 2 )
(d) Set the compressive strength equal to the stress to find the maximum weight the column can support.
F
= compressive strength, so F = (compressive strength)A = (2.0 ×108 Pa)(25 ×10−4 m 2 ) = 5.0 × 105 N .
A
18. (a) Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find the diameter and tensile stress of the wire.
Diameter:
∆L
4 FL
4(120 N)(3.0 m)
F
F
= 1
=Y
, so d =
=
= 1.3 mm .
π Y ∆L
A 4πd
L
π (120 × 109 Pa)(2.1× 10−3 m)
Tensile stress:
F
∆L
2.1× 10−3 m
=Y
= (120 × 109 Pa)
= 8.4 × 107 N/m 2
A
L
3.0 m
(b) Strategy Set the maximum stress equal to the tensile strength.
Solution Find the maximum weight the wire can support.
Wmax
1
= tensile strength, so Wmax = π (0.00135 m) 2 (4.0 × 108 N m 2 ) = 570 N .
A
4
19. Strategy Use Hooke’s law for volume deformations.
Solution Compute the fractional changes of the volume and radius of the sphere.
∆V
∆P 1.0 × 108 N m 2
Volume:
=
=
= 7.7 × 10−4
V
B
130 × 109 Pa
Radius:
4 3
3V
.
π r = V , so r = 3
3
4π
3V ′
3
V − ∆V
∆r r − r ′
r′
V′
∆V
3
=
= 1 − = 1 − 4π = 1 − 3
= 1− 3
= 1− 3 1−
= 1 − 1 − 7.7 × 10−4 = 2.6 × 10−4
3
V
r
r
r
V
V
V
3
4π
486
Physics
Chapter 10: Elasticity and Oscillations
20. Strategy Use ∆P = ρ gd and the relationship between volume, mass, and density.
Solution Find the percent increase in the density of water at a depth of 1.0 km.
∆ρ
ρ
=
ρ′ − ρ ρ′
= −1 =
ρ
ρ
Since V ′ ≈ V ,
100% ×
∆ρ
ρ
≈
∆ρ
ρ
≈−
m
V′
m
V
−1 =
∆V
V
V −V ′
−1 =
=−
V′
V′
V′
∆V
. Use Hooke’s law for volume deformations.
V
ρ gd
∆P
(1.0 × 103 kg m3 )(9.80 N kg)(1.0 × 103 m)
× 100% = w × 100% =
× 100% = 0.45%
B
B
2.2 × 109 Pa
21. Strategy Use Hooke’s law for volume deformations.
Solution Find the change in volume of the sphere.
∆V
V ∆P
(1.00 cm3 )(9.12 × 106 Pa)
∆P = − B
, so ∆V = −
=−
= −57 × 10−6 cm3 .
V
B
160 × 109 Pa
The volume of the steel sphere would decrease by 57 × 10−6 cm3 .
22. Strategy Use Hooke’s law for volume deformations. The pressure of the Moon is roughly 10−9 Pa.
Solution Find the change in volume of the aluminum.
∆V
V ∆P
(1.00 cm3 )(10−9 Pa − 1.013 × 105 Pa)
∆P = − B
, so ∆V = −
=−
= 1.4 × 10−6 cm3 .
V
B
70 × 109 Pa
The volume of the aluminum sphere would increase by 1.4 × 10−6 cm3 .
23. Strategy Set the shear stress equal to the total shear strength to find the maximum shearing force.
Solution Find the maximum shearing force F on the plates that the four bolts can withstand.
F
= shear strength, so F = 4π r 2 (shear strength) = 4π (0.010 m)2 (6.0 × 108 Pa) = 7.5 × 105 N .
4 Abolt
24. Strategy Use Hooke’s law for volume deformations.
Solution Find the change in the volume of the anchor.
∆V
∆P
V ∆P
(0.230 m3 )(1.75 × 106 Pa)
=−
=−
= − 6.71 cm3 .
, so ∆V = −
V
B
B
60.0 × 109 Pa
25. Strategy Use Hooke’s law for shear deformations.
Solution Find the magnitude of the tangential force.
∆x
F F
= 2 =S
, so F = S ∆xL = (940 Pa)(0.64 × 10−2 m)(0.050 m) = 0.30 N .
A L
L
487
Chapter 10: Elasticity and Oscillations
Physics
26. (a) Strategy The tangent of the deformation angle is equal to the relative displacement of the sponge divided by
its thickness.
Solution Compute the relative displacement, ∆x.
∆x = L tan γ = (2.0 cm) tan 8.0° = 2.8 mm
(b) Strategy Use Hooke’s law for shear deformations.
Solution Compute the shear modulus of the sponge.
FL
FL
F
12 N
S=
=
=
=
= 2.0 × 104 Pa
A∆x AL tan γ
A tan γ (42 × 10−4 m 2 ) tan 8.0°
27. Strategy At the maximum extension of the spring, x = A and the magnitude of the acceleration is maximum.
Use Eq. (10-22).
Solution Find the magnitude of the acceleration at the point of maximum extension of the spring.
4π 2 A 4π 2 (0.050 m)
am = ω 2 A =
=
= 7.9 m s 2
T2
(0.50 s) 2
28. Strategy The amplitude is half the maximum distance. Use Eq. (10-21).
Solution Find the maximum needle speed.
vm = ω A =
2π A 2π
=
T
( 8.42 ×10−3 m ) =
9.0 s
24
7.0 cm s
29. Strategy The amplitude is half the maximum distance. Use Eqs. (10-21) and (10-22).
Solution Find the maximum velocity and maximum acceleration of the prong.
⎛ 2.24
⎞
× 10−3 m ⎟ = 3.10 m s
vm = ω A = 2π fA = 2π (440.0 Hz) ⎜
2
⎝
⎠
⎞
2
2 2
2
2 ⎛ 2.24
am = ω A = 4π f A = 4π (440.0 Hz) ⎜
× 10−3 m ⎟ = 8560 m s 2
2
⎝
⎠
30. Strategy At the equilibrium point, the speed is at its maximum. Use Eq. (10-21).
Solution Find the speed at the equilibrium point.
2π
2π (0.050 m)
vm = ω A =
A=
= 0.63 m s
T
0.50 s
31. Strategy Replace each quantity with its SI units.
Solution a has units m s 2 . − ω 2 x has units s −2 ⋅ m = m s 2 . So, a = −ω 2 x is consistent for units.
ω has units s −1.
k m has units
Nm
=
kg
kg ⋅ m s 2
1
= 2 = s −1. So,
m ⋅ kg
s
488
k m has the same units as ω .
Physics
Chapter 10: Elasticity and Oscillations
32. (a) Strategy Solve for the spring constant in Eq. (10-20a).
Solution Find the spring constant.
k
ω=
, so k = ω 2 m = (3.00 Hz) 2 (2π rad cycle)2 (0.17 kg) = 60 N m .
m
(b) Strategy Use Eq. (10-17).
Solution The amplitude is A = 12.0 cm. Find the angular frequency.
ω = (2π rad cycle)(3.00 Hz) = 6.00π rad s
Thus, the equation that describes the position of the object as a function of time is
x(t ) = (12.0 cm) cos[(6.00π s −1 )t ] .
33. Strategy Use Eqs. (10-21) and (10-22).
Solution
(a) Find vm and am in terms of f. Then compare high- and low-frequency sounds.
vm = ω A = 2π fA ∝ f and am = ω 2 A = 4π 2 f 2 A ∝ f 2 , so vm and am are greatest for high frequency .
(b) vm = 2π (20.0 Hz)(1.0 × 10−8 m) = 1.3 × 10−6 m s
am = 4π 2 (20.0 Hz)2 (1.0 × 10−8 m) = 1.6 × 10−4 m s 2
(c) vm = 2π (20.0 × 103 Hz)(1.0 × 10−8 m) = 0.0013 m s
am = 4π 2 (20.0 × 103 Hz)2 (1.0 × 10−8 m) = 160 m s 2
34. Strategy Use Eqs. (10-21) and (10-22).
2
Solution For SHM, show that vm
= am A.
vm = ω A, so ω =
vm
v2
v2
2
. am = ω 2 A = m2 A = m , so vm
= am A.
A
A
A
35. Strategy The angular frequency of oscillation is inversely proportional to the square root of the mass. Form a
proportion.
Solution Find the new value of ω.
ω∝
1 mf
ω
1
=
, so f =
ωi
m
1 mi
mi
=
mf
1
1
ω
10.0 rad s
=
= 5.0 rad s .
. Therefore, ωf = i =
4.0 2.0
2.0
2.0
489
Chapter 10: Elasticity and Oscillations
Physics
36. (a) Strategy Use Hooke’s law.
Solution When the cart moves to the right, the spring on the
right compresses an amount x and the spring on the left stretches
an amount x. The spring on the right pushes on the cart with a
force of kx to the left, while the spring on the left pulls on the
cart with a force of kx to the left. When the cart moves to the
left, the situation is reversed. Thus, the forces are identical and
the magnitude of the net force on the cart is 2kx .
x
x
kx
kx
(b) Strategy Use Eq. (10-20a).
Solution The effective spring constant for the two springs (as if they were one) is 2k, so the angular
frequency of the cart is ω =
2k m .
37. Strategy Use Eqs. (10-20a) and (10-20b) and Newton’s second law.
Solution Find the spring constant.
m
g (24 kg)(9.80 m s 2 )
ΣFy = kx − mchild g = 0, so k = child =
= 840 N m.
0.28 m
x
Find the mass of the wooden horse.
k
k
ωchild =
, so mhorse =
− mchild .
mchild + mhorse
ωchild 2
Find the oscillation frequency of the spring when no one is sitting on the horse.
ω
1
k
1
k
1
840 N m
=
=
f horse = horse =
840 N m
k
π
2π
2π mhorse 2π
2
−
m
child
2
ωchild
(0.88 Hz) 2 (2π rad cycle)2
Fs
y
mchildg
− 24 kg
= 2.5 Hz
38. Strategy Use Eq. (10-22).
Solution Find the oscillation frequency of the bird’s wingtips.
am
am = ω 2 A = 4π 2 f 2 A, so f =
4π 2 A
=
12 m s 2
4π 2 (0.050 m)
= 2.5 Hz .
39. Strategy Use Eqs. (10-21) and (10-22) and Newton’s second law.
Solution Find the radio’s maximum displacement and maximum speed, and the maximum net force exerted on it.
a
(a) am = ω 2 A, so A = m2 =
ω
(b) vm = ω A = ω
am
ω2
=
am
ω
98 m s 2
2
4π (120 Hz)
=
2
= 1.7 × 10−4 m .
98 m s 2
= 0.13 m s
2π (120 Hz)
(c) According to Newton’s second law, Fm = mam = (5.24 kg)(98 m s 2 ) = 510 N .
490
Physics
Chapter 10: Elasticity and Oscillations
40. Strategy Use Eqs. (10-21) and (10-22).
Solution Compute the maximum speed and acceleration to which the pilot is subjected.
vm = ω A = 2π (25.0 Hz)(0.00100 m) = 0.157 m s
am = ω 2 A = 4π 2 (25.0 Hz)2 (0.00100 m) = 24.7 m s 2
41. (a) Strategy Use Newton’s second law and Eq. (10-22).
Solution Find the maximum force acting on the diaphragm.
Fm = mam = mω 2 A = 4π 2 (0.0500 kg)(2.0 × 103 Hz)2 (1.8 × 10−4 m) = 1.4 kN .
(b) Strategy The maximum elastic potential energy of the diaphragm is equal to the total mechanical energy.
Solution Find the mechanical energy of the diaphragm.
1
E = U = mω 2 A2 = 2π 2 (0.0500 kg)(2.0 × 103 Hz)2 (1.8 × 10−4 m)2 = 0.13 J .
2
42. Strategy Use Hooke’s law and Newton’s second law.
Solution
(a) According to Newton’s second law, at equilibrium, ΣFy = 0 = kd − mg , so
k = mg d . When the extension of the spring is a maximum,
2kd
kd
ΣFy = ma = k (2d ) − mg = 2(mg d )d − mg = mg. Therefore, a = g .
a
mg
mg
(b) At maximum extension, ΣFy = kx − mg = ma. Solve for x.
m
(1.0 kg)(9.80 N kg + 9.80 N kg)
= 0.78 m .
kx = mg + ma = m( g + a), so x = ( g + a) =
k
25 N m
43. (a) Strategy The speed is maximum when the spring and mass system is at its equilibrium point. Use Newton’s
second law.
Solution Find the extension of the spring.
mg (0.60 kg)(9.80 N kg)
ΣFy = kx − mg = 0, so x =
=
= 0.39 m .
k
15 N m
kx
mg
(b) Strategy Use Eqs. (10-20a) and (10-21).
Solution Find the maximum speed of the body.
k
15 N m
vm = ω A =
x=
(0.39 m) = 2.0 m s
m
0.60 kg
491
Chapter 10: Elasticity and Oscillations
Physics
44. Strategy Assuming SHM, the kinetic energy decreases by an amount equal to the increase in elastic potential
energy.
Solution Find the change in kinetic energy.
1
1
∆K = − kx 2 = − (25 N m)(0.050 m)2 = −0.031 J
2
2
45. Strategy Use Hooke’s law, Newton’s second law, and Eq. (10-20c).
Solution Find the “spring constant” of the boat. At equilibrium,
mperson g
.
ΣFy = kx − mperson g = 0, so k =
x
Compute the period of oscillation.
mtotal
mtotal
mtotal x
(47 kg + 92 kg)(0.080 m)
= 2π
= 2π
= 2π
= 0.70 s
T = 2π
k
mperson g x
mperson g
(92 kg)(9.80 m s 2 )
kx
mperson g
46. (a) Strategy Use Newton’s second law and Eq. (10-20a).
Solution Determine the spring constant of the cord.
mg
At equilibrium, ΣFy = 0 = kd − mg. So, k =
, where d = 0.20 m.
d
Calculate the period.
2π
2π
0.20 m
m
m
d
=
= 2π
= 2π
= 2π
= 2π
= 0.90 s
T=
ω
k
mg d
g
k m
9.80 m s 2
kd
mg
(b) Strategy Use Eqs. (10-21) and (10-20a).
Solution Find the maximum speed of the baby.
vm = ω A = A
k
mg
g
9.80 m s 2
=A
=A
= (0.080 m)
= 0.56 m s
m
dm
d
0.20 m
47. Strategy and Solution Since y(t ) = A sin ω t , f =
ω
2π
=
1.57 rad s
= 0.250 Hz .
2π
48. Strategy The maximum kinetic energy occurs at the equilibrium point where v = vm = ω A. For SHM,
ω = k m.
Solution Find the maximum kinetic energy of the body.
1 2 1
1 ⎛k⎞
1
1
K max = mvm
= mω 2 A2 = m ⎜ ⎟ A2 = kA2 = (2.5 N m)(0.040 m) 2 = 2.0 mJ
2
2
2 ⎝m⎠
2
2
492
Physics
Chapter 10: Elasticity and Oscillations
49. (a) Strategy The object will oscillate up and down with an amplitude determined by the spring constant and the
mass of the spring.
Solution Find the amplitude of the motion. At the equilibrium point, the net force on
the object is zero.
ΣFy = kA − mg = 0, so A = mg k = (0.306 kg)(9.80 m s 2 ) (25 N m) = 12 cm.
The object will move up and down a total vertical distance of 2 A = 24 cm. Thus, the
kA
pattern traced on the paper by the pen is a vertical straight line of length 24 cm .
(b) Strategy and Solution As the paper moves to the left at constant speed while the pen
oscillates vertically in SHM, the pen traces a pattern of
mg
a positive cosine plot of amplitude 12 cm .
24 cm
50. Strategy Graph (a) x, (b) vx , and (c) a x on the vertical axis and t on the horizontal axis. Analyze the slopes.
Solution
(a) x(t ) = A sin ω t
x
A
0
2π
ω
t
π
ω
(b) The slope of the x(t) graph is maximum at t = 0, so vx (t ) is max at t = 0. Since vx (t ) starts at its maximum
value and then is positive and decreasing, it is a cosine function, vx (t ) = vm cos ωt.
vx
vm
0
π
ω
2π
ω
t
(c) The slope of the vx (t ) graph is zero at t = 0 and becomes negative just after, so a x (t ) starts at zero and
decreases. Since a x (t ) starts at zero and then is negative and decreasing, it is a negative sine function,
a x (t ) = −am sin ω t.
ax
am
0
π
ω
2π
ω
t
(d) Compare minimums. x(t ) has its minimum at t =
3π
π
π
.
, vx (t ) has its at t = , and a x (t ) has its at t =
ω
2ω
2ω
π 3π π
π
π π
π
and −
.
=
, so vx (t ) is one quarter cycle ahead of x(t) and
− =
ω 2ω 2ω
2ω 2ω ω 2ω
a x (t ) is one quarter cycle ahead of vx (t ).
One quarter cycle is
493
Chapter 10: Elasticity and Oscillations
Physics
51. Strategy Use the definition of average speed and Eq. (10-21). In (d), graph vx on the vertical axis and t on the
horizontal axis.
Solution
(a) The average speed is the total distance traveled divided by the time of travel.
∆x 4 A
4A
2
ωA
vav =
=
=
=
π
∆t
2π ω
T
(b) The maximum speed for SHM is vm = ω A .
2
vav π ω A
=
=
ωA
π
vm
2
(c)
(d) Graph vx (t ) and a line from the origin to vm .
vx
ωA
2π
ω
0
t
π
ω
−ω A
If the acceleration were constant so that the speed varied linearly, the average speed would be 1/2
of the maximum velocity. Since the actual speed is always larger than what it would be for constant
acceleration, the average speed must be larger.
52. Strategy Use the equations of motion for constant acceleration. Graph y on the vertical axis and t on the
horizontal axis.
Solution Analyze the motion of the ball.
During the fall:
1
1
1
y − yi = y − h = viy ∆t − g (∆t ) 2 = (0)∆t − g (t − 0)2 , so y = h − gt 2. At y = 0, t =
2
2
2
2h
.
g
During the rise:
The speed of the ball just before and after it hits the ground is v = gt = g
ti =
2h
= 2 gh . So, at
g
1
2h
, viy = 2 gh , and y = yi = 0. If ti = 0 when y = yi = 0, then y = viy ∆t − g (∆t ) 2. The graph is shifted
2
g
2
⎛
2h
2h ⎞ 1 ⎛
2h ⎞
2h
− g ⎜t −
, so y = 2 gh ⎜ t −
.
⎟
⎟⎟ , and y = h when t = 2
⎜
⎟ 2 ⎜
g
g
g
g
⎝
⎠
⎝
⎠
to the right by
y
h
0
2h
g
2h
2 g
t
The motion is not SHM, since it is not a sine or cosine function . The graph is always nonnegative and has a
parabolic shape.
494
Physics
Chapter 10: Elasticity and Oscillations
53. (a) Strategy Use Eq. (10-20a) to find the spring constant. Then, find the elastic potential energy using 12 kx 2 .
Solution Find the spring constant.
k
ω=
, so k = ω 2 m = (2.00 Hz) 2 (2π rad cycle)2 (0.2300 kg) = 36.3 N m.
m
The equation for the elastic potential energy is
1
U (t ) = (36.3 N m)(0.0800 m)2 sin 2 [ (2.00 Hz)(2π rad cycle)t ] = (116 mJ) sin 2 ⎡ (4.00π s −1 )t ⎤ .
⎣
⎦
2
Since the sine function is squared, the period of U (t ) is half that of a sine function or
T=
π
π
=
= 250 ms. Graph U (t ).
ω 4.00π s −1
U (mJ)
116
0
0
500 t (ms)
250
(b) Strategy Find the kinetic energy using 12 mvx 2 .
Solution The equation for the kinetic energy is
2
K (t ) =
⎛ 2π rad ⎞
⎛ 2π rad ⎞ ⎤
1
2
2⎡
(0.2300 kg)(2.00 Hz) 2 ⎜
⎟ (0.0800 m) cos ⎢ (2.00 Hz) ⎜
⎟t⎥
2
cycle
⎝
⎠
⎝ cycle ⎠ ⎦
⎣
= (116 mJ) cos 2 ⎡(4.00π s −1 )t ⎤ .
⎣
⎦
Since the cosine function is squared, the period of K (t ) is half that of a cosine function or
T = π ω = π (4.00π s −1 ) = 250 ms, which is the same as U (t ). Graph K (t ).
K (mJ)
116
0
0
500 t (ms)
250
(c) Strategy Add U (t ) and K (t ) and graph the result.
Solution
E (t ) = U (t ) + K (t ) = (116 mJ) sin 2 ⎡(4.00π s −1 )t ⎤ + (116 mJ) cos 2 ⎡ (4.00π s −1 )t ⎤
⎣
⎦
⎣
⎦
2
−
1
2
−
1
⎡
⎤
⎡
⎤
= (116 mJ) sin (4.00π s )t + cos (4.00π s )t = (116 mJ)(1) = 116 mJ
⎣
⎦
⎣
⎦
Graph E (t ) = U (t ) + K (t ).
{
}
E (mJ)
116
0
0
250
500 t (ms)
(d) Strategy and Solution Friction does nonconservative work on the object, thus,
U , K , and E would gradually be reduced to zero .
495
Chapter 10: Elasticity and Oscillations
Physics
54. (a) Strategy Draw the velocity vector for point P in Fig. 10.17b and then find its x-component.
Solution Show that vx (t ) = −ω A sin ω t.
vm
y
θ
vy
x
vx
From the figure, vx = −vm sin θ . For SHM, vm = ω A and θ = ω t , so vx (t ) = −ω A sin ω t.
(b) Strategy Use conservation of energy and Eq. (10-20a).
Solution Verify that the expressions for x(t) and vx (t ) are consistent with energy conservation.
1
1
1
1
1
E = kA2 = K + U = mv 2 + kx 2 = mω 2 A2 sin 2 ωt + kA2 cos 2 ωt
2
2
2
2
2
2
1 ⎛ k ⎞ 2 2
1 2
1 2
1 2
1 2
2
2
2
= m⎜
⎟ A sin ωt + kA cos ωt = kA (sin ωt + cos ωt ) = kA (1) = kA
2 ⎜⎝ m ⎟⎠
2
2
2
2
The expressions for x(t) and vx (t ) are consistent with energy conservation.
55. Strategy Use Eq. (10-26b).
Solution Compute the period of the pendulum.
L
4.0 m
= 2π
= 4.0 s
T = 2π
g
9.80 m s 2
56. Strategy The total mechanical energy for a simple pendulum is E = 12 mω 2 A2. Use Eq. (10-26a).
Solution Find the amplitude of the pendulum.
2
1
1 ⎛ g ⎞ 2 mg 2
E = mω 2 A2 = m ⎜
A , so A =
⎟ A =
2
2 ⎜⎝ L ⎟⎠
2L
2 EL
=
mg
57. Strategy and Solution According to Eq. (10-26b), T = 2π
2(0.015 J)(0.75 m)
= 3.0 cm .
(2.5 kg)(9.80 N kg)
L
, which does not depend upon the mass.
g
Therefore, T = 1.5 s .
58. Strategy Use Eq. (10-26b) and form a proportion with the final and initial periods.
Solution Find the period of oscillation of the pendulum.
Tf 2π
=
Ti 2π
Lf
g
Li
g
=
Lf
=
Li
2L
= 2, so Tf = Ti 2 = (2.0 s) 2 = 2.8 s .
L
496
Physics
Chapter 10: Elasticity and Oscillations
59. (a) Strategy Graph vx on the vertical axis and t on the horizontal axis. Use Eq. (10-21).
Solution vx leads x by one quarter cycle [π (2ω )] and vm = ω A, so if x = A sin ωt , vx = ω A cos ωt .
vx
ωA
0
π
ω
2π
ω
t
−ω A
(b) Strategy Use Eqs. (6-6) and (10-21).
Solution Find the maximum kinetic energy.
1 2
1
K m = mvm
, so K m = mω 2 A2 .
2
2
60. Strategy Use Eq. (10-26b).
Solution Find the length of the pendulum.
L
gT 2 (9.80 m s 2 )(1.0 s) 2
T = 2π
, so L =
=
= 0.25 m .
g
4π 2
4π 2
61. Strategy Use Eq. (10-26b) to find the length of the pendulum. Then, form a ratio of the lengths.
Solution Solve for L.
L
gT 2
T = 2π
, so L =
.
g
4π 2
Form a proportion.
2
L2 T22 ⎛ 1.00 s ⎞
=
=
= 1.11
L1 T12 ⎜⎝ 0.950 s ⎟⎠
62. (a) Strategy and Solution Since the period of a pendulum is inversely proportional to gravitational field
strength, the greater period implies that the gravitational field strength on the other planet is less than
that on Earth.
(b) Strategy Use Eq. (10-26b). Form a proportion.
Solution Refer to the mystery planet as X.
T
L
L
L ⎛
L
TE = 2π
and TX = 2π
, so E = 2π
⎜⎜ 2π
gE
gX
TX
gE ⎝
gX
2
⎞
⎟⎟
⎠
−1
=
2
⎛T ⎞
⎛ 0.650 s ⎞
2
Thus, g X = g E ⎜⎜ E ⎟⎟ = (9.80 m s 2 ) ⎜
⎟ = 5.57 m s .
0.862
s
T
⎝
⎠
⎝ X⎠
497
gX
gE
.
Chapter 10: Elasticity and Oscillations
Physics
63. Strategy The amplitude is (20.0 mm) 2 = 10.0 mm and the period is 2.00 s. Use Eqs. (10-21) and (10-26b) and
conservation of energy.
Solution Find the maximum speed of the pendulum bob.
1st method:
2π
2π (10.0 × 10−3 m)
vm = ω A =
A=
= 3.14 cm s
T
2.00 s
2nd method:
Find the length L of the pendulum.
L
T 2 g (2.00 s) 2 (9.80 m s 2 )
T = 2π
, so L =
=
= 0.993 m.
g
4π 2
4π 2
θ max
L
A
Find θ max .
A 10.0 × 10−3 m
=
= 1.007 × 10−2 rad.
L
0.993 m
The height h of the pendulum bob above its lowest point is the difference between the
length of the pendulum L and the vertical distance from the axis to the bob when it is at its
maximum height, L cos θ . So, h = L(1 − cos θ ).
Find vm .
Since the displacement is small, θ max ≈ sin θ max =
∆K =
Axis
θ
L cos θ
L
1 2
mvm = −∆U = mgh, so
2
vm = 2 gh = 2 gL(1 − cos θ ) = 2(9.80 m s 2 )(0.993 m)(1 − cos1.007 × 10−2 ) = 3.14 cm s .
h
64. (a) Strategy Use Eq. (10-26b).
Solution Compute the period of the pendulum.
L
1.000 m
= 2π
= 2.01 s
T = 2π
g
9.80 m s 2
(b) Strategy Refer to the derivation of the physical pendulum in Section 10.8 of
College Physics.
Solution Let the mass of the point mass by m and the mass of the uniform thin rod
be mr . Find the net torque acting on the physical pendulum.
Axis
θ
L
L/2
m rg
L
gL
gL
Στ = F⊥ r = − mr g sin θ − mg sin θ L = −
sin θ ( mr + 2m) ≈ − θ ( mr + 2m),
2
2
2
mg
where the approximation for small amplitudes (sin θ ≈ θ ) was used. The net torque
is equal to the rotational inertia times the angular acceleration, so
gLθ (mr + 2m)
gL(mr + 2m)
gL
Στ = − θ (mr + 2m) = I α , and the angular acceleration is α = −
=−
θ , where
2
2I
2( I r + I )
I r and I are the rotational inertias for the rod and the point mass, respectively. Since we have SHM, our
equation for the angular acceleration is analogous to the equation for the linear acceleration of the oscillating
gL(mr + 2m)
spring, a x = −ω 2 x, where the angular frequency in our case is ω =
. Thus, the period of the
2( I r + I )
pendulum is T =
2π
ω
= 2π
2( I r + I )
. Now, the rotational inertias of the rod and point mass are
gL(mr + 2m)
498
Physics
Chapter 10: Elasticity and Oscillations
I r = 13 mr L2 and I = mL2 , respectively, so the period becomes T = 2π
compare the mass of the rod to the total mass.
(
(
2 L 1 + 3mm
2 L(mr + 3m)
r
T = 2π
= 2π
3 g (mr + 2m)
2
m
3g 1 + m
(
(
3m
mr
2
2L 1 +
⎛ T ⎞
=
⎜
⎟
⎝ 2π ⎠
3 g 1 + 2mm
1 + 3mm
r
1 + 2mm
r
r
)
)
r
2 L(mr + 3m)
. So, now we need to
3 g (mr + 2m)
)
)
2
⎛ T ⎞ 3 g 3T 2 g
=⎜
=
⎟
⎝ 2π ⎠ 2 L 8π 2 L
3m 3T 2 g ⎛ 2m ⎞ 3T 2 g 3T 2 g ⎛ 2m ⎞
=
+
1+
⎜1 +
⎟=
⎜
⎟
mr 8π 2 L ⎜⎝ mr ⎟⎠ 8π 2 L 8π 2 L ⎜⎝ mr ⎟⎠
m⎛
3T 2 g ⎞ 3T 2 g
⎜⎜ 3 − 2 ⎟⎟ = 2 − 1
mr ⎝
4π L ⎠ 8π L
3T 2 g
− 1 16π 2 L − 3T 2 g
2
m mr + m
1+
=
= 1 + 8π L
=
mr
mr
3T 2 g 24π 2 L − 6T 2 g
3−
4π 2 L
mr
mr + m
× 100% =
24π 2 L − 6T 2 g
16π 2 L − 3T 2 g
× 100% =
24π 2 (1.000 m) − 6(0.99 × 2.007 s) 2 (9.80 m s 2 )
16π 2 (1.000 m) − 3(0.99 × 2.007 s) 2 (9.80 m s 2 )
× 100% = 11.3%
The percentage of the total mass of the pendulum in the uniform thin rod is 11.3% .
65. Strategy The total mechanical energy of a pendulum is E = 12 mω 2 A2. Form a proportion.
Solution Find the mechanical energy of the pendulum.
2
2
⎛A ⎞
E2 A22
⎛ 3.0 cm ⎞
= 2 , so E2 = ⎜ 2 ⎟ E1 = ⎜
⎟ (5.0 mJ) = 11 mJ .
E1 A1
⎝ 2.0 cm ⎠
⎝ A1 ⎠
66. Strategy Use Eqs. (10-26a) and (10-28b).
Solution Show that the thin circular hoop oscillates with the same frequency as a simple pendulum of length
equal to the diameter of the hoop.
Hoop:
T = 2π
I
2mr 2
2r
D
= 2π
= 2π
= 2π
mgd
mgr
g
g
1
1 g
, where D is the diameter of the hoop.
=
T 2π D
Simple pendulum:
ω
1 g
fP =
=
= f H if L = D.
2π 2π L
So, f H =
499
Chapter 10: Elasticity and Oscillations
Physics
67. (a) Strategy The total mechanical energy of a pendulum is E = 12 mω 2 A2. Use Eq. (10-26a).
Solution Find the energy of the pendulum.
2
E=
1
1 ⎛ g ⎞ 2 mgA2 (0.50 kg)(9.80 m s 2 )(0.050 m)2
mω 2 A2 = m ⎜
=
= 6.1 mJ
⎟ A =
2
2 ⎜⎝ L ⎟⎠
2L
2(1.0 m)
(b) Strategy Use Eq. (10-26b) and the equation for the potential energy of an object in a uniform gravitational
field.
Solution Find the percentage of the pendulum’s energy lost during one cycle.
L
T = 2π
and U = m2 gh where m2 = 2.0 kg.
g
cycles per week =
U
cycles/wk
E
1 wk 604,800 s
=
T
2π
× 100% =
=
g
L
m2 gh
604,800 s
2π
g
L
(
1m g
2 1L
A2
)
× 100% =
4π m2h
(604,800 s)m1 A2
L3
× 100%
g
(1.0 m)3
4π (2.0 kg)(1.0 m)
(604,800 s)(0.50 kg)(0.050 m)2
× 100% = 1.1%
9.80 m s 2
68. Strategy E = 12 mω 2 A2 ∝ A2 for a pendulum. Form a proportion.
Solution Find by what factor the energy has decreased.
E2 A22 ( A1 20.0)2
1
1
=
=
=
=
2
2
E1 A12
400
A1
20.0
The energy has decreased by a factor of 400 .
69. Strategy E = 12 mω 2 A2 ∝ A2 for a pendulum.
Solution Find the percent decrease of the oscillator’s energy in ten cycles.
∆E
∆A 2
(1 − 0.0500)2 − 12
0.95002 − 12
× 100% = 2 × 100% =
×
=
× 100% = − 9.75%
100%
E
A
12
12
70. Strategy Use Eq. (10-20b).
Solution Find the spring constant.
1 k
f =
, so k = 4π 2 f12 m1 , where m1 is the combined mass of all four people and the car.
2π m
Compute the frequency when only the 45-kg person is present.
f2 =
1
2π
k
1
=
m2 2π
4π 2 f12 m1
(2.00 Hz)2 (1020 kg + 45 kg + 52 kg + 67 kg + 61 kg)
=
= 2.16 Hz
m2
1020 kg + 45 kg
500
Physics
Chapter 10: Elasticity and Oscillations
71. Strategy Use Eq. (10-21). Assume the amplitude of the pendulum is small.
Solution Find the period T of the pendulum.
2π
2π
0.50 m s
=
= 2.5 s .
vm = ω A = 0.50 m s and ω =
= 2.5 rad s. Thus, T =
ω 2.5 rad s
0.20 m
72. Strategy Use Eq. (10-26a).
Solution We must assume that the pendulum is located on Earth. Find its length.
ω=
g
g
9.80 m s 2
, so L =
=
= 0.994 m .
L
ω 2 (3.14 rad s)2
73. Strategy Use Eq. (10-20c).
Solution
(a) The period is directly proportional to the square root of the mass, and the period for the fish is longer than that
for the weight, so the fish weighs more than the weight.
(b) Form a proportion. Let f = fish and w = weight.
T
m
T = 2π
, so f =
k
Tw
mf
mw
⎛T
. Thus, Wf = ⎜⎜ f
=
Ww
⎝ Tw
Wf
2
2
⎞
⎛ 220 ⎞
⎟⎟ Ww = ⎜
⎟ (4.90 N) = 56 N .
⎝ 65 ⎠
⎠
74. Strategy Assume the weight of the cable is negligible compared to the weight of the aviator. Use Eq. (10-26b).
Solution Find the period for a pendulum assuming SHM.
45 m
L
= 2π
= 13 s
T = 2π
g
9.80 m s 2
75. Strategy Refer to Eqs. (10-20). Use conservation of energy and the fact that E ∝ A2 .
Solution Analyze the mass and spring system.
(a)
The frequency and period don’t vary with amplitude, they only vary with m and k . Since these two
values remain constant, so do the frequency and period.
(b) Since the total energy of a spring is directly proportional to the square of the amplitude,
the total energy for an amplitude of 2 D is four times that for an amplitude of D .
(c) The initial speed will essentially result in a greater initial displacement; therefore, it will have a greater
amplitude. Since the frequency and period don’t vary with amplitude, the frequency and period are still
the same.
(d) The energy is greater when given an initial push, since it has an amplitude greater than 2D. The increase in
energy is
1 mv 2 ,
i
2
due to the initial kinetic energy.
501
Chapter 10: Elasticity and Oscillations
Physics
76. Strategy Assume the web obeys Hooke’s law. Use Eq. (10-20b) and Newton’s second law.
Solution According to Newton’s second law, at equilibrium, ΣFy = kd − mg = 0, so k m = g d .
kd
Calculate the frequency of oscillation.
f =
1
2π
k
1
=
m 2π
g
1
=
d 2π
9.80 m s 2
0.030 × 10−3 m
= 91 Hz .
mg
77. Strategy Graph x on the vertical axis and t on the horizontal axis. Analyze the slope of the graph (the magnitude
of which is the speed) in terms of the distance between the dots to determine the fastest and slowest speeds of the
mass.
Solution Graph x(t ) = −(10 cm) cos[(1.57 s −1 )t ].
x (cm)
10
0
1.0
2.0
4.0 t (s)
3.0
−10
The distance between adjacent dots should be the least at the endpoints and greatest at the center,
so its speed is lowest at the endpoints and fastest at its equilibrium position.
78. Strategy Assume SHM. The amplitude of the motion is half the movement during one complete stroke, or 1.2
cm. Use Eqs. (10-20b), (10-21), and (10-22).
Solution Compute the maximum speed and maximum acceleration of the blade.
vm = ω A = 2π fA = 2π (28 Hz)(0.012 m) = 2.1 m s and am = ω 2 A = 4π 2 (28 Hz)2 (0.012 m) = 370 m s 2 .
79. Strategy I = mL2 for a simple pendulum of length L and mass m. Use Eq. (10-28a).
Solution
mgd
ω=
=
I
mgd
2
mL
=
gd
2
L
=
g ( L)
2
L
=
g
, which is the angular frequency of a simple pendulum.
L
502
Physics
Chapter 10: Elasticity and Oscillations
80. (a) Strategy Use Hooke’s law and Newton’s second law.
Solution Find the spring constant. At equilibrium, the spring has stretched a distance
d = 0.310 m − 0.200 m = 0.110 m.
ΣFy = kd − mg = 0, so k =
mg (1.10 kg)(9.80 m s 2 )
=
= 98.0 N m .
d
0.110 m
kd
mg
(b) Strategy Use conservation of energy.
Solution The total energy of the spring is equal to the maximum kinetic energy, as well as the maximum
potential energy. Let d = 0.0500 m.
K max =
1
1
mv 2 = U max = kd 2 , so vmax =
2 max
2
kd 2
(98.0 N m)(0.0500 m) 2
=
= 0.472 m s .
m
1.10 kg
(c) Strategy Use conservation of energy.
Solution Find the speed of the brick.
1
1
1
E = kd 2 = K + U = mv 2 + ky 2 , so
2
2
2
98.0 N m
k 2
2
(d − y ) =
[(0.0500 m)2 − (0.0250 m)2 ] = 0.409 m s .
v=
1.10 kg
m
(d) Strategy Use Eq. (10-20c).
Solution The time is will take for the brick to oscillate five times is five times the period.
⎛
m⎞
1.10 kg
= 3.33 s
5T = 5 ⎜⎜ 2π
⎟⎟ = 10π
k
98.0
N m
⎝
⎠
81. Strategy Since the body begins with its maximum amplitude at t = 0, the body oscillates according to a cosine
function (cos 0 = 1). Use Newton’s second law, Hooke’s law, and Eq. (10-20a).
Solution Find the amplitude A.
mg
4.0 N
ΣFy = kA − mg = 0, so A =
=
= 1.6 cm.
k
250 N m
Find the angular frequency ω .
ω=
k
=
m
kg
=
mg
kA
y
mg
(250 N m)(9.80 m s 2 )
= 25 rad s
4.0 N
Thus, the equation describing the motion of the body is y = (1.6 cm) cos [ (25 rad s )t ] .
82. Strategy Use dimensional analysis.
Solution [ A] = m, [k ] = kg s 2 , and [m] = kg; we need [ f ] = s −1. Only k has units which include seconds, so f
must be proportional to the square root of k to get s −1. kg must be eliminated and only m has units of kg;
therefore, f ∝ k m .
503
Chapter 10: Elasticity and Oscillations
Physics
83. (a) Strategy Use conservation of energy.
Solution Let the maximum displacement be d. Find d.
1
1
1
E = K + U = mv 2 + kx 2 = kd 2 , so
2
2
2
2
mv
(1.24 kg)(0.543 m s)2
d=
+ x2 =
+ (0.345 m) 2 = 0.395 m .
k
9.82 N m
(b) Strategy The maximum speed of the block occurs when the block is at it equilibrium position, x = 0.
Use conservation of energy.
Solution Find the maximum speed of the block.
1
1
1
E = K + U = mv 2 + kx 2 = kd 2 , so
2
2
2
2
2
k (d − x )
(9.82 N m)[(0.395 m) 2 − 02 ]
vmax =
=
= 1.11 m s .
m
1.24 kg
(c) Strategy Use the result from part (b).
Solution
v=
k (d 2 − x 2 )
(9.82 N m)[(0.3953 m) 2 − (0.200 m)2 ]
=
= 0.960 m s
m
1.24 kg
84. Strategy The angle θ through which the tuning pin must be turned is related to the extension of the wire ∆L by
the formula for arc length. Use Eq. (10-4).
Solution Relate θ , ∆L, and d p , the diameter of the tuning pin.
dp
s = rθ =
θ = ∆L
2
Find the angle through which the tuning pin must be turned to tune the piano wire.
(d p 2)θ
∆L
F
F
=
=Y
=Y
, so
2
A π (d w 2)
L
L
θ=
2 FL
2
d pπ (d w 2) Y
=
8FL
π d p d w2 Y
=
8(402 N − 381 N)(0.66 m)
2
11
π (0.0080 m)(0.00080 m) (2.0 × 10
s = ∆L
dp / 2 θ
⎛ 360° ⎞
⎜
⎟ = 2.0° .
Pa) ⎝ 2π ⎠
85. Strategy Use Eq. (10-2) to find the tensile stress. Then compare the tensile stress to the elastic limit of steel piano
wire.
Solution Find the tensile stress in the piano wire in Problem 90.
F
T
4T
4(402 N)
tensile stress = =
=
=
= 8.0 × 108 Pa < 8.26 × 108 Pa
A 1 π d 2 π d 2 π (0.80 × 10−3 m) 2
4
The tensile stress is 8.0 × 108 Pa; it is just under the elastic limit.
504
Physics
Chapter 10: Elasticity and Oscillations
86. Strategy Assume that the cable was horizontal prior to being stepped on by the tightrope walker. Neglect the
weight of the cable. Use Eqs. (10-1), (10-2), and (10-4), as well as Newton’s second law.
Solution
(a) Find the strain in the cable.
∆L L′ − L L′ − L′ cos θ
1
1
=
=
=
−1 =
− 1 = 8 × 10−4
strain =
L
L
L′ cos θ
cos θ
cos 0.040
L
θ
L′
(b) Find the tension in the cable.
ΣFy = 2T sin θ − mg = 0, so T =
mg
640 N
=
= 8.0 kN .
2sin θ 2sin 0.040
T
θ
θ
mg
(c) Find the cross-sectional area of the cable.
T
mg
∆L
⎛ 1
⎞
=
=Y
=Y⎜
− 1⎟ , so
A 2 A sin θ
L
⎝ cos θ
⎠
mg
640 N
A=
=
= 5 × 10−5 m 2 .
9
1
1
2Y sin θ cos θ − 1
2(200 × 10 Pa) sin 0.040 cos 0.040
−1
(
)
(
)
(d) Compute the stress.
T 8.0 × 103 N
=
= 1.6 × 108 Pa < 2.5 × 108 Pa
A 5 × 10−5 m 2
No , the cable has not been stretched beyond its elastic limit.
87. Strategy Treat the swinging gibbon as a physical pendulum. Use Eq. (10-28a).
Solution Estimate the frequency of oscillation of the gibbon.
f ≈
1
2π
mgd
1
=
I
2π
(9.80 m s 2 )(0.40 m)
0.25 m 2
= 0.63 Hz
88. (a) Strategy Approximate the tibia as two concentric circles.
Solution Find the average cross-sectional area of the tibia.
⎡⎛ 2.5 cm ⎞2 ⎛ 1.3 cm ⎞ 2 ⎤
2
A = π ro2 − π ri2 = π ⎢⎜
⎟ −⎜
⎟ ⎥ = 3.6 cm
⎢⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦
(b) Strategy Use Eq. (10-2).
Solution Find the compressive stress in the tibia.
F
2800 N
compressive stress = =
= 7.8 × 106 Pa
A 3.6 × 10−4 m 2
(c) Strategy Use Eq. (10-4) and the results of parts (a) and (b). Use Y for a femur in Table 10.1.
Solution Find the change in length for the tibia due to the compressive forces.
FL
∆L F
(2800 N)(0.40 m)
Y
= , so ∆L =
=
= 3.3 × 10−4 m .
4
2
9
−
L
A
AY (3.6 × 10 m )(9.4 × 10 Pa)
505
T
Chapter 10: Elasticity and Oscillations
Physics
89. Strategy The force on the column is its weight. Use Eq. (10-2) and the relationship between density, mass, and
volume.
Solution
(a) Calculate the compressive stress at the bottom of the column.
F mg ρVg ρ hAg
compressive stress = =
=
=
= ρ gh
A
A
A
A
(b) Find the absolute limit to the height of a cylindrical column, regardless of how wide it is.
compressive strength
2.0 × 108 Pa
hm =
=
= 7.6 km
ρg
(2.7 × 103 kg m3 )(9.80 m s 2 )
(c) It is unlikely that someone would want to build a marble column taller than 7.6 km. So, the answer is no; this
limit is of little practical concern. No beanstalk could ever reach a height of 7.6 km; its height is limited by
other means.
90. (a) Strategy Gravitational potential energy is converted into elastic potential energy in the bungee cord. Assume
the cord obeys Hooke’s law. Assume SHM and use Eq. (10-20c).
Solution Find k for the cord.
1 2
2mgh
.
ky = mgh, so k =
2
y2
Find the period of oscillations of the bungee cord.
m
m
2
2
T = 2π
= 2π
=πy
= π (50.0 m − 33.0 m)
= 3.42 s
2
k
gh
2mgh y
(9.78 m s 2 )(50.0 m)
(b) Strategy Use conservation of energy and the quadratic formula.
Solution Find the extension of the bungee cord y2.
1 2 1 ⎛ 2m1gh ⎞ 2
m y2
m y 2 (33.0 m)
ky2 = ⎜ 2 ⎟ y2 = m2 gh = m2 g ( y2 + 33.0 m), so 0 = y22 − 2 1 y2 − 2 1
.
2
2 ⎝ y1 ⎠
m1h
m1h
Solve for y2.
y2 =
m2 y12
m1h
2
⎛ m y2 ⎞
⎡ m y 2 (33.0 m) ⎤
± ⎜ − m2 h1 ⎟ − 4(1) ⎢ − 2 1 m h
1 ⎠
1
⎝
⎣
⎦⎥
2(1)
(80.0 kg)(17.0 m)2 1
=
±
2(60.0 kg)(50.0 m) 2
2
⎡ (80.0 kg)(17.0 m)2 ⎤
(80.0 kg)(17.0 m)2 (132 m)
⎢
⎥ +
(60.0 kg)(50.0 m)
⎢⎣ (60.0 kg)(50.0 m) ⎥⎦
= 20.3 m or − 12.6 m
y2 > 0, so y2 = 20.3 m, and 33.0 m + 20.3 m = 53.3 m > 50.0 m.
No, he should not use the same cord because his greater mass will stretch if too much and he will hit the
water.
506
Physics
Chapter 10: Elasticity and Oscillations
91. (a) Strategy Use Eq. (10-4) and the geometry of the web.
Solution Find the angle the web makes with the horizontal.
∆L
∆L 1.4 × 109 N m 2
Y
= maximum stress, so
=
= 0.35.
L
L
4.0 × 109 N m 2
r
θ
d
r + ∆L
L = r , so ∆L r = 0.35. The new length of a stretched web strand is the hypotenuse of
a right triangle.
r
1
1
1
cos θ =
, so θ = cos −1
=
= cos −1
= 42.2° .
L
L
∆
∆
r + ∆L 1 + r
1 + 0.35
1+ r
(b) Strategy Use Newton’s second law and Eq. (10-2).
Solution Find the tension.
θ
mg
.
ΣFy = T sin θ − mg = 0, so T =
sin θ
Determine the mass of the bug.
F T
mg
= =
= 1.4 × 109 N m 2 , so
A A A sin θ
m=
T
mg
A sin θ (1.4 × 109 N m 2 ) 50(1.0 × 10−11 m 2 ) sin 42.2°(1.4 × 109 N m 2 )
=
= 48 g .
g
9.80 m s 2
(c) Strategy The downward extension of the web is the leg of a right triangle opposite θ . The hypotenuse is
r + ∆L.
Solution Find the distance the web extends downward.
d = (r + ∆L) sin θ = r (1 + ∆L r ) sin θ = (0.10 m)(1.35) sin 42.2° = 9.1 cm
92. Strategy T = 2π I (mgd ) for a physical pendulum, where d is the distance from the axis to the center of mass.
I = 13 mL2 for a uniform rod with the axis through its end.
Solution Find the period of the pendulum for each horizontal axis.
(a) T = 2π
1 mL2
3
1
=
3gd
= 2π L
mgd
2π (1.00 m)
3(9.80 m s 2 )(0.500 m)
= 1.64 s
(b) Treat the meterstick as two rods with lengths 75 cm and 25 cm.
2
2
1 ⎛ m ⎞ ⎛ L ⎞ 1 ⎛ 3m ⎞⎛ 3L ⎞
1 2 ⎛ 1 27 ⎞ 1 2 ⎛ 28 ⎞ 7
2
I = ⎜ ⎟⎜ ⎟ + ⎜
⎟ = 3 mL ⎜ 64 + 64 ⎟ = 3 mL ⎜ 64 ⎟ = 48 mL
3 ⎝ 4 ⎠ ⎝ 4 ⎠ 3 ⎝ 4 ⎟⎜
4
⎠⎝ ⎠
⎝
⎠
⎝ ⎠
T = 2π
7 mL2
48
mgd
= 2π
7 L2
7L
7(1.00 m)
= 2π
= 2π
= 1.53 s
48 g ( L 4)
12 g
12(9.80 m s 2 )
(c) Treat the meterstick as two rods with lengths 60 cm and 40 cm.
2
2
1 ⎛ 4m ⎞⎛ 4 L ⎞ 1 ⎛ 6m ⎞ ⎛ 6 L ⎞
1
216 ⎞ 7
⎛ 64
I= ⎜
mL2
+ ⎜
= mL2 ⎜
+
⎟⎜
⎟
⎟
⎜
⎟
⎟=
3 ⎝ 10 ⎠⎝ 10 ⎠ 3 ⎝ 10 ⎠ ⎝ 10 ⎠
3
⎝ 1000 1000 ⎠ 75
T = 2π
7 mL2
75
mg ( L 10)
= 2π
14 L
14(1.00 m)
= 2π
= 1.94 s
15 g
15(9.80 m s 2 )
507
Chapter 10: Elasticity and Oscillations
Physics
93. (a) Strategy Use conservation of energy. Do not assume SHM.
Solution Find the speed of the pendulum bob at the bottom of its swing.
1
∆K = mv 2 = −∆U = mgL, so v = 2 gL .
2
(b) Strategy Assume (incorrectly, for such a large amplitude) that the motion is SHM. Use Eqs. (10-21) and
(10-26a).
Solution Find the speed of the pendulum bob at the bottom of its swing.
The amplitude A is a quarter of the circumference of a circle with radius L, or
Assuming SHM, vm = ω A =
g ⎛π ⎞ π
L =
L ⎜⎝ 2 ⎟⎠
2
gL . Since v ∝ ω ∝
2π L π
= L.
4
2
1
, a smaller speed implies a larger
T
π
gL
π
v
period. Since m = 2
=
> 1, the period of a pendulum for large amplitudes is larger than that
v
2 gL
2 2
given by Eq. (10-26b).
94. (a) Strategy Draw a diagram of a pendulum.
Solution From the figure below, we see that L cos θ + y = L, or y = L(1 − cos θ ).
θ
L
L cos θ
L
y
(b) Strategy Assume θ is small. Use the gravitational potential energy of a pendulum given in the problem
statement and the result of part (a).
Solution
⎡ 1 ⎛ x ⎞2 ⎤ 1 ⎛ mg ⎞
mg
1
U = mgy = mgL(1 − cos θ ) ≈ mgL ⎢ ⎜ ⎟ ⎥ = ⎜
x 2 = kx 2 with k =
.
⎟
L
2
⎢⎣ 2 ⎝ L ⎠ ⎥⎦ 2 ⎝ L ⎠
508
Physics
Chapter 10: Elasticity and Oscillations
95. Strategy The inertia of the system is I = 13 m1L2 + m2 L2. Use Eq. (10-28b) and the definition of center of mass.
Solution
(a) The distance to the center of mass from the rotation axis is
m1 L2 + m2 L
m
1
+ m2
= 2
L.
m1 + m2
m
Find the period of this physical pendulum.
d=
I
T = 2π
= 2π
mgd
1 m L2
3 1
+ m2 L2
mgd
= 2π
L2
mg ⎡
⎢⎣
(
(
m1
2
m1
3
+ m2
+ m2
)
) m⎤⎥⎦ L
= 2π
(
g(
L
) = 2π
+m )
m1
3
+ m2
m1
2
2
2 L(m1 + 3m2 )
3g (m1 + 2m2 )
(b) For each case, replace the smaller mass with zero. Then
for m1 >> m2 , T = 2π
2L
L
, and for m1 << m2 , T = 2π
.
3g
g
The former is the period for the uniform rod alone and the latter is the period for the block alone.
96. (a) Strategy Use Eq. (10-4).
Solution Find the force the wings must exert to extend the resilin.
F
∆L
YA∆L (1.7 × 106 N m 2 )(1.0 ×10−6 m 2 )(4.0 cm − 1.0 cm)
=Y
, so F =
=
= 5.1 N .
A
L
L
1.0 cm
(b) Strategy The energy stored in the resilin is elastic potential energy. Use Hooke’s law.
Solution Find the energy stored in the resilin.
1
1⎛F ⎞
1
1
U = kx 2 = ⎜ ⎟ x 2 = Fx = (5.1 N)(0.030 m) = 7.7 × 10−2 J
2
2⎝ x ⎠
2
2
509
Chapter 11
WAVES
Conceptual Questions
1. Wrapping a thick coil of copper wire around a piano string increases the string’s mass density and therefore
decreases the speed of waves traveling along it. The fundamental wavelength is fixed by the length of the string—
the decreased wave velocity must therefore be accompanied by a decrease in the frequency at which the string
vibrates.
2. Piano, guitar, and violin strings produce transverse waves when they are plucked or bowed—they do not produce
longitudinal sound waves. The transverse motion of the strings causes longitudinal sound waves to be produced in
the surrounding air.
3. The wavelength of the fundamental standing wave on a cello string depends only upon the length of the string.
The wavelength of the sound waves produced by the vibrating cello string is determined by the frequency of the
string’s vibration. This frequency is proportional to the string’s wave velocity, which depends upon both the mass
per unit length and the tension of the string.
4. (a) The wavelength of the fundamental will decrease.
(b) The frequency of the fundamental will increase.
(c) The wave velocity is constant, thus the time for a pulse to travel the length of the string will decrease.
(d) The maximum velocity of a point on the string is proportional to the frequency and will therefore increase.
(e) The maximum acceleration of a point on the string is proportional to the square of the frequency and will
therefore increase.
5. Words spoken by two people at the same time are comprehensible because sound waves travel through each
other—interfering while superimposed, but returning to their original waveform as they again separate.
6. (a) Increasing the tension of the string changes the speed of transverse waves and thus the frequency.
(b) Pressing her finger on the string changes the string length, which changes the wavelength and hence the
frequency.
(c) Each string has a different tension and mass per unit length; thus the speed of transverse waves differs from
string to string. The fundamental wavelength remains the same because the strings are all of the same length,
but the frequencies are different.
7. A transverse wave is produced by a disturbance that tends to shear the medium, separating layers at right angles to
the direction of the wave velocity.
510
Physics
Chapter 11: Waves
8. In the figure below, the shape of a wave traveling to the right is depicted at two times separated by a short
interval. Arrows indicate the direction of the string’s movement.
v
9. Since transverse waves do not travel through the core while longitudinal waves do, some part of the core is a
molten, viscous liquid that cannot support the transmission of a transverse wave. A longitudinal wave can create
compressions and rarefactions in the liquid and travel on through.
10. The electrical signal mimicking the noise is modified electronically to produce a wave that is out of phase with the
noise. When played in the headphones it interferes destructively with the noise, thereby canceling it.
11. If the wires going to one speaker are reversed, the sound waves emitted will be out of phase with those from the
other speaker. For a listener situated midway between the speakers, the sounds from the two speakers will
interfere destructively, and the sound level will be noticeably reduced. If the listener moves slightly, the difference
in the path length from the two speakers changes. If this change in path length is small compared to the
wavelength, the waves still interfere destructively to a large extent. For this reason there is a noticeably weaker
bass (long wavelength) over a large area between the speakers, while high frequencies (short wavelengths) are
less affected.
Problems
1. Strategy Form a proportion with the intensities, treating the Sun as an isotropic source. Use Eq. (11-1).
Solution Find the intensity of the sunlight that reaches Jupiter.
P
2
2
⎛r ⎞
IJ
r 2
4π rJ 2
⎛ 1 ⎞
=
= E , so I J = ⎜ E ⎟ I E = ⎜
(1400 W m 2 ) = 52 W m 2 .
⎟
P
2
IE
r
5.2
⎝
⎠
r
J
⎝
⎠
J
4π r 2
E
2. (a) Strategy It takes half the time for the sound to cross the valley as it does to make the round trip.
Solution The cliff is ∆x = v∆t = (343 m s)(0.75 s) = 260 m away.
(b) Strategy Treat the radio and echo as isotropic sources. Use Eq. (11-1) and form a proportion.
Solution Find the intensity of the music arriving at the cliff.
I cliff
I radio
=
P (4π rc 2 )
P (4π rr 2 )
=
rr 2
⎛r
, so I cliff = ⎜⎜ r
2
rc
⎝ rc
2
2
⎞
⎛ 1 ⎞
2
−5
−10 W m 2 .
⎟⎟ I radio = ⎜
⎟ (1.0 × 10 W m ) = 1.5 × 10
257.25
⎝
⎠
⎠
3. Strategy Form a proportion with the intensities, treating the jet airplane as an isotropic source. Use Eq. (11-1).
Solution Find the intensity of the sound waves at the ears of the person.
2
2
2
⎛r ⎞
⎛r ⎞
⎛ 5.0 m ⎞
(1.0 × 102 W m 2 ) = 170 mW m 2 .
=
= ⎜⎜ 1 ⎟⎟ , so I 2 = ⎜⎜ 1 ⎟⎟ I1 = ⎜
⎟
I1 P (4π r12 ) ⎝ r2 ⎠
⎝ 120 m ⎠
⎝ r2 ⎠
I2
P (4π r22 )
511
Chapter 11: Waves
Physics
4. Strategy The power equals the intensity times the area.
Solution Find the power radiated by the jet airplane in the form of sound waves.
2
⎛ 5.0 m ⎞
2
2
2
P = IA = ⎜
⎟ (1.0 × 10 W m )4π (120 m) = 31 kW
120
m
⎝
⎠
5. Strategy The power equals the intensity times the area.
Solution Find the rate at which the Sun emits electromagnetic waves.
P = IA = I (4π RE 2 ) = 4π (1.4 × 103 W m 2 )(1.50 × 1011 m)2 = 4.0 × 1026 W
6. Strategy Refer to the figure. Use the definition of average speed.
Solution
(a) vx =
∆x 1.80 m − 1.50 m
=
= 1.5 m s
∆t
0.20 s
(b) v y =
∆y 8.7 cm − 4.5 cm
=
= 21 cm s
∆t
0.20 s
7. Strategy Refer to the figure. Use the definition of average speed.
Solution
(a) Find the speed.
∆x 1.80 m − 1.50 m
vx =
=
= 1.5 m/s
∆t
0.20 s
Find the position.
xf = xi + v∆t = 1.80 m + (1.5 m s)(3.00 s − 0.20 s) = 6.0 m
x − xi
4.00 m − 1.80 m
+ ti =
+ 0.20 s = 1.7 s
(b) tf = f
vx
1.5 m s
8. Strategy Use Eq. (11-4).
Solution Find the linear mass density of the cord.
F
F
75 N
v=
, so µ =
=
= 3.8 g m .
2
µ
v
(140 m s) 2
9. Strategy Use Eq. (11-4).
Solution Find the speed of the transverse waves on the string.
90.0 N
F
=
= 168 m s
v=
µ
3.20 × 10−3 kg m
512
Physics
Chapter 11: Waves
10. Strategy Use Eq. (11-2) to find the wave speeds for both strings. Use the speeds and the lengths of the strings to
find the additional time required for the slower wave to reach the end of its string.
Solution
FL
(180.0 N)(15.0 m)
v1 = 1 1 =
= 186 m s and v2 =
m1
0.0780 kg
F2 L2
m2
=
(160.0 N)(15.0 m)
= 203 m s.
0.0580 kg
The pulse moves faster on the second string. Find the time difference.
⎛1 1 ⎞
⎛
⎞
∆x ∆x
1
1
∆x = v∆t , so ∆t1 − ∆t2 =
−
= ∆x ⎜⎜ − ⎟⎟ = (15.0 m) ⎜
−
⎟ = 6.9 ms .
v1 v2
⎝ 186.05 m s 203.42 m s ⎠
⎝ v1 v2 ⎠
11. Strategy Use Eq. (11-2) for the speed of the transverse waves.
Solution The weight of the string divided by the load is
0.25 N
= 2.5 × 10−4 = 0.025%.
3
1.00 × 10 N
The weight of the string is negligible since the result will be limited to two significant figures (by 0.25 N).
Find the time it takes the wave pulse to travel to the upper end of the string.
(0.25 N)(10.0 m)
∆y
L
mL
mgL
∆t =
=
=
=
=
= 16 ms
v
F
Fg
FL m
(1.00 × 103 N)(9.80 N kg)
12. Strategy Use Eq. (11-6).
Solution Find the speed of the wave.
v = f λ = (500.0 Hz)(0.500 m) = 250 m s
13. Strategy Use Eq. (11-5).
Solution Find the wavelength.
λ = vT = (75.0 m s)(5.00 × 10−3 s) = 0.375 m
14. Strategy Use Eq. (11-6).
Solution Find the frequency.
v
120 m s
= 400 Hz
f = =
λ 30.0 × 10−2 m
15. Strategy Use Eq. (11-6).
Solution Find the frequencies.
(a)
f =
(b) f =
v
λ
v
λ
=
340 m s
= 340 Hz
1.0 m
=
3.0 × 108 m s
= 3.0 × 108 Hz
1.0 m
513
Chapter 11: Waves
Physics
16. Strategy Use Eq. (11-6).
Solution Find the range of visible electromagnetic waves.
v 3.0 × 108 m s
v 3.0 × 108 m s
=
= 7.5 × 1014 Hz and f 2 =
=
= 4.3 × 1014 Hz.
f1 =
λ1 4.0 × 10−7 m
λ2 7.0 × 10−7 m
The frequency range is 4.3 × 1014 Hz to 7.5 × 1014 Hz .
17. Strategy Use Eq. (11-6).
Solution Find the frequency with which the buoy bobs up and down.
v 2.5 m s
= 0.33 Hz
f = =
7.5 m
λ
18. Strategy and Solution If another swimmer were 9.6 m away from you, your motions would be the same (in
phase). For periodic motion, the motion half a wavelength from any point along a line parallel to the motion of the
waves is opposite to the motion at that point, so the other swimmer should be half a wavelength away, or
9.6 m
= 4.8 m .
2
19. Strategy Use Eq. (11-7).
Solution Compute the wave speed.
ω
5.0 rad s
= 0.83 cm s
v= =
k 6.0 rad cm
20. Strategy Read the amplitude from the equation of the wave. Use Eq. (11-7) to find the wavelength.
Solution
⎧ π
y ( x, t ) = (3.5 cm) sin ⎨
[ x − (66 cm s)t ]⎫⎬ = A sin(kx − ωt ), so we have
⎩ 3.0 cm
⎭
(a) A = 3.5 cm
(b) λ =
2π
2π
= π
k
= 6.0 cm
3.0 cm
21. Strategy The wave on the string is of the form y ( x, t ) = A sin(ωt − kx). Use the equation, the given information,
Eq. (11-7), and the relationship between period and angular frequency to find the amplitude, wavelength, period,
and wave speed.
Solution
(a) A = 4.0 mm
(b) λ =
(c) T =
2π
2π
=
= 1.0 m
k
6.0 m −1
2π
ω
=
2π
= 0.010 s
6.0 × 102 s −1
514
Physics
Chapter 11: Waves
(d) v =
ω
k
=
6.0 × 102 s −1
6.0 m −1
= 100 m s
(e) Since the signs of ω t and kx are opposite, the wave travels in the +x-direction (to the right).
22. (a) Strategy Use Eq. (10-21).
Solution Find the maximum transverse speed of a point on the string.
vm = ω A = (130 rad s)(0.0220 m) = 2.9 m s
(b) Strategy Use Eq. (10-22).
Solution Find the maximum transverse acceleration of a point on the string.
am = ω 2 A = (130 rad s) 2 (0.0220 m) = 370 m s 2
(c) Strategy Use Eq. (11-7).
Solution Find the wave speed.
ω 130 rad s
v= =
= 8.7 m s
k 15 rad m
(d) Strategy Consider why transverse speed and wave speed are different.
Solution The answer to part (c) is different from the answer to part (a) because
the motion of the particles on the string is not the same as the motion of the wave along the string .
23. Strategy The equation for a transverse sinusoidal wave moving in the negative x-direction can be written in the
form y ( x, t ) = A sin(kx + ωt ). Use Eq. (11-7) to find the angular frequency and the wavenumber.
Solution A = 0.120 m, λ = 0.300 m, v = 6.40 m s, and y ( x, t ) = A sin(ω t + kx ).
ω=
2π v
λ
=
2π (6.40 m s)
2π
2π
= 134 s −1 and k =
=
= 20.9 m −1.
0.300 m
λ 0.300 m
Thus, the equation is
y ( x, t ) = (0.120 m) sin[(134 s−1 )t + (20.9 m −1 ) x ] .
24. Strategy The equation for a transverse sinusoidal wave moving in the positive x-direction and in the negative
y-direction in the next instant of time according to the situation given in the problem statement can be written in
the form y ( x, t ) = A sin(kx − ωt ). Use Eqs. (10-21) and (11-7).
Solution Find the wave speed v by finding the maximum speed of a point on the string.
v = 5.00vm = 5.00ω A
Find the wave number.
ω
ω
1
1
k= =
=
=
= 8.00 rad m
v 5.00ω A 5.00 A 5.00(0.0250 m)
The equation for the transverse sinusoidal wave is
y ( x, t ) = (2.50 cm) sin[(8.00 rad m) x − (2.90 rad s)t ] .
515
Chapter 11: Waves
Physics
25. Strategy The maximum y-value is the amplitude. Find the horizontal distance for which the wave repeats to find
the wavelength. The wave speed is the distance the wave travels divided by the time interval it took to travel that
distance. Use Eq. (11-6) to find the frequency of the wave. The period is the reciprocal of the frequency.
Solution
(a) ymax = 2.6 cm, so A = 2.6 cm .
(b) λ = ∆x = 16 m − 2 m = 14 m
(c) v =
(d) f =
(e) T =
∆x 7.5 m − 5.5 m
=
= 20 m s
∆t
0.10 s
v
λ
=
20 m s
= 1.4 Hz
14 m
1
14 m
=
= 0.70 s
f 20 m s
26. (a) Strategy The maximum y-value is the amplitude. Find the horizontal distance for which the wave repeats to
find the wavelength. The wave speed is the distance the wave travels divided by the time interval it took to
travel that distance.
Solution Plot y ( x, t ) = (4.0 cm) sin[(378 rad s)t − (314 rad cm) x] at the two given times.
y (cm)
4
t=0
2
0
0.01
t=
0.02
1
480 s
0.03 x (cm)
2
4
ymax = 4.0 cm, so A = 4.0 cm ; λ = ∆x = 0.020 cm − 0 = 0.020 cm ;
∆x 0.0025 cm − 0
v=
=
= 1.2 cm s
∆t
1 480 s
(b) Strategy Set x = 0; then the function to plot is y ( x, t ) = (4.0 cm) sin[(378 rad s)t ]. The period of the
vibration is equal to the time interval for which the wave repeats. Multiply the wave speed by the period of
the vibration and compare to the wavelength found in part (a).
Solution
y (cm)
4
x=0
2
0
2
π
378
π
189
t (s)
4
T=
π
⎛ π
s ≈ 16.6 ms and vT = (1.2 cm s) ⎜
189
⎝ 189
⎞
s ⎟ = 0.020 cm = λ.
⎠
516
Physics
Chapter 11: Waves
27. Strategy Use Eqs. (10-21) and (10-22). Plot the graphs.
Solution Find the maximum speed and maximum acceleration of a point on the string.
vm = ω A = (4.0π rad s)(0.0050 m) = 0.063 m s and am = ω 2 A = (4.0π rad s)2 (0.0050 m) = 0.79 m s 2 .
Plot the graphs.
y(0, t ) = (0.0050 m) cos[(4.0π s −1)t ]
y (m)
0.0050
x=0
0
0.25
0.50 t (s)
0.0050
v y leads y by
1
4
vy (m/s)
0.063
0
cycle; v y (0, t ) = −(0.063 m/s) sin[(4.0π s −1)t ]
x=0
0.50 t (s)
0.25
0.063
a y leads v y by
ay (m/s2)
0.79
0
1
4
cycle; a y (0, t ) = −(0.79 m/s 2 ) cos[(4.0π s −1)t ]
x=0
0.50 t (s)
0.25
0.79
28. Strategy Use Eqs. (10-21) and (10-22). v y leads y by 1/4 cycle, so v y is a cosine function. Plot the graphs.
Solution y ( x, t ) = (1.2 mm) sin[(2.0π s −1 )t − (0.50π m −1 ) x]; calculate the maximum speed.
vm = ω A = (2.0π s −1 )(1.2 mm) = 7.5 mm/s, so v y (0, t ) = (7.5 mm s) cos(2.0π s −1 )t at x = 0.
y (mm)
vy (mm/s)
7.5
1.2
x=0
x=0
1.0
0
0.5
0
t (s)
7.5
1.2
517
0.5
1.0
t (s)
Chapter 11: Waves
Physics
29. (a) Strategy Substitute t = 0, 0.96 s, and 1.92 s into y ( x, t ) = (0.80 mm) sin[(π 5.0 cm −1 ) x − (π 6.0 s −1 )t ] and
graph the resulting equations.
Solution The three equations are:
y ( x, 0) = (0.80 mm) sin[(π 5.0 cm −1 ) x], y ( x, 0.96 s) = (0.80 mm) sin[(π 5.0 cm −1 ) x − 0.50], and
y ( x, 1.92 s) = (0.80 mm) sin[(π 5.0 cm −1 ) x − 1.0].
Find the wavelength.
2π
2π
=
= 10 cm
λ=
π (5.0 cm)
k
The amplitude of the wave is A = 0.80 mm. The first graph (solid) begins at the origin. The second graph
(dashed) is shifted to the right by (5.0 cm × 0.50) π = 0.80 cm. The third graph (dotted) is shifted to the right
twice as far as the second graph, or 1.6 cm.
The graphs are shown:
y (mm)
t = 0.96 s
0.80
t = 1.92 s
10
0
x (cm)
5.0
t=0
0.80
(b) Strategy Substitute t = 0, 0.96 s, and 1.96 s into y ( x, t ) = (0.50 mm) sin[(π 5.0 cm −1 ) x + (π 6.0 s −1 )t ] and
graph the resulting equations.
Solution The three equations are:
y ( x, 0) = (0.50 mm) sin[(π 5.0 cm −1 ) x], y ( x, 0.96 s) = (0.50 mm) sin[(π 5.0 cm −1 ) x + 0.50], and
y ( x, 1.92 s) = (0.50 mm) sin[(π 5.0 cm −1 ) x + 1.0].
Find the wavelength.
2π
2π
=
= 10 cm
λ=
π (5.0 cm)
k
The amplitude of the wave is A = 0.50 mm. The first graph (solid) begins at the origin. The second graph
(dashed) is shifted to the left by (5.0 cm × 0.50) π = 0.80 cm. The third graph (dotted) is shifted to the left
twice as far as the second graph, or 1.6 cm. The graphs are shown:
y (mm)
t = 0.96 s
0.50
t=0
5.0
t = 1.92 s
0
10
x (cm)
0.50
(c) Strategy Refer to the results of parts (a) and (b).
Solution The graphs obtained in part (a) move to the right as time progresses, so
y ( x, t ) = (0.80 mm) sin (kx − ωt ) represents a wave traveling in the +x-direction. The graphs obtained in part
(b) move to the left as time progresses, so y ( x, t ) = (0.50 mm) sin (kx + ωt ) represents a wave traveling in the
–x-direction.
518
Physics
Chapter 11: Waves
30. (a) Strategy and Solution Since the argument of the cosine function is ω t + kx (both terms are positive), the
wave is moving to the left.
(b) Strategy The maximum y-value is the amplitude. The wave repeats every 4.0 cm, so the wavelength is
0.040 m. Use Eq. (11-7) to find the angular frequency and wavenumber.
Solution
ymax = 2.0 mm, so
2π
k=
λ
=
A = 2.0 mm ; ω =
2π
= 160 rad m
0.040 m
2π v
λ
=
2π (10.0 m s)
= 1600 rad s ;
0.040 m
(c) Strategy Choose the point (x, y) = (0, 0). 0 = A cos[ωt + k (0)], so 0 = cos ωt. cos ωt = 0 when ωt = nπ 2,
where n is an odd integer. The smallest nonnegative n is 1 and it will give the smallest nonnegative time. Use
Eq. (11-6).
Solution
t=
π
2ω
=
π
4π f
=
1
λ
0.040 m
=
=
= 1.0 ms
4 f 4v 4 (10.0 m s )
The period is T =
1 λ 0.040 m
= =
= 4.0 ms, so the three times are 1.0 ms, 5.0 ms, and 9.0 ms.
f v 10.0 m s
31. Strategy Compute the positions of the peaks of each pulse for the given times. Then use the principle of
superposition to graph the shape of the cord for each time.
Solution
t (s)
Short Pulse Position
Tall Pulse Position
0.15
10 cm + (40 cm s)(0.15 s) = 16 cm
30 cm − (40 cm s)(0.15 s) = 24 cm
0.25
10 cm + (40 cm s)(0.25 s) = 20 cm
30 cm − (40 cm s)(0.25 s) = 20 cm
0.30
10 cm + (40 cm s)(0.30 s) = 22 cm
30 cm − (40 cm s)(0.30 s) = 18 cm
y (cm)
1.5
y (cm)
1.5
1.0
1.0
t = 0.15 s
0.5
0
y (cm)
1.5
0.5
10
20
30
40 x (cm)
20
30
40 x (cm)
0
t = 0.30 s
1.0
0.5
0
t = 0.25 s
10
519
10
20
30
40 x (cm)
Chapter 11: Waves
Physics
32. Strategy Compute the positions of the peaks of each pulse for the given times. Then use the principle of
superposition to graph the shape of the cord for each time.
Solution
t (s)
Positive Pulse Position
Negative Pulse Position
0.60
2.0 m + (2.5 m s)(0.60 s) = 3.5 m
6.0 m − (2.5 m s)(0.60 s) = 4.5 m
0.80
2.0 m + (2.5 m s)(0.80 s) = 4.0 m
6.0 m − (2.5 m s)(0.80 s) = 4.0 m
0.90
2.0 m + (2.5 m s)(0.90 s) = 4.3 m
6.0 m − (2.5 m s)(0.90 s) = 3.8 m
y (cm)
10.0
y (cm)
10.0
t = 0.60 s
5.0
0
2
5.0
4
6
t = 0.80 s
5.0
8
0
x (m)
5.0
2
4
6
8
x (m)
10.0
10.0
y (cm)
10.0
t = 0.90 s
5.0
0
2
5.0
6
8
x (m)
10.0
33. Strategy Sketch the sine waves. Use the principle of superposition to find the amplitudes. Let y1 = A sin(ω t + kx)
⎛α + β ⎞
⎛α − β ⎞
and y2 = A sin(ω t + kx − φ ) and use the trigonometric identity sin α + sin β = 2sin ⎜
⎟ cos ⎜ 2 ⎟ .
2
⎝
⎠
⎝
⎠
Solution
(a) y (cm)
4
2
0
2
60°
180°
300°
420°
θ
4
Use the principle of superposition.
φ⎞ φ
φ⎞
⎛
⎛
y = y1 + y2 = A sin(ω t + kx ) + A sin(ω t + kx − φ ) = 2 A sin ⎜ ω t + kx − ⎟ cos = A′ sin ⎜ ω t + kx − ⎟
2⎠
2
2⎠
⎝
⎝
φ
60.0°
where A′ = 2 A cos = 2(4.0 cm) cos
= 6.9 cm .
2
2
520
Physics
Chapter 11: Waves
(b) y (cm)
4
2
0
2
60°
180°
4
A′ = 2(4.0 cm) cos
300°
420°
θ
90.0°
= 5.7 cm
2
34. Strategy Let y1 = A sin(ω t + kx) and y2 = A sin(ω t + kx − φ ) and use the trigonometric identity
⎛α + β ⎞
⎛α − β ⎞
sin α + sin β = 2sin ⎜
⎟ cos ⎜ 2 ⎟ . Use the principle of superposition.
2
⎝
⎠
⎝
⎠
Solution Find the traveling wave.
φ⎞
φ
φ⎞
⎛
⎛
y = y1 + y2 = A sin(ω t + kx) + A sin(ω t + kx − φ ) = 2 A sin ⎜ ω t + kx − ⎟ cos = A′ sin ⎜ ω t + kx − ⎟
2
2
2⎠
⎝
⎠
⎝
where A′ = 2 A cos
2 A cos
φ
2
φ
2
= A. Find φ .
= A, so φ = 2 cos −1
1
= 120° .
2
35. Strategy Use the principle of superposition and the trigonometric identity
⎛α + β ⎞
⎛α − β ⎞
sin α + sin β = 2sin ⎜
⎟ cos ⎜ 2 ⎟ .
2
⎝
⎠
⎝
⎠
Solution Find the traveling sine wave.
φ⎞
φ
φ⎞
⎛
⎛
y = y1 + y2 = A sin(ω t + kx) + A sin(ω t + kx − φ ) = 2 A sin ⎜ ω t + kx − ⎟ cos = A′ sin ⎜ ω t + kx − ⎟
2⎠
2
2⎠
⎝
⎝
where A′ = 2 A cos
cos
φ
2
=
φ
2
= 6.69 cm. Find φ .
A′
A′
6.69 cm
= 2 cos −1
= 96.0° .
, so φ = 2 cos −1
2A
2A
2(5.00 cm)
36. Strategy f = v λ and the frequency is the same in both mediums.
Solution Find the wavelength of the light in water.
va vw
v
=
, so λw = w λa = 0.750(0.500 × 10−6 m) = 375 nm .
λa λw
va
37. Strategy Refer to the figure. Use ∆x = vx ∆t and the principle of superposition.
Solution The pulse moves 1.80 m − 1.50 m = 0.30 m in 0.20 s. So, the speed of the wave is
0.30 m
= 1.5 m s. When the pulse reaches the right endpoint, it is reflected and inverted. When exactly half
v=
0.20 s
of the pulse has been reflected and inverted, the superposition of the incident and reflected waves results in the
x 4.0 m − 1.5 m
= 1.7 s .
cancellation of the waves ( y1 + y2 = 0). Thus, the string looks flat at t = =
v
1.5 m s
521
Chapter 11: Waves
Physics
38. Strategy Refer to the figure. Use ∆x = vx ∆t.
Solution The pulse moves 1.80 m − 1.50 m = 0.30 m in 0.20 s. So, the speed of the wave is
0.30 m
= 1.5 m s. The pulse travels to the right until it reaches the endpoint; it is then reflected and inverted.
v=
0.20 s
It then travels to the left until it hits the left endpoint; it is again reflected and inverted. The pulse travels to the
right until it reaches x = 1.5 m. The total distance traveled is 2.5 m + 4.0 m + 1.5 m = 8.0 m. The elapsed time is
x
8.0 m
= 5.3 s .
t= =
v 1.5 m s
39. Strategy The waves are coherent. Use the principle of superposition.
Solution
(a) The resulting wave will have its largest amplitude if the waves interfere constructively. The phase difference
is 0° , and the amplitude is A1 + A2 = 5.0 cm + 3.0 cm = 8.0 cm .
(b) The resulting wave will have its smallest amplitude if the waves interfere destructively. The phase difference
is 180° , and the amplitude is A1 − A2 = 5.0 cm − 3.0 cm = 2.0 cm .
(c) 8.0 cm : 2.0 cm = 4 :1
40. Strategy The waves are coherent. Use the principle of superposition. Intensity is proportional to the square of the
amplitude.
Solution
(a) The resulting wave will have its highest intensity when the waves interfere constructively. The phase
difference is 0° , and the amplitude is A = A1 + A2 = 6.0 cm + 3.0 cm = 9.0 cm .
(b) The resulting wave will have its lowest intensity when the waves interfere destructively. The phase difference
is 180° , and the amplitude is A = A1 − A2 = 6.0 cm − 3.0 cm = 3.0 cm .
(c) Form a proportion.
2
2
I1 ⎛ A1 ⎞
⎛ 9.0 cm ⎞
=⎜
⎟ =⎜
⎟ = 9.0
I 2 ⎝ A2 ⎠
⎝ 3.0 cm ⎠
The ratio is 9 :1 .
522
Physics
Chapter 11: Waves
41. Strategy Intensity is proportional to the amplitude squared. For constructive interference, the amplitude of the
superposition is the sum of the original amplitudes.
Solution Find A1/A2.
A1
=
A2
I1
25
5.0
=
=
I2
15
3.0
Find the amplitude of the superposition.
⎛
5.0
5.0 ⎞
+ A2 = A2 ⎜⎜1 +
A = A1 + A2 = A2
⎟
3.0
3.0 ⎟⎠
⎝
Find the intensity of the superposition.
2
2
⎛
⎛
I
A
5.0
5.0 ⎞
5.0 ⎞
2
2
=
= 1+
, so I = ⎜⎜ 1 +
⎟⎟ I 2 = ⎜⎜ 1 +
⎟⎟ (15 mW m ) = 79 mW m .
I 2 A2
3.0
3.0
3.0
⎝
⎠
⎝
⎠
42. Strategy Intensity is proportional to the amplitude squared. For destructive interference, the amplitude of the
superposition is the absolute value of the difference of the original amplitudes.
Solution Find A1/A2.
A1
=
A2
I1
=
I2
25
28
Find the amplitude of the superposition.
⎛
25 ⎞
A = A1 − A2 = A2 ⎜⎜1 −
⎟
28 ⎟⎠
⎝
Find the intensity of the superposition.
2
2
⎛
⎛
I
A
25
25 ⎞
25 ⎞
2
2
−3
=
= 1−
, so I = ⎜⎜ 1 −
⎟⎟ I 2 = ⎜⎜ 1 −
⎟⎟ (28 × 10 W m ) = 80 µW m .
I 2 A2
28
28
28
⎝
⎠
⎝
⎠
43. Strategy Intensity is proportional to the amplitude squared. For constructive interference, the amplitude of the
superposition is the sum of the original amplitudes. For destructive interference, the amplitude of the
superposition is the absolute value of the difference of the original amplitudes. For incoherent waves, the
intensities add.
Solution
(a) Find A1/A2.
A1
=
A2
I1
0.040 2.0
=
=
I2
0.090 3.0
Find the amplitude of the superposition.
⎛ 2.0 ⎞
⎛ 2.0 ⎞
A = A1 + A2 = A2 ⎜
⎟ + A2 = A2 ⎜ 1 +
⎟
3.0
⎝
⎠
⎝ 3.0 ⎠
Find the intensity of the superposition.
2
2
I
A
2.0
⎛ 2.0 ⎞
⎛ 2.0 ⎞
2
2
=
= 1+
, so I = ⎜1 +
⎟ I 2 = ⎜1 +
⎟ (0.090 W m ) = 0.25 W m .
I 2 A2
3.0
⎝ 3.0 ⎠
⎝ 3.0 ⎠
523
Chapter 11: Waves
Physics
(b) Find the amplitude of the superposition.
⎛ 2.0 ⎞
A = A1 − A2 = A2 ⎜1 −
⎟
⎝ 3.0 ⎠
Find the intensity of the superposition.
2
2
⎛ 2.0 ⎞
⎛ 2.0 ⎞
2
2
I = ⎜1 −
⎟ I 2 = ⎜1 −
⎟ (0.090 W m ) = 0.010 W m
⎝ 3.0 ⎠
⎝ 3.0 ⎠
(c) Find the intensity of the superposition.
I = I1 + I 2 = 0.040 W m 2 + 0.090 W m 2 = 0.130 W m 2
44. Strategy The intensity minimums imply that the distance 37.1 m − 25.8 m = 11.3 m is equal to a whole number
of wavelengths m plus one-half wavelength. Determine the number (or numbers) of wavelengths m that gives a
frequency (or frequencies) between 100 Hz and 150 Hz. Use Eq. (11-6).
Solution The number of wavelengths m is related to the distance between intensity minimums by
1⎞
v ⎛
1 ⎞ 343 m s
⎛
= (30.35 Hz)m + 15.18 Hz.
⎜ m + ⎟ λ = 11.3 m. In terms of frequency, we have f = = ⎜ m + ⎟
2
2 ⎠ 11.3 m
λ
⎝
⎠
⎝
Substitute 100 Hz and 150 Hz for f and solve for m to find the range of possible values.
(30.35 Hz)m + 15.18 Hz = 100 Hz, so m > 2.8 and (30.35 Hz)m + 15.18 Hz = 150 Hz, so m < 4.4.
The two possible values of m are 3 and 4. Try them both.
1 ⎞ 343 m s
1 ⎞ 343 m s
⎛
⎛
f (3) = ⎜ 3 + ⎟
= 106 Hz and f (4) = ⎜ 4 + ⎟
= 137 Hz.
2 ⎠ 11.3 m
2 ⎠ 11.3 m
⎝
⎝
Both values are within the range of allowed frequencies, so the possible frequencies of the sound waves coming
from the speakers are 106 Hz and 137 Hz.
45. Strategy Use Eqs. (11-2) and (11-13).
Solution
2
⎛ f ⎞
T1
T
, so ⎜ 2 ⎟ = 2 .
4 Lm
T1
⎝ f1 ⎠
Calculate the percentage reduction in the tension.
f1 =
1 T1L
v
TL
=
and v =
. f1 =
2L
m
2L m
T1 − T2
× 100% =
T1
T1 −
( ) T ×100% = ⎢⎡1 − ⎛⎜ f
f2 2
1
f1
T1
⎢
⎣
⎞
⎟
f
⎝ 1⎠
2
2⎤
2
⎡
1 − 0.040 ⎞ ⎤
⎥ × 100% = ⎢1 − ⎛⎜
⎟ ⎥ × 100% = 7.8%
1
⎥
⎠ ⎥⎦
⎢⎣ ⎝
⎦
46. Strategy According to Eq. (11-13), the fundamental frequency of a guitar string is directly proportional to the
speed of a wave on the string. According to Eq. (11-2), the speed of a wave on the string in directly proportional
to the square root of the tension in the string.
Solution f ∝ v and v ∝ T , so f ∝ T , where T is the tension. Therefore, since
the frequency increases by 7% .
47. Strategy Nodes are separated by a distance of λ /2. Use Eq. (11-7).
Solution Find the distance between adjacent nodes.
λ 2π
π
distance = =
=
= 0.016 m
2 2k 2.0 × 102 rad m
524
1.15 = 1.07,
Physics
Chapter 11: Waves
48. Strategy The frequencies are given by f n = nv (2 L). The speed of the transverse waves is related to the tension
by v = T µ .
Solution
(a) Find the speed of the transverse waves.
v
= f1 , so v = 2 Lf1 = 2(1.50 m)(450.0 Hz) = 1350 m s .
2L
(b) Find the tension.
v
1 T
=
f1 =
, so T = 4 µ L2 f12 = 4(25.0 × 10−6 kg m)(1.50 m) 2 (450.0 Hz)2 = 45.6 N .
2L 2L µ
(c) The frequencies are the same for both mediums, but the wavelength depends upon the wave speed.
v 340 m s
f = f1 = 450.0 Hz, so λ = =
= 0.76 m.
f 450.0 Hz
The wavelength and frequency are 0.76 m and 450.0 Hz , respectively.
49. Strategy The frequencies are given by f n = nv (2 L). The speed of the transverse waves is related to the tension
by v = T µ .
Solution
(a) Find the frequency of the fundamental oscillation.
v
1 T
1
12 N
f1 =
=
=
= 33 Hz
2 L 2 L µ 2(1.5 m) 1.2 × 10−3 kg m
(b) Find the tension.
4µ L2 f32 4(1.2 × 10−3 kg m)(1.5 m) 2 (0.50 × 103 Hz)2
3v
3 T
, so T =
f3 =
=
=
= 300 N .
2L 2L µ
9
9
50. Strategy Use Newton’s second law and Eqs. (11-4) and (11-13).
Solution Find the tension in the string.
ΣFy = T − mg = 0, so T = mg .
T
Find the speed of waves on the string.
T
mg
=
v=
µ
2.00 m
µ
mg
Find the fundamental frequency.
f1 =
T
v
1 mg
1
(2.20 kg)(9.80 m s 2 )
=
=
= 616 Hz
2L 2L µ
2(2.00 m)
3.55 × 10−6 kg m
525
Chapter 11: Waves
Physics
51. Strategy The frequencies are given by f n = nv (2 L). The speed of the transverse waves is related to the mass per
unit length by v = T µ .
Solution Find the mass per unit length of the guitar string.
v
T
1 T
82 N
f1 =
=
= 4.5 × 10−4 kg m .
, so µ = 2 2 =
2
2
2L 2L µ
4 L f1
4(0.65 m) (329.63 Hz)
52. Strategy The fundamental frequency is given by f n = nv (2 L) with n = 1. Use the definition of average speed.
Solution Find the fundamental frequency.
2.0 m
f1 =
v
= 0.050 s = 10 Hz
2 L 2(2.0 m)
53. (a) Strategy and Solution All frequencies higher than the fundamental are integral multiples of the
fundamental. Since there are no other frequencies between the two given, the fundamental is the difference
between those two. Thus, the fundamental frequency is 1040 Hz − 780 Hz = 260 Hz .
(b) Strategy Use Eqs. (11-2) and (11-13).
Solution Find the total mass of the string.
v
F
F
1 FL
1200 N
=
=
=
= 2.8 g .
f1 =
, so m =
2
2L 2L m
4mL
4 f1 L 4(260 Hz) 2 (1.6 m)
54. Strategy The weight of the weight equals the tension in the string. Use Eqs. (11-4) and (11-13) to find the
required weight of the weight.
Solution Find the weight F.
fn =
F=
nv
n
=
2L 2L
4 µ L2 f n 2
n2
F
µ
=
=
n2 F
4 µ L2
, so
4(0.120 × 10−3 kg m)(0.42 m) 2 (110 Hz) 2
n2
= 1.0 n 2 N, where n = 1, 2, 3, ... .
55. Strategy Use Eqs. (11-2) and (11-13).
Solution
nv
TL
fn =
and v =
. Find the total mass of the wire.
2L
m
f1 =
1 TL
v
=
=
2L 2L m
300.0 N
T
T
, so m =
=
= 0.050 kg .
2
4 Lm
4 Lf1
4(2.0 m)(27.5 Hz)2
526
Physics
Chapter 11: Waves
56. (a) Strategy Replace each quantity with its SI units.
Solution Show that
FL
has units
m
FL m has units of speed.
N⋅m
=
kg
kg ⋅ m s 2 ⋅ m
= m s , which are the units of speed.
kg
(b) Strategy Set [ F ]a [ L]b [m]c equal to m s , the units of speed, and determine a, b, and c.
FL m has units of speed.
Solution Show that no combination of L, m, and F other than
[ F ]a [ L]b [m]c = m ⋅ s −1
N a mb kg c =
kg a ⋅ m a
s
2a
mb kg c =
kg a + c m a +b s −2a = kg 0 ⋅ m1 ⋅ s −1
Equate exponents.
a + c = 0, so a = −c. a + b = 1, so b = 1 − a = 1 + c. −2a = −1, so a =
1
1
1
, c = − , and b = .
2
2
2
The only combination of F, L, and m that gives units of speed is F 1/ 2 L1/ 2 m −1/ 2 = FL / m ; therefore, the
FL / m times some dimensionless constant.
speed of transverse waves on the string can only be
57. Strategy The wave speed for the 1.0-Hz waves is twice that for the 2.0-Hz waves, so it takes the 1.0-Hz waves
120 s to reach you. (120 s + 120 s = 240 s is the time it takes the 2.0-Hz waves to reach you; twice as long.)
Solution Compute the distance to the boat.
∆x = v∆t = (1.56 m s)(120 s) = 190 m
58. Strategy The wave is harmonic; y ( x, t ) = (1.2 cm) sin[(0.50π rad/s)t − (1.00π rad/m)x]. Use Eqs. (10-21) and
(10-22). Plot the graphs.
Solution Find the maximum velocity and maximum acceleration for a point on the string.
vm = ω A = (0.50π rad s)(1.2 cm) = 1.9 cm s and am = ω 2 A = (0.50π rad s)2 (1.2 cm) = 3.0 cm s 2 .
v y leads y by a quarter cycle, so v y is a cosine function.
v y = (1.9 cm s) cos[(0.50π rad s)t − (1.00π rad m) x]
a y leads v y by a quarter cycle, so a y leads y by a half cycle; a y is a negative sine function.
a y = −(3.0 cm s 2 ) sin[(0.50π rad s)t − (1.00π rad m) x]
The period is T =
y (cm)
1.2
0
1.2
2π
ω
=
2π
= 4.0 s.
0.50π rad s
vy (cm/s)
1.9
x=0
3
6
t (s)
ay (cm/s2)
3.0
x=0
0
3
6
t (s)
0
3.0
1.9
527
x=0
3
6
t (s)
Chapter 11: Waves
Physics
59. Strategy Use Eq. (11-6).
Solution Find the wavelength of the radio waves.
v 3.0 × 108 m s
= 3.3 m
λ= =
f
90 × 106 Hz
60. (a) Strategy In a longitudinal wave, the motion of particles in the medium is along the same line as the direction
of propagation of the wave.
Solution
A given particle will oscillate sinusoidally in the ± y -direction about its equilibrium position with an
amplitude of 5.0 cm and a period of 1/8 s.
[T = λ v = 10 cm (80 cm s ) = 1 8 s].
(b) Strategy In a transverse wave, the motion of particles in the medium is perpendicular to the direction of
propagation of the wave.
Solution If the wave under consideration were transverse,
the particles would oscillate along a direction perpendicular to the y -axis .
61. Strategy Speed is inversely proportional to the time of travel. Form a proportion and use ∆x = v∆t.
Solution Relate the speeds to the times of travel.
vP 10.0 km s 5.00 tS
=
=
=
vS
8.0 km s
4.0 tP
Find the time for the S wave to travel from the source to the detector.
2.0 s
⎛ 4.0 ⎞
∆t = tS − tP = tS − tS ⎜
⎟ = tS (1 − 0.80), so tS = 0.20 = 10 s.
5.00
⎝
⎠
Calculate the distance between the source and the detector.
d = vStS = (8.0 km s)(10 s) = 80 km
62. Strategy Refer to the figure to determine the frequency; count the number of cycles during the indicated time
period. Use Eq. (11-6) and the definition of average speed.
Solution Estimate the wavelength of the seismic waves.
λ=
v
=
f
180,000 m
30.0 s
4.0 cycles
2.6 s −1.0 s
= 2.4 km
528
Physics
Chapter 11: Waves
63. Strategy According to Eq. (11-13), the frequency of the string is inversely proportional to the length of the string.
Solution Form a proportion.
1
L
f
f ∝ , so 2 = 1 .
L
L1
f2
Relate the distance between frets to the frequencies and the length of the string.
⎛
f
f ⎞
∆L = L1 − L2 = L1 − 1 L1 = L1 ⎜1 − 1 ⎟
f2
f
2⎠
⎝
1 ⎞
⎛
First fret: ∆L = (64.8 cm) ⎜1 −
⎟ = 3.64 cm
1.0595
⎝
⎠
1 ⎞
⎛
Second fret: 3.64 cm + (64.8 cm − 3.64 cm) ⎜ 1 −
⎟ = 7.07 cm
1.0595
⎝
⎠
1 ⎞
⎛
Third fret: 7.074 cm + (64.8 cm − 7.074 cm) ⎜ 1 −
⎟ = 10.32 cm
⎝ 1.0595 ⎠
64. Strategy Use Eq. (11-13).
Solution Find the lowest standing wave frequencies in each situation.
(a) Since f n =
v
nv
, f n = nf1 = n(300.0 Hz).
and f1 =
2L
2L
n
nf1
2
600.0 Hz
3
900.0 Hz
4
1.200 kHz
(b) The lowest frequency is now f 2 , and only even harmonics are allowed (always a node at the center).
n
nf1
2
600.0 Hz
4
1.200 kHz
6
1.800 kHz
8
2.400 kHz
(c) The effective length of the string is now half of the original length.
nv
nv
nv
, so f n′ =
fn =
=
= 2 f n.
2L
2( L 2) L
n
f n′ = 2 f n
1
600.0 Hz
2
1.200 kHz
3
1.800 kHz
4
2.400 kHz
529
Chapter 11: Waves
Physics
65. Strategy Even though the wire is cut in two, the linear mass density does not change (half the length and half the
mass). According to Eq. (11-4), the speed of waves on a wire is directly proportional to the square root of the
tension. According to Eq. (11-13), the frequency of the waves on a wire is directly proportional to the speed of the
waves. Therefore, the frequency is directly proportional to the square root of the tension in a wire. Use Newton’s
second law.
Solution For the single wire:
ΣFy = T1 − mg = 0, so T1 = mg .
T1
T2
For the two wires:
ΣFy = 2T2 − mg = 0, so T2 = mg 2.
Therefore, T1 = 2T2 . Form a proportion.
f2
f1
=
T2
T1
=
T2
2T2
=
1
2
, so f 2 =
f1
2
mg
T2
mg
.
Thus, the new fundamental frequency of each wire is 660 Hz
2 = 470 Hz .
66. (a) Strategy For a harmonic wave moving to the left (−x-direction),
2π ⎞
⎛ 2π
y = A sin(ωt + kx) = A sin ⎜
t+
x .
λ ⎟⎠
T
⎝
Solution Write an equation for the surface seismic waves.
2π
⎛ 2π
⎞
y( x, t ) = (0.020 m) sin ⎜
t+
x ⎟ , so y( x, t ) = (0.020 m) sin[(1.6 rad/s)t + (0.0016 rad/m) x] .
3
4.0
s
4.0 × 10 m ⎠
⎝
(b) Strategy Use Eq. (10-21).
Solution Find the maximum speed of the ground as the waves move by.
2π A 2π (0.020 m)
=
= 0.031 m s
vm = ω A =
T
4.0 s
(c) Strategy Use Eq. (11-6).
Solution Find the wave speed.
λ 4.0 km
= 1.0 km s
v=λf = =
T
4.0 s
67. (a) Strategy Use Hooke’s law.
Solution Explain why the tension in the spring is approximately proportional to the length.
Hooke’s law: T = k ( x − x0 ) ≈ kx for x x0.
(b) Strategy Use Eq. (11-2) and the result of part (a).
Solution Find the time it takes the wave to travel the length of the spring.
v=
TL
≈
m
kL2
L
since T ≈ kx = kL, and L = v∆t , so
=
m
∆t
kL2
or ∆t =
m
m
.
k
Since neither m nor k change, the increase in length does not affect the time of travel, so ∆t = 4.00 s .
530
Physics
Chapter 11: Waves
68. Strategy Use dimensional analysis.
Solution λ has units m. g has units m s 2 . λ ⋅ g has units m 2 s 2 .
λ g has units m s. So, v ∝ λ g .
69. Strategy Use dimensional analysis.
Solution γ has units N m = kg s 2 . ρ has units kg m3 . λ has units m. γ (λ ⋅ ρ ) has units
kg s 2
m ⋅ kg m
3
=
m2
s
2
. [γ (λ ⋅ ρ )]1/ 2 has units m s. So, v ∝
γ
. Since v depends upon λ , surface waves are
λρ
dispersive.
70. Strategy Use the properties of traveling waves to answer the questions concerning the seismic wave described by
the equation y ( x, t ) = (7.00 cm) cos[(6.00π rad cm) x + (20.0π rad s)t ].
Solution
(a) Because kx and ωt have the same sign, the wave is moving to the left .
(b) The particles in the medium move a distance from their equilibrium positions equal to the amplitude of the
wave. Thus, they move 7.00 cm .
(c) The frequency of this wave is f =
ω 20.0π s −1
=
= 10.0 Hz .
2π
2π
(d) The wavelength of this wave is λ =
(e) The wave speed is v =
ω
k
=
2π
2π
=
= 0.333 cm .
k
6.00π cm −1
20.0π s −1
6.00π cm −1
= 3.33 cm s .
(f) A particle that is at y = 7.00 cm and x = 0 when t = 0
oscillates sinusoidally along the y -axis about y = 0 with an amplitude of 7.00 cm .
(g) Since the motion of particles in the medium is perpendicular to the direction of propagation of the wave, the
wave is a transverse wave.
71. Strategy and Solution Refer to the figure.
(a) Since the wave is moving to the left, the peak of the pulse has not yet reached the position of point A.
So, point A is moving upward.
(b) The peak has passed point B, so point B is moving downward.
(c) The slope of the string is larger at point A than at point B, so the speed of the string segment is larger at
point A.
531
Chapter 11: Waves
Physics
72. Strategy The position of the particle will follow the shape of the wave. The velocity is a bit more complicated.
As the wave passes the point under consideration, the particle moves upward rapidly until it reaches the “top” of
the wave. When it reaches the top, its velocity is instantaneously zero. The point then moves downward at an
average velocity less than that with which it rose. The velocity is equal to the slope of the position graph for any
time t.
Solution The plot of the position as a function of time:
x
t
The plot of the velocity as a function of time:
vx
t
73. Strategy Destructive interference occurs when the path length difference of the two sound waves is an odd
multiple of half of the wavelength.
Solution The wavelength is λ = v / f = (340 m/s)/(680 Hz) = 0.50 m. The largest possible path length difference
is equal to the distance between the speakers, 1.5 m. λ / 2 = 0.25 m, so the path length differences that cause
destructive interference are 0.25 m, 0.75 m, and 1.25 m. Let the speakers lie along the x-axis at x = ±0.75 m.
Then the path length difference is zero along the y-axis and 1.5 m along the x-axis. As the listener walks along the
circle of radius 1 m, the path length difference varies from 0 to 1.5 m. The path length difference equals 0.25 m,
0.75 m, and 1.25 m once for each quadrant of the circle (three occurrences of destructive interference). There are
four quadrants, so the listener observes destructive interference at 12 points along the circle.
74. (a) Strategy Substitute t = 0, 1.0 s, and 2.0 s into
{
}
y ( x, t ) = (5.0 cm) sin[(π 5.0 cm −1 ) x − (π 6.0 rad s)t ] + sin[(π 5.0 cm −1 ) x + (π 6.0 rad s)t ]
and graph the resulting equations. Use Eq. (11-7) to find the wavelength.
Solution The three equations are:
{
}
y ( x, 0) = (5.0 cm) sin[(π 5.0 cm −1 ) x] + sin[(π 5.0 cm −1 ) x] = (10 cm) sin[(π 5.0 cm −1 ) x],
{
}
−
1
−
1
y ( x, 2.0 s) = (5.0 cm) {sin[(π 5.0 cm ) x − 1.05] + sin[(π 5.0 cm ) x + 1.05]} .
y ( x, 1.0 s) = (5.0 cm) sin[(π 5.0 cm −1 ) x − 0.52] + sin[(π 5.0 cm −1 ) x + 0.52] , and
Find the wavelength.
2π
2π
λ=
=
= 10 cm
π (5.0 cm)
k
The amplitude of the first (t = 0) wave is A = 10 cm. The second wave (t = 1.0 s) has a smaller amplitude than
the first, due to the opposite phase shifts. The third wave is even smaller in amplitude. The graphs are shown.
y (cm)
10
t=0
t = 1.0 s
0
t = 2.0 s
5.0
10
x (cm)
10
(b) Strategy and Solution The wave doesn’t “travel” anywhere—it just oscillates up and down—so this is a
standing wave.
532
Physics
Chapter 11: Waves
75. Strategy Refer to Problem 78. The function is y = A[sin( kx − ωt ) + sin( kx + ωt )]. Use the trigonometric identity
sin α + sin β = 2sin[(α + β ) 2]cos[(α − β ) 2] and the principle of superposition.
Solution Use the identity.
y = A[sin(kx − ωt ) + sin(kx + ωt )] = A sin(ωt − kx) + A sin(ω t + kx)
⎛ kx − ωt + kx + ωt ⎞
⎡ kx − ωt − (kx + ωt ) ⎤
= 2 A sin ⎜
⎟ cos ⎢
⎥ = 2 A sin( kx) cos(−ωt ) = [2 A cos(ωt )]sin(kx) = A′ sin( kx)
2
2
⎝
⎠
⎣
⎦
Therefore, A′ = 2 A cos(ωt ). Prove this holds for each of the amplitudes of the graphs from Problem 79.
{
}
y ( x, t ) = (5.0 cm) sin[(π 5.0 cm −1 ) x − (π 6.0 rad s)t ] + sin[(π 5.0 cm −1 ) x + (π 6.0 rad s)t ]
cm −1 ) x]cos[(π
= 2(5.0 cm) sin[(π 5.0
6.0 rad s)t ]
−
1
= (10 cm) sin[(π 5.0 cm ) x]cos[(π 6.0 rad s)t ]
So the maximum amplitude is 10 cm for x = 2.5 cm.
At t = 1.0 s,
y (2.5 cm, 1.0 s) = (10 cm) sin[(π 5.0 cm −1 )(2.5 cm)]cos[(π 6.0 rad s)(1.0 s)] = (10 cm)(1) cos(π 6.0) = 8.7 cm.
At t = 2.0 s,
y (2.5 cm, 2.0 s) = (10 cm) sin[(π 5.0 cm −1 )(2.5 cm)]cos[(π 6.0 rad s)(2.0 s)] = (10 cm)(1) cos(π 3.0) = 5.0 cm.
The values correspond nicely with those shown in the graphs in the solution for Problem 79.
Using the original function, we have
y (2.5 cm, 1.0 s) = (5.0 cm){sin[(π 5.0 cm−1 )(2.5 cm) − (π 6.0 rad s)(1.0 s)]
+ sin[(π 5.0 cm −1 )(2.5 cm) + (π 6.0 rad s)(1.0 s)]} = 8.7 cm
and
y (2.5 cm, 2.0 s) = (5.0 cm){sin[(π 5.0 cm −1 )(2.5 cm) − (π 6.0 rad s)(2.0 s)]
+ sin[(π 5.0 cm −1 )(2.5 cm) + (π 6.0 rad s)(2.0 s)]} = 5.0 cm.
Therefore, the amplitudes of the graphs of Problem 79 satisfy the equation A′ = 2 A cos(ωt ), where A′ is the
amplitude of the wave plotted and A is 5.0 cm.
533
Chapter 11: Waves
Physics
76. Strategy Use the principle of superposition.
Solution ∆x = 1.80 m − 1.50 m = 0.30 m in ∆t = 0.20 s, so v = 0.30 m (0.20 s) = 1.5 m s.
Find the position of the peak at t = 2.2 s.
xpeak = 1.5 m + (1.5 m s)(2.2 s) = 4.8 m (3.2 m; 4.8 m − 4.0 m = 0.8 m to the left)
The peak of the pulse is now inverted due to reflection.
y (cm)
10
5
0
5
x = 4.0 m
x=0
10
x = 3.2 m
77. Strategy Use the principle of superposition.
Solution ∆x = 1.80 m − 1.50 m = 0.30 m in ∆t = 0.20 s, so v = 0.30 m (0.20 s) = 1.5 m s.
Find the position of the peak at t = 1.6 s.
xpeak = xi + vt = 1.5 m + (1.5 m s)(1.6 s) = 3.9 m
The peak of the pulse is nearly to the end of the string. The reflected pulse is below the string, so most of the
height of the original pulse is cancelled.
y (cm)
10
5
0
x = 3.9 m
x=0
x = 4.0 m
534
Chapter 12
SOUND
Conceptual Questions
1. The wavelength of the standing waves inside a bassoon is determined by the length of its air chamber. With a
fixed fundamental wavelength, the frequency of these waves depends only upon the speed of sound in air, which
itself depends significantly upon the air temperature. As a result of thermal contraction and expansion, changes in
air temperature also affect the frequency of waves generated on a cello string—albeit much less significantly than
for the bassoon.
2. The piano’s tuning will change by only a very small amount due to thermal contraction of the strings, so we can
ignore it and assume it stays constant. When the air in the room is colder, the speed of sound in air will be less
than before. Therefore, the frequency of the sound waves vibrating in the organ pipes will be less, so the organ
will be lower in pitch than the piano.
3. The range-finder sends out an ultrasonic sound pulse toward the opposite wall and measures how long it takes for
the pulse to be reflected and return. Assuming the air in the room is around room temperature, it then uses the
known speed of sound to calculate the distance to the opposite wall.
4. Diffraction around an object becomes significant when the wavelength is about the same size as the object or
larger. Lower-frequency sounds have longer wavelengths, so they tend to diffract more around the head. The
intensity of the sound received by each ear is therefore nearly identical, so this method doesn’t work well for lowfrequency sounds.
5. For high-frequency sounds the wavelength is relatively small. The phase difference of the sounds arriving at each
ear is then very sensitive to small variations in head position, wind speed, and several other factors. This makes
the phase difference method unreliable for high-frequency sounds.
6. Yes, the threshold of pain is at 110 dB, but hearing is affected starting at 85 dB. You should not remain in the
vicinity of such noise without protection for your ears.
7. When you hear your own voice, the sound reaches your eardrums by traveling partly through the air to your ears
and partly through the inside of your head. The quality of a sound can be affected by the medium through which it
travels. Other people hear the sound of your voice through the air only.
8. When a wave arrives at a place where there is an abrupt change in wave speed, there is a large reflection of the
wave. Without the gel spread over the skin, there would be a small air gap between the skin and the imaging
apparatus, resulting in a significant amount of reflection of the ultrasonic pulses. Since the speed of sound in the
gel is not much different than in the body, the gel greatly reduces the amount of reflection.
9. The maximum loudness of a stereo is proportional to the maximum intensity level that it can produce. Doubling
the power of the stereo’s amplifier also doubles its intensity, but it increases the intensity level by only 3 dB.
10. We first view the wall as a stationary observer receiving a Doppler shifted sound wave from a moving source. The
sound wave is absorbed and reemitted by the wall which now acts as a stationary source providing a signal to a
moving observer. The observer finds once again that the sound wave has been Doppler shifted—this time due to
his own motion. The wall does not change the sound wave but serves only to provide a stationary reference frame.
The situation is therefore independent of the existence of the wall, so the Doppler shifts due to the motion of the
source and the observer may be combined.
535
Chapter 12: Sound
Physics
11. Air elements move from regions of high pressure surrounding condensations toward regions of low pressure
surrounding rarefactions. At locations where the pressure is either a maximum or a minimum, the displacement of
air elements is zero. This occurs because the pressure gradient is identical in either direction at these locations. If
the displacement was anything other than zero, the air elements would have to “choose” in which direction to
travel.
12. The speed of sound in a medium depends directly upon the restoring force of the medium (as measured by its bulk
modulus) and inversely with the medium’s inertia (as measured by its density). Although a solid has a higher
density than air, its bulk modulus is far greater—enough so that the greater restoring force produces a higher
sound speed.
13. Doubling the pressure amplitude of a sound wave doubles the displacement amplitude, quadruples the intensity,
and increases the intensity level by 6 dB.
14. By changing from a reference frame in which the source and observer are at rest into one in which the air is at
rest, it becomes apparent that a non-zero wind velocity in one frame is equivalent to the observer and source
moving with identical velocities in another frame. The latter does not alter the frequency of the sound wave—the
same must be true of the former.
15. Depressing a valve on some instruments increases the total length of the air column and thus increases the
fundamental wavelength. The fundamental frequency must therefore decrease so that the pitch is lowered when
the valve is depressed.
16. The sound of six violas playing together is not six times as loud as the sound of one because the difference in
loudness between two sounds is proportional to the logarithm of the ratio of the intensities. Because the sound
waves from the six violas are incoherent, the total intensity from all of them is equal to the sum of their individual
intensities. Thus, the intensity is increased by a factor of six and the intensity level increases by approximately 8
dB.
17. Although the fundamental frequency of the highest note on a piano is approximately 4 kHz, the harmonics of that
note occur at higher frequencies. An instrument’s timbre is determined by the presence or absence of these
harmonics. High quality audio equipment must therefore be able to reproduce frequencies up to the limit of the
human audible range in order to faithfully reproduce the sound of an instrument.
Problems
1. Strategy Use Eqs. (11-6) and (12-3).
Solution Find the wavelength of the ultrasonic waves.
v
T
v v
T
331 m s
, so λ = = 0
=
λ = and v = v0
f
T0
f
f T0 1.0 × 105 Hz
273.15 K + 15 K
= 3.4 mm .
273.15
2. Strategy Use Eq. (11-6).
Solution Find the wavelength of the ultrasonic waves.
v
1533 m s
λ= =
= 6.1 mm
f 2.5 × 105 Hz
3. Strategy Use ∆x = v∆t and Eq. (12-3).
Solution Find the time it takes the sound of the bat connecting to the ball to travel to the spectator’s ears.
∆x
T
∆x ∆x T0
60.0 m
273.15 K
∆t =
and v = v0
=
=
= 173 ms .
, so ∆t =
331 m s 273.15 K+27.0 K
v
T0
v
v0 T
536
Physics
Chapter 12: Sound
4. (a) Strategy Use Eqs. (12-2) and (12-3).
Solution Compute the speed of sound at T = 12°C.
v = v0
T
273.15 K + 12 K
= (331 m s)
= 338 m s
273.15 K
T0
(b) Strategy The speed of light is so much faster than the speed of sound in air that the time it takes for light to
reach an observer is negligible in this case. Use ∆x = v∆t.
Solution Find the distance to the lightning strike.
∆x = v∆t = (338 m s)(8.2 s) = 2.8 km
5. Strategy Use ∆x = vx ∆t and the speeds of light and sound. For T = 20.0°C, v = 343 m s.
Solution Verify the rule of thumb.
∆x 1.6 × 103 m
∆x 1.6 × 103 m
∆tlight =
=
=
µ
is
negligible;
5
s
∆
t
=
=
= 4.7 s .
sound
v
343 m s
c
3 × 108 m s
4.7 s = 5 s to one significant figure. The rule of thumb is approximately correct.
6. Strategy Use Eq. (12-5).
Solution Find the speed of sound in the thin rod.
v=
Y
ρ
=
1.1× 1011 Pa
8.92 × 103 kg m3
= 3.5 km s
The copper alloy has a slightly lower speed of sound than that listed in Table 12.1 for copper, 3560 m s.
7. Strategy Use Eq. (12-1).
Solution Find the speed of sound in mercury.
v=
B
ρ
=
2.8 × 1010 Pa
1.36 × 104 kg m3
= 1.4 km s
8. Strategy Use Eqs. (12-2) and (12-3) and the binomial approximation.
Solution Derive (12-4).
(a) v = v0
T + 273.15
TC
T
= v0 C
= v0 1 +
273.15
273.15
T0
(b) Use the binomial approximation (1 + x)1/ 2 ≈ 1 +
v = v0 1 +
1
x for x << 1.
2
⎡
⎤
TC
TC
≈ (331 m s) ⎢1 +
⎥ = (331 + 0.606TC ) m s
273.15
⎣ 2(273.15) ⎦
537
Chapter 12: Sound
Physics
9. Strategy Replace each quantity with its SI units and simplify. In (a), use Eq. (12-1). In (b), analyze each
combination of ρ and B.
Solution
(a) Show that Eq. (12-1) gives the speed of sound in m s.
v=
B
ρ
, so
N m2
kg m
3
=
(kg ⋅ m s 2 ) m 2
kg m
3
=
1 (m ⋅ s 2 )
1m
(b) Show that no other combination of B and ρ of than
ρ
B
2
has units
3
=
m2
s2
= m s.
B ρ can give dimensions of speed.
2
kg m
kg
s
kg N
kg 2 ⋅ m s 2
kg 2
⋅
=
=
⋅
=
=
; and
has
units
;
ρ
B
m3 m 2
m5
m4 ⋅ s2
m3 N
m ⋅ kg ⋅ m s 2 m 2
1
m4 ⋅ s2
.
has units
ρB
kg 2
No power of the above three combinations (other than −1 2, which gives B ρ ) will give the dimensions
of speed; therefore, Eq. (12-1) must be correct except for the possibility of a dimensionless constant.
10. Strategy Let the distance to the train from Stan and Ollie be d. Then, d = vsteel ∆tsteel = vair ∆tair . Also,
∆tair = ∆tsteel + 2.1 s.
Solution Find ∆tsteel .
vsteel ∆tsteel = vair ∆tair = vair (∆tsteel + 2.1 s) = vair ∆tsteel + vair (2.1 s), so ∆tsteel =
vair (2.1 s)
vsteel − vair
=
2.1 s
vsteel
vair
Compute d.
d = vsteel ∆tsteel = (5790 m s)
2.1 s
5790
343
−1
= 770 m
11. Strategy Intensity is power per unit area and power is energy per unit time. Use Eq. (12-8).
Solution Solve for the intensity.
I
β = (10 dB) log
I0
β
I
10 10 dB =
I0
β
I = I 010 10 dB
Compute the energy absorbed by the eardrum.
P
E
I= =
, so
A A∆t
β
80.0 dB
⎛ 60 s ⎞
E = IA∆t = I 010 10 dB A∆t = (10−12 W m 2 )10 10 dB (0.600 × 10−4 m 2 )(3.0 min) ⎜
⎟ = 1.1 µJ .
⎝ 1 min ⎠
538
−1
.
Physics
Chapter 12: Sound
12. Strategy Intensity is power per unit area. Power is the rate at which energy is produced. Assume the loudspeaker
to be an isotropic source of sound waves. Use Eq. (12-8).
Solution Solve for the intensity.
I
β = (10 dB) log
I0
β
I
10 10 dB =
I0
β
I = I 010 10 dB
Find the rate at which sound energy is produced by the loudspeaker.
β
71 dB
P = IA = I 010 10 dB (4π r 2 ) = 4π (10−12 W m 2 )10 10 dB (25 m)2 = 0.099 W
13. Strategy Solve for the intensity in Eq. (12-8). The sound is incoherent, so add the three intensities; then solve for
the combined intensity level of the three machines.
Solution Solve for the intensity.
β
β
I
I
β = (10 dB) log10 , so 1010 dB =
and I = I 01010 dB .
I0
I0
Find the combined intensity level.
β1
β3
β2
I total = I1 + I 2 + I 3 = I 01010 dB + I 0 1010 dB + I 01010 dB , so
β total = (10 dB) log10
I total
I0
(
β1
I total
I0
β1
β3
β2
= 1010 dB + 1010 dB + 1010 dB . Thus,
β3
β2
)
(
85 dB
90 dB
93 dB
= (10 dB) log10 1010 dB + 1010 dB + 1010 dB = (10 dB) log10 10 10 dB + 10 10 dB + 1010 dB
)
= 95 dB .
Since 95 dB is comparable to 93 dB, the intensity level of all three machines running is not much different than
with only one machine running.
14. Strategy Solve for the intensity in Eq. (12-8). The sound is incoherent, so add the eight intensities; then solve for
the combined intensity level of the eight cars at point P.
Solution Solve for the intensity.
β
β
I
I
β = (10 dB) log10 , so 1010 dB =
and I = I 01010 dB .
I0
I0
Find the combined intensity level.
β1
β8
I total = I1 + ⋅⋅⋅ + I8 = I 01010 dB + ⋅⋅⋅ + I 01010 dB , so
β total = (10 dB) log10
I total
I0
(
I total
I0
β8
β1
β8
β1
= 1010 dB + ⋅⋅⋅ + 1010 dB . Thus,
)
(
98.0 dB
98.0 dB
= (10 dB) log10 1010 dB + ⋅⋅⋅ + 1010 dB = (10 dB) log10 10 10 dB + ⋅⋅⋅ + 10 10 dB
= 107 dB .
539
)
Chapter 12: Sound
Physics
15. (a) Strategy Use Eqs. (12-7) and (12-8). For T = 20.0°C, v = 343 m s.
Solution Solve for the intensity.
I
β = (10 dB) log
I0
β
I
10 10 dB =
I0
β
I = I 010 10 dB
Find the pressure amplitude, p0.
β
p02
= I = I 010 10 dB , so
2ρ v
β
p0 = 2 ρ vI 010 10 dB = 2(1.20 kg m3 )(343 m s)(10−12 W m 2 )10
120.0 dB
10 dB
= 28.7 N m 2 .
(b) Strategy The force exerted on the eardrum is equal to the pressure amplitude times the area of the eardrum.
Solution Compute the force exerted on the eardrum.
F = p0 A = (28.7 N m 2 )(0.550 × 10−4 m 2 ) = 1.58 mN
16. Strategy Use Eq. (12-8).
Solution Solve for the intensity.
I
β = (10 dB) log
I0
β
I
10 10 dB =
I0
β
I = I 010 10 dB
Compute the percentage change in intensity.
β
β
1.00 dB
⎛ β +1.00 dB
⎞
I 0 ⎜10 10 dB − 10 10 dB ⎟
10 10 dB ⎛⎜10 10 dB − 1⎞⎟
∆I
⎞
⎠ × 100% =
⎝
⎠ × 100% = ⎛ 10 1.00
× 100% = ⎝
⎜ 10 − 1⎟ × 100% = 26%
β
β
I
⎝
⎠
I 10 10 dB
10 10 dB
0
540
Physics
Chapter 12: Sound
17. Strategy Use Eq. (12-8) to find an expression for the intensity. Then use each relationship given for the
intensities to obtain the relationships for the intensity levels.
Solution Solve for the intensity.
I
β = (10 dB) log
I0
β
I
10 10 dB =
I0
β
I = I 010 10 dB
(a) Show that if I 2 = 10.0 I1, β 2 = β1 + 10.0 dB.
I 2 = 10.0I1
β2
β1
I 010 10 dB = 10.0 I 010 10 dB
β2
β1
10 10 dB = (10.0)10 10 dB
β2
β1
β1
⎡
⎤
log10 10 dB = log ⎢ (10.0)10 10 dB ⎥ = log10.0 + log10 10 dB
⎣
⎦
β2
10 dB
= 1.00 +
β1
10 dB
β 2 = β1 + 10.0 dB
(b) Show that if I 2 = 2.0 I1, β 2 = β1 + 3.0 dB.
I 2 = 2.0 I1
β2
β1
I 010 10 dB = 2.0 I 010 10 dB
β2
β1
10 10 dB = (2.0)10 10 dB
β2
β1
β1
⎡
⎤
log10 10 dB = log ⎢ (2.0)10 10 dB ⎥ = log 2.0 + log10 10 dB
⎣
⎦
β2
10 dB
= 0.30 +
β1
10 dB
β 2 = β1 + 3.0 dB
18. (a) Strategy Intensity is proportional to amplitude squared.
Solution Find the percent increase in the intensity.
2
⎡⎛ ⎞ 2 ⎤
⎡
⎤
I − I0
A2 − A02
∆I
⎢⎜ A ⎟ − 1⎥ × 100% = ⎢⎜⎛ 1.500 ⎟⎞ − 1⎥ × 100% = 125.0% .
100%
× 100% =
× 100% =
×
=
2
⎢⎝ A0 ⎠
⎥
I0
I0
A0
⎢⎣⎝ 1 ⎠
⎥⎦
⎣
⎦
(b) Strategy Intensity is proportional to amplitude squared. Use Eq. (12-9).
Solution Find the ratio of the intensities.
I2
I
A2
=
= 2 = 2.250
I1 I 0 A0
Find the change in the intensity level.
I
∆β = (10 dB) log 2 = (10 dB) log 2.250 = 3.522 dB
I1
541
Chapter 12: Sound
Physics
19. Strategy The longest possible length of the tube is the longest length that produces a frequency between
20.0 kHz and 40.0 kHz. Use Eq. (11-13).
Solution Solve for the length of the whistle.
v
v
1
f1 =
, so L =
. Since L ∝ , the longest length is produced by the lowest frequency, 20.0 kHz.
f1
2L
2 f1
343 m s
L=
= 8.58 mm
2(20.0 × 103 Hz)
20. (a) Strategy f n = nv (4 L) for a pipe closed at one end and v = 343 m s for T = 20.0°C.
Solution Find the length of the organ pipe.
v
v
343 m s
, so L =
f1 =
=
= 32.8 cm
4L
4 f1 4(261.5 Hz)
(b) Strategy The frequency of the organ pipe is proportional to the speed of the waves and the speed is
proportional to the square root of temperature, so f ∝ v ∝ T .
Solution Find the fundamental frequency after the temperature drop.
T
273.15 K + 0.0 K
f 0.0° = f 20° 0.0° = (261.5 Hz)
= 252.4 Hz
T20°
273.15 K + 20.0 K
21. (a) Strategy f n = nv (2 L) for a pipe open at both ends and v = 343 m s for T = 20.0°C.
Solution Find the length of the organ pipe.
v
v
343 m s
f1 =
=
= 65.6 cm
, so L =
2L
2 f1 2(261.5 Hz)
(b) Strategy The frequency of the organ pipe is proportional to the speed of the waves and the speed is
proportional to the square root of temperature, so f ∝ v ∝ T .
Solution Find the fundamental frequency after the temperature drop.
T
273.15 K + 0.0 K
f 0.0° = f 20° 0.0° = (261.5 Hz)
= 252.4 Hz
T20°
273.15 K + 20.0 K
22. Strategy The frequency of the organ pipe is proportional to the speed of the waves and the speed is proportional
to the square root of temperature, so f ∝ v ∝ T .
Solution Find the fundamental frequency after the temperature increase.
T
273.15 K + 20.0 K
f 20.0° = f 0.0° 20.0° = (382 Hz)
= 396 Hz
T0.0°
273.15 K + 0.0 K
23. Strategy f n = nv (2 L) for a pipe open at both ends.
Solution Find the length of the organ pipe.
v
v
331 m s
f1 =
, so L =
=
= 43.3 cm .
2L
2 f1 2(382 Hz)
542
Physics
Chapter 12: Sound
24. (a) Strategy Determine the fundamental frequency and compare it to the resonant frequencies given to
determine whether they are produced by a pipe open at both ends or closed at one end.
Solution Compute ratios of these frequencies.
546 Hz 7 546 Hz 7
390 Hz 5
= ,
= , and
= . Resonant frequencies for a pipe open at both ends are integral
390 Hz 5 234 Hz 3
234 Hz 3
multiples of the fundamental. Resonant frequencies for a pipe closed at one end are odd integral multiples.
Therefore, this pipe is closed at one end, since the numbers in the ratios of these frequencies are all odd.
(b) Strategy and Solution Using the results from part (a), we find that the fundamental frequency is
546 Hz
= 78.0 Hz .
7
(c) Strategy Use Eq. (12-10b).
Solution Find the length of the pipe.
v
v
343 m s
, so L =
f1 =
=
= 1.10 m .
4L
4 f1 4(78.0 Hz)
25. Strategy f n = nv (2 L) for a pipe open at both ends and v = v0 T T0 .
Solution Find an expression for the temperature in terms of n.
2
2
⎛ 2 Lf n ⎞
273.15 K ⎡ 2(2.0 m)(702 Hz) ⎤
19, 658 K
T
, so T = T0 ⎜
.
⎟ =
⎢
⎥ =
2
T0
nv
331
m
s
n
n2
⎣
⎦
⎝ 0 ⎠
The assumed temperature range is 293 K (20°C) to 308 K (35°C). We need to find n such that T falls within this
fn =
nv nv0
=
2L 2L
range. By trial and error, n is found to be 8. So, T = 19, 658 K 82 = 307 K = 34°C .
26. Strategy If the displacement antinode is at the end of the tube, L = 30.0 cm. For a pipe closed at one end,
λn = 4L n , where n = 1, 3, 5, …. The resonance produced by the smallest value of L corresponds to the
fundamental. In (c), use Eq. (11-6).
Solution
(a) Compute the wavelength of the sound.
λ1 = 4L = 4(30.0 cm) = 1.20 m
(b) The next larger value of L corresponds to n = 3.
3λ
3(1.20 m)
= 90.0 cm
L= 1 =
4
4
(c) Compute the speed of sound in the tube.
v = λ f = (1.20 m)(282 Hz) = 338 m s
27. Strategy Use Eq. (12-10b) with n = 3 for A and Eq. (11-13) with n = 2 for B.
Solution For tuning fork A, f3 = 3v (4 L). For tuning fork B, f 2 = v L . The ratio of A’s frequency to B’s
3v v
frequency is
= 34 .
4L L
543
Chapter 12: Sound
Physics
28. Strategy Assume that the tuba is a long straight pipe open at both ends. f n = nv (2 L) for a pipe open at both
ends and v = 343 m s for T = 20.0°C.
Solution Find the length of the tuba.
343 m s
v
v
, so L =
f1 =
=
= 1.31 m .
2L
2 f1 2(130.8 Hz)
29. (a) Strategy The rod is analogous to a pipe open at both ends.
Solution There is a displacement node (pressure antinode) at the center of the rod and displacement
antinodes (pressure nodes) at the ends.
(b) Strategy For a pipe open at both ends, λ1 = 2 L. Use Eq. (11-6).
Solution Calculate the speed of sound in aluminum.
v = λ1 f1 = 2 Lf1 = 2(1.0 m)(2.55 × 103 Hz) = 5100 m s .
(c) Strategy The frequency of the sound wave in the rod is the same as the frequency of the sound wave in the
air. Use Eq. (11-6).
Solution Find the wavelength of the sound wave in the air.
v
334 m s
λ= =
= 13.1 cm
f 2.55 × 103 Hz
(d) Strategy and Solution The longitudinal motion of the rod is symmetrical about a central axis, so the ends
move in opposite directions and, thus, they are out of phase.
30. (a) Strategy Use Eq. (12-11).
Solution Find the frequency of Brian’s string.
f beat = ∆f = f 2 − f1, so f1 = f 2 − ∆f = 440.0 Hz − 2.0 Hz = 438.0 Hz .
(b) Strategy and Solution Since the frequency of the string is directly proportional to the square root of its
tension. Brian needs to tighten his A string to increase its frequency.
31. (a) Strategy The beat frequency decreased as the tension was increased, so the original frequency was low. Use
Eq. (12-11).
Solution Find the original frequency of the untuned string.
f beat = ∆f = f 2 − f1, so f1 = f 2 − ∆f = 293.0 Hz − 3.0 Hz = 290.0 Hz .
(b) Strategy Since the frequency of the string is directly proportional to the square root of its tension, the tension
is directly proportional to the square of the frequency.
Solution Compute the percent increase in the tension.
⎡⎛ ⎞ 2 ⎤
⎡⎛ 293.0 − 1.0 ⎞2 ⎤
⎛ F
⎞
F − F0
f
∆F
⎢
× 100% =
× 100% = ⎜
− 1⎟ × 100% = ⎜ ⎟ − 1⎥ × 100% = ⎢⎜
⎟ − 1⎥ × 100%
⎢⎝ f 0 ⎠
⎥
F0
F0
⎢⎣⎝ 290.0 ⎠
⎥⎦
⎝ F0 ⎠
⎣
⎦
= 1.4%
544
Physics
Chapter 12: Sound
32. Strategy The frequencies of the organ pipes are proportional to the speeds of the waves and the speeds are
proportional to the square root of the temperatures, so f ∝ v ∝ T . Form a proportion.
Solution Find the beat frequency of the two pipes.
⎛ 273.15 K+25.0 K
⎞
f 25
T
T
= 25 , so f beat = ∆f = f 25 − f 20 = f 20 25 − f 20 = (264.0 Hz) ⎜
− 1⎟ = 2 Hz .
⎜
⎟
f 20
T20
T20
⎝ 273.15 K+20.0 K
⎠
33. (a) Strategy Use Eqs. (11-2) and (11-13).
Solution Solve for the tension in the string.
FL
v
v=
and f1 =
, so F = 4mLf12 = 4(0.300 × 10−3 kg)(0.655 m)(330.0 Hz)2 = 85.6 N .
m
2L
(b) Strategy and Solution The waves travel on the string with a speed of
v = 2 Lf1 = 2(0.655 m)(330.0 Hz) = 432 m s .
(c) Strategy and Solution Since the other musician is lowering the frequency of the whistle, the frequency
being played is 330.0 Hz + 5 Hz = 335 Hz when beats are first heard.
(d) Strategy Use Eq. (12-10b) for a pipe closed at one end.
Solution Find the length of the slide whistle.
v
v
343 m s
f =
, so L =
=
= 0.256 m .
4L
4 f 4(335 Hz)
34. Strategy The third harmonic of the cello string has a frequency of three times the fundamental. Use Eq. (12-11).
Solution Find the beat frequency between the cello and the violin.
f beat = ∆f = 3(65.40 Hz) − 196.0 Hz = 0.2 Hz
35. Strategy Since both source and observer are in motion, use Eq. (12-14).
Solution vo and vs are positive, since they are both traveling in the direction of the propagation of the sound
wave. Compute the frequency heard by the driver of the car (the observer).
fo =
vo
v
vs
1− v
1−
28 m s
fs =
1 − 343 m s
44 m s
1 − 343 m s
(550 Hz) = 580 Hz
36. Strategy Since the observer is moving and the source is stationary, use Eq. (12-13).
Solution As the driver approaches the factory, vo < 0. As the driver recedes from the factory, vo > 0.
Find the change in frequency heard by the driver.
⎛
⎛
v ⎞
v ⎞
2v f
2(85 × 103 m h)(500 Hz) ⎛ 1 h ⎞
f f − fi = ⎜ 1 − o ⎟ fs − ⎜ 1 + o ⎟ fs = − o s = −
⎜
⎟ = −69 Hz
⎜
⎜
v ⎟⎠
v ⎟⎠
v
343 m s
⎝ 3600 s ⎠
⎝
⎝
545
Chapter 12: Sound
Physics
37. Strategy Since the observer is moving and the source is stationary, use Eq. (12-13).
Solution As Mandy walks toward one siren (1), vo < 0. As she recedes from the other siren (2), vo > 0.
Find the beat frequency heard by Mandy.
⎛
⎛
v ⎞
v ⎞
2 vo fs 2(1.56 m s)(698 Hz)
f1 − f 2 = ⎜ 1 + o ⎟ fs − ⎜1 − o ⎟ fs =
=
= 6.35 Hz
⎜
⎜
v ⎟⎠
v ⎟⎠
v
343 m s
⎝
⎝
38. Strategy Since the source is moving and the observer is stationary, use Eq. (12-12).
Solution Compute the frequencies of the sound received by the stationary observer.
(a) The source is moving toward the stationary observer (vs > 0).
f
1.0 kHz
f o = sv =
= 2.0 kHz
1 − s 1 − 0.50
v
(b) The source is now moving away (vs < 0).
1.0 kHz
fo =
= 670 Hz
1 + 0.50
39. Strategy Since the observer is moving and the source is stationary, use Eq. (12-13).
Solution Compute the frequencies of the sound observed by the moving observer.
(a) The observer is moving toward a stationary source (vo < 0).
⎛ v ⎞
f o = ⎜ 1 − o ⎟ fs = [1 − (−0.50)](1.0 kHz) = 1.5 kHz
v ⎠
⎝
(b) The observer is now moving away from the source (vo > 0).
f o = (1 − 0.50)(1.0 kHz) = 500 Hz
40. Strategy Since the observer is moving and the source is stationary, use Eq. (12-13).
Solution As the child swings toward the whistle (1), vo < 0. As she swings away (2), vo > 0.
Find the speed of the child at the bottom of the swing.
−1
⎛
v ⎞ ⎡⎛
v ⎞ ⎤
v + vo
f
1.050 = 1 = ⎜ 1 + o ⎟ fs ⎢⎜ 1 − o ⎟ fs ⎥ =
, so
f 2 ⎝⎜
v ⎠⎟ ⎢⎣⎝⎜
v ⎠⎟ ⎦⎥
v − vo
0.050v 0.050(343 m s)
1.050(v − vo ) = v + vo , or vo =
=
= 8.4 m s .
2.050
2.050
546
Physics
Chapter 12: Sound
41. Strategy Since both source and observer are moving, use Eq. (12-14).
Solution Compute the frequencies of the sound observed by the moving observer.
(a) A source and an observer are traveling toward each other (vs > 0, vo < 0).
vo
v
v
1 − vs
1−
fo =
fs =
1 − (−0.50)
(1.0 kHz) = 3.0 kHz
1 − 0.50
(b) A source and an observer are traveling away from each other (vs < 0, vo > 0).
1 − 0.50
fo =
(1.0 kHz) = 330 Hz
1 + 0.50
(c) A source and an observer are traveling in the same direction (vs < 0, vo < 0).
1 + 0.50
fo =
(1.0 kHz) = 1.0 kHz
1 + 0.50
42. Strategy First treat the cell as the observer moving toward a source; then treat the cell as a source moving toward
an observer. Use Eqs. (12-12) and (12-13).
Solution Cell as observer moving toward a source (vo < 0):
⎛
v ⎞
⎛ v⎞
f1 = ⎜ 1 − o ⎟ fs = ⎜ 1 + ⎟ f
v
⎝ u⎠
sound ⎠
⎝
Cell as source moving toward an observer (vs > 0):
fo =
fs
1− v
vs
sound
= fr =
(
)
1 + uv f 1 + v
f1
u f
=
=
1 − uv
1 − uv
1 − uv
43. Strategy Since the source is moving, use Eq. (12-12). Assume the source’s speed is small compared to the speed
of sound and use the binomial approximation from Appendix A.5.
Solution Show that the fractional shift in the frequency of sound from a moving source observed by a stationary
observer is vs v .
For a moving source, f o =
fs
. Using the binomial approximation (1 − x) −1 ≈ 1 + x for x << 1, f o becomes
1 − vs v
v
v
∆f vs
v
⎛ v ⎞
f o ≈ fs ⎜1 + s ⎟ assuming s << 1. Thus, f o = fs + s fs , so f o − fs = ∆f = s fs , or
= . Therefore, the
v⎠
v
v
fs
v
v
⎝
∆f
f o − fs
vs
=
fractional shift in observed frequency
is equal to .
fs
fs
v
547
Chapter 12: Sound
Physics
44. Strategy Since the race car is in motion and the spectators are stationary, use Eq. (12-12).
Solution Find the ratios of the frequencies observed to the frequency of the source.
f
1
Toward the spectators: o1 =
(vs > 0)
fs 1 − vs
v
f
1
(vs < 0)
Away from the spectators: o2 =
fs 1 + vs
v
The observed frequency of the sound as the car recedes from the spectators is 0.75 times that observed during the
car’s approach, so f o2 = 0.75 f o1. Find vs , the speed of the racecar. v = 343 m s for T = 20.0°C.
0.75
1−
vs
v
=
1
v
1 + vs
v
⎛ v ⎞
0.75 ⎜1 + s ⎟ = 1 − s
v
v
⎝
⎠
vs
(0.75 + 1) = 1 − 0.75
v
1.75vs = 0.25v
v
343 m s
vs =
=
= 49 m s
7.0
7.0
The race car is moving with a speed of 49 m s .
45. (a) Strategy The distance traveled (round trip) by the sound of the firing pistol in time ∆t is v∆t. The distance
between the ship and one side of the fjord is half this distance. Use Eq. (12-3).
Solution Find the distance between the ship and one side of the fjord.
1
1
T
1
273.15 K + 5.0 K
d = v∆t = ∆tv0
= (4.0 s)(331 m s)
= 670 m
2
2
T0 2
273.15 K
(b) Strategy Let the distance to the closer side of the fjord be d1 = 12 vt1; then the distance to the other side of
the fjord is d 2 = 12 vt2.
Solution Find the time interval between the two echoes.
d
2d
2d 2
⎛ 1.80 km − 0.668 km ⎞
∆t = t2 − t1 = 2 − t1 =
− t1 = 2 t1 − t1 = ⎜
− 1⎟ (4.0 s) = 2.8 s
v
d1
2d1 t1
0.668 km
⎝
⎠
548
Physics
Chapter 12: Sound
46. (a) Strategy The distance traveled (round trip) by the sound in time ∆t is v∆t. The depth of the water is half this
distance.
Solution Find the depth of the water.
1
1
d = v∆t = (1533 m s)(0.68 s) = 520 m
2
2
(b) Strategy Use Eq. (11-6).
Solution Find the wavelength of the sound wave in water.
v
1533 m s
λ= =
= 4.0 cm
f 38 × 103 Hz
(c) Strategy The frequency is the same in air as it is in water. Use Eq. (11-6).
Solution Find the wavelength of the sound wave in air.
v
350 m s
λ= =
= 9.2 mm
f 38 × 103 Hz
47. Strategy The distance traveled (round trip) by the sound wave in time ∆t is v∆t. The depth d of the lake is half
this distance.
Solution Find the depth of the lake.
1
1
d = v∆t = (0.540 s)(1493 m s) = 403 m
2
2
48. Strategy The distance traveled (round trip) by the sound wave in time ∆t is v∆t. The depth d of the ocean at the
location is half this distance.
Solution Find the depth of the ocean.
1
1
d = v∆t = (1533 m s)(7.07 s) = 5.42 km .
2
2
49. Strategy First treat the moth as a receiver moving away from the source (the bat); then treat the moth as the
source moving away from an observer (the bat). Use Eqs. (12-3) and (12-14).
Solution Moth as receiver (vm and vb > 0; in the direction of propagation):
f1 =
vm
v
v
1 − vb
1−
fs
Moth as source (vm and vb < 0; opposite the direction of propagation):
(
(
v
)(
)(
v
)
)
v
1 + vb 1 − vm
1 + vb
(v + vb )(v − vm )
f2 =
f =
fs =
fs
vm 1
vb
vm
(v + vm )(v − vb )
1+ v
1+ v 1− v
Find v at 10.0°C.
T
273.15 K + 10.0 K
v = v0
= (331 m s)
= 337 m s
T0
273.15 K
Calculate the frequency, f 2.
(337 m s + 4.40 m s)(337 m s − 1.20 m s)
f2 =
(82.0 kHz) = 83.6 kHz
(337 m s + 1.20 m s)(337 m s − 4.40 m s)
549
Chapter 12: Sound
Physics
50. Strategy The distance traveled (round trip) by the chirp in time ∆t is v∆t. The distance between the bat and the
moth is half this distance. The speed of sound in air when T = 10.0°C is
T
273.15 K + 10.0 K
v = v0
= (331 m s)
= 337 m s.
T0
273.15 K
337 m s >> 4.40 m s > 1.20 m s; the relative speed between the bat and moth is less than 1% of the speed of
sound, so we can ignore the fact that the bat and moth are moving.
Solution The echo is heard by the bat just after the outgoing chirp is finished, so the sound wave traveled round
trip in a time of ∆t ≈ 2.0 ms. The distance between the bat and the moth is
1
1
d = v∆t = (337 m s)(2.0 × 10−3 s) = 34 cm .
2
2
51. Strategy Use the result of Problem 42 for the frequency of reflected waves during angiodynography and Eq.
(12-11).
Solution Find the beat frequency.
v
f beat = ∆f = f r − f =
1 + v cell
sound
v
1 − v cell
sound
⎛ 1 + 0.10 m s
⎞
1570 m s
⎟ = 640 Hz
f − f = (5.0 × 10 Hz) ⎜
−
1
⎜ 1 − 0.10 m s
⎟
⎜ 1570 m s
⎟
⎝
⎠
6
52. Strategy Use the result of Problem 42 for the frequency of reflected waves during angiodynography and Eq.
(12-11). In (b), use the binomial approximation from Appendix A.5.
Solution
(a) Find the beat frequency.
f beat = ∆f = f r − f =
(
)
2v
⎛ 1+ v
⎞
1 + uv − 1 − uv
u − 1⎟ f =
u f = 2 fv
⎜
−
=
=
f
f
f
v
v
⎜ 1− v
⎟
u −v
−
−
1 − uv
1
1
u
u
u
⎝
⎠
1 + uv
(b) From Appendix A.5, the binomial approximation for (1 − x)−1 is 1 + x for x << 1, so the beat frequency is
f beat = ∆f =
2v
u
1−
v
u
f ≈
⎛ 2v 2v 2 ⎞
2v ⎛ v ⎞
2v
⎛2f ⎞
1+ ⎟ f = ⎜ + 2 ⎟ f ≈
f =⎜
⎜
⎟v
⎜ u
⎟
u ⎝ u⎠
u
u ⎠
⎝ u ⎠
⎝
(since 2v 2 u 2 is extremely small for v << u ). Therefore, ∆f ∝ v if v << u.
53. (a) Strategy Use Eqs. (11-2) and (11-13).
Solution Find f 2 .
v=
FL
v
and f 2 = , so f 2 =
m
L
F
7.00 N
=
= 319 Hz .
mL
(0.230 × 10−3 kg)(0.300 m)
(b) Strategy The frequency in air is the same as the frequency for the string. Use Eq. (11-6).
Solution The frequency of the sound in the surrounding air is 319 Hz. The wavelength of the sound in the
v 350 m s
surrounding air is λ =
=
= 1.1 m .
f2
319 Hz
550
Physics
Chapter 12: Sound
54. Strategy Find the time is takes for Kyle’s voice to travel the 5.00 m to the surface of the water. Then, use this
time, the given time of 0.0210 s, and the speed of sound in water to find Rob’s depth below the boat.
Solution Find ∆tair .
d air = vair ∆tair , so ∆tair =
dair
vair
. The time of travel of the sound in the water is ∆twater = 0.0210 s − ∆tair .
Find Rob’s depth below the boat, d water .
⎛
d
d water = vwater ∆twater = vwater (0.0210 s − ∆tair ) = vwater ⎜⎜ 0.0210 s − air
vair
⎝
= 9.8 m
⎞
⎛
5.00 m ⎞
⎟⎟ = (1533 m s) ⎜ 0.0210 s −
⎟
343 m s ⎠
⎝
⎠
55. Strategy f n = nv (4 L) for a pipe closed at one end, where n = 1, 3, 5, …. The speed of sound at T = 20.0°C is
v = 343 m s.
Solution Calculate the four lowest standing-wave frequencies for the organ pipe.
343 m s
3(343 m s)
5(343 m s)
f1 =
= 17.9 Hz , f3 =
= 53.6 Hz , f5 =
= 89.3 Hz , and
4(4.80 m)
4(4.80 m)
4 ( 4.80 m )
f7 =
7(343 m s)
= 125 Hz .
4(4.80 m)
56. Strategy f n = nv (4 L) for a pipe open at one end, where n = 1, 3, 5, …. The speed of sound at T = 20.0°C is
v = 343 m s. Refer to Fig. 12.12 to determine the effect resonance has on the sensitivity of the ear at various
frequencies.
Solution Calculate the three lowest standing-wave frequencies for a pipe open at one end.
343 m s
3(343 m s)
5(343 m s)
f1 =
= 3.4 kHz , f3 =
= 10 kHz , and f5 =
= 17 kHz .
4(0.025 m)
4(0.025 m)
4(0.025 m)
Resonance at 3.4 kHz (2 < 3.4 < 5) enhances the sensitivity of the ear for frequencies critical to
speech recognition.
57. (a) Strategy and Solution The intensity is inversely proportional to distance, so I1 ∝ I 0 / d 2 , where I1 and I 0
are the intensity at the object causing the reflection and the original intensity, respectively, and d is the
distance between the bat and the object. Similarly, for the intensity of the echo, I 2 ∝ I1 / d 2 where I 2 is the
intensity of the echo at the bat. Therefore, I 2 ∝
I1
d
2
∝
I0 / d 2
d
2
∝
1
d4
.
(b) Strategy Use the result of part (a).
Solution Find the percent increase in the intensity of the echo in terms of the distances.
1 − 1
⎡⎛ d ⎞ 4 ⎤
∆I
I − I0
d 4 d04
⎢⎜ 0 ⎟ − 1⎥ × 100%
× 100% =
× 100% =
×
100%
=
1
I0
I0
⎢⎣⎝ d ⎠
⎥⎦
d04
Compute the percent increase for each object.
⎡⎛ 0.60 ⎞4 ⎤
⎡⎛ 1.10 ⎞4 ⎤
− 1⎥ × 100% = 110% ; second object: ⎢⎜
First object: ⎢⎜
⎟
⎟ − 1⎥ × 100% = 46%
⎢⎣⎝ 0.50 ⎠
⎥⎦
⎢⎣⎝ 1.00 ⎠
⎥⎦
551
Chapter 12: Sound
Physics
58. Strategy The distance traveled (round trip) by the sound waves in time ∆t is v∆t. The distance between the bat
and the object is half this distance.
Solution
(a) Find the distance to the object.
1
1
d = v∆t = (331 m s)(0.060 s) = 9.9 m
2
2
(b) Find the time interval between the emission of the signal and the arrival of the echo.
2d 2(0.30 m)
∆t =
=
= 1.8 ms
v
331 m s
(c) No; the 3-ms sound pulse is longer than the 1.8-ms time interval between the beginning of the emission of the
sound pulse and the arrival of the echo, so the bat will still be emitting the first signal.
59. Strategy Use Eq. (11-6).
Solution Compute the frequency of the sound wave.
v 343 m s
= 2.3 kHz
f = =
0.15 m
λ
60. Strategy Assume that the object causing the echo is stationary or moving slowly. First treat the bat as a moving
source. Then treat the bat as a moving receiver. Use Eqs. (12-11), (12-12), and (12-13). The speed of sound at
T = 20.0°C is v = 343 m s.
Solution Bat as source (emits sound) (vb > 0; in the direction of propagation):
1
f1 =
f
v
1 − vb
Bat as receiver (of the echo) (vb < 0; opposite the direction of propagation):
v
1 + vb
v + vb
⎛ v ⎞
f 2 = ⎜ 1 + b ⎟ f1 =
f =
f
v
b
v ⎠
v − vb
⎝
1−
v
Calculate the beat frequency.
⎛ v + vb ⎞ v + vb − v + vb
2v f
2(15 m s)(35 kHz)
f beat = ∆f = f 2 − f = f ⎜
− 1⎟ =
f = b =
= 3.2 kHz
v − vb
v − vb
343 m s − 15 m s
⎝ v − vb
⎠
61. Strategy The distance traveled (round trip) by the sound pulse in time ∆t is v∆t. The distance to the ocean floor
is half this distance.
Solution Find the elapsed time between an emitted pulse and the return of its echo at the correct depth d.
2d = v∆t , so ∆t =
(
2d 2(40.0 fathoms) 1.83
=
v
1533 m s
m
fathom
)=
552
0.0955 s .
Physics
Chapter 12: Sound
62. Strategy Use Eq. (12-8) and P = IA.
Solution Compute the sound intensity levels for each distance.
(a) β = (10 dB) log
I
P
0.800 W
= (10 dB) log
= (10 dB) log −12
= 138 dB
I0
I0 A
(10
W m 2 )π (0.127 m 2) 2
(b) Assume that the trumpet is an isotropic source.
0.800 W
β = (10 dB) log −12
= 88.0 dB
(10
W m 2 )4π (10.0 m)2
63. (a) Strategy The fundamental frequency produced by the chimney when both of its ends are open is
3
(261.6 Hz) (1 2 ) = 32.70 Hz. Use Eq. (11-13).
Solution Find the height of the chimney.
v
v
330 m s
f1 =
, so L =
=
= 5.05 m .
2L
2 f1 2(32.70 Hz)
(b) Strategy Use Eq. (12-10b).
Solution Find the fundamental frequency produced by the chimney when its bottom is closed.
f1 =
v
330 m s ⎡ 330 m s ⎤
=
⎢ 2(32.70 Hz) ⎥
4L
4
⎣
⎦
−1
=
32.70 Hz
= 16.35 Hz
2
64. Strategy Find the greatest common factor of the frequencies to determine the fundamental. Use Eq. (11-6).
Solution Factor the numerical value of each frequency into its prime factors.
36 = 2 × 2 × 3 × 3; 60 = 2 × 2 × 3 × 5; 84 = 2 × 2 × 3 × 7
The factors in common are 2, 2, and 3. The greatest common factor is 2 × 2 × 3 = 12, so the fundamental frequency
is 12 Hz. Compute the wavelength of the wave.
v 180 m s
λ= =
= 15 m
f
12 Hz
65. Strategy The greatest common factor of the frequencies is the fundamental. Compute ratios of the frequencies
and use these ratios to determine the greatest common factor of the frequencies.
Solution Compute the ratios.
392
588 3
980 5
= 196 × 1, 196 × 3 = 588, and 196 × 5 = 980.
= and
= , so
2
392 2
392 2
The fundamental frequency is 196 Hz.
553
Chapter 12: Sound
Physics
66. Strategy Use Eqs. (12-3) and (12-10b) to find the fundamental frequency of the tube. Then use Eq. (11-13) to
find the appropriate wave speed for the wire and Eq. (11-4) to find the necessary tension, which is equal to the
weight that should be attached to the wire.
Solution Find the fundamental frequency of the tube.
v
v0
T
f1 = air =
4 Ltube 4 Ltube T0
Find the appropriate wave speed for the wire.
v
f1 = wire , so vwire = 2 Lwire f1.
2 Lwire
Find the linear mass density of the wire.
m
µ = wire = ρ wire Awire
Lwire
Find the weight to be hung from the wire.
F
F
vwire =
=
, so
µ
ρ wire Awire
2
⎛L
⎞ v 2T
F = ρ wire Awire vwire = ρ wire Awire (2 Lwire f1 ) = ρ wire Awire ⎜⎜ wire ⎟⎟ 0
⎝ Ltube ⎠ 4T0
2
2
⎛ 1.00 m ⎞ (331 m s) (18.0 K + 273.15 K)
= (7860 kg m3 )π (0.00200 m) 2 ⎜
= 1280 N .
⎟
4(273.15 K)
⎝ 1.50 m ⎠
2
2
67. Strategy The frequency in the fluid is the same as for the wave in air. Use Eq. (12-7).
Solution Find the ratio of the pressure amplitude of the wave in air to that in the fluid.
0.80 pair 2
pfluid 2
p
ρair vair
(1.20)(343)
0.80 I air =
, so air =
= I fluid =
=
= 0.019.
2 ρair vair
2 ρfluid vfluid
pfluid
0.80 ρfluid vfluid
0.80(1001.80)(1493)
The ratio is 0.019.
68. (a) Strategy The maximum speed of an element of air in the sound wave is given by vm = ω A = ω s0. Use Eqs.
(12-6) and (12-7).
Solution Find the maximum speed.
I=
p02
(ω v ρ s0 )2 (ω s0 )2 ρ v vm 2 ρ v
, so vm =
=
=
=
2ρ v
2ρ v
2
2
2I
=
ρv
2(1.0 × 10−12 W m 2 )
(1.3 kg m3 )(340 m s)
= 6.7 × 10−8 m s .
(b) Strategy The average kinetic energy of the eardrum is equal to half the kinetic energy of an air element with
the maximum speed.
Solution Find the maximum kinetic energy.
K
1 2 1
K av = m = mvm
= (0.0001 kg)(6.7 × 10−8 m s)2 = 1× 10−19 J
2
4
4
(c) Strategy and Solution The average kinetic energy of the eardrum due to collisions with air molecules in the
presence of a sound wave at the threshold of human hearing is about ten times that in the absence of a sound
wave. The ear is about as sensitive as it can be.
554
Physics
Chapter 12: Sound
69. Strategy Since the combined sound intensity is incoherent, the sound intensity of one of the violins is 1 8 of the
combined total ( I 2 = I1 / 8). Use Eq. (12-9).
Solution Find the sound intensity level for one violin.
I
1
β 2 − β1 = (10 dB) log 2 = (10 dB) log = − (10 dB) log 8, so
I1
8
β 2 = β1 − (10 dB) log 8 = 38.0 dB − (10 dB) log 8 = 29.0 dB .
555
REVIEW AND SYNTHESIS: CHAPTERS 9–12
Review Exercises
1. Strategy The magnitude of the buoyant force on an object in water is equal to the weight of the water displaced
by the object.
Solution
(a) Lead is much denser than aluminum, so for the same mass, its volume is much less. Therefore, aluminum has
the larger buoyant force acting on it; since it is less dense it occupies more volume.
(b) Steel is denser than wood. Even though the wood is floating, it displaces more water than does the steel.
Therefore, wood has the larger buoyant force acting on it; since it displaces more water than the steel.
m
(c) Lead: ρ w gVPb = ρ w g Pb = (1.00 × 103 kg m3 )(9.80 m s 2 )
ρ
Pb
Aluminum: ρ w gVAl = ρ w g
Steel: ρ w gVSteel = ρ w g
mAl
ρ Al
mSteel
ρSteel
1.0 kg
11,300 kg m3
= (1.00 × 103 kg m3 )(9.80 m s 2 )
= (1.00 × 103 kg m3 )(9.80 m s 2 )
= 0.87 N
1.0 kg
2702 kg m3
1.0 kg
7860 kg m3
= 3.6 N
= 1.2 N
Wood: mg = (1.0 kg)(9.80 m s 2 ) = 9.8 N (Since the wood is floating, the buoyant force is equal to its
weight.)
2. (a) Strategy The relationship between the fraction of a floating object’s volume that is submerged to the ratio of
the object’s density to the fluid in which it floats is Vf Vo = ρo ρf .
Solution Find the percentage of the plastic that is submerged in the water.
Vsubmerged ρ plastic
890 kg m3
=
=
= 0.89, or 89%
ρ water 1.00 × 103 kg m3
Vplastic
(b) Strategy and Solution
In Figure C, the plastic will actually rise. Think about it this way. In Figure B, the part of the plastic
above the water has a buoyant force from the air lifting it up. Since air is not very dense, this force is
not very strong. Diagram C is correct. The oil is very dense so the buoyant force from the oil is strong
and the plastic rises.
(c) Strategy The buoyant forces of the water and oil on the plastic are equal to the total buoyant force on the
plastic.
Solution Let Vw and Vo be the volume of the plastic within the water and oil, respectively. Then, the total
volume is Vp = Vo + Vw . Find the percentage of the plastic submerged in the water.
ρ w gVw + ρo gVo = ρ p gVp
ρ wVw + ρo (Vp − Vw ) = ρ pVp
ρ wVw − ρoVw = ρ pVp − ρoVp
ρ p − ρo
Vw
890 − 830
Vp
=
ρ w − ρo
=
1000 − 830
= 0.35, or 35%
556
Physics
Review and Synthesis: Chapters 9–12
3. Strategy Use the continuity equation for incompressible fluids and Bernoulli’s equation.
Solution Let the entrance point be 1 and the faucet be 2. Find v1 , the speed of the water in the main pipe.
A
A1v1 = A2 v2 , so v2 = 1 v1 =
A2
P1 + ρ gy1 +
r12
r22
v1.
1
1
ρ v 2 = P2 + ρ gy2 + ρ v22
2 1
2
v12 =
2( P2 − P1 )
ρ
⎛ r2 ⎞
+ 2 g ( y2 − y1 ) + ⎜ 1 v1 ⎟
⎜r 2 ⎟
⎝ 2
⎠
2
2
⎛ r2 ⎞
2( P1 − P2 )
− 2 g ( y2 − y1 )
⎜ 1 2 v1 ⎟ − v12 =
⎜r
⎟
ρ
⎝ 2
⎠
4
⎤
⎡ 2( P1 − P2 )
⎤ ⎡⎢⎛ r1 ⎞
v1 = ⎢
− 2 g ( y2 − y1 ) ⎥ ⎜⎜ ⎟⎟ − 1⎥
⎥
ρ
⎣
⎦ ⎢⎝ r2 ⎠
⎣
⎦
−1
⎡ 2(52.0 × 103 Pa)
⎤ ⎡⎛ 5.00 ⎞4 ⎤
− 2(9.80 m s 2 )(4.20 m + 0.90 m) ⎥ ⎢⎜
v1 = ⎢
⎟ − 1⎥
⎥⎦
⎢⎣1.00 × 103 kg m3
⎥⎦ ⎣⎢⎝ 1.20 ⎠
−1
= 0.116 m s
4. (a) Strategy As Arnold falls, gravity does mgh = (82 kg)(9.80 m s 2 )(10 m) = 8.0 kJ of work on him, so each
of his legs must absorb about 4.0 kJ of energy when he lands; that is, each leg must do work W = Fav ∆y to
bring him to rest.
Solution Compute the compressive stress on Arnold’s legs.
W
4000 J
=
= 8.0 × 105 N. So, the compressive stress on each
The average force on each femur is Fav =
∆y 0.005 m
Fav
8.0 × 105 N
= 2 × 109 N m 2 . Since the compressive stress on each femur exceeds the
A
5 × 10−4 m 2
maximum ultimate strength for compression, Arnold’s femur will break.
femur is
=
(b) Strategy and Solution This time, Arnold has 30 cm instead of 5 mm to come to a stop, and it is the snow
4000 J
= 1.3 × 104 N and the compressive stress is
that does the work. So, the average force is
0.30 m
1.3 × 104 N
= 3 × 107 N m 2 . This time, the compressive stress is less than the maximum ultimate strength
5 × 10−4 m 2
for compression, therefore, Arnold’s femur will not break.
557
Review and Synthesis: Chapters 9–12
Physics
5. Strategy Use Eqs. (10-20a), (10-21), and (10-22), and Newton’s second law.
Solution The normal force on the 1.0-kg block m1 is N = m1 g . So, the force of friction
on m1 is f = µ N = µ m1 g .
Find the maximum acceleration that the top block can experience before it starts to slip.
µ m1 g
f
ΣF = f = m1a, so a =
=
= µ g.
m1
m1
N
1.0 kg
m1g
k
For SHM, the maximum acceleration is am = ω 2 A = A, which in this case is equal to
m
kA
. Equate the accelerations and solve for A.
m1 + m2
µ (m1 + m2 ) g
kA
= µ g , so A =
.
m1 + m2
k
m1
m2
The maximum speed is vm = ω A. Compute the maximum speed that this set of blocks
can have without the top block slipping.
m + m2
1.0 kg + 5.0 kg
k
⎡ µ (m1 + m2 ) g ⎤
vm = ω A =
= µg 1
= 0.45(9.80 m s 2 )
= 0.88 m s
⎢
⎥
150 N m
m1 + m2 ⎣
k
k
⎦
6. Strategy Use Eq. (12-13) and conservation of energy.
Solution Find the maximum speed of the child, which occurs at the lowest point of her swing.
As the child swings toward the whistle (1), vo < 0. As she swings away (2), vo > 0.
−1
v ⎞ ⎡⎛
v ⎞ ⎤
v + vo
f1 ⎛
, so
= ⎜ 1 + o ⎟ f s ⎢⎜ 1 − o ⎟ f s ⎥ =
f 2 ⎝⎜
v ⎠⎟ ⎢⎣⎝⎜
v ⎠⎟ ⎥⎦
v − vo
0.040v 0.040(343 m s)
1.040(v − vo ) = v + vo , or vo =
=
= 6.7 m s.
2.040
2.040
At the lowest point of the swing, the kinetic energy is at its maximum and the potential energy is at its minimum.
Find how high the child is swinging; that is, her maximum height.
v 2
1
(6.7 m s) 2
mghmax = mvmax 2 , so hmax = max =
= 2.3 m .
2
2g
2(9.80 m s 2 )
1.040 =
7. (a) Strategy Use Eq. (11-7).
Solution Compute the wave speed for each traveling wave.
ω
ω
6.00 s −1
3.00 s −1
= 1.50 cm s and vII = II =
= 1.00 cm s.
vI = I =
1
−
kI 4.00 cm
kII 3.00 cm −1
Eq. I has the fastest wave speed of 1.50 cm s .
(b) Strategy Use Eq. (11-7).
Solution Compute the wavelengths for each traveling wave.
2π
2π
2π
2π
λI =
=
= 1.57 cm and λII =
=
= 2.09 cm.
kI 4.00 cm −1
kII 3.00 cm −1
Eq. II
has the longest wavelength of 2.09 cm .
558
Physics
Review and Synthesis: Chapters 9–12
(c) Strategy Use Eq. (10-21).
Solution Compute the maximum speed of a point in the medium for each traveling wave.
vIm = ωI AI = (6.00 s −1 )(1.50 cm) = 9.00 cm s and vIIm = ωII AII = (3.00 s −1 )(4.50 cm) = 13.5 cm s.
Eq. II
has the fastest maximum speed of a point in the medium of 13.5 cm s .
(d) Strategy and Solution kx − ωt indicates a wave is moving in the positive x-direction and kx + ωt indicates
a wave is moving in the negative x-direction. So, Eq. II
is moving in the positive x-direction.
8. (a) Strategy Use Eq. (10-26a) and ω = 2π f .
Solution Compute the frequency of the Foucault pendulum.
f =
ω
1
=
2π 2π
g
1
=
L 2π
9.80 m s 2
= 0.133 Hz
14.0 m
(b) Strategy Find the amplitude of the oscillation and use it and the frequency to find the
maximum speed. Use Eq. (10-21).
θ
L
Solution The amplitude of the oscillation is A = L sin θ . The maximum speed of the
pendulum is
g
vm = ω A =
L sin θ = gL sin θ = (9.80 m s 2 )(14.0 m) sin 6.10° = 1.24 m s .
L
A
(c) Strategy The maximum speed and tension occur at the equilibrium position. Use Newton’s second law.
Solution Find the maximum tension.
v 2
v2
ΣFy = F − mg = mar = m
= m m , so
r
L
2
⎛
⎞
⎡
v
(1.24 m s) 2 ⎤
F = m ⎜ g + m ⎟ = (15.0 kg) ⎢9.80 m s 2 +
⎥ = 149 N .
⎜
14.0 m ⎦⎥
L ⎟⎠
⎣⎢
⎝
F
mg
(d) Strategy Use Eqs. (11-2) and (11-13).
Solution Find the fundamental frequency of the wire.
v
1 FL
F
149 N
f1 =
=
=
=
= 16.3 Hz
2L 2L m
4 Lm
4(14.0 m)(0.0100 kg)
9. (a) Strategy Use Eqs. (11-2) and (11-13).
Solution Find the tension in the guitar string.
v
1 FL
F
f1 =
=
=
, so F = 4 Lmf12 = 4(0.655 m)(0.00331 kg)(82 Hz) 2 = 58 N .
2L 2L m
4 Lm
(b) Strategy Use Eqs. (11-4) and (11-13).
Solution Find the length of the lowest frequency string when it is fingered at the fifth fret.
v
1 F
1 F
1
58 N
=
=
= 49 cm .
, so L =
f1 =
2L 2L µ
2 f1 µ 2(110 Hz) 0.00331 kg (0.655 m)
559
Review and Synthesis: Chapters 9–12
Physics
10. Strategy Use Eq. (11-4) and ∆x = v∆t.
Solution Find the time it takes for the wave pulse to travel from one child to the other.
∆x
µ
0.0013 kg m
∆t =
= ∆x
= (12 m)
= 0.15 s
v
F
8.0 N
11. Strategy For two adjacent steps, the extra distance traveled by the wave reflected from the upper step is twice the
tread depth. For the waves to cancel, they must be one-half wavelength out of phase or an odd multiple of onehalf wavelength. Therefore, the minimum tread depth is half of one-half wavelength or one-quarter wavelength.
Solution Find the minimum tread depth d min .
v
λ
v
343 m s
λ = , so d min = =
=
= 21.4 cm .
f
4 4 f 4(400.0 Hz)
12. (a) Strategy First, treat the brick wall as the observer. Next, treat the wall as the source. Use Eq. (12-12).
Solution The motion of Akiko is in the direction of sound propagation, so vAkiko > 0.
f wall = f o =
1
1−
vs
v
fs =
1
1−
vAkiko
v
f Akiko
Now, since the wall and Haruki are stationary, there is no further Doppler effect. So,
1
1
f Haruki = f wall =
f
=
(512.0 Hz) = 522.7 Hz .
7.00 m s
vAkiko Akiko
1 − 343 m s
1−
v
(b) Strategy The situation is similar to that of part (a), but instead of a stationary Haruki, we have Junichi as an
observer moving in the direction of propagation of the sound reflected from the wall (vJunichi > 0). Use Eq.
(12-13).
Solution
⎛ 2.00 m s ⎞
⎛ v ⎞
⎛ v
⎞
f Junichi = fo = ⎜1 − o ⎟ fs = ⎜ 1 − Junichi ⎟ f wall = ⎜1 −
⎟ (522.67 Hz) = 519.6 Hz
v
v
343 m s ⎠
⎝
⎠
⎝
⎠
⎝
13. Strategy The frequency of the sound is increased by a factor equal to the number of holes in the disk. Use Eq.
(11-6).
Solution The frequency of the sound is
f = 25(60.0 Hz) = 1500 Hz .
Compute the wavelength that corresponds to this frequency.
v
343 m s
λ= =
= 22.9 cm
f 1.50 × 103 Hz
560
Physics
Review and Synthesis: Chapters 9–12
14. Strategy Use Eqs. (11-4) and (11-13).
Solution Find the frequency in terms of the tension.
v
1 F
=
f1 =
2L 2L µ
Form a proportion to find the new fundamental frequency.
f2
1 3F ⎛
µ⎞
=
⎜⎜ 2 L
⎟ = 3, so f 2 = 3 f1 = 3(847 Hz) = 1470 Hz .
f1 2 L µ ⎝
F ⎟⎠
15. Strategy and Solution There are 1000 cm3 in 1 L. The heart pumps blood at a rate of
80 cm3 60 s
×
= 4800 cm3 min . Since 5 L = 5000 cm3 ≈ 4800 cm3 , it takes about 1 minute for the medicine
s
1 min
to travel throughout the body.
16. Strategy Use Eq. (11-6).
Solution
(a) The initial wavelength is λ =
v 341 m s
=
= 27.7 cm .
f 1231 Hz
(b) The frequency of the sound in the wall is the same as it was in the air. Therefore, the wavelength of the sound
v 620 m s
= 50 cm .
in the wall is λ = =
f 1231 Hz
(c) The frequency and the speed of sound are the same as before they entered the wall, so the wavelength is
27.7 cm.
17. Strategy The source is moving in the direction of propagation of the sound, so vs > 0. The observer is moving in
the direction opposite the propagation of the sound, so vo < 0. Use Eq. (12-14).
Solution Find the frequency heard by the passenger in the oncoming boat.
v − vo
343 m s − (−15.6 m s)
fo =
fs =
(312 Hz) = 346 Hz
v − vs
343 m s − 20.1 m s
18. (a) Strategy The wavelength of the diaphragm is 0.20 m. The frequency of the gas in the tube is the same as that
of the diaphragm. Use Eq. (11-6).
Solution Find the speed of sound in the gas.
v = f λ = (1457 Hz)(0.20 m) = 290 m s
(b) Strategy and Solution
The piles of sawdust represent displacement nodes: regions where the air remains at rest. They also
represent pressure antinodes.
561
Review and Synthesis: Chapters 9–12
Physics
19. (a) Strategy Use the continuity equation for incompressible fluids and Bernoulli’s equation.
Solution Let the lower end be 1 and the upper end be 2. Find the speed of the water as it exits the pipe.
A1
2
r12
⎛ 10.0 ⎞
v1 = ⎜
⎟ (15.0 cm s) = 41.7 cm s .
2
A2
⎝ 6.00 ⎠
r2
Find the pressure at the lower end.
1
1
P1 + ρ gy1 + ρ v12 = P2 + ρ gy2 + ρ v22
2
2
1
P1 = P2 + ρ g ( y2 − y1 ) + ρ (v22 − v12 )
2
4
1⎡
⎧
⎫
⎤
− 1⎥ (0.150 m s) 2 ⎬
P1 = 101.3 kPa + (1.00 × 103 kg m3 ) ⎨(9.80 m s 2 )(1.70 m) + ⎢ 10.0
6.00
2
⎣
⎦
⎩
⎭
P1 = 118 kPa
A1v1 = A2 v2 , so v2 =
v1 =
( )
(b) Strategy Use the equations of motion for a changing velocity.
Solution Find the time it takes for the water to fall to the ground.
1
∆y = vi sin θ∆t − g (∆t ) 2
2
0 = g (∆t ) 2 − 2vi sin θ∆t + 2∆y
∆t =
∆t =
2vi sin θ ± 4vi 2 sin 2 θ − 8 g ∆y
2g
2(0.4167 m s) sin 60.0° ± 4(0.4167 m s)2 sin 2 60.0° − 8(9.80 m s 2 )(−0.300 m)
2(9.80 m s 2 )
∆t = 0.287 s ( − 0.213 s is rejected, since time is positive.)
The horizontal distance from the pipe outlet where the water lands is
∆x = vi cos θ∆t = (41.7 cm s) cos 60.0°(0.287 s) = 5.98 cm .
20. Strategy The standing wave is the first harmonic, f 2 . The wave is in position B one-quarter period after it is in
position A. It is in position C one-half period after it is in position A. Use Eqs. (11-4) and (11-13).
Solution
The speed of the wave on the string is v =
F
µ
v 1
, so f 2 = =
L L
F
µ
.
1
µ
0.200 × 10−3 kg m
=L
= (0.720 m)
= 7.20 ms.
f2
F
2.00 N
Since the string is at A when t = 0, it will be at least one period before a photo of the string can be taken at
position A; then it will be another period before the second photo can be taken. Therefore, the two earliest times
after t = 0 that the string can be photographed in position A are 7.20 ms and 14.4 ms. Unlike positions A and C,
which only appear once per period, position B occurs twice per period. The string is at position B one-quarter
period (T 4) after it is at position A, and it will be at position B again an additional one-half period later
The period is T =
(T 4 + T 2 = 3T 4); therefore, the two earliest times after t = 0 that the string can be photographed in position B
are 1.80 ms and 5.40 ms. The string is at position C one-half period (T 2) after it is at position A, and it will be
at position C again an additional period later (T 2 + T = 3T 2); therefore, the two earliest times after t = 0 that the
string can be photographed in position C are 3.60 ms and 10.8 ms.
562
Physics
Review and Synthesis: Chapters 9–12
21. Strategy Use the equations describing waves on a string fixed at both ends.
Solution
(a) The wavelength of the fundamental mode is λ = 2 L = 2(0.640 m) = 1.28 m .
(b) The wave speed on the string is v = λ f = (1.28 m)(110.0 Hz) = 141 m s .
(c) Find the linear mass density of the string.
133 N
F
F
=
= 6.71 g m .
, so µ =
v=
2
µ
(140.8 m s) 2
v
(d) Find the maximum speed of a point on the string.
v = ω A = 2π fA = 2π (110.0 Hz)(0.00230 m) = 0.253 m s
(e) The frequency in air is that same as that in the bridge and body of the guitar, which is the same as the
frequency of the string: 110.0 Hz .
(f) Find the speed of sound in air.
v = 331 m s + (0.60 m s)(20) = 343 m s
Find the wavelength of the sound wave in air.
v 343 m s
λ= =
= 3.12 m
f 110.0 Hz
22. Strategy Use conservation of energy, conservation of momentum, and Newton’s second law.
Solution
(a) Find the speed of the oranges just before they land on the pan.
1 2
mv = mgh, so v = 2 gh = 2(9.80 m s 2 )(0.300 m) = 2.425 m s.
2
Find the speed of the oranges and the pan immediately after the oranges land on the pan.
m
2.20 kg
mvoi = (m + M )v, so v =
voi =
(2.425 m s) = 2.18 m s .
m+M
2.20 kg + 0.250 kg
(b) Find the new equilibrium point.
mg (2.20 kg)(9.80 m s 2 )
ΣF = k ∆y − mg , so ∆y =
=
= 4.8 cm .
k
450 N m
(c) Find the amplitude of the oscillations.
1
1
1
mtot g ∆y + k (∆y ) 2 + mv 2 = kA2 , so
2
2
2
A=
=
2mtot g ∆y + k (∆y )2 + mtot v 2
k
2(2.45 kg)(9.80 m s 2 )(0.0479 m) + (450 N m)(0.0479 m) 2 + (2.45 kg)(2.18 m s) 2
= 0.18 m
450 N m
(d) Find the frequency of the oscillations.
1 k
1 450 N m
=
= 2.2 Hz .
f =
2π m 2π
2.45 kg
563
Review and Synthesis: Chapters 9–12
Physics
23. (a) Strategy Find the mass of the balloon and helium. Then use Newton’s second law.
Solution Find the mass.
4
m = mHe + mb = ρ HeV + mb = (0.179 kg m3 ) π (0.120 m)3 + 2.80 × 10−3 kg = 0.00410 kg
3
Find the tension in the ribbon.
ΣF = FB − mg − T = 0, so
4
⎡
⎤
T = FB − mg = ρairVg − mg = ⎢(1.29 kg m3 ) π (0.120 m)3 − 0.004096 kg ⎥ (9.80 m s 2 ) = 5.13 × 10−2 N .
3
⎣
⎦
(b) Strategy Use the small angle approximation for sin θ. Consider Hooke’s law.
Solution Find the period of oscillation.
F
∆x
, which implies that k ≈ T . Thus, we can find the period using
F = FT sin θ ≈ FTθ ≈ FT
L
L
T = 2π
m
mL
(4.10 × 10−3 kg)(2.30 m)
= 2π
= 2π
= 2.69 s .
k
FT
5.13 × 10−2 N
24. Strategy Use Bernoulli’s equation and the pressure difference of a static liquid.
Solution We have the two relations:
1
P2 − P1 = ρliquid gh and P1 + ρair v 2 = P2 .
2
Eliminate the pressure difference and find the minimum speed of the air.
2 ρliquid gh
1
2(800 kg m3 )(9.8 m s 2 )(0.030 m)
=
= 19 m s .
ρair v 2 = ρliquid gh, so v =
ρair
2
1.29 kg m3
25. Strategy Use Newton’s second law and Hooke’s law.
Solution
(a) Find the tension in the string when hanging straight down.
ΣF = T1 − mg = 0, so T1 = mg .
Find the tension in the string while it is swinging.
mg
ΣFy = T2 cos θ − mg = 0, so T2 =
.
cos θ
Find the stretch of the string.
∆T A
Y=
, so
∆L L
1
L∆T Lmg ⎛ 1
⎞ (2.200 m)(0.411 kg)(9.8 m s 2 ) ⎛
⎞
∆L =
=
− 1⎟ =
− 1⎟ = 6.17 × 10−4 m .
⎜
⎜
2
9
cos
65.0
AY
AY ⎝ cos θ
°
⎠ π (0.00125 m) (4.00 × 10 Pa) ⎝
⎠
(b) Find the kinetic energy.
ΣFx = T2 sin θ = ma x =
mv 2 2 K
, so
=
r
r
mg
rT2 sin θ L sin θ cos θ sin θ mgL tan θ sin θ (0.411 kg)(9.80 m s 2 )(2.200 m) tan 65.0° sin 65.0°
=
=
=
2
2
2
2
= 8.61 J
K=
564
Physics
Review and Synthesis: Chapters 9–12
(c) Find the time it takes a transverse wave pulse to travel the length of the string. Neglect the small change due
to the stretch when finding the linear mass density.
T
T2
v= 2 =
and L = vt , so
µ
ρA
t=
L + ∆L
ρA
ρ A cos θ
(1150 kg m3 )π (0.00125 m)2 cos 65.0°
= ( L + ∆L )
= ( L + ∆L )
= (2.2006 m)
v
T2
mg
(0.411 kg)(9.80 m s 2 )
= 0.0536 s
26. Strategy The tension is 0.93 of the tensile strength. Use Eqs. (10-2), (11-2), and (11-13).
Solution Set the stress equal to the strength.
T
TL
F
nv
. Find f1.
= strength = 0.93 , so T = 0.93 A(strength). f n =
and v =
m
A
A
2L
f1 =
=
v
1 TL 1
=
=
2L 2L m
2
TL
mL2
=
1 0.93 A(strength) L 1 0.93(strength)V 1 0.93(strength)
=
=
2
2
2
mL2
mL2
ρ L2
1
0.93(6.3 × 108 Pa)
= 1.4 kHz
2 (8500 kg m3 )(0.094 m)2
MCAT Review
1. Strategy The buoyant force on the brick is equal in magnitude to the weight of the volume of water it displaces.
Solution The brick is completely submerged, so its volume is equal to that of the displaced water. The weight of
the displaced water is 30 N − 20 N = 10 N. Find the volume of the brick.
10 N
10 N
ρ w gVw = ρ w gVbrick = 10 N, so Vbrick =
=
= 1× 10−3 m3
ρ w g (1000 kg m3 )(10 m s 2 )
The correct answer is A .
2. Strategy The expansion of the cable obeys F = k ∆L.
Solution Compute the expansion of the cable.
F
5000 N
∆L = =
= 10−3 m
k 5.0 × 106 N m
The correct answer is A .
3. Strategy Solve for the intensity in the definition of sound level.
Solution Find the intensity of the fire siren.
I
SL = 10 log10 , so I = I 010SL 10 = (1.0 × 10−12 W m 2 )10100 10 = 1.0 × 10−2 W m 2 .
I0
The correct answer is D .
4. Strategy and Solution The two centimeters of liquid with a specific gravity of 0.5 are equivalent to one
centimeter of water; that is, the 6-cm column of liquid is equivalent to a 5-cm column of water. Therefore, the
new gauge pressure at the base of the column is five-fourths the original. The correct answer is C .
565
Review and Synthesis: Chapters 9–12
Physics
5. Strategy The wavelength is inversely proportional to the index n. Let λn = 8 m and λn + 2 = 4.8 m.
Solution Find n.
L=
nλn
4
=
(n + 2)λn + 2
4
−1
, so
⎛ λ
⎞
λn
n+2
2
=
= 1 + , or n = 2 ⎜⎜ n − 1⎟⎟ .
λn + 2
n
n
⎝ λn + 2 ⎠
Compute L.
L=
nλn
⎛ λ
⎞
= 2 ⎜⎜ n − 1⎟⎟
4
⎝ λn + 2
⎠
−1
λn
4
=
1⎛ 8 m
⎞
− 1⎟
⎜
2 ⎝ 4.8 m ⎠
−1
(8 m) = 6 m
The correct answer is C .
6. Strategy The wave may interfere within the range of possibility of totally constructive or totally destructive
interference.
Solution For totally constructive interference, the amplitude of the combined waves is 5 + 3 = 8 units. For totally
destructive interference, the amplitude of the combined waves is 5 − 3 = 2 units. The correct answer is B .
7. Strategy and Solution As the bob repeatedly swings to and fro, it speeds up and slows down, as well as changes
direction. Therefore, its linear acceleration must change in both magnitude and direction.
The correct answer is D .
8. Strategy Since K ∝ v 2 and v ∝ r −4 , K ∝ r −8 .
Solution Compute the ratio of kinetic energies.
K2
⎛2⎞
=⎜ ⎟
K1 ⎝ 1 ⎠
−8
=
1
1
or K 2 : K1 = 1: 44.
=
256 44
The correct answer is B .
9. Strategy The buoyant force on a ball is equal to the weight of the volume of water displaced by that ball.
Solution Since B1 is not fully submerged and B2 and B3 are, the buoyant force on B1 is less than the buoyant
forces on the other two. Since B2 and B3 are fully submerged, the buoyant forces on each are the same.
The correct answer is B .
10. Strategy and Solution Ball 1 is floating, so its density is less than that of the water. Since Ball 2 is submerged
within the water, its density is the same as that of the water. Ball 3 is sitting on the bottom of the tank, so its
density is greater than that of the water. The correct answer is A .
11. Strategy The supporting force of the bottom of the tank is equal to the weight of Ball 3 less the buoyant force of
the water.
Solution Compute the supporting force on Ball 3.
ρ Ball 3 gVBall 3 − ρ water gVBall 3 = gVBall 3 ( ρ Ball 3 − ρ water )
= (9.80 m s 2 )(1.0 × 10−6 m3 )(7.8 × 103 kg m3 − 1.0 × 103 kg m3 ) = 6.7 × 10−2 N
The correct answer is B .
566
Physics
Review and Synthesis: Chapters 9–12
12. Strategy The relationship between the fraction of a floating object’s volume that is submerged to the ratio of the
object’s density to the fluid in which it floats is Vf Vo = ρo ρf . So, the fraction of the object that is not
submerged is Vns = 1 − Vf Vo = 1 − ρo ρf .
Solution Find the fraction of the volume of Ball 1 that is above the surface of the water.
ρ
8.0 × 102 kg m3 1
1 − Ball 1 = 1 −
=
ρ water
1.0 × 103 kg m3 5
The correct answer is D .
13. Strategy The pressure difference at a depth d in water is given by ρ water gd .
Solution Compute the approximate difference in pressure between the two balls.
ρ water gd = (1.0 × 103 kg m3 )(9.80 m s 2 )(0.20 m) = 2.0 × 103 N m 2
The correct answer is C .
14. Strategy For the force exerted on Ball 3 by the bottom of the tank to be zero, Ball 3 must have the same density
as the water.
Solution Find the volume of the hollow portion of Ball 3, VH .
ρ Ball 3 =
mBall 3
VBall 3
VH = VBall 3 −
=
ρ FeVFe
VBall 3
= ρ water and VH = VBall 3 − VFe , so
⎛ ρ
ρ water
VBall 3 = VBall 3 ⎜⎜1 − water
ρ Fe
ρ Fe
⎝
⎛ 1.0 × 103 kg m3 ⎞
⎞
−6 3
−6 3
⎟⎟ = (1.0 × 10 m ) ⎜⎜ 1 − 7.8 × 103 kg m3 ⎟⎟ = 0.87 × 10 m .
⎠
⎝
⎠
The correct answer is C .
567
Chapter 13
TEMPERATURE AND THE IDEAL GAS
Conceptual Questions
1. The development of standard temperature scales requires use of the zeroth law of thermodynamics. This law tells
us that if a thermometer is in thermodynamic equilibrium with both a test object and an object used to define a
standard, then the test and standard objects must be in thermodynamic equilibrium with each other. If this law did
not hold, the temperature scale on a thermometer would have no relation to the actual temperature of the test
object, and therefore, no standard could be defined.
2. Absolute zero is the temperature at which the motion of atoms in a substance is a minimum and the temperature
can decrease no further. There is no fundamental significance to the zero degree point of the Celsius and
Fahrenheit temperature scales—they are simply of historical origin.
3. A temperature difference of 1 K is the same as a temperature change of 1°C. Thus, either scale may be used in
applications dealing only with temperature differences.
4. Imagine a circle drawn on the plate instead of a hole cut into it. The drawn circle must expand in the same manner
as the cut hole. Thus, the center of the plate expands outward whether or not a hole exists and the hole must
therefore grow larger.
5. The thermal expansion coefficients of silver and brass only differ by about 5%. Thus, a bimetallic strip made from
these materials would bend only slightly during expansion, contrary to its intended purpose.
6. Metals have thermal expansion coefficients several orders of magnitude larger than glass. Running the jar under
hot water therefore facilitates its opening since the lid expands more than the jar as its temperature increases.
7. According to the ideal gas law, if the temperature of an ideal gas is negative, then either the pressure or the
volume must likewise have a negative value. This requirement is nonsensical since absolute pressure and volume
must be positive quantities. Unlike the Celsius and Fahrenheit temperature scales, temperatures in the Kelvin scale
are always positive and therefore avoid the aforementioned problem.
8. Conversion of the price per cubic foot of natural gas into the price per mole requires knowledge of the number of
moles contained within the given volume. From the macroscopic ideal gas law, the number of moles in a sample
is proportional to the volume, pressure, and temperature of the gas. Using the known volume and the additional
quantities of pressure and temperature, the number of moles may be calculated and the price may be converted.
The conversion could also be made if either the number density or the mass density and the mass per molecule
were known. The latter method works whether or not the gas is ideal.
9. The SI units of mass density and number density are kg/m3 and m−3, respectively. An equal number density does
not imply an equal mass density because the mass of an individual atom may be different in each gas.
10. From the ideal gas law, two gases at equal temperature and pressure must have identical number densities. The
mass of a nitrogen atom is greater than the mass of a helium atom—the mass density of nitrogen is therefore
greater.
11. One mole of aluminum atoms has a mass of 27.0 g.
12. The pressure of the air inside the ball increases as it is heated, pushing outward on the dents.
568
Physics
Chapter 13: Temperature and the Ideal Gas
13. The pressure of the air outside the balloon decreases as its distance above the Earth increases. The balloon
expands until the pressures inside and outside are equal.
14. Hydrogen and helium molecules in the high-energy tail of the Maxwell-Boltzman distribution have enough kinetic
energy to escape from Earth’s atmosphere. Other molecules are gradually boosted into the vacated high-energy
region and eventually escape themselves. Only a negligible number of hydrogen and helium molecules have
enough kinetic energy to escape from Jupiter.
15. The kinetic energy is proportional to the velocity squared, so the molecules with large velocities count more in the
average than those with small velocities.
16. Since the diameter of the molecules is considerably smaller than the average intermolecular distance, there is a lot
of empty space for the molecules to move around in without hitting one another.
17. We could use any two of the quantities to decide whether the gas is dilute. The gas is dilute if the number density
is low enough, or equivalently if the average intermolecular distance is large enough compared to the molecular
diameter. For example, if the mean free path is much larger than the diameter, then we know that the
intermolecular distance is large enough and the number density small enough so that the gas is dilute.
18. In this case the air would be a liquid or solid so there would be no sense in speaking of a mean free path; the
molecules are so close together that they are never free of interactions with other molecules.
19. The passenger’s velocity changes more gradually as he sinks into the bag than if he comes into contact with a hard
surface (dashboard or high-pressure bag). If the air pressure in the bag is too low, the passenger is not sufficiently
slowed as the bag collapses.
20. Water boils at a temperature below 100°C at altitudes above sea level because the air pressure is lower. Since the
water temperature is less, the reactions responsible for cooking the egg proceed more slowly.
Problems
1. Strategy Use Eqs. (13-2b) and (13-3).
Solution Convert the temperature.
T − 32°F 84°F − 32°F
=
= 29°C
(a) TC = F
1.8°F °C
1.8°F °C
(b) T = 29 K + 273.15 K = 302 K
2. Strategy Use Eqs. (13-2a) and (13-3).
Solution Convert the temperature.
(a) TC = T − 273.15 K = 77 K − 273.15 K = −196°C
(b) TF = (1.8°F °C)TC + 32°F = (1.8°F °C)(−196.15°C) + 32°F = −321°F
569
Chapter 13: Temperature and the Ideal Gas
Physics
3. (a) Strategy Set TC = TF = T in Eq. (13-2a) and solve for T.
Solution Find the temperature.
32
T = 1.8T + 32, so T =
= − 40 .
1 − 1.8
(b) Strategy Set TC = TK − 273.15 and TK = TF = T in Eq. (13-2a) and solve for T.
Solution Find the temperature.
T = 1.8(T − 273.15) + 32 = 1.8T − 459.67, so T =
−459.67
= 575 .
1 − 1.8
4. Strategy Use Eq. (13-2a) and the fact that the kelvin has the same degree size as the Celsius scale.
Solution Find the temperature changes.
(a) ∆T = ∆TC = −6.0 K
(b) ∆TF = (1.8°F °C)∆TC = (1.8°F °C)(−6.0°C) = −11°F
5. Strategy and Solution There are 78 + 114 = 192 degrees C and 144 degrees J between the freezing and boiling
points of ethyl alcohol. Thus, the conversion factor is 144 192 = 0.750. So, TJ = (0.750 °J °C)TC + A, where A is
the offset to be determined. Find A by setting both temperatures equal to their respective boiling temperatures.
144°J = (0.750 °J °C)(78°C) + A, so A = 144°J − (0.750 °J °C)(78°C) = 85.5°J.
Thus, the conversion from °J to °C is given by TJ = (0.750 °J °C)TC + 85.5°J .
6. Strategy Use Eq. (13-4).
Solution Find the amount the faucet rises.
∆L
= α∆T , so ∆L = L0α∆T = (2.4 m)(16 × 10−6 K −1 )(90.0°C − 20.0°C) = 2.7 mm .
L0
7. Strategy Use Eq. (13-4).
Solution Find the expansion of the rods to determine how far the unfixed end moves.
∆L
= α∆T , so ∆L = L0α∆T . For the system of rods, we have
L0
∆LCu + ∆LAl = L0α Cu ∆T + L0α Al ∆T = L0 ∆T (α Cu + α Al )
= (0.350 m)(16 × 10−6 K −1 + 22.5 × 10−6 K −1 )(150°C − 0.0°C) = 2.0 mm .
8. Strategy Since each section of track expands along its entire length, only half of the expansion is considered for a
particular gap. Two sections meet at a gap, so the gap should be as wide as the expansion of one section of track.
Use Eq. (13-4).
Solution Find the amount of space that should be left between the track sections.
∆L
= α∆T , so ∆L = L0α∆T = (18.30 m)(12 × 10−6 K −1)(50.0°C − 10.0°C) = 8.8 mm .
L0
570
Physics
Chapter 13: Temperature and the Ideal Gas
9. Strategy Since each concrete slab expands along its entire length, only half of the expansion is considered for a
particular gap. Two sections meet at a gap, so the gap should be as wide as the expansion of one concrete slab.
Use Eq. (13-4).
Solution Find the sizes of the expansion gaps.
(a) ∆L = L0α∆T = (15 m)(12 × 10−6 K −1)(40.0°C − 20.0°C) = 3.6 mm
(b) ∆L = L0α∆T = (15 m)(12 × 10−6 K −1)(−20.0°C − 20.0°C) = −7.2 mm
gap width = 7.2 mm + 3.6 mm = 10.8 mm
10. Strategy Form a proportion with ∆LPb and ∆Lglass (which are set equal) to solve for Tglass when TPb = 50.0°C.
Use Eq. (13-4).
Solution Find the final temperature of the glass.
L0α Pb ∆TPb
∆LPb
α Pb ∆TPb
1=
, so
=
=
∆Lglass L0α glass ∆Tglass α glass (Tglass − 20.0°C)
Tglass =
29 × 10−6 K −1
9.4 × 10−6 K −1
(50.0°C − 20.0°C) + 20.0°C = 113°C .
11. Strategy The hole expands just as if it were a solid brass disk. Use Eq. (13-6).
Solution Find the increase in area of the hole.
∆A = 2α A0∆T = 2(1.9 × 10−5 °C−1)(1.00 mm 2 )(30.0°C − 20.0°C) = 3.8 × 10−4 mm 2
12. Strategy According to Eq. (13-4), the change in diameter is given by ∆d = d 0α∆T = d − d0.
Solution Find the diameter of the rivets.
d = d0 + ∆d = d0 + d0α∆T = d0 (1 + α∆T ), so
d
0.6350 cm
d0 =
=
= 0.6364 cm .
1 + α∆T 1 + (22.5 × 10−6 K −1)(−78.5°C − 20.5°C)
13. Strategy A decrease in volume for a fixed mass increases the density, and vice versa. Use Eq. (13-7).
Solution
(a)
∆V
∆ρ
∆V
∆ρ
= − β∆T , so ∆ρ = − βρ∆T .
. Thus,
= β∆T and
=−
V0
ρ
ρ
V0
(b) Compute the fractional change in density.
∆ρ
= −β∆T = −(57 × 10−6 K −1)(−10.0°C − 32°C) = 2.4 × 10−3
ρ
571
Chapter 13: Temperature and the Ideal Gas
Physics
14. Strategy Determine how much more the volume of the water expands than that of the brass container. Use Eq.
(13-7).
Solution Find the amount of water that overflows.
∆V
= β ∆T , so ∆Vwater − ∆Vbrass = V0 ( β water − β brass )∆T
V0
= (75.0 cm 2 )(20.0 cm)(207 × 10−6 K −1 − 57 × 10−6 K −1 )(95.0°C − 25.0°C) = 15.8 cm3 .
15. Strategy Determine how much the volume of the water expands. Use Eq. (13-7).
Solution Find the amount of water that will have spilled.
∆V
= β ∆T , so ∆V = V0 β∆T = (268.4 mL)(207 × 10−6 K −1 )(32.0°C − 2.0°C) = 1.67 mL .
V0
16. (a) Strategy Both the glass and the water expand, but the water expands much more than the glass. Use Eq.
(13-7).
Solution Find the difference in volume expansions to find the amount of water spilled.
∆VH O − ∆Vglass = V0 β H O ∆T − V0 β glass ∆TV0 = V0 ( β H O − β glass )∆T
2
2
2
= (268.4 mL)(207 × 10−6 K −1 − 28.2 × 10−6 K −1 )(32.0°C − 2.0°C) = 1.44 mL
Less water spilled compared to the situation in Problem 15.
(b) Strategy Refer to Problem 15 and part (a).
Solution Compute the percent change.
1.44 mL − 1.67 mL
× 100% = −14%
1.67 mL
So, 14% less water was spilled than when we don’t consider the expansion of the glass .
17. Strategy Use Eq. (13-4). The internal radius of the ring expands as if it were a solid piece of brass.
Solution Find the temperature at which the internal radius of the ring is 1.0010 cm.
∆L
∆L
= α∆T , so
= T − T0 . Compute the temperature.
α L0
L0
T=
∆L
1.0010 cm − 1.0000 cm
+ T0 =
+ 22.0°C = 75°C
α L0
(19 × 10−6 K −1 )(1.0000 cm)
18. Strategy For a physical pendulum, T ∝ L . Form a proportion. Use Eq. (13-5).
Solution Find the fractional change in the period of the oscillating steel rod.
Tnew
Lnew
L0 (1 + α∆T )
=
=
= 1 + α∆T , so
T0
L0
L0
∆T Tnew − T0 Tnew
=
=
− 1 = 1 + α∆T − 1 = 1 + (12 × 10−6 K −1)(0°C − 25°C) − 1 = −1.5 × 10−4
T0
T0
T0
572
Physics
Chapter 13: Temperature and the Ideal Gas
19. Strategy Use Eq. (13-4).
Solution Find the maximum change in the length of the span over an entire year.
⎛ 1K ⎞
∆L = α L0∆T = (12 × 10−6 K −1)(1.6 × 103 m)[105°F − (−15°F)] ⎜
⎟ = 1.3 m
⎝ 1.8°F ⎠
20. Strategy The area of the hole expands as if it were a solid disk of brass. Use Eq. (13-6).
Solution Find the temperature at which the area of the hole is 4.91000 cm 2.
∆A
∆A
4.91000 cm 2 − 4.90874 cm 2
= 2α ∆T = 2α (T − T0 ), so T =
+ T0 =
+ 20.0°C = 26.8°C .
A0
2α A0
2(19 × 10−6 K −1)(4.90874 cm 2 )
21. Strategy The diameter of the hole expands as if it were a solid piece of copper. Use Eq. (13-4).
Solution Find the temperature at which the diameter of the hole is 1.0000 cm.
∆L
∆L
1.0000 cm − 0.9980 cm
= α∆T = α (T − T0 ), so T =
+ T0 =
+ 20.0°C = 150°C .
L0
α L0
(16 × 10−6 K −1)(0.9980 cm)
22. Strategy Find the temperature at which the diameters of the washer and bolt are the same. Use Eq. (13-5).
Solution Set the final diameters equal and solve for the final temperature.
Lc0 + Lc0α c ∆T = Ls0 + Ls0α s ∆T
Lc0α c (Tf − Ti ) = Ls0 − Lc0 + Ls0α s (Tf − Ti )
Tf ( Lc0α c − Ls0α s ) = Ls0 − Lc0 + Ti ( Lc0α c − Ls0α s )
Ls0 − Lc0
Tf =
+ Ti
Lc0α c − Ls0α s
1.0000 cm − 0.9980 cm
Tf =
+ 20.0°C = 520°C
(0.9980 cm)(16 × 10−6 K −1 ) − (1.0000 cm)(12 × 10−6 K −1 )
23. Strategy Use Eqs. (13-4) and (13-5).
Solution Find the new scale of the rule.
∆Lr Lr − Lr0
L
L
=
= r − 1 = α r ∆T , so r = 1 + α r ∆T .
Lr0
Lr0
Lr0
Lr0
Divide the new length of the brick by the new scale.
Lb
L (1 + α b ∆T ) (25.00 cm)[1 + (0.75 × 10−6 K −1)(60.00 K)]
= b0
=
= 24.98 cm
1 + α r ∆T
1 + α r ∆T
1 + (12 × 10−6 K −1)(60.00 K)
24. Strategy Find the temperature change required to increase the circumference of the A340 by 26 cm. Use Eq.
(13-4).
Solution Find the required increase in temperature.
∆L
∆L
0.26 m
= α∆T , so ∆T =
=
= 650 K .
−
L0
α L0 (22.5 × 10 6 K −1 )(17.72 m)
The melting point of aluminum is 660°C.
573
Chapter 13: Temperature and the Ideal Gas
Physics
25. Strategy Use Eqs. (13-4) and (13-6) and the given initial and final areas.
Solution Find the fractional change.
∆A A − A0 (s0 + ∆s)2 − s02 s02 + 2s0∆s + (∆s)2 − s02 2s0∆s + (∆s)2 ∆s(2s0 + ∆s)
=
=
=
=
=
A0
A0
s02
s02
s02
s02
Now, since s0 >> ∆s, we have
∆s
∆A ∆s(2s0 + 0) 2s0∆s 2∆s
= α∆T .
≈
=
=
= 2α∆T since
2
2
s0
A0
s0
s0
s0
26. Strategy Use Eqs. (13-4) and (13-7) and the given initial volume.
Solution Find the fractional change.
∆V V − V0 (s0 + ∆s)3 − s03 s03 + 2s02∆s + s0 (∆s)2 + s02∆s + 2s0 (∆s) 2 + (∆s)3 − s03 3s02∆s + 3s0 (∆s) 2 + (∆s)3
=
=
=
=
V0
V0
s03
s03
s03
Now, since ∆s << s0 , 3s02∆s >> s0 (∆s)2 and (∆s)3. Thus, we have
∆s
∆V 3s02∆s
∆s
= α∆T .
≈
=3
= 3α∆T since
3
s0
V0
s0
s0
27. Strategy Use the definition of one mol and Avogadro’s number.
Solution Find the conversion between atomic mass units u and kg.
12 g
12 mol
NA
=
0.001 kg
6.022 × 1023
, so 1 u = 1.66 × 10−27 kg.
28. Strategy Add the molar masses of each element in ammonia.
Solution Find the molar mass.
mNH3 = mN + 3mH = 14.00674 g mol + 3(1.00794 g mol) = 17.03056 g mol
29. Strategy Add the molecular masses of each element in carbon dioxide.
Solution Find the mass of carbon dioxide in kg.
mass of CO 2 in kg = mC + 2mO = [12.011 u + 2(15.9994 u)](1.6605 × 10−27 kg u) = 7.31× 10−26 kg
30. (a) Strategy and Solution By the definition of the atomic mass unit, 13.003 g mol = 13.003 u .
(b) Strategy Multiply the mass of one carbon-13 atom by the conversion factor 1.6605 × 10−27 kg u .
Solution Find the mass in u.
m
mass of 13C in kg =
= (13.003 u)(1.6605 × 10−27 kg u) = 2.1591× 10−26 kg
NA
31. Strategy Divide the mass of the water in the human by the mass of a water molecule.
Solution Estimate the number of water molecules.
mass of water
0.62(80.2 kg)
N=
=
= 1.7 × 1027
molecular mass [2(1.0 u) + 16.0 u](1.66 × 10−27 kg u)
574
Physics
Chapter 13: Temperature and the Ideal Gas
32. Strategy Use Eq. (13-10).
Solution Find the number density of carbon atoms.
N ρ (3500 kg m3 )(10−6 m3 cm3 )
number per unit volume =
= =
= 1.8 × 1023 atoms/cm3
V m (12.011 u)(1.66 × 10−27 kg u)
33. Strategy Divide the total mass by the molar mass of sucrose to find the number of moles. Then use Eq. (13-11) to
find the number of hydrogen atoms.
Solution mC12 H 22O11 = 12(12.011 g mol ) + 22(1.00794 g mol ) + 11(15.9994 g mol ) = 342.30 g mol
684.6 g
= 2.000 mol of sucrose.
342.30 g mol
There are 2.000(22) = 44.00 moles of hydrogen. Find the number of hydrogen atoms.
There are 342.30 grams of sucrose per mole, so there are
N = nN A = (44.00 mol)(6.022 × 1023 mol−1) = 2.650 × 1025 atoms
34. Strategy Divide the total mass of He by its molar mass.
Solution Find the number of moles.
mass of He
13 g
=
= 3.2 mol
molar mass of He 4.00260 g mol
35. Strategy Divide the total mass of methane by its molar mass.
Solution Find the number of moles.
mass of CH 4
144.36 g
nCH 4 =
=
= 8.9985 mol
molar mass of CH 4 12.011 g mol + 4(1.00794 g mol)
36. Strategy Divide the molar mass of gold by Avogadro’s number.
Solution Find the mass of one gold atom.
molar mass
1 mol
⎛
⎞
mass of one gold (Au) atom =
= (196.96654 g mol)(10−3 kg g) ⎜
⎟
23
NA
⎝ 6.022 × 10 atoms ⎠
= 3.271× 10−25 kg
37. Strategy Use Eq. (13-10).
Solution Find the number of air molecules.
⎛ 1 m3 ⎞ ⎛ 1 ⎞ ⎛
⎞
ρV
1u
= (1.2 kg m3 )(1.0 cm3 ) ⎜ 6
= 2.5 × 1019 molecules
N=
⎟
⎜
⎜
⎟
⎜ 10 cm3 ⎟ ⎝ 29.0 u ⎠ ⎜ 1.66 × 10−27 kg ⎟⎟
m
⎝
⎠
⎝
⎠
575
Chapter 13: Temperature and the Ideal Gas
Physics
38. (a) Strategy Use Eq. (13-11).
Solution Find the number density.
N nN A (1.00 mol)(6.022 × 1023 mol−1)
=
=
= 2.69 × 1025 m −3
V
V
0.0224 m3
(b) Strategy Assume that each molecule is at the center of a sphere of radius r.
Solution The volume of the sphere is
V
1
1
= N =
= 3.72 × 10−26 m3 per molecule.
N V
2.69 × 1025 atoms
3
m
V 4 3
Then
= π r ≈ 4r 3, since π ≈ 3.
N 3
The distance separating molecules is approximately the diameter of the spheres.
Solve for d = 2r.
1/ 3
⎛ V ⎞
d = 2r = 2 ⎜
⎟
⎝ 4N ⎠
1/ 3
⎛ 3.72 × 10−26 m3 ⎞
= 2⎜
⎟
⎜
⎟
4
⎝
⎠
≈ 4 nm
(c) Strategy Multiply the number of moles by the molar mass and use the definition of mass density.
Solution Find the total mass and mass density.
M = (1.00 mol)[2(14.00674 g mol)] = 28.0 g and ρ =
M
28.0 g ⎛ 1 kg ⎞
3
=
⎜
⎟ = 1.25 kg m .
V
0.0224 m3 ⎜⎝ 103 g ⎟⎠
39. Strategy The number of SiO 2 molecules (and the number of Si atoms) N is roughly equal to the volume of a
sand grain Vg divided by the volume of a SiO 2 molecule Vm .
Solution Find the order of magnitude of the number of silicon atoms in a grain of sand.
N=
Vg
Vm
=
4 π r3
3 g
4 π r3
3 m
3
3
⎛ dg ⎞ ⎛ 0.5 × 10−3 m ⎞
=⎜
⎟ = 1018 atoms
⎟ = ⎜⎜
−9
⎟
d
⎝ m ⎠ ⎝ 0.5 × 10 m ⎠
40. Strategy Use Gay-Lussac’s law.
Solution Compute the fractional change in pressure required to increase the cabin temperature from
18°C to 24°C.
P − Pi Tf − Ti
24 − 18
P ∝ T , so f
=
=
= 0.02 .
Pi
Ti
18 + 273.15
41. Strategy Use the macroscopic form of the ideal gas law, Eq. (13-16).
Solution Find the new temperature of the air.
PV = nRT , so
Tf =
Tf
Ti
=
Pf Vf
nR
PV
i i
nR
=
Pf Vf
PV
i i
, or
Pf Vf Ti 20.0 Pi (0.111Vi )Ti
=
= 20.0(0.111)(30 K + 273.15 K) = 673 K = 400°C .
PV
PV
i i
i i
576
Physics
Chapter 13: Temperature and the Ideal Gas
42. Strategy Use the macroscopic form of the ideal gas law, Eq. (13-16).
Solution Find the final volume in terms of the initial volume.
PV = nRT , so
Vf
Vi
=
nRT
Pf
nRT
Pi
P
P
= i , or Vf = i Vi .
Pf
Pf
The volume outside of the tire is equal to the difference between the total final volume and the initial volume.
⎛P
⎞
P
⎛ 36.0
⎞
− 1⎟ = 0.0362 m3
Vf − Vi = i Vi − Vi = Vi ⎜⎜ i − 1⎟⎟ = (0.0250 m3 ) ⎜
Pf
P
14.70
⎝
⎠
⎝ f
⎠
43. Strategy Use the macroscopic form of the ideal gas law, Eq. (13-16).
Solution
PV = nRT , so V =
nRT (1.00 mol)[8.314 J (mol ⋅ K) ](273.15 K + 0.0 K)
=
= 0.0224 m3.
P
(1.00 atm)(1.013 × 105 Pa atm)
The result is verified.
44. Strategy Replace each quantity with its SI units.
Solution PV has SI units Pa ⋅ m3 = (N m 2 ) ⋅ m3 = N ⋅ m = J.
45. Strategy The volume and moles of the gas are constant. Use Gay-Lussac’s law.
Solution Find the pressure at the higher temperature.
T
70.0 K + 273.15 K
P ∝ T , so Pf = f Pi =
(115 kPa) = 135 kPa .
Ti
20.0 K + 273.15 K
46. Strategy The pressure does not change. Use the microscopic form of the ideal gas law, Eq. (13-13).
Solution Find the fraction of air molecules that must be pushed outside.
Initially: PV = Ni kTi
Finally: PV = N f kTf
The fraction pushed out is f =
Ni − N f
N
T
16.0 K + 273.15 K
= 1− f = 1− i = 1−
= 0.0136 .
Ni
Ni
Tf
20.0 K + 273.15 K
47. Strategy Use Eq. (13-10) and the microscopic form of the ideal gas law, Eq. (13-13).
Solution Compute the mass density for each temperature.
mN mP
ρ=
=
since PV = NkT.
V
kT
(a) ρ =
(b) ρ =
(29 u)(1.6605 × 10−27 kg u)(1.0 atm)(1.013 × 105 Pa atm)
(1.38 × 10−23 J K)(273.15 K − 10 K)
(29 u)(1.6605 × 10−27 kg u)(1.0 atm)(1.013 × 105 Pa atm)
(1.38 × 10
−23
J K)(273.15 K + 30 K)
577
= 1.3 kg m3
= 1.2 kg m3
Chapter 13: Temperature and the Ideal Gas
Physics
48. Strategy The volume and moles of the gas are constant. Use Gay-Lussac’s law.
Solution Divide the final by the initial pressure.
273.15 K + 100.0 K
P T
P ∝ T , so f = f =
= 1.55 .
273.15 K − 33 K
Pi Ti
49. Strategy The number of moles of the gas is constant. Use the ideal gas law.
Solution Find the volume of the hydrogen.
(1.00 × 105 N m 2 )(5.0 m3 )(273.15 K − 13 K)
Pf Vf PV
PV T
= i i , so Vf = i i f =
= 1.3 × 103 m3 .
Tf
Ti
Pf Ti
(0.33 × 103 N m 2 )(273.15 K + 27 K)
50. Strategy The number of moles of the gas is constant. Use the ideal gas law.
Solution Find the new pressure of the ideal gas.
Pf Vf PV
PV T
(1.0 × 105 Pa)(1.2 m3 )(273.15 K + 227 K)
= i i , so Pf = i i f =
= 3.3 × 105 Pa .
Tf
Ti
Vf Ti
(0.60 m3 )(273.15 K + 27 K)
51. Strategy The temperature and the number of moles of the gas are constant. Use Boyle’s law and Eq. (9-3).
Solution Find the factor by which the diver’s lungs expand.
1
V
P P + ρ gd
ρ gd
(1.03 × 103 kg m3 )(9.80 m s 2 )(5.0 m)
P ∝ , so f = i = f
= 1+
= 1+
= 1.50 .
V
Vi Pf
Pf
Pf
1.013 × 105 Pa
52. Strategy Use the microscopic form of the ideal gas law, Eq. (13-13).
Solution Find the absolute pressure of intergalactic space.
NkT 1 atom 6
=
(10 cm3 m3 )(1.38 × 10−23 J K)(3 K) = 4 × 10−17 Pa
P=
3
V
cm
53. Strategy Use the microscopic form of the ideal gas law, Eq. (13-13).
Solution Find the number of air molecules released.
PV
N=
, and V, k, and T are constant, so
kT
∆N =
V ∆P (1.0 m3 )(15.0 atm − 20.0 atm)(1.013 × 105 Pa atm)
=
= −1.3 × 1026.
−
23
kT
(1.38 × 10
J K)(273 K)
1.3 × 1026 air molecules were released.
54. Strategy Find the number of moles of molecular oxygen by dividing the total mass by the molar mass. Use the
macroscopic form of the ideal gas law, Eq. (13-16).
Solution Compute the volume occupied by the gas.
nRT (0.532 kg)(103 g kg ) [8.314 J (mol ⋅ K) ](0.0 K + 273.15 K)
V=
=
×
= 0.38 m3
P
2(15.9994 g mol)
1.0 × 105 Pa
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Physics
Chapter 13: Temperature and the Ideal Gas
55. Strategy The number of moles of the gas is constant and V = 16 π d 3. Use the ideal gas law and Eq. (9-3).
Solution Find the diameter of the bubble when it reaches the surface.
Pf Vf PV
1
PT
( P + ρ gh)Tf ⎛ 1 3 ⎞
= i i , so Vf = π df3 = i f Vi = f
⎜ 6 π di ⎟ . Solve for d f .
6
Tf
Ti
Pf Ti
Pf Ti
⎝
⎠
d f = di 3
( Pf + ρ gh)Tf
Pf Ti
= (1.00 mm) 3
[(1.0 atm)(1.013 × 105 Pa atm) + (1.0 × 103 kg m3 )(9.80 m s 2 )(80.0 m)](273.15 K + 18 K)
(1.0 atm)(1.013 × 105 Pa atm)(273.15 K + 4 K)
= 2.1 mm
56. Strategy The number of moles of the gas is constant. Use the ideal gas law and Eq. (9-3).
Solution Find the volume of the bubble just before it breaks the surface of the water.
Pf Vf PV
= i i , so
Tf
Ti
(
)
⎡ (1.00×103
ρ gd
V
T
1
+
⎢⎣1 +
i
f
Pf
( Pf + ρ gd )ViTf
Vf =
=
=
Pf Ti
Ti
kg m3 )(9.80 m s 2 )(20.0 m) ⎤
⎥⎦ (1.00
1.013×105 Pa
cm3 )(273.15 K + 25.0 K)
273.15 K + 10.0 K
3
= 3.09 cm .
57. Strategy The number of moles of air is constant. Assume that the temperature is constant.
∆V
∆V
= 0.500 L s = 5.00 × 102 cm3 s and Pf = Pi + ρ gd . Use Boyle’s law.
Vf =
∆t where
∆t
∆t
Solution Find how long the tank of air will last for each depth.
(a)
∆V ∆t
P
Ptank
Vf
, so
= ∆t
= tank =
Vi
Vi
Pf
Patm + ρ gd
∆t =
=
∆V
∆t
PtankVi
( Patm + ρ gd )
(5.00 × 102 cm3 s)(10−6
(1.0 × 107 Pa)(0.010 m3 )
m3 cm3 )(60 s min) ⎡1.013 × 105 Pa + (1.0 × 103 kg m3 )(9.80 m s 2 )(2.0 m) ⎤
⎣
⎦
= 28 min
(1.0 × 107 Pa)(0.010 m3 )
(5.00 × 102 cm3 s)(10−6 m3 cm3 )(60 s min) ⎡1.013 × 105 Pa + (1.0 × 103 kg m3 )(9.80 m s 2 )(20.0 m) ⎤
⎣
⎦
= 11 min
(b) ∆t =
579
Chapter 13: Temperature and the Ideal Gas
Physics
58. (a) Strategy The temperature and the number of moles of the gas are constant. Use Boyle’s law.
Solution Find the volume of the oxygen at atmospheric pressure.
1
PV (2200 lb in 2 )(0.60 ft 3 )
P ∝ , so Vf = i i =
= 90 ft 3 .
V
Pf
14.70 lb in 2
(b) Strategy Divide the volume of the oxygen at atmospheric pressure by the volume flow rate.
Solution Find how long the cylinder of oxygen will last.
V
9.0 × 101 ft 3 ⎛
1L
⎞⎛ 1 h ⎞
∆t = ∆V =
= 5.3 h
⎜
3 ⎟ ⎜ 60 min ⎟
8
L
min
0.0353
ft
⎠
⎝
⎠⎝
∆t
59. (a) Strategy The pressure and number of moles is constant. Use the ideal gas law.
Solution Show that ∆V V0 = β∆T , where β = 1 T0 .
V
T
=
V0 T0
V
T
−1 =
−1
V0
T0
V − V0 T − T0
=
V0
T0
∆V ⎛ 1 ⎞
= ⎜ ⎟ ∆T = β∆T
V0 ⎝ T0 ⎠
(b) Strategy Use the result of part (a).
Solution Compute the coefficient of volume expansion for an ideal gas. Refer to Table 13.2.
1
1
βideal =
=
= 3410 × 10−6 K −1
T0 273.15 K + 20 K
From Table 13.2, β air = 3340 × 10−6 K −1 and the range for liquids is 182 × 10−6 K −1 for mercury to
1240 × 10−6 K −1 for benzene (excepting the negative value for water). So, β ideal ≈ β air and β ideal is 3 to 19
times larger than the values for liquids.
60. Strategy Use Eq. (13-20).
Solution Find the temperature of the ideal gas.
2 K tr
3
2(3.20 × 10−20 J)
K tr = kT , so T =
=
= 1550 K .
2
3k
3(1.38 × 10−23 J K)
61. Strategy The total translational kinetic energy of the gas molecules is equal to the number of molecules times the
average translational kinetic energy per molecule. Use Eqs. (13-13) and (13-20).
Solution Find the total translational kinetic energy of the gas molecules.
3
⎛3 ⎞ 3
K total = N K tr = N ⎜ kT ⎟ = PV = (1.013 × 105 Pa)(0.00100 m3 ) = 152 J
2
⎝2 ⎠ 2
580
Physics
Chapter 13: Temperature and the Ideal Gas
62. Strategy The total translational kinetic energy of the gas molecules is equal to the number of molecules times the
average translational kinetic energy per molecule. Use Eqs. (13-13) and (13-20).
Solution Divide the total translational kinetic energy by the volume for each pressure.
(a)
N 32 kT
K total N K tr
3
3
=
=
= P = (1.00 atm)(1.013 × 105 Pa atm) = 1.52 × 105 J/m3
V
V
V
2
2
(b)
K total 3
= (300.0 atm)(1.013 × 105 Pa atm) = 4.559 × 107 J/m3
V
2
1 2
1 2
Nm 2
, and from Eq. (13−17), P =
具v 典 = vrms
具vx 典. So,
3
3
V
Nm ⎛ 1 2 ⎞ 1 ⎛ Nm ⎞ 2
1⎛ M ⎞ 2
1 2
P=
vrms ⎟ = ⎜
vrms = ⎜ ⎟ vrms
.
= ρ vrms
⎜
⎟
V ⎝3
3⎝ V ⎠
3
⎠ 3⎝ V ⎠
63. Strategy and Solution 具vx2 典 =
64. Strategy The speed of sound at 0.0°C and 1.00 atm is 331 m/s. Refer to Fig 13.13.
Solution According to Fig. 13.13, approximately 50% of the O 2 molecules are moving faster than the speed
of sound.
65. Strategy The total internal kinetic energy of the ideal gas is equal to the number of molecules times the average
kinetic energy per molecule. Use Eq. (13-20).
Solution Find the total internal kinetic energy of the ideal gas.
3
3
3
K total = N K tr = N
kT = nRT = (1.0 mol)[8.314 J (mol ⋅ K) ](273.15 K + 0.0 K) = 3.4 kJ
2
2
2
66. Strategy Use the ideal gas law and Eq. (13-22).
Solution Find the temperature of the nitrogen gas.
PV
PV = nRT , so T =
.
nR
Compute the rms speed of the nitrogen molecules.
vrms =
3kT
3kPV
3(1.381× 10−23 J K)(1.6 atm)(1.013 × 105 Pa atm)(0.25 m)3
=
=
= 370 m s
m
mnR
2(14.0 u)(1.6605 × 10−27 kg u)(2.0 mol)[8.314 J (mol ⋅ K)]
67. Strategy Use Eq. (13-22).
Solution Find the ratio of the rms speeds.
vrms, Ar
vrms, Ne
=
3kT
mAr
3kT
mNe
=
mNe
=
mAr
1
=
2
1
2
68. Strategy Use Eq. (13-22).
Solution Find the rms speed of the particle.
vrms =
3kT
=
m
3(1.38 × 10−23 J K)(273.15 K + 27 K)
1.38 × 10−17 kg
581
= 3.00 cm s
Chapter 13: Temperature and the Ideal Gas
Physics
69. Strategy Use Eq. (13-22).
Solution Find the rms speeds of the molecules.
(a) vrms =
(b) vrms =
(c) vrms =
3kT
=
m
3(1.38 × 10−23 J K)(273.15 K + 0.0 K)
2(14.00674 u)(1.66 × 10−27 kg u)
3(1.38 × 10−23 J K)(273.15 K + 0.0 K)
= 493 m s
= 461 m s
2(15.9994 u)(1.66 × 10−27 kg u)
3(1.38 × 10−23 J K)(273.15 K + 0.0 K)
[12.011 u + 2(15.9994 u)](1.66 × 10−27 kg u)
= 393 m s
70. Strategy Use Eq. (13-22).
Solution Find the rms speeds of the atom and molecules.
He: vrms =
N 2: vrms =
H 2: vrms =
O 2: vrms =
3kT
3(1.38 × 10−23 J K)(273.15 K + 25 K)
=
= 1360 m s
m
(4.00260 u)(1.66 × 10−27 kg u)
3(1.38 × 10−23 J K)(273.15 K + 25 K)
= 515 m s
2(14.00674 u)(1.66 × 10−27 kg u)
3(1.38 × 10−23 J K)(273.15 K + 25 K)
= 1920 m s
2(1.00794 u)(1.66 × 10−27 kg u)
3(1.38 × 10−23 J K)(273.15 K + 25 K)
= 482 m s
2(15.9994 u)(1.66 × 10−27 kg u)
71. Strategy Compare the rms speeds to the escape speed. Use Eq. (13-22).
Solution Find the rms speeds for atomic and molecular hydrogen.
H: vrms =
H 2: vrms =
3kT
=
m
3(1.38 × 10−23 J K)(160 K)
(1.00794 u)(1.66 × 10−27 kg u)
3(1.38 × 10−23 J K)(160 K)
2(1.00794 u)(1.66 × 10−27 kg u)
= 2.0 km s
= 1.4 km s
60 km/s >> 2.0 km/s > 1.4 km/s; so yes , the astronaut should expect to find lots of hydrogen there.
72. Strategy Use Eqs. (13-13) and (13-22).
Solution Find the rms speed for an ideal gas.
3kT
3k ⎛ PV ⎞
3PV
=
=
⎜
⎟
m
m ⎝ Nk ⎠
Nm
Form a proportion with the initial and final rms speeds and solve for the pressure P.
vrms =
v0
1
=
=
0.90v0 0.90
3P0V
Nm
3PV
Nm
=
P0
, so P = 0.902 P0 = 0.902 (2.0 atm) = 1.6 atm .
P
582
Physics
Chapter 13: Temperature and the Ideal Gas
73. Strategy Use Eq. (13-20).
Solution Find the temperature of the ideal gas.
2 K tr
3
2(4.60 × 10−20 J)
=
= 2220 K .
K tr = kT , so T =
2
3k
3(1.38 × 10−23 J K)
74. Strategy and Solution The average translational kinetic energy of a molecule in an ideal gas is K tr =
The rms speed is vrms = 具v 2 典 , so
1 2
3PV
, and using
mvrms = K tr . From Eq. (13-19), we know that K tr =
2
2N
the ideal gas law, PV = NkT, we have
1 2
3 ⎛ PV
mvrms = ⎜
2
2⎝ N
⎞ 3
⎟ = 2 kT , so vrms =
⎠
3kT
.
m
75. Strategy and Solution The average translational energy of a molecule in an ideal gas is K tr =
rms speed is vrms = 具v 2 典 , so
1
m具v 2 典.
2
1
m具v 2 典. The
2
1 2
3PV
mvrms = K tr . From Eq. (13-19), we know that K tr =
, and using the
2
2N
ideal gas law, PV = nRT, we have
1 2
3 ⎛ PV
mvrms = ⎜
2
2⎝ N
⎞ 3 ⎛ nRT
⎟= 2⎜ N
⎠
⎝
⎞
⎛ n
⎟ , so vrms = 3 ⎜ mN
⎠
⎝
3RT
⎞
⎟ RT = M .
⎠
76. Strategy Form a proportion with the two reaction rates and solve for the activation energy. Use Eq. (13-24).
Solution Find the activation energy.
−
Ea
E
(
a 1− 1
1.878 e kT2
= Ea = e k T1 T2
−
1
e kT1
E ⎛1 1 ⎞
ln1.878 = a ⎜ − ⎟
k ⎝ T1 T2 ⎠
Ea =
)
(1.38 × 10−23 J K) ln1.878
k ln1.878
=
= 1.4 × 10−19 J
1 − 1
1
1
−
T
T
273.15 K + 5.0 K 273.15 K +10.0
1
2
583
Chapter 13: Temperature and the Ideal Gas
Physics
77. Strategy Form a proportion with the two reaction rates and solve for the temperature increase. Use Eq. (13-24).
Solution Find the temperature increase.
−
Ea
E
(
a 1− 1
1.035 e kT2
= Ea = e k T1 T2
−
1
e kT1
E ⎛1 1 ⎞
ln1.035 = a ⎜ − ⎟
k ⎝ T1 T2 ⎠
k ln1.035 1 1
= −
Ea
T1 T2
1
1 k ln1.035
= −
T2 T1
Ea
⎛ 1 k ln1.035 ⎞
T2 = ⎜ −
⎟
Ea
⎝ T1
⎠
⎛ 1 k ln1.035 ⎞
∆T = ⎜ −
⎟
Ea
⎝ T1
⎠
)
−1
−1
− T1
⎡
1
(1.38 × 10−23 J K) ln1.035 ⎤
=⎢
−
⎥
2.81× 10−19 J
⎣⎢ 273.15 K + 10.00 K
⎦⎥
−1
− (273.15 K + 10.00 K) = 0.14°C
78. Strategy Form a proportion with the two reaction rates and solve for the activation energy. Use Eq. (13-24).
Solution Find the activation energy implied by the rule.
−
Ea
E
(
a 1− 1
2 e kT2
= Ea = e k T1 T2
−
1
e kT1
E ⎛1 1 ⎞
ln 2 = a ⎜ − ⎟
k ⎝ T1 T2 ⎠
Ea =
k ln 2
=
− T1
1
T1
2
)
(1.38 × 10−23 J K) ln 2
= 1.3 × 10−19 J
1
1
−
273.15 K +90.0 K 273.15 K +100.0 k
79. Strategy Use Eqs. (13-13) and (13-25).
Solution Estimate the mean free path for a nitrogen molecule is each situation.
(a) Λ =
(b) Λ =
(c) Λ =
1
2π d 2
N
V
=
1
2π d 2
( kTP )
=
kT
2π d 2 P
(1.38 × 10−23 J K)(230 K)
2π (3 × 10−10 m)2 (5.0 × 104 Pa)
(1.38 × 10−23 )(230 K)
2π (3 × 10−10 m)2 (1× 103 Pa)
=
(1.38 × 10−23 J K)(290 K)
2π (3 × 10−10 m)2 (1.0 × 105 Pa)
= 200 nm
= 8 µm
584
= 100 nm
Physics
Chapter 13: Temperature and the Ideal Gas
80. Strategy Use Eq. (13-26).
Solution Find the time for the perfume molecule to diffuse 5.00 m in one direction.
x2
(5.00 m)2
= 1.25 × 106 s .
xrms = 2 Dt , so t = rms =
2D
2(1.00 × 10−5 m 2 s)
81. Strategy Use Eq. (13-26).
Solution Estimate the time it takes a sucrose molecule to move 5.00 mm in one direction.
x2
(5.00 × 10−3 m)2
xrms = 2 Dt , so t = rms =
= 2.5 × 104 s .
2D
2(5.0 × 10−10 m 2 s)
82. Strategy Use Eq. (13-26). Use a proportion to find the time.
Solution Find the time for a perfume molecule to diffuse 6.0 m in one direction.
2
t
x2
x6.0
(6.0 m)2
and
=
=
(20 s) = 80 s .
xrms = 2 Dt , so 6.0 = 6.0
t
t
6.0
2
2 3.0
t3.0 x3.0
x3.0
(3.0 m)2
83. Strategy Use Eq. (10-10) for the change in pressure and Eq. (13-7) for the fractional change in volume of the
gasoline.
Solution The gasoline expands outward against the can, so the negative sign in Eq. (10-10) is inappropriate.
∆V
∆P = Pf − Pi = B
= B β∆T , so
V0
1 atm
∆V
⎛
⎞
= 1.0 atm + (1.00 × 109 N m 2 )(950 × 10−6 K −1 )(15.0 K) ⎜
Pf = Pi + B
⎟ = 140 atm .
5
V0
⎝ 1.013 × 10 Pa ⎠
84. Strategy Use Eqs. (10-4) and (13-4).
Solution Find the stress.
∆L
stress = Y
= Y α∆T = (2.0 × 1011 N m 2 )(12 × 10−6 K −1)(20.0 K) = 4.8 × 107 N/m 2
L0
85. Strategy Find the temperature at which the radius of the steel sphere and the internal radius of the brass ring are
the same. Use Eq. (13-5).
Solution Set the final lengths of the radii equal and solve for the final temperature.
Lb0 + Lb0α b ∆T = Ls0 + Ls0α s ∆T
Lb0α b (Tf − Ti ) = Ls0 − Lb0 + Ls0α s (Tf − Ti )
Tf ( Lb0α b − Ls0α s ) = Ls0 − Lb0 + Ti ( Lb0α b − Ls0α s )
Ls0 − Lb0
+ Ti
Tf =
Lb0α b − Ls0α s
1.0010 cm − 1.0000 cm
+ 22.0°C = 165°C
Tf =
(1.0000 cm)(19 × 10−6 K −1 ) − (1.0010 cm)(12 × 10−6 K −1 )
585
Chapter 13: Temperature and the Ideal Gas
Physics
86. (a) Strategy Sum the molar masses of the elements that make up the molecule.
Solution Find the molar mass of oleic acid.
18(12.011 g mol) + 34(1.00794 g mol) + 2(15.9994 g mol) = 282.47 g mol
(b) Strategy Divide the mass by the molar mass.
Solution Find the number of moles of oleic acid in one drop.
2.3 × 10−5 g
= 8.1× 10−8 mol
282.47 g mol
(c) Strategy The volume is equal to the area times the height of the spread-out drop.
Solution Find d.
V = Ah = A(7d ), so d =
V
2.6 × 10−5 cm3
=
= 5.3 × 10−8 cm .
7A
7(70.0 cm 2 )
(d) Strategy Divide the total area by the area of one drop.
Solution Find the number of oleic acid molecules in one drop.
total area
70.0 cm 2
=
= 2.5 × 1016 molecules
area of one molecule (5.3 × 10−8 cm)2
(e) Strategy Divide the number of molecules by the number of moles.
Solution Estimate Avogadro’s number.
2.5 × 1016
= 3.1× 1023 mol−1
8.1× 10−8 mol
87. Strategy The relative number of atoms of an element contained within the molecule is equal to the molecular
mass of the molecule times the percentage of the molecular mass of the molecule that is the mass of the atoms of
element divided by the molecular mass of the element.
Solution Find the chemical formula.
(63 u)(0.016)
(63 u)(0.222)
(63 u)(0.762)
= 1.0 H;
= 1.0 N;
= 3.0 O
1.00794 u
14.00674 u
15.9994 u
The chemical formula is HNO3 .
586
Physics
Chapter 13: Temperature and the Ideal Gas
88. Strategy Plot the pressure on the vertical axis and the temperature on the horizontal axis. Estimate the value of
absolute zero by drawing a best-fit line through the data points and finding the temperature at which the line
intersects the T-axis (where the pressure is zero).
Solution The graph is shown.
P (atm)
1.50
1.00
0.50
T (°C)
300
250
200
150
100
50
0
50
100
According to the graph, absolute zero is approximately −270°C .
89. (a) Strategy Use Eq. (13-20).
Solution Compute the average kinetic energy of the air molecules.
3
3
⎡5
⎤
K tr = kT = (1.38 × 10−23 J K) ⎢ (98.6 − 32.0) K + 273.15 K ⎥ = 6.42 × 10−21 J
2
2
⎣9
⎦
(b) Strategy Since the average kinetic energy is directly proportional to the absolute temperature, find the
percent change in the absolute temperature.
Solution Find the percentage by which the kinetic energy of the molecules increased.
5 (100.0 − 98.6) K
∆K
∆T
9
× 100% =
× 100% = 5
× 100% = 0.25%
K0
T0
(98.6
−
32.0)
K
+
273.15
K
9
90. Strategy Use the ideal gas law with the number of moles constant to find Va , the volume of air at standard
temperature and pressure.
Solution Find the number of breaths taken per day.
PaVa PLVL
PTV
=
, so Va = L a L .
Ta
TL
PaTL
volume per day
volume per day (volume per day)PaTL
=
=
volume per breath
Va
PLTaVL
) (1 atm)(1.013 ×10 Pa atm)(39 K + 273.15 K) = 4000 breaths
(450 mm Hg) (133.3
) (0 K + 273.15 K)(100 cm )
( 210 ) (
=
L
day
103 cm3
L
5
Pa
mm Hg
587
3
Chapter 13: Temperature and the Ideal Gas
Physics
91. Strategy Use the ideal gas law.
Solution
(a) Find the number of moles in terms of the pressure.
PV
PV = nRT , so n =
.
RT
Find the percent change in the number of moles of air in the cabin.
nf − ni
× 100% =
ni
Pf V
RT
−
PV
i
RT
PV
i
RT
× 100% =
Pf − Pi
7.62 × 104 Pa − 1.01× 105 Pa
× 100% =
× 100% = −25%
Pi
1.01× 105 Pa
The number of moles decreases by 25% .
(b) Find the final temperature in terms of the pressure.
Tf
=
Ti
Pf V
nR
PV
i
nR
=
Pf
P
7.62 × 104 Pa
, so Tf = f Ti =
(25.0 K + 273.15 K) = 225 K = − 48°C .
Pi
Pi
1.01× 105 Pa
92. Strategy The extra volume is equal to the difference between the final and initial volumes. Let the initial radius
be of a cannonball be r0 and the final radius be r0 + ∆r. Use Eq. (13-4).
Solution Find the extra volume required.
Vextra
3
⎡
⎤
⎛4 ⎞
⎛ 4 ⎞ 3 ⎢⎛ ∆r ⎞
⎛4 ⎞ 3
3
3
3
= 500 ⎜ π ⎟ [(r0 + ∆r ) − r0 ] = 500 ⎜ π ⎟ r0 ⎜1 +
⎟ − 1⎥ = 500 ⎜ π ⎟ r0 [(1 + α∆T ) − 1]
⎥
r0 ⎠
⎝3 ⎠
⎝ 3 ⎠ ⎢⎝
⎝3 ⎠
⎣
⎦
4
⎛
⎞
= 500 ⎜ π ⎟ (0.05 m)3{[1 + (12 × 10−6 K −1)(8 K)]3 − 1} = 80 cm3
⎝3 ⎠
93. Strategy and Solution The average of the test scores is
The rms value is
83 + 62 + 81 + 77 + 68 + 92 + 88 + 83 + 72 + 75
= 78.1 .
10
832 + 622 + 812 + 77 2 + 682 + 922 + 882 + 832 + 722 + 752
= 78.6 . The most probable value
10
is 83 , since it appears twice as often as any other score.
94. Strategy Use the ideal gas law. Let L be the length of the cylinder and d be the distance the piston must be
pushed down before air will flow into the tire.
Solution The volume of the pump just as air begins to flow into the tire is Vf = A( L − d ).
PV = nRT , so V =
Find d.
nRT
.
P
nRT
Vf
P
P
P
P
= f = i , so Vf = A( L − d ) = AL − Ad = i Vi = i AL, or
nRT
Vi
Pf
Pf
Pf
Pi
⎛
P
d = L ⎜⎜1 − i
⎝ Pf
⎞
14.70
⎛
⎞
⎟⎟ = (18.0 in) ⎜1 −
⎟ = 13.2 inches .
⎝ 14.70 + 40.0 ⎠
⎠
588
Physics
Chapter 13: Temperature and the Ideal Gas
95. (a) Strategy The slope of a graph is rise over run.
Solution Find the slope of pressure versus temperature.
∆P 8.00 mm
=
= 0.400 mm Hg °C
∆T
20.0°C
(b) Strategy Use the ideal gas law and the result of part (a).
Solution Find the number of moles of gas present.
(∆P )V = nR (∆T ), so
V ∆P
0.500 L
10−3 m3
n=
=
(0.400 mm Hg °C)
(1.333 × 102 Pa mm Hg) = 3.21× 10−3 mol .
R ∆T 8.314 J (mol ⋅ K)
1L
96. Strategy Use the ideal gas law.
Solution
(a) The top pushes on the gas with a force of magnitude F = mg . So, the total pressure of the gas at equilibrium
is Ptotal =
mg
+ Patm .
A
Find the volume of the gas.
PV = nRT , so
nRT
nRT
(2.25 × 10−3 mol)[8.314 J (mol ⋅ K)](23.0 K + 273.15 K)
V=
=
=
= 50.9 cm3 .
2)
mg
(5.40
kg)(9.80
m
s
P
+P
+ 1.013 × 105 Pa
A
atm
70.0×10−4 m 2
(b) When the gas is again in equilibrium, the new pressure is the same as the old.
nRTf
Vf
T
223.0 + 273.15
= P = f =
= 1.675
nRT
i
23.0 + 273.15
Vi
Ti
P
The volume of the gas increases by a factor of 1.675 .
97. Strategy Assume that each air molecule is at the center of a sphere (with volume V/N) of diameter d. Then the
average distance between air molecules is approximately d. Use the microscopic form of the ideal gas law, Eq.
(13-13).
Solution Estimate the average distance between air molecules.
1/ 3
⎛ 1
⎞
⎛ 6kT ⎞
PV = P ⎜ N π d 3 ⎟ = NkT , so d = ⎜
⎟
6
⎝
⎠
⎝ πP ⎠
1/ 3
⎡ 6(1.38 × 10−23 J K)(273.15 K + 0.0 K) ⎤
=⎢
⎥
5
⎢⎣ π (1.00 atm)(1.013 × 10 Pa atm) ⎥⎦
589
≈ 4 nm .
Chapter 13: Temperature and the Ideal Gas
Physics
98. (a) Strategy Find the distance between N 2 molecules and multiply by the scale factor
0.0375 m
= 1.25 × 108 to get the distance between ping pong balls. Let each N 2 molecule be at the
0.30 × 10−9 m
center of a sphere with diameter d. Then the average distance between N 2 molecules is approximately d. The
volume of each sphere is V/N with V = 0.0224 m3 for N = 6.02 × 1023 molecules, P = 1.00 atm, and T =
0.0°C.
Solution Find the average distance between the ping-pong balls.
V 1 3
6V
= π d , so d = 3
.
N 6
πN
Therefore, the distance between ping-pong balls is (1.25 × 108 ) 3
6(0.0224 m3 )
π (6.02 × 1023 )
= 52 cm .
(b) Strategy Find the mean free path for an N 2 molecule and multiply by the scale factor to find the average
distance between ping pong ball collisions. Use Eq. (13-25) and the results of part (a).
Solution Find the mean free path.
1
kT
Λ=
=
2
2π d ( N / V )
2π d 2 P
So, the average distance between ping pong ball collisions is
(1.25 × 108 )(1.38 × 10−23 J K)(273.15 K + 0.0 K)
= 12 m .
2π (0.30 × 10−9 m)2 (1.00 atm)(1.013 × 105 Pa atm)
99. (a) Strategy Use Eq. (13-10) to find the number of air molecules per m3 ; multiply by 0.21 to find the number
of O 2 molecules.
Solution Find the number of oxygen molecules per cubic meter.
N O2 0.21ρair
0.21(1.20 × 103 g m3 )(6.022 × 1023 mol−1)
=
=
V
mair
0.78(2)(14.00674 g mol) + 0.21(2)(15.9994 g mol) + 0.01(39.948 g mol)
= 5.2 × 1024 m −3
(b) Strategy The ratio of the surface volume of air to the volume of the air at a depth of 100.0 m multiplied by
21% gives the percentage of O 2 molecules. Use the ideal gas law with N, k, and T constant and Eq. (9-3).
Solution Find the appropriate percentage of oxygen molecules in the tank.
Patm
V2
P
(1.013 × 105 Pa)(21%)
(21%) = 1 (21%) =
(21%) =
= 1.9%
V1
P2
Patm + ρ gh
1.013 × 105 Pa + (1025 kg m3 )(9.80 m s 2 )(100.0 m)
100. Strategy Use Eq. (13-22). Form a proportion.
Solution Show that, in two gases at the same temperature, the rms speeds are inversely proportional to the square
root of the molecular masses.
(v )
3kT
3kT
and (vrms ) 2 =
(vrms )1 =
, so rms 1 =
(vrms )2
m1
m2
3kT
m1
3kT
m2
590
=
m2
.
m1
Physics
Chapter 13: Temperature and the Ideal Gas
101. Strategy Use Eq. (13-4).
Solution Find the approximate temperature for the SR-71 while it is in flight.
∆L
∆L
0.20 m
= α∆T , so Tf = Ti +
= 20°C +
= 630°C .
−
L0
α L0
(10.1× 10 6 K −1 )(32.70 m)
102. Strategy Use Eq. (13-26).
Solution Find the time it takes for a water molecule to diffuse out through the pore.
x2
(2.5 × 10−5 m)2
xrms = 2 Dt , so t = rms =
= 1.3 × 10−5 s .
2D
2(2.4 × 10−5 m 2 s)
103. Strategy Use the microscopic form of the ideal gas law, Eq. (13-13).
Solution Find the number of air molecules N.
5
−3
3
4
PV (1.00 × 10 Pa) 3 π (0.125 × 10 m)
NkT = PV , so N =
=
= 1.9 × 1014 molecules .
kT
(1.38 × 10−23 J K)(310 K)
104. (a) Strategy If V = 10.0 L and N is the number of N 2 molecules, V/N is approximately the volume of a sphere
centered on one molecule with diameter d. d is approximately the nearest-neighbor distance.
Solution Estimate the nearest-neighbor distance.
1/ 3
1/ 3
1 3 V
⎛ 6V ⎞
π d = , so d = ⎜
⎟
6
N
⎝πN ⎠
⎡
⎤
⎢ 6(10.0 L)(10−3 m3 L) ⎥
=⎢
23
−1 ⎥
⎢ π (12 g)(6.022×10 mol ) ⎥
⎣ 2(14.00674 g mol ) ⎦
= 4.2 nm .
(b) Strategy and Solution Since the nearest neighbor distance is significantly larger than the diameter of an N 2
molecule (4.2 nm/0.3 nm = 14), the gas is dilute .
105. Strategy Use Eq. (13-22).
Solution Find the decrease in the rms speed of the air molecules.
3kTf
3kTi
∆vrms =
−
m
m
=
3(1.38 × 10−23 J K)(10.0 K + 273.15 K)
[0.750(2)(14.00674 u) + 0.250(2)(15.9994 u)](1.66 × 10−27 kg u)
−
3(1.38 × 10−23 J K)(40.0 K + 273.15 K)
[0.750(2)(14.00674 u) + 0.250(2)(15.9994 u)](1.66 × 10−27 kg u)
= −25 m/s
The rms speed will have decreased by 25 m/s.
591
Chapter 13: Temperature and the Ideal Gas
Physics
106. Strategy Use Eqs. (13-4) and (13-5).
Solution
(a) Find T such that ∆L = 0.00200 cm.
∆L
0.00200 cm
∆L = α L0∆T , so T =
+ T0 =
+ 20.0°C = 44°C .
−
α L0
(12 × 10 6 K −1)(7.00000 cm)
(b) Set the diameters ( Ls and Lb ) equal and solve for T.
Ls = Lb
Ls0 + Ls0α s ∆T = Lb0 + Lb0α b∆T
∆T ( Ls0α s − Lb0α b ) = Lb0 − Ls0
Lb0 − Ls0
T = T0 +
Ls0α s − Lb0α b
7.00200 cm − 7.00000 cm
= 20.0°C +
= −21°C
(7.00000 cm)(12 × 10−6 K −1) − (7.00200 cm)(19 × 10−6 K −1)
107. Strategy The increase in volume of the mercury minus the increase in volume of the glass bulb equals the
volume of mercury that moves up the tube. Use Eqs. (13-6) and (13-7).
Solution Find the how far the thread of mercury moves h.
∆VHg − ∆Vg = Atubeh
V0 β Hg ∆T − V0 β g ∆T = ( A0 + 2α g A0∆T )h
h=
V0 (β Hg − β g )∆T
A0 (1 + 2α g ∆T )
=
(0.200 × 10−6 m3 )[(182 − 9.75) × 10−6 K −1](1.00 K)
1 π (0.120 × 10−3
4
m)2[1 + 2(3.25 × 10−6 K −1)(1.00 K)]
= 3.05 mm
108. (a) Strategy Use Eq. (13-4).
Solution Find the temperature to which the band must be heated.
∆L
∆L
134.460 cm − 134.448 cm
= α∆T , so T = T0 +
= 20.0°C +
= 27.4°C .
L0
α L0
(12 × 10−6 K −1)(134.448 cm)
(b) Strategy Use Eq. (10-4).
Solution Find the tension in the band when it cools.
F
∆L
=Y
, so
A
L0
∆L
134.460 cm − 134.448 cm
F = AY
= (0.0500 m)(0.00500 m)(2.0 × 1011 N m 2 )
= 4.5 kN .
L0
134.448 cm
592
Physics
Chapter 13: Temperature and the Ideal Gas
109. Strategy Use ideal gas law and Hooke’s law.
Solution Find the pressure of the gas.
nRT
PV = nRT , so Pgas =
. The force with which the piston pushes on the spring is equal to
V
F = ( Pgas − Patm ) Apiston . Set this equal to F = k ∆x to find the spring constant.
k ∆x = ( Pgas − Patm ) Apiston =
− Patm ) Apiston , so
( nRT
V
2
⎡ (6.50×10−2 mol)[8.314 J (mol⋅K)](20.0 K + 273.15 K)
5 Pa ⎤ π ( 0.0800 2 m) = 7.4 × 103 N m .
k=⎢
−
1.013
×
10
⎥⎦
(0.120 m + 0.0540 m)π (0.0800 2 m)2
0.0540 m
⎣
110. Strategy Use Eq. (13-4) and the formula for arc length.
Solution
(a) Find the relation for the radius of the strip.
∆L1
∆L2
= α1∆T and
= α 2∆T , α 2 > α1.
L0
L0
L0 + ∆L1 = Rθ and L0 + ∆L2 = ( R + s)θ (arc length).
Solve for R.
L + ∆L1
L0 + ∆L2 = ( R + s) 0
R
R( L0 + L0α 2∆T ) = ( R + s)( L0 + L0α1∆T )
R(1 + α 2∆T ) = R(1 + α1∆T ) + s(1 + α1∆T )
R(α 2 − α1)∆T = s(1 + α1∆T )
≈s
(α1∆T << 1)
s
R≈
(α 2 − α1)∆T
(b) Compute the radius.
0.1× 10−3 m
R=
= 0.7 m
[(19 − 12) × 10−6 K −1](20.0 K)
593
Chapter 14
HEAT
Conceptual Questions
1. Heat flows from the hotter to the colder object.
2. The moisture forms an insulating layer between the hot iron and the skin, so that brief contact is safe. The water
temperature will not go over 100°C and evaporating the water takes a lot of heat due to its large heat of
vaporization. When the iron is sufficiently hot, it vaporizes some of the water on the finger upon contact, forming
an additional insulating layer of vapor between the iron and moistened finger.
3. Water is most dense at 3.98°C and becomes less dense as the temperature is lowered further. The coldest water
(below 3.98°C) therefore floats to the surface, which is why ice forms there.
4. The water can then cool by evaporation.
5. The air between the layers acts as very good thermal insulation.
6. The water is slow to cool in the winter due to the large specific heat of water, so it helps keep the vineyard warmer
than if there were no large body of water nearby.
7. No, the metal has a higher thermal conductivity than the wood, so the rate at which it takes in heat from the hand
is greater.
8. The breeze blows from the cooler land toward the warmer water.
9. The fins increase the area of the surface responsible for radiating heat from the engine to the air so that the engine
cools more efficiently.
10. The air under the bridge cools faster than the earth under the roadway. Thus, the bridge cools faster than the
roadway.
11. At higher elevations the atmospheric pressure is lower so that water boils at a lower temperature and cooking
times need to be lengthened.
12. The boiling temperature of water increases as its pressure increases. Food therefore cooks more quickly in a
pressure cooker because the increased pressure heightens the temperature at which the food cooks.
13. Water contained in the sauce and cheese of a pizza has a higher specific heat and thermal conductivity than the
crust. Thus, for a given time interval, the amount of heat energy transferred to the roof of the mouth by the sauce
and cheese is greater than that which is transferred to the hand by the crust.
14. The internal energy of a gas of monatomic molecules is a function of the average molecular translational energy.
In addition to this translational mode, the internal energy of a gas of diatomic molecules is also affected by the
average molecular rotational and vibrational energies. Some of the heat energy put into a diatomic gas goes into
exciting these rotational and vibrational modes and not into the translational modes. Since temperature is only a
function of the average translational energy, the energy put into these other modes does not affect the temperature.
As a result, the molar specific heat of a diatomic gas is greater than that for a monatomic gas.
594
Physics
Chapter 14: Heat
15. At very low temperatures, the internal energy of a diatomic gas is too low to excite the rotational and vibrational
energy modes of its constituent molecules. At room temperature, some of the energy put into the system goes into
exciting these modes and not toward increasing the temperature of the gas. Thus, the molar specific heat is greater
for higher temperatures.
16. No, Stefan’s law may be used to compare the radiation rates of two bodies only if the absolute temperature (in
kelvins) of each body is used.
17. The rate at which heat flows via conduction, convection, and radiation depends upon the temperature difference
between the hot and cold bodies. By adding the milk to the coffee immediately, the temperature difference
between the coffee and the surrounding air is minimized, so the rate of heat flow is smaller. Additionally, when
the color of the coffee is lighter, the radiative heat loss will be further reduced.
18. The air gap inside a double-paned window is designed to reduce the flow of heat through the window via
conduction. If the gap size were larger, convective currents could be set up that would increase the rate of energy
flow via convection.
19. A greenhouse effect is set up within the packages of meat that causes the temperature inside to be larger than that
of the surrounding air. This could be remedied by turning the clear plastic wrapping face-down away from the
lights.
20. The average translational kinetic energy of a single molecule in the pot of hot tea is larger than that of a molecule
in the partially frozen lake. The total internal energy of the water in the lake is however much greater than that of
the water in the teapot because the mass of the water in the lake is orders of magnitude larger than the mass of
water in the teapot.
21. In winter, the walls of a room are at a lower temperature than the air within the room. In summer, the walls are
warmer than the inside air. The amount of heat lost from a person standing within the room via conduction and
convection is constant throughout the year since the temperature of the air is held fixed. The temperature
difference between the body and the walls is greatest in winter. As a result, more heat is lost via radiative transfer
in winter and the room therefore feels cooler.
22. A dark-toned and flat-finished object has a higher emissivity than a light-toned and shiny-finished object. Stefan’s
law therefore tells us that a radiator painted with a dark finish is a more efficient radiator of heat than one finished
with a light tone.
23. The two objects must be in thermal equilibrium with each other and the walls of the evacuated chamber in which
they reside. If an object were to absorb more heat energy than it emitted, its temperature would change and the
system would no longer be in thermal equilibrium. Each body must therefore emit exactly as much radiation as it
absorbs—a good absorber must be a good emitter.
24. (a) In fluid flow, the pressure difference between ends of pipe = volume flow rate × fluid flow resistance; in heat
conduction, the temperature difference between ends of thermal conductor = rate of heat flow × thermal
resistance.
(b) Yes; the pressure at the end of pipe 1 is equal to the pressure at the beginning of pipe 2, and so forth, so the
same reasoning that leads to Eq. (14-13) holds for fluid flow.
595
Chapter 14: Heat
Physics
Problems
1. (a) Strategy The gravitational potential energy of the 1.4 kg of water is converted to internal energy in the
6.4-kg system.
Solution Compute the increase in internal energy.
U = mgh = (1.4 kg)(9.80 m s 2 )(2.5 m) = 34 J
(b) Strategy and Solution Yes; the increase in internal energy increases the average kinetic energy of the water
molecules, thus the temperature is slightly increased.
2. Strategy The decrease in gravitational potential energy per unit mass is equal to the increase in internal energy
per unit mass.
Solution Compute the internal energy produced per kg.
U
= gh = (9.80 N kg)(105 m) = 1.03 kJ kg
m
3. Strategy The amount of internal energy generated is equal to the decrease in kinetic energy of the bullet.
Solution Compute the amount of internal energy generated.
1
1
∆K = mvi2 = (0.0200 kg)(7.00 × 102 m s)2 = 4.90 kJ
2
2
4. Strategy The amount of internal energy generated is equal to the decrease in kinetic energy of the ball.
Solution Compute the amount of internal energy generated.
2
∆K =
1 2 1
⎛ 1h ⎞
mvi = (0.1475 kg)(162,000 m h)2 ⎜
⎟ = 149 J
2
2
⎝ 3600 s ⎠
5. (a) Strategy The decrease in gravitational potential energy of the child is equal to the amount of internal energy
generated.
Solution Compute the amount of internal energy generated.
U = mgh = (15 kg)(9.80 m s 2 )(1.7 m) = 250 J
(b) Strategy and Solution Friction warms the slide and the child, and the air molecules are deflected by the
child’s body. The energy goes into all three .
6. Strategy According to conservation of energy, ∆E = 0.
Solution Find the amount of energy dissipated by air resistance.
1
1
0 = ∆U + ∆K + U air = mg ∆y + m(vf2 − vi2 ) + U air = −mgh + mvf2 + U air , so
2
2
1 2
1
⎡
⎤
2
3
U air = mgh − mvf = (64 kg) ⎢ (9.80 m s )(0.90 × 10 m) − (5.8 m s)2 ⎥ = 560 kJ .
2
2
⎣
⎦
596
Physics
Chapter 14: Heat
7. Strategy The total change in energy is zero. The total energy of the system is conserved.
Solution Find the amount of energy dissipated by air drag.
0 = ∆U + ∆K + U air
1
U air = −∆U − ∆K = −mg ∆y − m(vf2 − vi2 )
2
1
⎧
⎫
2
= −(0.60 kg) ⎨(9.80 m s )(3.0 m − 2.0 m) + [(4.5 m s) 2 − (7.6 m s) 2 ]⎬ = 5.4 J
2
⎩
⎭
8. Strategy The potential energy of the mass is converted into heat in the water.
Solution Determine the descending mass.
∆U g
1.00 × 103 J
∆U g = mg ∆y, so m =
=
= 2.72 kg .
g ∆y (9.80 m s 2 )(30.0)(1.25 m)
9. Strategy The conversion factor is 1 kW ⋅ h = 3.600 MJ.
Solution Convert 1.00 kJ to kilowatt-hours.
1 kW ⋅ h
(1.00 × 103 J)
= 2.78 × 10−4 kW ⋅ h
3.600 × 106 J
10. Strategy The heat capacity of an object is equal to its mass times its specific heat.
Solution Find the heat capacity of 20.0 kg of silver.
C = mc = (20.0 kg)[0.235 kJ (kg ⋅ K)] = 4.70 kJ K
11. Strategy The heat capacity of an object is equal to its mass times its specific heat.
Solution Find the heat capacity of the 5.00-g gold ring.
C = mc = (0.00500 kg)[0.128 kJ (kg ⋅ K)] = 6.40 × 10−4 kJ K
12. Strategy Solve Eq. (14-4) for the final temperature.
Solution Find the final temperature of the water.
Q
125.6 kJ
+ Ti =
+ 22°C = 82°C .
Q = mc∆T = mc(Tf − Ti ), so Tf =
(0.500 kg)[4.186 kJ (kg ⋅ K)]
mc
13. Strategy Use Eq. (14-4).
Solution Find the amount of heat that must flow into the water.
Q = mc∆T = (2.0 × 10−3 m3 )(1.0 × 103 kg m3 )[4186 J (kg ⋅ K) ](80.0 − 20.0) K = 0.50 MJ
14. Strategy The final kinetic energy equals the energy content of the banana.
Solution Find the man’s speed.
1 2
mv = U , so vf =
2 f
2U
=
m
2(1.00 × 102 kcal)(4186 J kcal)
= 100 m s .
83 kg
597
Chapter 14: Heat
Physics
15. Strategy The 3.3% of the energy from the food is converted to gravitational potential energy of the high jumper.
Solution Find the height the athlete could jump.
U
(3.00 × 106 cal)(4.186 J cal)(0.033)
U = mgh, so h =
=
= 700 m .
mg
(60.0 kg)(9.80 m s 2 )
16. Strategy The heat capacity of an object is equal to its mass times its specific heat.
Solution Find the heat capacity of the 30.0-kg block of ice.
C = mc = (30.0 kg)[2.1 kJ (kg ⋅ K)] = 63 kJ K
17. Strategy The heat capacity of an object is equal to its mass times its specific heat. The mass of an object is equal
to its density times its volume.
Solution Find the heat capacities.
(a) C = mc = ρVc = (2702 kg m3 )(1.00 m3 )[0.900 kJ (kg ⋅ K)] = 2430 kJ K
(b) C = mc = ρVc = (7860 kg m3 )(1.00 m3 )[0.44 kJ (kg ⋅ K)] = 3500 kJ K
18. Strategy The heat capacity of an object is equal to its mass times its specific heat. Find the heat capacities of the
systems by adding the individual heat capacities.
Solution Find the heat capacities.
(a) C = mb cb + mw cw = (0.450 kg)[0.384 kJ (kg ⋅ K) ] + (0.050 kg)[4.186 kJ (kg ⋅ K)] = 0.38 kJ K
(b) C = mw cw + mAl cAl = (7.5 kg)[4.186 kJ (kg ⋅ K)] + (0.75 kg)[0.900 kJ (kg ⋅ K)] = 32 kJ K
19. Strategy Use Eq. (14-4) to find the heat required.
Solution The heat capacity of the system is C = mAl cAl + mw cw .
Q = mc∆T = {(0.400 kg)[0.900 kJ (kg ⋅ K)] + (2.00 kg)[4.186 kJ (kg ⋅ K)]}(100.0°C − 15.0°C) = 742 kJ
20. Strategy Use Eq. (14-4).
Solution Find the heat required to raise the woman’s body temperature.
Q = mc∆T = (50.0 kg)[3.5 kJ (kg ⋅ K)](38.4°C − 37.0°C) = 250 kJ
21. Strategy Use Eq. (14-4).
Solution Find the specific heat of lead.
Q
0.88 kJ
c=
=
= 0.13 kJ (kg ⋅ K)
m∆T (0.35 kg)(20.0 K)
598
Physics
Chapter 14: Heat
22. Strategy The gravitational potential energy of the falling water is converted into internal energy of the system
(total water). Neglect heat flow into the vessel. Use Eq. (14-4).
Solution Find the mass of water that was in the vessel.
m1c∆T + mvc∆T = c∆T (m1 + mv ) = m1gh, so
⎡ (9.80 m s 2 )(1.00 × 102 m) ⎤
m gh
− 1⎥ = 1.34 kg .
mv = 1 − m1 = (1.00 kg) ⎢
c ∆T
⎣⎢ [4186 J (kg ⋅ K) ](0.100 K) ⎦⎥
23. Strategy The decrease in the internal energy of the mercury is equal to the amount of heat that flowed out of it.
Use Eq. (14-4).
Solution Find the amount of energy lost.
∆U = Q = mc∆T = (0.10 ×10−3 kg)[0.139 kJ (kg ⋅ K) ](8.5°C − 15.0°C) = −0.090 J
0.090 J of energy was lost by the mercury.
24. Strategy Use the definition of average power and Eq. (14-4).
Solution Solve for ∆t using P = ∆E ∆t = ∆Q ∆t .
∆Q mc∆T (0.50 kg)[4186 J (kg ⋅ K) ](100.0°C − 20.0°C)
∆t =
=
=
= 80 s .
P
P
2.1× 103 W
25. Strategy The energy required to increase the internal energy of the gas is equal to nCV ∆T where
CV = 20.4 J (mol ⋅ K) for H 2. Use the ideal gas law to find the number of moles of H 2.
Solution Find the energy required.
PV
(10.0 atm)(1.013 × 105 Pa atm)(250 L)(10−3 m3 L)[20.4 J (mol ⋅ K)](25.0 K)
CV ∆T =
∆U = nCV ∆T =
RT
[8.314 J (mol ⋅ K) ](273.15 K + 0.0 K)
= 57 kJ
26. Strategy Use the ideal gas law to find the number of moles of nitrogen gas in the container. Then, solve for the
final temperature of the gas in Eq. (14-6).
Solution Find the number of moles of gas in the container.
PV
PV = nRT , so n =
.
RT
Find the new temperature of the gas after the heat is added.
PV
Q = nCV ∆T =
C (T − Ti ), so
RTi V f
Tf = Ti +
RTi Q
PVCV
= 23°C +
[8.314 J (mol ⋅ K) ](23 K + 273.15 K)(26.6 × 103 J)
(3.5 atm)(1.013 × 105 Pa atm)(425 L)(10−3 m3 L)[20.8 J (mol ⋅ K)]
599
= 44°C .
Chapter 14: Heat
Physics
27. Strategy The number of moles of air is given by the ideal gas law, n = PV ( RT ) , where these are the initial
quantities. The power generated by the crowd is 501Pav = ∆Q ∆t and ∆Q = 52 nR∆T for a diatomic gas.
Solution Find ∆T .
5
5 ⎛ PV ⎞
nR∆T = ⎜
R∆T = ∆Q = 501Pav ∆t , so
2
2 ⎝ RT ⎟⎠
2(501)TPav ∆t 2(501)(273.15 K + 20.0 K)(110 W)(2.0 h)(3600 s h)
∆T =
=
= 58°C .
5PV
5(1.01× 105 Pa)(8.00 × 103 m3 )
28. Strategy The gas is monatomic, so the molar specific heat is
3
2
Q=
3
2
R = 3Nk (2n). Therefore, the heat is
Nk ∆T .
Solution Find N.
3
2Q
2(10.0 J)
Nk ∆T = Q, so N =
=
= 4.83 × 1021 molecules .
23
−
2
3k ∆T 3(1.38 × 10
J K)(1.00 × 102 K)
29. (a) Strategy Phase changes are indicated by the graph where the temperature is constant while heat is added.
Solution Initially, the substance is solid. As the temperature increases, the substance changes from the solid
to the liquid phase, then from the liquid phase to the gas phase. There are two phase changes shown by the
graph: from B to C, solid to liquid ; and from D to E, liquid to gas .
(b) Strategy and Solution The beginning of the phase change from solid to liquid indicates the melting point of
the substance. That point is labeled by the letter B .
(c) Strategy and Solution The beginning of the phase change from liquid to gas indicates the boiling point of
the substance. That point is labeled by the letter D .
30. Strategy The sum of the heat flows is zero. Use Eqs. (14-4) and (14-9).
Solution Find the heat of vaporization of water.
0 = Qs + Qw + Qc = −ms Lv + mscw ∆Ts + mw cw ∆Tw + mccc∆Tc , so
c (m ∆T + mw ∆Tw ) + mccc∆Tc
Lv = w s s
ms
[4.186 J (g ⋅ K)][(18.5 g)(−38.0 K) + (2.00 × 102 g)(47.0 K)] + (3.00 × 102 g)[0.380 J (g ⋅ K)](47.0 K)
18.5 g
= 2260 J g
=
600
Physics
Chapter 14: Heat
31. Strategy The sum of the heat flows is zero. Use Eqs. (14-4) and (14-9).
Solution Find the heat of fusion of water.
0 = Qice + Qw + Qc = mice Lf + micecw ∆Tice + mw cw ∆Tw + mccc∆Tc , so
c (m ∆T + mw ∆Tw ) + mccc∆Tc
Lf = − w ice ice
mice
[4.186 J (g ⋅ K)][(30.0 g)(8.5 K) + (2.00 × 102 g)( − 11.5 K)]+ (3.00 × 102 g)[0.380 J (g ⋅ K)](−11.5 K)
30.0 g
= 330 J g
=−
32. Strategy The sum of the heat flows is zero. Use Eqs. (14-4) and (14-9).
Solution Find the mass of the liquid nitrogen that is converted to gas.
0 = Qb + QN = mb cb ∆Tb + mN Lv , so
m c ∆T
(25.0 g)[0.384 kJ (kg ⋅ K)][(77.2 − 293) K]
mN = − b b b = −
= 10.4 g .
Lv
199.1 kJ kg
33. Strategy The sum of the heat flows is zero. Find the mass of the water required to melt the ice and leave the
temperature of the drink at 5.0°C. Use Eqs. (14-4) and (14-9).
Solution Find the mass of water to be added to the cup.
0 = Qw + Qice
0 = mw cw ∆Tw + mice Lf + mice cice ∆T1 + mice cw ∆T2
− mw cw ∆Tw = mice ( Lf + cice ∆T1 + cw ∆T2 )
m ( L + cice ∆T1 + cw ∆T2 )
mw = ice f
−cw ∆Tw
(50.0 g + 50.0 g){333.7 J g + [2.1 J (g ⋅ K)](15.0 K) + [4.186 J (g ⋅ K)](5.0 K)}
= 461 g
mw = −
[4.186 J (g ⋅ K)](−20.0 K)
34. Strategy Find the sum of the heats required to raise the temperature of the ice to 0.0°C, melt the ice, raise the
resulting water to 100.0°C, evaporate the water, and raise the temperature of the resulting steam to 110.0°C. Use
Eqs. (14-4) and (14-9).
Solution Find the required heat.
Q = Qice + Qw + Qs
= mLf + mcice ∆Tice + mcw ∆Tw + mLv + mcs ∆Ts
= (1.0 kg){333.7 kJ kg + [2.1 kJ (kg ⋅ K)](20.0 K) + [4.186 kJ (kg ⋅ K)](100.0 K) + 2256 kJ kg
+ [2.01 kJ (kg ⋅ K)](10.0 K)}
= 3100 kJ
35. Strategy Heat flows from the water to the ice, melting some of it. Find the mass of ice required to lower the
temperature of the water to 0.0°C. Use Eqs. (14-4) and (14-9).
Solution Find the required mass of ice.
0 = Qice + Qw = mice Lf + mw cw ∆Tw , so
mice = −
mw cw ∆Tw
Lf
=−
(5.00 × 102 mL)(1.00 g mL)[4.186 J (g ⋅ K)](−25.0 K)
= 157 g .
333.7 J g
601
Chapter 14: Heat
Physics
36. Strategy The sum of the heat flows is zero. The tea is basically water. The mass of the tea is found by
multiplying the density of water by the volume of the tea. Neglect the temperature change of the glass.
Use Eqs. (14-4) and (14-9).
Solution Find the mass of the ice required to cool the tea to 10.0°C.
0 = Qt + Qice = ρ wVt cw ∆Tt + mice Lf + mice cice ∆T1 + mice cw ∆T2 = ρ wVt cw ∆Tt + mice ( Lf + cice ∆T1 + cw ∆T2 ), so
mice =
− ρ wVt cw ∆Tt
Lf + cice ∆T1 + cw ∆T2
=
−(1.00 × 103 kg m3 )(2.00 × 10−4 m3 )[4.186 kJ (kg ⋅ K)](−85.0 K)
= 179 g .
333.7 kJ kg + [2.1 kJ (kg ⋅ K)](10.0 K) + [4.186 kJ (kg ⋅ K)](10.0 K)
37. Strategy The sum of the heat flows is zero. The tea is basically water. The mass of the tea is found by
multiplying the density of water by the volume of the tea. Do not neglect the temperature change of the
glass. Use Eqs. (14-4) and (14-9).
Solution Find the mass of the ice required to cool the tea to 10.0°C. Let ∆T be the temperature change of the tea
and the glass.
0 = Qt + Qice + Qg
0 = ρ wVt cw ∆T + mice Lf + mice cice ∆T1 + mice cw ∆T2 + mg cg ∆T
0 = ( ρ wVt cw + mg cg )∆T + mice ( Lf + cice ∆T1 + cw ∆T2 )
mice =
−( ρ wVt cw + mg cg )∆T
Lf + cice ∆T1 + cw ∆T2
−{(1.00 × 103 kg m3 )(2.00 × 10−4 m3 )[4.186 kJ (kg ⋅ K)] + (0.35 kg)[0.837 kJ (kg ⋅ K)]}(−85.0 K)
333.7 kJ kg + [2.1 kJ (kg ⋅ K)](10.0 K) + [4.186 kJ (kg ⋅ K)](10.0 K)
= 242 g
=
The percentage change from the answer for Problem 40 is
242 g − 179 g
× 100% = 35% .
179 g
38. Strategy Use Eq. (14-9) for fusion.
Solution The heat required to melt the ice is 12.0 kJ − 4.0 kJ = 8.0 kJ. Find the mass of the ice.
Q
8.0 kJ
m=
=
= 24 g
Lf 333.7 kJ kg
39. Strategy Heat flows from the aluminum into the ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of aluminum required to melt 10.0 g of ice.
QAl + Qice = mAlcAl∆TAl + mice Lf = 0, so
m L
(10.0 g)(333.7 J g)
mAl = − ice f = −
= 46.3 g .
cAl∆TAl
[0.900 J (g ⋅ K)](0.0°C − 80.0°C)
40. Strategy Heat flows from the aluminum into the ice. Use Eqs. (14-4) and (14-9).
Solution Find the temperature of the aluminum at which it will melt a mass of the ice equal to its own.
L
333.7 J g
+ 0.0°C = 371°C .
mAlcAl∆TAl + mice Lf = mAlcAl (Tf − Ti ) + mice Lf = 0, so Ti = f + Tf =
cAl
0.900 J (g ⋅ K)
Since 371°C < 660°C (the melting point of aluminum), the answer is yes .
602
Physics
Chapter 14: Heat
41. Strategy Use Eq. (14-4) for vaporization. Q is equal to the rate of heat loss per square meter times the area times
the time (1 h = 3600 s).
Solution Find the mass of water lost through transpiration.
Q (250 W m 2 )(0.005 m 2 )(3600 s)
m=
=
= 2g
Lv
2256 J g
42. Strategy The rate of heat loss is ∆Q ∆t = Lv ∆m ∆t since Q = mLv for evaporation and where ∆m represents
the mass of water evaporated.
Solution Compute the rate of heat lost through transpiration.
∆Q
⎛ 1 min ⎞
= (0.618 g min) ⎜
⎟ (2256 J g) = 23.2 W
∆t
⎝ 60 s ⎠
43. Strategy The heat flows from the coffee to the ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of ice required to cool the coffee.
Qice + Qc = mice Lf + micec∆Tice + mcc∆Tc = mice ( Lf + c∆Tice ) + mcc∆Tc = 0, so
−mcc∆Tc
(0.25 kg)[4186 J (kg ⋅ K) ](−20.0 K)
=−
= 36 g .
mice =
Lf + c∆Tice
333.7 × 103 J kg + [4186 J (kg ⋅ K) ](60.0 K)
44. (a) Strategy Follow each line segment and consider what happens to the substance represented by the diagram.
Solution For each segment, the pressure, temperature, and phase changes are given in the table.
Segment
Pressure
Temperature
Phase Change
AB
Decreases
Constant
Fluid to solid
BC
Constant
Increases
Solid to liquid
CD
Decreases
Constant
Liquid to vapor
DE
Constant
Decreases
Vapor to solid
(b) Strategy and Solution a is the critical point: If the path for changing a liquid to a gas goes around the
critical point without crossing the vapor pressure curve, no phase change occurs. At temperatures above the
critical temperature or pressures above the critical pressure, it is impossible to make a clear distinction
between the liquid and gas phases. b is the triple point: The three states―solid, liquid, and vapor―can
coexist in equilibrium.
45. Strategy The heat supplied heats the substance to its melting point, melts it, then raises the temperature of the
resulting liquid to 327°C. Use Eqs. (14-4) and (14-9).
Solution Compute the heat of fusion.
Q
31.15 kJ
Q = mLf + mc∆T , so Lf = −c∆T + = −[0.129 kJ (kg ⋅ K)](327 − 21) K +
= 22.8 kJ kg .
m
0.500 kg
603
Chapter 14: Heat
Physics
46. Strategy The rate of heat loss is ∆Q ∆t = Lv ∆m ∆t since Q = mLv for evaporation and where ∆m represents
the mass of water evaporated.
Solution Compute the rate of heat lost by the dog through panting.
∆Q
⎛ 1 min ⎞
= (670 min −1)(0.010 g)(2256 J g) ⎜
⎟ = 250 W
∆t
⎝ 60 s ⎠
47. Strategy The R-factor is equal to the quotient of the length of a material d and its thermal conductivity κ .
Solution
(a) Form a proportion.
Rcork
Rair
dcork
=1=
κ cork
dair
, so dcork =
κ air
κ cork
0.046 W (m ⋅ K)
dair =
(1.0 cm) = 2.0 cm .
κ air
0.023 W (m ⋅ K)
(b) Use the result from part (a). Replace cork with tin.
κ
66.8 W (m ⋅ K)
d tin = tin dair =
(0.010 m) = 29 m
κ air
0.023 W (m ⋅ K)
48. Strategy Use the latent heat of fusion for water to determine the rate of heat flow. Then, use Fourier’s law of heat
conduction to find the thermal conductivity of the rod and identify the metal.
Solution Determine the rate of heat flow. The temperature difference of 180°F is equal to 100 K.
Q mLv
=
∆t
∆t
Find the thermal conductivity.
dmLv
Q
∆T mLv
(1.10 m)(0.00132 kg)(333.7 × 103 J kg)
=κA
=
, so κ =
=
= 66.6 W (m ⋅ K) .
∆t
d
∆t
A∆T ∆t
π (0.0230 m 2) 2 (100 K)(175 s)
Referring to Table 14.5, we see that the thermal conductivity of tin is 66.8 W (m ⋅ K). No other value is close to
66.6 W (m ⋅ K) , so the metal rod is tin .
49. Strategy Use Eq. (14-12).
Solution Compute the thermal resistance for each material.
(a) R =
(b) R =
(c) R =
d
2.0 × 10−2 m
=
= 0.12 K W
κ A [0.17 W (m ⋅ K)](1.0 m 2 )
2.0 × 10−2 m
[80.2 W (m ⋅ K)](1.0 m 2 )
2.0 × 10−2 m
[401 W (m ⋅ K)](1.0 m 2 )
= 2.5 × 10−4 K W
= 5.0 × 10−5 K W
604
Physics
Chapter 14: Heat
50. Strategy Add the thermal resistances. Use Eqs. (14-12) and (14-13).
Solution Find the equivalent thermal resistance of the rods.
d
d
d⎛ 2
1 ⎞
ΣRn = Cu + Fe = ⎜
+
⎟ where d = d Fe .
κ Cu A κ Fe A A ⎝ κ Cu κ Fe ⎠
Find the rate of heat transfer.
∆T
A∆T
(6.0 × 10−6 m 2 )(100.0 K)
ᏼ=
=
=
= 0.14 W
2
1
⎡
⎤
ΣRn d 2 + 1
(0.25
m)
+
κ Cu κ Fe
⎢⎣ 401 W (m⋅K) 80.2 W (m⋅K) ⎥⎦
(
)
51. Strategy From the given information, ∆T = ᏼ1R1 = ᏼ 2 R2 = ᏼ( R1 + R2 ).
Solution Find the rate of heat flow per unit area.
R1 ᏼ 2
ᏼ R
ᏼ
ᏼ
20.0 W m 2
= 6.67 W m 2 .
=
and ᏼ = 2 2 . Thus, ᏼ = R 2 = ᏼ 2 =
20.0 + 1
1
2
R1 + R2
R2 ᏼ1
1
1
+
+
10.0
R
ᏼ
2
1
52. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution
(a) Find the temperature at the interface between the wood and cork if the cork is on the inside and the wood on
the outside. Set ᏼ w = ᏼ c.
∆T
∆T
κ w A w = κc A c
d
d
κ w (0.0°C − T ) = κ c (T − 20.0°C)
(0.0°C)κ w + (20.0°C)κ c = T (κ c + κ w )
(20.0°C)(0.046)
= 5.2°C
T=
0.046 + 0.13
(b) Find the temperature at the interface between the wood and cork if the wood is inside and the cork is outside.
κ w (20.0°C − T ) = κ c (T − 0.0°C)
(20.0°C)κ w = T (κ c + κ w )
(20.0°C)(0.13)
= 15°C
T=
0.046 + 0.13
(c)
The temperature at the interface differs for the two cases, but the total thermal resistance is
the same either way, so it doesn’t matter whether the cork is placed on the inside or the
outside of the wooden wall.
53. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Find the lowest temperature the dog can withstand without increasing its heat output.
T −T
dᏼ
(0.050 m)(51 W)
∆T
ᏼ=κA
= κ A i o , so To = Ti −
= 38°C −
= −37°C .
κA
d
d
[0.026 W (m ⋅ K)](1.31 m 2 )
54. Strategy Use Eq. (14-11).
Solution Find the required heat output.
37°C
∆T
ᏼ=
=
= 110 W
R
0.33 K W
605
Chapter 14: Heat
Physics
55. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Compute the rate of heat conduction for each situation.
(a) ᏼ = κ A
∆T
35°C − 4.0°C
= [0.040 W (m ⋅ K)](1.2 m 2 )
= 300 W
0.0050 m
d
(b) ᏼ = [0.60 W (m ⋅ K)](1.2 m 2 )
35°C − 4.0°C
= 4500 W
0.0050 m
56. Strategy Use Fourier’s law of heat conduction, Eq. (14-10), and Eqs. (14-11) and (14-12).
Solution The relevant quantities are
d
d
∆T
ᏼ g = κg A
, Rg =
, and Rf =
.
κg A
κf A
d
Find the effective thermal resistance.
d
d
d⎛ 1
1 ⎞
ΣRn =
+
= ⎜
+
⎟
κ g A κ f A A ⎜⎝ κ g κ f ⎟⎠
Form a proportion with the initial and final rates of heat flow.
ᏼ gf
ᏼg
=
∆T
d⎛ 1 1
⎜ +
A ⎜⎝ κ g κ f
κ g A ∆dT
⎞
⎟
⎟
⎠
=
1
κ g ⎜⎛ κ1 + κ1 ⎟⎞
⎝
g
f
=
⎠
1
κ
1 + κg
f
=
1
1.0
1 + 0.025
= 0.024
The conduction rate has been reduced by a factor of 0.024 .
57. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution
(a) ᏼ = κ A
(b)
∆T
104°C − 24°C
= [401 W (m ⋅ K)](1.0 × 10−6 m 2 )
= 0.32 W
d
0.10 m
∆T 104°C − 24°C
=
= 800 K m
d
0.10 m
(c) The effective length has doubled.
ᏼ = [401 W (m ⋅ K)](1.0 × 10−6 m 2 )
(d) The effective area has doubled.
ᏼ = [401 W (m ⋅ K)](2.0 × 10−6 m2 )
104°C − 24°C
= 0.16 W
0.20 m
104°C − 24°C
= 0.64 W
0.10 m
(e) Since the bars are identical, the temperature at the junction will be midway between the temperatures of the
baths.
104°C + 24°C
= 64°C
2
606
Physics
Chapter 14: Heat
58. (a) Strategy and Solution
The skier with the down jacket will stay warmer longer, since the jacket’s thermal
conductivity is lower and the jacket is thicker.
(b) Strategy Use Fourier’s law of heat conduction, Eq. (14-10). Form a proportion.
Solution
∆T
ᏼ down κ d A dd
κ d
(0.025)(0.50)
1
=
= d w =
=
∆
T
κ w dd (0.040)(2.0) 6.4
ᏼ wool κ w A d
w
The person with the down jacket can stay outside 6.4 times longer.
59. Strategy Use Wien’s law, Eq. (14-17).
Solution Compute the wavelength of the maximum intensity.
2.898 × 10−3 m ⋅ K
= 1.76 µm .
λmax T = 2.898 × 10−3 m ⋅ K, so λmax =
1650 K
60. Strategy Find a mathematical relationship that fits the set of data.
Solution When the temperature is 2000 K, the wavelength is 1.45 µm. When the temperature is 1500 K, the
wavelength is 1.9 µm. So, as the temperature decreases, the wavelength increases. This seems to imply an inverse
relationship, such as λ =
k
, where k is a constant given by k = λ1T1 = λ2T2 . Compute k.
T
λ1T1 = (1.45 × 10−6 m)(2000 K) = 2.9 × 10−3 m ⋅ K and λ2T2 = (1.9 × 10−6 m)(1500 K) = 2.9 × 10−3 m ⋅ K. The
products are the same, so the constant predicted by Wien is 2.9 × 10−3 m ⋅ K .
61. Strategy Use Stefan’s law of radiation, Eq. (14-16).
Solution Compute the power radiated by the bulb.
ᏼ = eσ AT 4 = 0.32[5.670 × 10−8 W (m 2 ⋅ K 4 )](1.00 × 10−4 m 2 )(3.00 × 103 K) 4 = 150 W
62. Strategy Use Stefan’s law of radiation, Eq. (14-16).
Solution Find the surface area of the filament.
ᏼ
40.0 W
=
= 4.8 × 10−5 m 2 .
ᏼ = eσ AT 4 , so A =
4
−8
2
4
3
4
0.32[5.670 × 10 W (m ⋅ K )](2.6 × 10 K)
eσ T
63. Strategy The person must lose heat at the same rate as it is produced and absorbed to maintain a constant body
temperature. Power is equal to intensity times area.
Solution Compute the rate of heat loss required.
Produced: 90 W
Absorbed: ᏼ = IA = (7.00 × 102 W m 2 )(0.57)(1.80 m 2 )(0.42) = 300 W
Rate of heat loss: 90 W + 300 W = 390 W
607
Chapter 14: Heat
Physics
64. Strategy Use Eq. (14-18).
Solution Find the rate at which the student “burns” calories.
ᏼ net = eσ A(T 4 − Ts 4 )
⎛ 1 kcal ⎞ ⎛ 3600 s ⎞
= 1.0[5.670 × 10−8 W (m 2 ⋅ K 4 )](1.7 m 2 )[(35 K + 273.15 K) 4 − (16 K + 273.15 K) 4 ] ⎜
⎟⎜
⎟
⎝ 4186 J ⎠ ⎝ 1 h ⎠
= 170 kcal h
65. Strategy Use Stefan’s law of radiation, Eq. (14-16). Form a proportion. Assume the cross-sectional area A is
constant.
Solution Find the temperature of the filament.
14
ᏼ = eσ AT 4 , so
4
⎛ᏼ ⎞
ᏼ 58 T58
= 4 and T58 = ⎜ 58 ⎟
ᏼ 60 T60
⎝ ᏼ 60 ⎠
14
⎛ 58.0 ⎞
T60 = ⎜
⎟
⎝ 60.0 ⎠
(2820 K) = 2800 K .
66. Strategy Use Wien’s law, Eq. (14-17).
Solution Compute the temperature of the blackbody.
2.898 × 10−3 m ⋅ K
λmax T = 2.898 × 10−3 m ⋅ K, so T =
= 1090 K .
2.65 × 10−6 m
67. Strategy Assume that the black wood stove is a blackbody (e = 1). Use Eq. (14-18).
Solution Find the net rate at which heat is radiated into the room.
ᏼ net = eσ A(T 4 − Ts4 ) = (1)[5.670 × 10−8 W (m 2 ⋅ K 4 )](1.20 m 2 )[(175 K + 273.15 K)4 − (20 K + 273.15 K)4 ]
= 2.24 kW
68. (a) Strategy Power is equal to intensity times area and the heat absorbed is given by Eq. (14-4).
Solution Find the rate of increase of the lizard’s temperature ∆T ∆t .
Q mc∆T
IA
∆T
=
= IA, so
=
=
∆t
∆t
∆t mc
1 (1.4 × 103
2
W m 2 )(1.6 × 10−4 m 2 )
(3.0 g)[4.2 J (g ⋅ °C)]
= 8.9 × 10−3 °C s .
(b) Strategy The time required to raise the temperature of the lizard is equal to the temperature difference
divided by the rate of temperature increase.
Solution
∆T1
5.0°C
∆t1 =
=
= 9.4 min
∆T ∆t 8.9 × 10−3 °C s
608
Physics
Chapter 14: Heat
69. Strategy Approximate the pots as cubes of similar volume. Use Eq. (14-18).
Solution Find the net rate of radiative heat loss from the two pots.
s3 = V , so s = V 1/ 3 and 6s 2 = A = 6V 2 / 3.
Coffeepot:
ᏼ net = eσ A(T 4 − Ts4 )
= 0.12[5.670 × 10−8 W (m 2 ⋅ K 4 )][6(1.00 L)2/3 (10−3 m3 L) 2 / 3 ][(98 K + 273.15 K) 4 − (25 K + 273.15 K)4 ]
= 4.5 W
Teapot:
ᏼ net = 0.65[5.670 × 10−8 W (m 2 ⋅ K 4 )][6(1.00 L) 2/3 (10−3 m3 L) 2 / 3 ][(98 K + 273.15 K) 4 − (25 K + 273.15 K)4 ]
= 24 W
70. Strategy Power is equal to intensity times area and the heat absorbed is given by Eq. (14-4).
Solution Find the rate of increase of the leaf’s temperature ∆T ∆t .
Q mc∆T
∆T
IA 0.700(9.00 × 102 W m 2 )(5.00 × 10−3 m 2 )
=
= IA, so
=
=
= 1.70 °C s
∆t
∆t
∆t mc
(0.500 g)[3.70 J (g ⋅ °C)]
71. (a) Strategy The power absorbed by the leaf must equal that radiated away. Power is equal to intensity times
area. Use Stefan’s law of radiation, Eq. (14-16).
Solution Absorbed:
I top A + eσ ATs4 = 0.700(9.00 × 102 W m 2 )(5.00 × 10−3 m 2 ) +
(1)[5.670 × 10−8 W (m 2 ⋅ K 4 )](5.00 ×10−3 m 2 )(273.15 K + 25.0 K) 4
= 5.39 W
Find the temperature of the leaf. The area is now 2(5.00 × 10−3 m 2 ) = 10.0 × 10−3 m 2 (both sides of the leaf).
ᏼ = eσ AT 4 , so
1/ 4
⎛ ᏼ ⎞
T =⎜
⎟
⎝ eσ A ⎠
1/ 4
⎡
⎤
5.39 W
=⎢
−8
−3 2 ⎥
2
4
⎣⎢ (1)[5.670 × 10 W (m ⋅ K )](10.0 × 10 m ) ⎦⎥
= 312 K − 273 K = 39°C .
(b) Strategy Since the bottom of the leaf absorbs and emits at the same rate, it can be ignored.
Solution Find the power per unit area that must be lost by other methods.
ᏼ abs,Sun ᏼ rad ᏼ other
ᏼ
=
+
= eσ T 4 + other , so
A
A
A
A
ᏼ
ᏼ other
abs,Sun
=
− eσ T 4 = 0.700(9.00 × 102 W m 2 ) − (1)[5.670 × 10−8 W (m 2 ⋅ K 4 )](273.15 K + 25.0 K)4
A
A
= 182 W m 2
Thus, the power per unit area that must be lost by other methods is 182 W m 2 .
72. Strategy Heat must be added to the room at the same rate that it is lost. Use Fourier’s law of heat conduction, Eq.
(14-10).
Solution Find the approximate rate at which heat must be added to the room.
∆T
75°F − 32°F ⎛ 1 K ⎞
ᏼ=κA
= [0.63 W (m ⋅ K)](1.56 m)(0.762 m)
⎜
⎟ = 1.38 kW
d
0.0130 m ⎝ 1.8°F ⎠
609
Chapter 14: Heat
Physics
73. Strategy Use Eq. (14-4) and the relationship between power and intensity.
Solution The energy provided by the sunlight is converted to heat in the water. The energy provided is
ᏼ∆t = IA∆t. Compute the time to heat the water.
mc∆T (1.0 L)(1000 g L)[4.186 J (g ⋅ K) ](100.0 − 15.0) K
Q = mc∆T = IA∆t , so ∆t =
=
= 320 s .
IA
(750 W m 2 )(1.5 m 2 )
74. Strategy The temperature of the ice must be raised to 0°C. Next, is must be melted. Then, the resulting liquid
water must be raised to 100°C. Finally, the water must be vaporized. Use Eqs. (14-4) and (14-9).
Solution Find the heat energy required to convert the ice to steam.
Q = mcice ∆Tice + mLf + mcwater ∆Twater + mLv = m(cice ∆Tice + Lf + cwater ∆Twater + Lv )
= 5(0.0220 kg){[2.1 kJ (kg ⋅ K)](50.0 K) + 333.7 kJ kg + [4.186 kJ (kg ⋅ K)](100.0 K) + 2256 kJ kg}
= 342 kJ
75. (a) Strategy The change in the internal energy of the bullet is equal to the initial kinetic energy of the bullet. Use
Eq. (14-4).
Solution Calculate the temperature increase of the bullet.
1
(4.00 × 102 m s) 2
v2
mc∆T = mv 2 , so ∆T =
=
= 180°C .
2
2c 2[0.44 × 103 J (kg ⋅ K)]
(b) Strategy The change in the internal energy of the bullet and block system is equal to the initial kinetic
energy of the bullet. Use Eq. (14-4).
Solution Calculate the equilibrium temperature T.
1
mFecFe ∆T + mw cw ∆T = (mFecFe + mw cw )∆T = mFev 2 , so
2
Tf =
=
mFev 2
+ Ti
2(mFecFe + mw cw )
(10.0 × 10−3 kg)(4.00 × 102 m s)2
2{(10.0 × 10−3 kg)[0.44 × 103 J (kg ⋅ K)] + (0.500 kg)[1680 J (kg ⋅ K)]}
+ 20.0°C = 20.9°C
76. Strategy The person must lose heat at the same rate as it is produced and absorbed to maintain a constant body
temperature. Refer to Problem 75. Use Eq. (14-9).
Solution Find the perspiration rate.
Produced: 90 W
Absorbed: ᏼ = IA = (7.00 × 102 W m 2 )(0.57)(1.80 m 2 )(0.42) = 300 W
Rate of heat loss: 90 W + 300 W = 390 W
So, heat must be carried away from the body by perspiration at a rate of 390 W.
⎞⎛ 1 L ⎞
390 J s ⎛ 3600 s ⎞ ⎛
1
Q mLv
m ᏼ
=
=
=
= 0.58 L h .
Q = mLv , so ᏼ =
, or
⎟⎟ ⎜ 3
⎜
⎟ ⎜⎜
3
∆t
∆t
∆t Lv 2430 J g ⎝ 1 h ⎠ ⎝ 1.0 g cm ⎠ ⎝ 10 cm3 ⎠⎟
610
Physics
Chapter 14: Heat
77. (a) Strategy Use Eqs. (14-4) and (14-9).
Solution Find the heat given up by the steam.
Q = − mcw ∆T + mLv = (4.0 g){ − [4.186 J (g ⋅ K)](45.0 − 100.0) K + 2256 J g} = 9.9 kJ
(b) Strategy Use Eq. (14-4).
Solution Compute the mass of the tissue.
Q
9945 J
m=
=
= 360 g
c∆T [3.5 J (g ⋅ K)](45.0 − 37.0) K
78. (a) Strategy Use Eq. (14-4).
Solution Find the heat given up by the water.
Q = − mc∆T = −(4.0 g)[4.186 J (g ⋅ K)](45.0 − 100.0) K = 920 J
(b) Strategy Use Eq. (14-4).
Solution Compute the mass of the tissue.
Q
920 J
m=
=
= 33 g
c∆T [3.5 J (g ⋅ K)](45.0 − 37.0) K
From Problem 93b, m = 360 g.
33 g
1
= 0.093 < , so less than 1 10 as much skin was involved .
355 g
10
79. Strategy Use Eq. (14-4), substituting for Q the expression given in the problem statement.
Solution Compute the temperature rise.
Q 0.544 × 10−3 J + (1.46 × 10−3 J cm)(1.5 cm)
Q = mc∆T , so ∆T =
=
= 6.5 × 10−3°C .
mc
(0.10 g)[4.186 J (g ⋅ K)]
80. Strategy The heat loss is given by ᏼ∆t and it is equal to Q = mLv .
Solution Find the mass of water required to replenish the fluid loss.
ᏼ∆t (650 W)(30.0 min)(60 s min)
m=
=
= 480 g
Lv
2430 J g
81. Strategy The heat loss is proportional to the temperature difference.
Solution Compute the increase in heat loss.
∆T
−8.0°C − (−18°C)
ᏼ 2 = 2 ᏼ1 =
ᏼ1 = 0.25ᏼ1
∆T1
22°C − (−18°C)
Therefore, ᏼ1 = 4.0ᏼ 2 ; thus, the heat loss would be 4.0 times higher without the insulation.
611
Chapter 14: Heat
Physics
82. Strategy Use Eqs. (14-4) and (14-9).
Solution First check to see if the ice completely melts.
Water:
Qw = mc∆T = (186 g)[4.186 J (g ⋅ K) ](0°C − 24°C) = −19 kJ
Ice:
mc∆T = 2(62 g)[2.1 J (g ⋅ K) ][0°C − (−15°C)] = 3.9 kJ
mLf = 2(62 g)(333.7 J g) = 41 kJ
Since 41 kJ + 3.9 kJ = 45 kJ > 19 kJ, the ice does not completely melt, so the final temperature is 0°C .
83. Strategy Use Eq. (14-4).
Solution Find the specific heat of granite.
Q
2.93 kJ
Q = mc∆T , so c =
=
= 0.792 kJ (kg ⋅ K) .
m∆T (0.500 kg)(7.40 K)
84. Strategy The potential energy of the spring is equal to 12 kx 2. Use Eq. (14-4).
Solution Find the temperature change of the water.
1
kx 2 (8.4 × 103 N m)(0.10 m)2
Q = mc∆T = ∆U = kx 2 , so ∆T =
=
= 0.010°C .
2
2mc 2(1.0 kg)[4186 J (kg ⋅ K) ]
85. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Determine the rate of heat flow.
∆T
327 K − 37 K
ᏼ=κA
= [67.5 W (m ⋅ K)]π (0.0130 m)2
= 10.4 W
d
1.00 m
86. Strategy Heat flows from the iron to the water. Use Eq. (14-4).
Solution Find the mass of 20.0°C water required to cool the iron to 23.0°C.
0 = Qw + QFe = mw cw ∆Tw + mFecFe∆TFe , so
m c ∆T
(0.38 kg)[0.44 kJ (kg ⋅ K) ](23.0°C − 498°C)
= 6.3 kg .
mw = − Fe Fe Fe = −
[4.186 kJ (kg ⋅ K) ](23.0°C − 20.0°C)
cw ∆Tw
87. Strategy Use Fourier’s law of heat conduction, Eq. (14-10). The rate the student “burns” calories is equal to the
rate of heat conduction through the 3.0-mm thickness of water right next to his skin.
Solution Compute the rate of heat flow.
∆T
⎡ (35 − 16) K ⎤ ⎛ 1 kcal ⎞ ⎛ 3600 s ⎞
ᏼ=κA
= [0.58 W (m ⋅ K)](1.7 m 2 ) ⎢
⎟⎜
⎟ = 5400 kcal h
⎥⎜
d
⎣ 0.0030 m ⎦ ⎝ 4186 J ⎠ ⎝ 1 h ⎠
612
Physics
Chapter 14: Heat
88. (a) Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Find the temperature at the copper-steel interface.
Set ᏼ st = ᏼ Cu .
κ st A
∆Tst
∆T
κ d
(401)(0.350)
= κ Cu A Cu , so ∆Tst = Cu st ∆TCu =
∆TCu = 20.3∆TCu .
κ st dCu
dst
dCu
(46.0)(0.150)
Since ∆Tst + ∆TCu = ∆T = 4.00°C, ∆TCu (20.3406 + 1) = 4.00°C, or ∆TCu = 0.187°C.
Thus, T = 104.00°C − 0.1874°C = 103.81°C.
∆m
. The rate at which heat enters the
∆t
∆Q ∆mLv
∆m κ A∆T
∆Q
∆T
⎛ ∆m ⎞
=
= Lv ⎜
water is
=ᏼ=κA
, and
⎟ . So, ∆t = dL . Use the values for steel.
∆t
∆t
∆t
d
⎝ ∆t ⎠
v
(b) Strategy and Solution The rate at which the water evaporates is
(
)
2
0.180 m (3.8126°C)
∆m [46.0 W (m ⋅ K)]π
2
=
= 0.565 g s
∆t
(0.00350 m) ( 2256 J g )
89. Strategy Gravitational potential energy is converted into internal energy. Use Eq. (14-9) and U = mgh.
Solution Find the mass of the ice melted by friction.
0.75mgh 0.75(75 kg)(9.80 m s 2 )(2.43 m)
Q = mm Lf = 0.75U = 0.75mgh, so mm =
=
= 4.0 g .
Lf
333,700 J kg
90. Strategy The basal metabolic rate is the minimal energy intake necessary to sustain life in a state of complete
inactivity.
Solution Calculate the BMR kg of body mass and BMR m 2 of surface area for each animal.
Animal
(a) BMR/kg
(b) BMR/m 2
Mouse
210
1200
Dog
51
1000
Human
32
1000
Pig
18
1000
Horse
11
960
(a) According to the table, since BMR kg is larger for smaller animals, it is true that smaller animals must
consume more food per kilogram of body mass.
(c) When an animal is resting, the food energy metabolized must be shed as heat (no work). Since
radiative loss depends upon surface area , BMR m 2 must be approximately the same for different-sized
animals.
613
Chapter 14: Heat
Physics
91. Strategy Use Eq. (14-18).
Solution
(a) Compute the net rate of heat loss through radiation.
ᏼ net = eσ A(T 4 − Ts4 )
= 0.97[5.670 × 10−8 W (m 2 ⋅ K 4 )](2.20 m 2 )[(273.15 K + 37.0 K)4 − (273.15 K + 23.0 K)4 ] = 190 W
(b) Find the skin temperature such that the net heat loss due to radiation is equal to the basal metabolic rate.
eσ A(T 4 − Ts4 ) = BMR, so
1/ 4
⎛ BMR
⎞
T =⎜
+ Ts4 ⎟
⎝ eσ A
⎠
(
)
1/ 4
⎡ (2167 kcal day)(4186 J kcal) 1 day
⎤
86,400 s
⎢
4⎥
=⎢
+ (273.15 K + 23.0 K) ⎥
2
4
2
−8
⎢⎣ 0.97[5.670 × 10 W (m ⋅ K )](2.20 m )
⎥⎦
− 273.15 K
= 31°C
(c) Wearing clothing slows heat loss by radiation because air layers trapped between clothing layers act as
insulation and thus reduce the net radiative heat loss. (T 4 − Ts4 is reduced.)
92. Strategy Set the rates of heat conduction equal. Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Compute the ratio of the depths.
∆T
∆T
d
κ
3.1
= κs A
ᏼ = κb A
, so b = b =
= 1.3, or d b = 1.3ds .
db
ds
ds κ s 2.4
93. Strategy W = Fd (work) and P = Fv (power) so W = Pd v . Use Eq. (14-4).
Solution Find the distance that the cheetah can run before it overheats.
0.700W = Q
0.700 Pd
= mc∆T
v
3
1h
vmc∆T (110 × 10 m h) 3600 s (50.0 kg)[3500 J (kg ⋅ °C)](41.0°C − 38.0°C)
d=
=
= 140 m
0.700P
0.700(160,000 W)
(
)
94. Strategy The shaking will heat the water, although very slowly. The work done by the scientist is converted to
heat in the water. The rate at which energy is supplied by the shaking is equal to the gravitational potential energy
lost by the water during each fall times the frequency of the shaking. Use Eq. (14-4).
Solution Find the time it will take to heat the water.
mc∆T
c∆T
[4186 J (kg ⋅ K) ](87°C − 12°C)
= mghf , so ∆t =
=
= 2 days .
∆t
ghf
(9.80 m s 2 )(0.333 m)(30 min −1)(1440 min day)
614
Physics
Chapter 14: Heat
95. Strategy Heat flows from the copper block to the water and iron pot. Use Eq. (14-4).
Solution Find the final temperature of the system.
0 = Qw + QCu + QFe
Qw = −QCu − QFe
mw cw (Tf − Ti ) = −mCu cCu (Tf − TCu ) − mFecFe (Tf − Ti )
Tf (mw cw + mCu cCu + mFecFe ) = mCu cCuTCu + (mFecFe + mw cw )Ti
Solve for Tf .
mCu cCuTCu + (mFecFe + mw cw )Ti
mw cw + mCu cCu + mFecFe
(2.0 kg)[385 J (kg ⋅ K) ](100.0°C) + {(2.0 kg)[440 J (kg ⋅ K) ] + (1.0 kg)[4186 J (kg ⋅ K) ]}(25.0°C)
=
= 35°C
(1.0 kg)[4186 J (kg ⋅ K) ] + (2.0 kg)[385 J (kg ⋅ K) ] + (2.0 kg)[440 J (kg ⋅ K) ]
Tf =
96. Strategy Heat flows from the gold to the water and copper pot. Use Eq. (14-4).
Solution Find the final temperature of the system.
0 = Qw + QAu + QCu
= mw cw (Tf − Ti ) + mAu cAu (Tf − TAu ) + mCu cCu (Tf − Ti )
= Tf (mw cw + mAu cAu + mCu cCu ) − mAu cAuTAu − (mw cw + mCu cCu )Ti
m c T + (mw cw + mCu cCu )Ti
Tf = Au Au Au
mw cw + mAu cAu + mCu cCu
(0.250 kg)[128 J (kg ⋅ K) ](75.0°C) +
{(0.500 L)(1.00 kg L)[4186 J (kg ⋅ K) ] + (1.500 kg)[385 J (kg ⋅ K) ]}(22.0°C)
= 22.6°C
=
(0.500 L)(1.00 kg L)[4186 J (kg ⋅ K) ] + (0.250 kg)[128 J (kg ⋅ K) ] + (1.500 kg)[385 J (kg ⋅ K) ]
97. (a) Strategy The work done by each animal is proportional to the heat generated. Use Eq. (14-4). Form a
proportion.
Solution Compare the temperature changes.
Qc 2.00mc∆Tc
0.700
=
=
= 14.0, so ∆Tc = 7.00∆Td .
Qd
mc∆Td
0.0500
The temperature change of the cheetah is 7.00 times higher than that of the dog.
(b) Strategy Use the result from part (a).
Solution Find the final temperature of the dog.
Tc − Ti = 7.00(Td − Ti )
Tc + 6.00Ti = 7.00Td
T + 6.00Ti 40.0°C + 6.00(35.0°C)
Td = c
=
= 35.7°C
7.00
7.00
The dog is a much better regulator of temperature and, as a result, has more endurance.
615
Chapter 14: Heat
Physics
98. Strategy If all the kinetic energy of the bullet is converted into heat, then Q = 12 mv 2 , where Q is the minimum
heat required to raise the temperature of the bullet to its melting point and melt it and, therefore, v is the minimum
speed of the bullet. Use Eqs. (14-4) and (14-9).
Solution Find the minimum required speed.
1
Q = mLf + mc∆T = mv 2 , so
2
v = 2( Lf + c∆T ) = 2{22.9 J g + [0.13 J (g ⋅ K)](327°C − 87.0°C)}(103 g kg) = 330 m s .
99. Strategy Heat flows from the tetrachloromethane to the water. Use Eq. (14-4).
Solution Find the specific heat of CCl4 , ct .
0 = Qw + Qt = mw cw ∆Tw + mt ct ∆Tt , so
m c ∆T
(2.00 kg)[4.186 kJ (kg ⋅ K)](18.54 − 18.00) K
ct = − w w w = −
= 0.84 kJ/(kg ⋅ K) .
mt ∆Tt
(2.50 × 10−1 kg)(18.54 − 40.00) K
100. Strategy Heat flows from the drink to the ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of the drink required to just melt the ice.
0 = Qdrink + Qice = mdrink c∆T + mice Lf , so
m L
(0.10 kg)(333.7 kJ kg)
mdrink = − ice f = −
= 0.32 kg .
c∆T
[4.186 kJ (kg ⋅ K) ](0°C − 25°C)
101. Strategy Multiply the energy required to melt the urethane by the molar mass and divide by the total mass to find
the latent heat.
Solution Find the latent heat of fusion of urethane.
(17.10 kJ)[3(12.011) + 7(1.00794) + 2(15.9994) + 14.00674] g mol
1.00 × 102 g
= 15.2 kJ mol
102. Strategy Heat flows from the bullet to the ice. Assume that all the kinetic energy goes into heating the bullet and
ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of ice that melts.
1
0 = mice Lf + mPbcPb∆T + mPb (vf2 − vi2 ), so
2
mPb ⎛ 1 2
0.0200 kg ⎡ 1
⎞
⎤
mice =
vi − cPb ∆T ⎟ =
(5.00 × 102 m s)2 − [130 J (kg ⋅ K)](0°C − 47.0°C) ⎥ = 7.86 g .
⎜
⎢
Lf ⎝ 2
⎠ 333,700 J kg ⎣ 2
⎦
616
Chapter 15
THERMODYNAMICS
Conceptual Questions
1. Yes, but it wouldn’t be a very good heat pump. Like an electric heater, the heat output would be equal to the work
input, with no heat being taken from the cold reservoir.
2. That one process uses fuel more efficiently than another does not guarantee that it is a “better” process. In
comparing electrical and gas heating systems, one has to consider not only the efficiency of the heating apparatus,
but also the efficiency of the process by which fuel is produced. The most common method of producing electrical
energy is via the burning of fossil fuels. The electrical energy must then be transported from the plant at which it
was produced over inefficient transmission lines. A gas furnace may therefore be a “better” system despite the
fact that the furnace itself is less efficient.
3. In terms of a heat engine, “win” would mean that you can get more work out than the heat input. According to the
first law, the work output must be less than or equal to the heat input. “Break even” means that the work output is
equal to the heat input. An engine that “breaks even” has 100% efficiency, which is only possible at T = 0
according to the second law.
4. Imagine that two reversible heat engines are operating between the same two reservoirs with the efficiency of the
first engine greater than the second. If the second engine is reversed so that it acts as a heat pump, there is a net
flow of heat out of the cold reservoir and into the hot one. Without any external work input, heat is flowing from
cold to hot, violating the second law of thermodynamics. Therefore, all reversible engines operating between the
same two reservoirs must have identical efficiencies.
5. Energy is always conserved, so there would be just as much energy as before when the fossil fuels are exhausted.
There wouldn’t be as much high quality or useful energy though, so it would be better to call it something like a
“high quality energy crisis.”
6. The kitchen would get warmer. The total energy of the room increases because the refrigerator draws electrical
power. Without a way for energy to leave the room, the temperature must increase.
7. The coefficient of performance for a heat pump decreases as the temperature of the cold reservoir (the outside air)
decreases. For real, non-ideal heat pumps, the coefficient of performance eventually falls below one, meaning that
not as much heat is delivered to the inside as work is input to the pump. At this point it is more efficient to use an
electric heater, which delivers an amount of heat exactly equal to the amount of work input.
8. Heat pumps are most efficient when the difference in temperature between the inside and outside is not too great.
If the outside temperature is extremely low, it becomes more efficient to use an electric heater than a heat pump.
9. No, entropy changes don’t require a flow of heat. For example, when a gas expands freely into a vacuum, its
entropy increases but there is no heat flow. Beating an egg is another example of a process that increases entropy
with no flow of heat.
10. No, there is always some heat output to the cold reservoir, making it warmer eventually. The only way to have no
heat going into the cold reservoir would be if its temperature were absolute zero, but it is impossible to cool
anything completely down to absolute zero.
11. The entropy of the lemonade decreases while the entropy of the ice increases. The increase in entropy of the ice is
greater than the decrease in entropy of the lemonade, so the total entropy for the process increases.
617
Chapter 15: Thermodynamics
Physics
12. The claim cannot be true. Such units would violate the 2nd law. Work is done on the air conditioner and it
removes heat from the room. There must be a hot reservoir into which heat is exhausted from the room. If the
room were used as the reservoir, the room would not be cooled.
13. This is not a violation. Although the salt crystals are in a more ordered state, the surroundings of the bucket are in
a more disordered state. The gaseous state of the evaporated sea water is less ordered than the water.
14. A gas held at constant volume cannot do any work on its surroundings, while at constant pressure a gas does work
as it expands during heating, for example. In the latter case, some of the added heat energy must go into the work
done by the gas, in order to conserve energy. This is why the two specific heats are not the same. You need to add
more heat to a gas at constant pressure to achieve a given temperature increase than to a gas a constant volume.
For solids and liquids, the change in volume (hence, the work done) upon heating or cooling is usually very small,
so the two kinds of specific heats are not very different.
Problems
1. Strategy The work done by Ming is equal to the magnitude of the force of friction f = µ N times the total
“rubbing” distance. Use the first law of thermodynamics.
Solution Find the change in internal energy.
∆U = Q + W = 0 + µ Nd = 0.45(5.0 N)[8(0.16 m)] = 2.9 J
2. Strategy Use the first law of thermodynamics. Q > 0 and W < 0.
Solution Find the change in internal energy.
∆U = Q + W = 550 J − 840 J = −290 J
3. Strategy Use the first law of thermodynamics. ∆U > 0 and W > 0.
Solution Find the heat flow.
Q = ∆U − W = 400 J − 500 J = −100 J
100 J of heat flows out of the system.
4. Strategy Use the first law of thermodynamics and the work-kinetic energy theorem. Q > 0 and W < 0.
Solution Find the change in internal energy.
1
1
∆U = Q + W = Q − ∆K = Q − mvf2 = 135 J − (9.00 kg)(3.00 m s)2 = 95 J
2
2
5. Strategy No work is done during the constant volume process, but work is done during the constant pressure
process. Use Eq. (15-3).
Solution Compute the total work done by the gas.
W = Pi ∆V = (2.000 atm)(1.013 × 105 Pa atm)(2.000 L − 1.000 L)(10−3 m3 L) = 202.6 J
6. Strategy No work is done during the constant volume process, but work is done during the constant pressure
process. Use Eq. (15-3).
Solution Compute the total work done by the gas.
W = Pi ∆V = (1.000 atm)(1.013 × 105 Pa atm)(2.000 L − 1.000 L)(10−3 m3 L) = 101.3 J
618
Physics
Chapter 15: Thermodynamics
7. (a) Strategy Use the ideal gas law. Refer to the figure.
Solution According to the graph, the pressure at point C is 98.0 kPa . Find the temperature at point C.
PV = nRT , so T =
PV (98.0 × 103 Pa)(2.00 L)(10−3 m3 L)
=
= 1180 K .
nR
(0.0200 mol)[8.314 J (mol ⋅ K) ]
(b) Strategy This is a constant volume process. Use Eq. (15-6) and the ideal gas law.
Solution Find the change in internal energy of the gas.
⎛ 3 ⎞ ⎛ P V PV ⎞ 3V
∆U = nCv ∆T = n ⎜ R ⎟ ⎜ f − i ⎟ =
( Pf − Pi )
⎝ 2 ⎠ ⎝ nR nR ⎠ 2
=
3(1.00 L)(10−3 m3 L)
(98.0 × 103 Pa − 230 × 103 Pa) = −200 J
2
(c) Strategy The work done per cycle is equal to the area contained within the curve.
Solution Compute the work done per cycle.
1
W = (230 × 103 Pa − 98.0 × 103 Pa)(2.00 L − 1.00 L)(10−3 m3 L) = 66 J
2
(d) Strategy and Solution At the beginning and end of a complete cycle, regardless of the starting point, the
temperature is the same, so ∆U = 0 because ∆T = 0 in a cycle.
8. Strategy For an isothermal process, ∆U = 0. 5.00 kJ of work is done on the gas. Use the first law of
thermodynamics.
Solution Find the heat flow.
Q = ∆U − W = 0 − W = −5.00 kJ
Since Q < 0, the heat flows out of the gas and into the reservoir .
9. (a) Strategy Oxygen gas is diatomic. Use Eqs. (15-7) and (15-9).
Solution Find the heat absorbed by the gas.
7
7
⎛5
⎞
Q = nCp ∆T = n(Cv + R)∆T = n ⎜ R + R ⎟ ∆T = nR∆T = (1.00 mol)[8.314 J (mol ⋅ K) ](25.0°C − 10.0°C)
2
2
2
⎝
⎠
= 436 J
(b) Strategy Use the ideal gas law.
Solution Find the change in volume of the gas.
nR∆T (1.00 mol)[8.314 J (mol ⋅ K) ](25.0°C − 10.0°C) ⎛ 3 L ⎞
∆V =
=
⎜ 10
⎟ = 1.23 L
P
(1.00 atm)(1.013 × 105 Pa atm)
m3 ⎠
⎝
(c) Strategy Use Eq. (15-2).
Solution Find the work done by the gas.
W = P∆V = (1.00 atm)(1.013 × 105 Pa atm)(1.23 × 10−3 m3 ) = 125 J
619
Chapter 15: Thermodynamics
Physics
(d) Strategy Use the first law of thermodynamics.
Solution Calculate the change in internal energy of the gas.
∆U = Q + W = 436.5 J − 124.7 J = 312 J
10. (a) Strategy No work is done during the constant volume process, but work is done during the constant pressure
process. Use Eq. (15-3).
Solution Compute the total work done on the gas.
W = − Pi ∆V = −(1.000 atm)(1.013 × 105 Pa atm)(16.00 L − 4.00 L)(10−3 m3 L) = −1216 J
(b) Strategy Use the first law of thermodynamics and the ideal gas law, as well as Eqs. (15-3), (15-6), and
(15-7).
Solution Calculate the total change in internal energy of the gas and the total heat flow into the gas during
the entire process.
A to B (constant volume):
3
∆U1 = nCV ∆T1 = nR∆T1
2
3
∆PV
, so ∆U1 = ∆PV .
Using the ideal gas law, ∆T1 =
2
nR
B to C to D (constant pressure):
5
∆U 2 = Q + W = nCP ∆T2 − Pi ∆V = nR∆T2 − Pi∆V
2
5
3
Pi ∆V
Using the ideal gas law, ∆T2 =
, so ∆U 2 = Pi ∆V − Pi ∆V = Pi ∆V .
2
2
nR
The total change in internal energy is
3
3
∆U = ∆U1 + ∆U 2 = ∆PV + Pi∆V
2
2
3
= [(1.000 atm − 2.000 atm)(4.000 L) + (1.000 atm)(16.00 L − 4.000 L)](1.013 × 105 Pa atm)(10−3 m3 L)
2
= 1216 J
The total heat flow is Q = ∆U − W = 1215.6 J + 1215.6 J = 2431 J .
11. (a) Strategy For A–C (constant temperature), W = nRT ln Vi Vf , and for C –D (constant pressure),
W = − Pi ∆V . Use the ideal gas law to find T.
Solution
Vi
⎛PV ⎞ V
− P∆V = nR ⎜ A A ⎟ ln A − Pi∆V
Vf
⎝ nR ⎠ VC
4.000 L
⎡
⎤
= ⎢ (2.000 atm)(4.000 L) ln
− (1.000 atm)(16.000 L − 8.000 L) ⎥ (1.013 × 105 Pa atm)(10−3 m3 L)
8.000 L
⎣
⎦
= −1372 J
Wtotal = nRT ln
620
Physics
Chapter 15: Thermodynamics
(b) Strategy For constant temperature, ∆U = 0. For constant pressure,
5 ⎛ P ∆V ⎞
3
∆U = Q + W = nCP ∆T − Pi ∆V = nR ⎜ i
− Pi ∆V = Pi∆V .
2 ⎝ nR ⎟⎠
2
Solution
3
∆U = (1.000 atm)(16.000 L − 8.000 L)(1.013 × 105 Pa atm)(10−3 m3 L) = 1216 J .
2
The total heat flow is Q = ∆U − W = 1216 J + 1372 J = 2588 J .
12. (a) Strategy For A–E (constant pressure), W = − Pi ∆V , and for E –D (constant temperature),
W = nRT ln Vi Vf . Use the ideal gas law to find T.
Solution
Wtotal = − Pi ∆V + nRT ln
Vi
⎛PV
= − Pi∆V + nR ⎜ E E
Vf
⎝ nR
⎞ VE
⎟ ln V
⎠
D
8.000 L ⎤
⎡
5
3
−3
= ⎢ −(2.000 atm)(8.000 L − 4.000 L) + (2.000 atm)(8.000 L) ln
⎥ (1.013 × 10 Pa atm)(10 m L)
16.00
L
⎣
⎦
= −1934 J
(b) Strategy For constant temperature, ∆U = 0. For constant pressure,
5 ⎛ P ∆V ⎞
3
∆U = Q + W = nCP ∆T − Pi ∆V = nR ⎜ i
− Pi ∆V = Pi∆V .
⎟
2 ⎝ nR ⎠
2
Solution
3
∆U = (2.000 atm)(8.000 L − 4.000 L)(1.013 × 105 Pa atm)(10−3 m3 L) = 1216 J .
2
The total heat flow is Q = ∆U − W = 1215.6 J + 1933.85 J = 3149 J .
13. (a) Strategy The net work done in one cycle is equal to the area inside the graph.
Solution Compute the net work done per cycle.
W = (4.00 atm − 1.00 atm)(1.013 × 105 P a atm)(0.800 m3 − 0.200 m3 ) = 182 kJ
(b) Strategy and Solution The net heat flow into the engine is equal to the work done per cycle, so
Qnet = 182 kJ .
14. Strategy Use Eq. (15-12).
Solution Find the efficiency of the generator.
W
1.17 kW ⋅ h ⎛ 3.600 × 106 J ⎞
e = net =
⎜
⎟ = 0.628
Qin
6.71× 106 J ⎜⎝ 1 kW ⋅ h ⎟⎠
621
Chapter 15: Thermodynamics
Physics
15. Strategy Use Eq. (15-15) and conservation of energy.
Solution Find the amount of heat taken from the cold reservoir.
Q
Q
K p = H , so Wnet = H . By conservation of energy, Wnet = QH − QC , or
Wnet
Kp
QC = QH − Wnet = QH −
⎛
1
= QH ⎜ 1 −
⎜
Kp
⎝ Kp
QH
⎞
1 ⎞
⎛ 3600 s ⎞⎛
⎟ = (7.81× 103 W)(10.0 h) ⎜
⎟⎜ 1 −
⎟ = 240 MJ .
⎟
⎝ 1 h ⎠⎝ 6.85 ⎠
⎠
16. (a) Strategy Use Eq. (15-12).
Solution Find the heat absorbed by the engine.
W
1.00 × 103 J
QH = net =
= 3.00 kJ
e
0.333
(b) Strategy The net work done by an engine during one cycle is equal to the net heat flow into the engine
during the cycle.
Solution Find the heat exhausted by the engine.
Wnet = QH − QC , so QC = QH − Wnet = 3.00 kJ − 1.00 kJ = 2.00 kJ .
17. (a) Strategy Use Eq. (15-12).
Solution Find the net work done by the engine.
Wnet = eQH = 0.21(1.00 kJ) = 210 J
(b) Strategy The net work done by an engine during one cycle is equal to the net heat flow into the engine
during the cycle.
Solution Find the heat released by the engine.
Wnet = QH − QC , so QC = QH − Wnet = 1.00 × 103 J − 210 J = 790 J .
18. Strategy The net rate of work done by the engine is ∆K ∆t .
Solution The net rate of work done is
2
1
Wnet ∆K 2 mv
mv 2
=
=
=
.
∆t
∆t
∆t
2 ∆t
According to the definition of efficiency, the rate of heat flow in at the high temperature is
Qin 1 Wnet mv 2 (1800 kg)(27 m s)2
=
=
=
= 2.6 × 105 W .
∆t e ∆t
2 e ∆t
2(0.27)(9.5 s)
Referring to the result of Example 15.5, the rate of heat flow out at the low temperature is
Qout Wnet ⎛ 1 ⎞ (1800 kg)(27 m s) 2 ⎛ 1
⎞
=
− 1⎟ = 1.9 × 105 W .
⎜ − 1⎟ =
⎜
∆t
∆t ⎝ e ⎠
2(9.5 s)
⎝ 0.27 ⎠
622
Physics
Chapter 15: Thermodynamics
19. (a) Strategy Since 5.0 × 1016 J of electric energy is generated by power plants with an average efficiency of
5.0
0.30, Qin for the power plants is
× 1016 J. The heat dumped into the environment daily is Qout . Use
0.30
conservation of energy.
Solution
⎛ 5.0
Qout = Qin − Wnet = Qin − eQin = Qin (1 − e) = ⎜
× 1016
0.30
⎝
⎞
J ⎟ (1 − 0.30) = 1.2 × 1017 J
⎠
(b) Strategy Use Eq. (14-4).
Solution
Q
1.2 × 1017 J
m=
=
= 1.4 × 1013 kg
c∆T [4186 J (kg ⋅ K) ](2.0°C)
20. Strategy The power of the solar power plant is equal to the intensity of the sunlight times the area of the
collectors times the efficiency of the collectors.
Solution Find the area required.
P
1.0 × 109 W
eIA = P, so A =
=
= 2.5 × 107 m 2 or 25 km 2 .
eI 0.200(0.20 × 103 W m 2 )
21. Strategy Use Eq. (15-12). The net work done by an engine during one cycle is equal to the net heat flow into the
engine during the cycle.
Solution Find the efficiency of the engine.
W
Wnet
1
1
e = net =
=
=
= 0.182
0.450
kJ
Q
out
Qin Wnet + Qout 1 +
1+
Wnet
0.100 kJ
22. Strategy The work done by the engine is equal to the increase in gravitational potential energy of the crate plus
the increase in kinetic energy. Use Eq. (15-12).
Solution Find the required heat input.
2
1
Wnet mgh + 2 mv
m⎛
v 2 ⎞ 5.00 kg ⎡
(4.00 m s) 2 ⎤
=
= ⎜ gh + ⎟ =
Qin =
(9.80 m s 2 )(10.0 m) +
⎢
⎥ = 1770 J
e
e
e ⎜⎝
2 ⎟⎠
0.300 ⎣⎢
2
⎦⎥
23. Strategy The net work is equal to the electricity supplied. Use Eq. (15-15).
Solution Find the heat delivered by the heat pump.
Q
K p = H , so QH = K pWnet = (3.0)(1.00 kJ) = 3.0 kJ .
Wnet
24. Strategy Use Eq. (15-16).
Solution Find the daily cost to run the air conditioning unit.
Q
Wnet = C
Kr
⎛ $0.10 ⎞
⎛ $0.10 ⎞ QC ⎛ $0.10 ⎞ 1.73 × 108 J ⎛ 1 kW ⋅ h ⎞
⎜ kW ⋅ h ⎟ Wnet = ⎜ kW ⋅ h ⎟ K = ⎜ kW ⋅ h ⎟
⎜
⎟ = $2.40
2.00
⎝
⎠
⎝
⎠ r ⎝
⎠
⎝ 3.6 × 106 J ⎠
623
Chapter 15: Thermodynamics
Physics
25. Strategy Use Eq. (15-17).
Solution Find the temperature of the cold reservoir.
T
er = 1 − C , so TC = TH (1 − er ) = (622 K)(1 − 0.725) = 171 K .
TH
26. (a) Strategy Use Eq. (15-12). The net work done by an engine during one cycle is equal to the net heat flow into
the engine during the cycle.
Solution Compute the efficiency of the engine.
W
Q − Qout
Q
82
e = net = in
= 1 − out = 1 −
= 0.34
Qin
Qin
Qin
125
(b) Strategy Use Eq. (15-17).
Solution Compute the efficiency of an ideal engine.
T
293 K
er = 1 − C = 1 −
= 0.640
TH
815 K
27. Strategy The minimum amount of heat is discharged when the steam engine is reversible. Use Eqs. (15-12) and
(15-17), and Wnet = QH − QC .
Solution Compute the efficiency of a reversible engine.
T
273.15 K + 27 K
er = 1 − C = 1 −
= 0.250
TH
273.15 K + 127 K
Compute the minimum amount of heat discharged.
⎛1
⎞
W
⎛ 1
⎞
QC = QH − Wnet = net − Wnet = Wnet ⎜⎜ − 1⎟⎟ = (8.34 kJ) ⎜
− 1⎟ = 25.0 kJ
er
e
0.250
⎝
⎠
⎝ r
⎠
28. Strategy The maximum efficiency is that of a reversible heat engine. Use Eq. (15-17).
Solution Calculate the maximum possible efficiency.
T
273.15 K + 4.0 K
er = 1 − C = 1 −
= 0.0481
TH
273.15 K + 18.0 K
29. Strategy Assume constant rates and reversibility. Use Eq. (15-17) and conservation of energy. P = Wnet ∆t .
Solution Compute the efficiency.
T
273.15 K + 2.0 K
er = 1 − C = 1 −
= 0.1213
TH
273.15 K + 40.0 K
Find the power used.
Q ∆t 0.10 × 103 W
W
Q
W ⎛1 ⎞
⎛1 ⎞
=
= 14 W .
QC = QH − Wnet = net − Wnet = Wnet ⎜ − 1⎟ , so C = net ⎜ − 1⎟ , and P = 1C
1 −1
∆t
∆t ⎝ e ⎠
e
−
1
⎝e ⎠
e
0.1213
30. Strategy The pump requires the minimum possible work if it is reversible. Use Eqs. (15-15) and (15-19).
Solution
Wnet =
⎛ T ⎞
QH
273.15 K − 10.0 K ⎞
⎛
= QH ⎜ 1 − C ⎟ = (1.0 × 103 J) ⎜ 1 −
⎟ = 100 J
K p,rev
⎝ 273.15 K + 20.0 K ⎠
⎝ TH ⎠
624
Physics
Chapter 15: Thermodynamics
31. Strategy Use Eq. (15-17) and Wnet = QH − QC .
Solution Find the waste heat exhausted.
Coal:
T
273.15 K + 27 K
er = 1 − C = 1 −
= 0.700 and
TH
273.15 K + 727 K
QC = QH − Wnet =
Wnet
⎛1 ⎞
⎛ 1
⎞
− Wnet = Wnet ⎜ − 1⎟ = (1.00 MJ) ⎜
− 1⎟ = 0.43 MJ.
e
⎝e ⎠
⎝ 0.700 ⎠
Nuclear:
T
273.15 K + 27 K
⎛ 1
⎞
er = 1 − C = 1 −
= 0.625 and QC = (1.00 MJ) ⎜
− 1⎟ = 0.60 MJ.
TH
273.15 K + 527 K
⎝ 0.625 ⎠
The coal-fired plant and the nuclear plant exhaust 0.43 MJ and 0.60 MJ of heat, respectively.
32. Strategy The power output is directly proportional to the efficiency. Use Eq. (15-17). Form a proportion to find
the power output of the other engine.
Solution Find the power output.
⎛ T ⎞
P
e
= = e ⎜⎜ 1 − C ⎟⎟
Pr er
⎝ TH ⎠
−1
⎛ T ⎞
, so P = ePr ⎜⎜ 1 − C ⎟⎟
⎝ TH ⎠
−1
⎛ 350 ⎞
= 0.42(2.3 × 104 W) ⎜1 −
⎟
⎝ 750 ⎠
−1
= 1.8 × 104 W .
33. (a) Strategy Use Eq. (15-17).
Solution Find the efficiency of the reversible engine.
T
273.15 K + 300.0 K
er = 1 − C = 1 −
= 0.3436
TH
273.15 K + 600.0 K
(b) Strategy Use Eq. (15-18).
Solution Find the amount of heat exhausted to the cold reservoir.
QC TC
T
273.15 K + 300.0 K
=
, so QC = C QH =
(420.0 kJ) = 275.7 kJ .
QH TH
TH
273.15 K + 600.0 K
34. (a) Strategy Use Eq. (15-17).
Solution Find the temperature of the hot reservoir.
T
T
310.0 K
er = 1 − C , so TH = C =
= 443 K .
TH
1 − er 1 − 0.300
(b) Strategy Use Wnet = QH − QC and the definition of efficiency of an engine.
Solution Find the amount of heat exhausted to the cold reservoir.
⎛1
⎞
W
⎛ 1
⎞
Wnet = QH − QC = net − QC , so QC = Wnet ⎜⎜ − 1⎟⎟ = (0.100 × 103 J) ⎜
− 1⎟ = 233 J .
er
e
0.300
⎝
⎠
⎝ r
⎠
625
Chapter 15: Thermodynamics
Physics
35. Strategy For maximum efficiency, assume reversibility. Use Eq. (15-17).
Solution Find the percent decrease in theoretical maximum efficiency.
T
(
T
)
T
T
1 − TCf − 1 − TCi
− TCf + TCi
27°C − 47°C
∆er
T −T
H
H
H
H
× 100% =
× 100% =
× 100% = Ci Cf × 100% =
× 100%
T
T
Ci
Ci
500.0°C − 27°C
er
TH − TCi
1− T
1− T
H
H
= −4.2%
The theoretical maximum efficiency would decrease by 4.2% .
36. Strategy Use Eq. (15-17).
Solution Find the theoretical maximum efficiency.
T
373 K
er = 1 − C = 1 −
= 0.517
TH
773 K
37. Strategy The maximum possible efficiency occurs if the engine is reversible. Use Eq. (15-17).
Solution Find the maximum possible efficiency.
T
273.15 K + 10.0 K
er = 1 − C = 1 −
= 0.0174
TH
273.15 K + 15.0 K
38. Strategy The maximum possible efficiency occurs if the engine is reversible. Use Eq. (15-17).
Solution Find the maximum possible efficiency.
T
273.15 K + 25 K
er = 1 − C = 1 −
= 0.039
TH
273.15 K + 37 K
39. Strategy Use Eq. (15-16). Since the water is initially at 0°C, QC = mLf is the amount of heat that must be
removed from the water to freeze it.
Solution Find the work required to freeze the water.
Q
Q
mLf (1.0 kg)(333.7 kJ kg )
K r = C , so Wnet = C =
=
= 110 kJ .
Wnet
Kr
Kr
3.0
40. (a) Strategy The maximum efficiency is that of a reversible heat engine. Use Eq. (15-17).
Solution Compute the efficiency.
⎛ T ⎞
⎛ 350 ⎞
e = 0.650er = 0.650 ⎜⎜ 1 − C ⎟⎟ = 0.650 ⎜1 −
⎟ = 0.30
T
⎝ 650 ⎠
⎝
H ⎠
(b) Strategy Use energy conservation and the definition of efficiency of an engine.
Solution Find the amount of work done by the engine.
−1
−1
W
⎛1 ⎞
⎛ 1
⎞
Wnet = QH − QC = net − QC , so Wnet = QC ⎜ − 1⎟ = (6.3 × 103 J) ⎜
− 1⎟ = 2.7 kJ .
e
⎝e ⎠
⎝ 0.30 ⎠
626
Physics
Chapter 15: Thermodynamics
41. Strategy The maximum rate at which the river can carry away heat is QC ∆t = mc∆T ∆t . Use energy
conservation and the definition of efficiency of an engine.
Solution Find the maximum possible power the plant can produce.
Wnet QH − QC Wnet mc∆T
, so
=
=
−
e ∆t
∆t
∆t
∆t
−1
Wnet ⎛ 1 ⎞−1 m
⎛ 1
⎞
c∆T = ⎜
= ⎜ − 1⎟
− 1⎟ (5.0 × 106 kg s)[4186 J (kg ⋅ K)](0.50 K) = 4.5 GW .
∆t
⎝ e ⎠ ∆t
⎝ 0.300 ⎠
42. Strategy Use Eqs. (15-15) and (15-17).
Solution The coefficient of performance of a heat pump is given by
Q
heat delivered
Kp =
= H .
net work input Wnet
The net work is equal to the heat from the hot reservoir times the efficiency of the pump. So,
Q
Q
1
Kp = H = H = .
Wnet er QH er
The efficiency of a reversible engine is given by er = 1 − TC TH , so the coefficient of performance for a
reversible heat pump is
1
Kp =
.
1 − TC TH
43. Strategy The refrigerator is reversible and the electric motor does 250 J of work per second. Use Eqs. (15-12)
and (15-17), and Wnet = QH − QC .
Solution Calculate er .
T
256 K
er = 1 − C = 1 −
= 0.20
320 K
TH
Calculate QC QH for the refrigerator.
QC QH − Wnet
W
=
= 1 − net = 1 − er = 1 − 0.20 = 0.80
QH
QH
QH
Since QC QH < 1, more heat is added to the room than is removed. Assuming constant rates, the net rate is
QH QC QH ⎛ QC ⎞ Wnet er
0.20 Wnet Wnet
−
=
(1 − 0.80) =
×
=
= +250 W .
⎜1 −
⎟=
∆t ∆t
∆t ⎝ QH ⎠
∆t
er
∆t
∆t
Or, more simply, since the room is insulated, and the electric motor is ideal (250 W of electrical energy is
converted to 250 W of work), by conservation of energy, the net rate of heat added to the room must be +250 W.
627
Chapter 15: Thermodynamics
Physics
44. Strategy Use Eqs. (15-16) and (15-17).
Solution The coefficient of performance of a refrigerator is given by
Q
heat removed
Kr =
= C .
net work input Wnet
The heat removed from the cold reservoir is equal to the difference between the heat exhausted into the hot
reservoir and the net work done. So,
Q
Q − Wnet
Q
Kr = C = H
= H − 1.
Wnet
Wnet
Wnet
The net work is equal to the product of the efficiency and the heat exhausted to the hot reservoir, so
Q
Q
1
K r = H − 1 = H − 1 = − 1.
Wnet
er QH
er
The efficiency of a reversible engine is given by er = 1 − TC TH , so the coefficient of performance for a
reversible refrigerator is
T T
1 − 1 + TC TH
1
1
−1 =
= C H =
Kr =
.
1 − TC TH
1 − TC TH
1 − TC TH TH TC − 1
45. Strategy Use Wnet = QH − QC , Eq. (15-17), and the definition of efficiency of an engine.
Solution
QC = QH − Wnet =
⎛
⎞
⎛ TH
⎛1
⎞
Wnet
T −T ⎞
TC
1
Wnet
− Wnet = Wnet ⎜ − 1⎟ = Wnet ⎜
− 1⎟ = Wnet ⎜
− H C⎟=
er
⎝ er
⎠
⎝ 1 − TC TH
⎠
⎝ TH − TC TH − TC ⎠ TH − TC
46. Strategy and Solution The number of moles is the same for each case. For an equal number of moles, gas has
more entropy than liquid, and the more diffuse the gas, the greater the entropy, so the order is (c), (a), (b) .
47. Strategy and Solution The mass is the same for each case. For equal masses, water has more entropy than ice,
and warmer water has more entropy than cooler water, so the order is (b), (a), (c), (d) .
48. Strategy The temperature is constant and the heat entering the system is Q = mLf . Use Eq. (15-20).
Solution Find the change in the ice cube’s entropy.
Q mLf (1.00 g)(333.7 J g)
∆S = =
=
= 1.22 J K
T
T
273.15 K + 0.0 K
49. Strategy The temperature is constant and the heat entering the system is Q = mLv . Use Eq. (15-20).
Solution Find the change in the entropy of the water.
Q mLv (1.00 kg)(2256 kJ kg )
∆S = =
=
= +6.05 kJ K
T
T
273.15 K + 100.0 K
Gas is more disordered than liquid, so the entropy increases.
628
Physics
Chapter 15: Thermodynamics
50. Strategy The temperature is constant and the heat leaving the system is Q = − mLv . Use Eq. (15-20).
Solution Find the change in the entropy of the steam.
Q − mLv −(0.010 kg)(2256 kJ kg )
∆S = =
=
= −60 J K
T
T
273.15 K + 100.0 K
As the steam condenses to liquid, heat is given off. Whatever object that absorbs the heat that is released when
this process occurs, increases in entropy by an amount greater than 60 J K ; therefore, the entropy of the universe
increases by an amount > 60 J K .
51. Strategy Use Eq. (15-20).
Solution
(a) Compute the change in entropy of the block.
Q
1.0 J
∆S C =
=
= 3.4 × 10−3 J K
TC 273.15 K + 20.0 K
(b) Compute the change in entropy for the water.
Q
−1.0 J
∆S H =
=
= −2.8 × 10−3 J K
TH 273.15 K + 80.0 K
(c) Calculate the change in entropy of the universe.
Q Q
∆S tot = ∆SH + ∆SC =
+
= −2.8 × 10−3 J K + 3.4 × 10−3 J K = 6 × 10−4 J K
TH TC
52. Strategy A small amount of heat is transferred from the water to the iron, but the temperatures change little
during the 10.0 s. Use Eq. (15-20).
Solution Calculate the change in entropy of the system.
⎛ 1
Q Q
1 ⎞
1
1
⎛
⎞
3
∆S = −
+
= Q ⎜⎜
−
−
⎟⎟ = (41.86 × 10 J) ⎜
⎟ = +41.1 J K
TH TC
T
T
273.15
K
0.0
K
273.15
K
100.0
K
+
+
⎝
⎠
H ⎠
⎝ C
53. Strategy The rate at which the entropy of the universe is changing is equal to the total change in entropy per unit
time. Use Eq. (15-20).
Solution
∆S Q ⎛ 1
1 ⎞
1
1
⎛
⎞
=
−
−
⎜
⎟ = (220.0 W) ⎜
⎟ = 0.102 J (K ⋅ s)
273.15
K
15.0
K
273.15
K
20.0
K
∆t ∆t ⎝ TC TH ⎠
−
+
⎝
⎠
54. Strategy An amount of heat is transferred from the hot reservoir to the cold reservoir, but their temperatures
change little. Use Eq. (15-20).
Solution Calculate the total change in entropy.
⎛ 1
1 ⎞
1
1
⎛
⎞
∆S = Q ⎜⎜
−
−
⎟⎟ = (418.6 kJ) ⎜
⎟ = 237 J K
T
T
273.15
K
100.0
K
273.15
K
200.0
K
+
+
⎝
⎠
H ⎠
⎝ C
629
Chapter 15: Thermodynamics
Physics
55. (a) Strategy Convert the rate of energy consumption from kcal per day to watts.
Solution
⎛ 1 day ⎞
∆Q
= (2, 000, 000 cal day)(4.186 J cal) ⎜
⎟ = 97 W
∆t
⎝ 86,400 s ⎠
(b) Strategy The rate of change of entropy is equal to the rate at which the heat is released divided by the
temperature of the room.
Solution
∆S ∆Q ∆t
97 W
=
=
= 0.33 W K
∆t
T
273.15 K + 20 K
56. Strategy Use Eqs. (14-4), (14-9), and (15-17), as well as Wnet = QH − QC and the definition of power.
Solution Calculate the efficiency.
T
273.15 K − 5.0 K
er = 1 − C = 1 −
= 0.0853
TH
273.15 K + 20.0 K
Wnet
⎛1 ⎞
− Wnet = Wnet ⎜ − 1⎟ where Wnet = P∆t. QC is equal to the heat removed from the
e
⎝e ⎠
water, mLf − mc∆T .
QC = QH − Wnet =
⎛1 ⎞
QC = mLf − mc∆T = P∆t ⎜ − 1⎟ , so
⎝e ⎠
s
1
1 −1
(148 W)(2.0 h) 3600
P∆t e − 1
h
0.0853
=
= 31 kg .
m=
333, 700 J kg − [4186 J (kg ⋅ K) ](−8.0°C)
Lf − c∆T
(
)
57. Strategy and Solution
(
)(
)
The engine will not work. There is no energy available to do the work necessary to
extract the water’s internal energy.
58. Strategy The molar specific heat of a diatomic ideal gas at constant pressure is CP = CV + R =
5
2
R+R =
Q = nCP ∆T is equal to the energy required and n = PVi ( RTi ) according to the ideal gas law.
Solution Find the energy required.
7
7 ⎛ PV ⎞
7
20.0 K
⎛
⎞
Q = nCP ∆T = nR∆T = ⎜ i ⎟ R∆T = (1.013 × 105 Pa)(200.0 L)(10−3 m3 L) ⎜
⎟
2
2 ⎝ RTi ⎠
2
⎝ 273.15 K + 20.0 K ⎠
= 4.84 kJ
59. Strategy At constant pressure, the work done by the expanding gas is W = − P∆V . ∆U = Q + W , so
∆U = Q − P∆V . According to the ideal gas law, Vf = Tf Vi Ti (n, R, and P are constant).
Solution Find the increase in internal energy of the gas.
⎛T
⎞
⎛ 273.15 K + 27 K ⎞
∆U = Q − P(Vf − Vi ) = Q − PVi ⎜ f − 1⎟ = 25 kJ − (2.0 × 105 Pa)(0.10 m3 ) ⎜
− 1⎟ = 15 kJ
⎝ 273.15 K − 73 K ⎠
⎝ Ti
⎠
630
7
2
R.
Physics
Chapter 15: Thermodynamics
60. Strategy Use Eq. (15-11).
Solution Find the work done on the air in the swim bladder.
V
Vi
8.16 mL
W = nRT ln i = PV
= (1.1 atm)(1.013 × 105 Pa atm)(8.16 × 10−3 L)(10−3 m3 L) ln
i i ln
Vf
Vf
7.48 mL
= 0.079 J
61. (a) Strategy The work done per cycle is equal to the area contained within the curve.
Solution Compute the work done per cycle.
1
W = (5.00 atm − 1.00 atm)(1.013 × 105 Pa atm)(2.00 m3 − 0.500 m3 ) = 304 kJ
2
(b) Strategy Use the ideal gas law to compute the temperatures at the upper left and lower right points on the
curve.
Solution Compute the temperatures.
PV
T
PV = nRT , so 2 2 = 2 .
PV
T1
1 1
Tul =
P2V2T1
PV
1 1
=
(5.00)(0.500)(470.0 K)
= 2350 K
(1.00)(0.500)
P2V2T1
(1.00)(2.00)(470.0 K)
=
= 1880 K
(1.00)(0.500)
PV
1 1
The maximum temperature is 2350 K.
Tlr =
(c) Strategy Use the ideal gas law at the lower-left corner of the diagram.
Solution Find the number of moles of gas used in the engine.
PV (1.00 atm)(1.013 × 105 Pa atm)(0.500 m3 )
n=
=
= 13.0 mol
RT
[8.314 J (mol ⋅ K) ](470.0 K)
62. Strategy Use Eq. (15-17). Form a proportion. Let f be the ratio of the low-temperature reservoir decrease to the
high-temperature reservoir increase.
Solution Compute f.
T −∆T
T −T +∆T
H
C
1 − CT
T + ∆T
decrease low T
∆T
TH
H
= f =
=
= H
= 1+
>1
+∆
−
T
T
T
T
C
H
C
TH
TH
increase high T
1 − T +∆
T
TH +∆T
H
Therefore, decreasing the low temperature reservoir by ∆T will result in a greater efficiency than increasing the
high temperature reservoir by ∆T .
631
Chapter 15: Thermodynamics
Physics
63. (a) Strategy The change in internal energy of the iron is equal to the heat that flows out of the iron
(−mcAl∆TAl ) and into the aluminum plus the work done on the iron (zero). Use the first law of
thermodynamics.
Solution Calculate the final temperature of the iron.
∆U = mcFe∆TFe = mcFe (Tf − Ti ) = Q + W = −mcAl∆TAl + 0, so
c
0.900
Tf = − Al ∆TAl + Ti = −
(2.0°C) + 20.0°C = 15.9°C .
cFe
0.44
(b) Strategy Since the temperatures vary significantly, use average values to estimate the entropy change. Use
Eq. (15-20).
Solution Compute the average temperatures.
20.0°C + 22.0°C
= 21.0°C = TH
Aluminum: Tav =
2
20.0°C + 15.91°C
= 17.955°C = TC
Iron: Tav =
2
Estimate the change in entropy of the system.
⎛ 1
⎛ 1
1 ⎞
1 ⎞
∆S = Q ⎜
−
−
⎟ = −mcAl∆TAl ⎜
⎟
T
T
T
T
H⎠
H⎠
⎝ C
⎝ C
1
1
⎛
⎞
= −(0.50 kg)[0.900 × 103 J (kg ⋅ K) ](2.0°C) ⎜
−
⎟ = −0.03 J K
⎝ 273.15 K + 17.955 K 273.15 K + 21.0 K ⎠
(c) Strategy and Solution ∆Ssystem = ∆Suniverse in this case, and the second law of thermodynamics states that
the entropy of the universe never decreases. Therefore, ∆Suniverse < 0 is impossible, and so is this process.
64. Strategy and Solution A mixture of two substances is more disordered than the same amount of the two
substances at the same temperature, but separate. Thus, (b) has a higher entropy than (a). (a) has a higher entropy
than (c), since (a) has twice the amount of the substances as (c), but all other quantities are the same. (d) has the
highest entropy, since it is a mixture and it is at the highest temperature. The list in order from least to greatest
entropy is (c), (a), (b), (d) .
65. Strategy The energy of the mixed state, U, must equal the sum of the original (unmixed) states, U1 and U 2. The
energy for a monatomic ideal gas is related to the temperature by U = 32 NkT .
Solution Find the final temperature T of the mixture.
U = U1 + U 2
3
3
3
( N1 + N 2 )kT = N1kT1 + N 2kT2
2
2
2
3
3
3
(n1 + n2 ) RT = n1RT1 + n2 RT2
2
2
2
n1T1 + n2T2 (4.0 mol)(20.0°C) + (3.0 mol)(30.0°C)
T=
=
= 24°C
n1 + n2
4.0 mol + 3.0 mol
632
Physics
Chapter 15: Thermodynamics
66. Strategy For an isobaric expansion, Q = nCP ∆T where CP = 72 R for a diatomic gas.
Solution Find the energy required to the raise the temperature of the nitrogen.
7
7 ⎛ PV ⎞
7(1.0 atm)(1.013 × 105 Pa atm)(160 L)(10−3 m3 L)(45°C − 25°C)
Q = nCP ∆T = nR∆T = ⎜ i ⎟ R∆T =
2
2 ⎝ Ti R ⎠
2(273.15 K + 25 K)
= 3.8 kJ
67. Strategy The efficiency is equal to the work done lifting the weight divided by the sum of the work done and the
internal energy dissipated in the muscle. The work done is equal to the increase in gravitational potential energy
of the weight.
Solution Determine the efficiency of the muscle.
e=
⎡
⎤
W
1
1
139 J
=
=
= ⎢1 +
⎥
∆
∆
U
U
W + ∆U 1 +
1 + mgh ⎣ (161 N)(0.577 m) ⎦
W
−1
= 0.401 or 40.1%
68. (a) Strategy The temperature is constant and the heat entering the system is Q = mLf . Use Eq. (15-20).
Solution Find the change in the entropy.
Q mLf (1.00 mol)(15.9994 g mol + 2 × 1.00794 g mol)(333.7 J g)
∆S = =
=
= 22.0 J K
T
T
273.15 K + 0.0 K
(b) Strategy Use Eq. (15-20).
Solution Compute the entropy change of the universe when the ice melts.
⎛ 1
1 ⎞
∆S = Q ⎜⎜
−
⎟⎟
⎝ TC TH ⎠
1
1
⎛
⎞
= (1.00 mol ) (15.9994 g mol + 2 × 1.00794 g mol)(333.7 J g) ⎜
−
⎟
⎝ 273.15 K + 0.0 K 273.15 K + 10.0 K ⎠
= 0.777 J K
69. Strategy The amount of heat that flows into the water is given by Q = mc∆T . Assuming the heat flows at an
average temperature gives T = (20.0°C + 50.0°C)/2 = 35.0°C. Use Eq. (15-20).
Solution Estimate the entropy change.
Q mc∆T (0.85 kg)[4186 J (kg ⋅ K) ](50.0 − 20.0) K
∆S = =
≈
= 350 J K
T
T
273.15 K + 35.0 K
633
Chapter 15: Thermodynamics
Physics
70. Strategy Assume that a heat pump takes in heat from outdoors at 10°C below the ambient outdoor temperature,
and that its output must be 10°C hotter than the house. Use Eq. (15-19).
Solution Compute the coefficient of performance.
1
1
1
K p,rev =
=
.
, so K realistic =
TC −10 K
TC −10 K
TC
1 − T +10 K 1 − 273.15 K + 20 K +10 K
1− T
H
H
Convert TC from kelvins to °C.
1
1
K realistic =
=
TC + 273.15 K −10 K
TC + 263.15 K
1−
1 − 303.15 K
303.15 K
Graph K p = K realistic.
Kp
15
10
5
15
10
5
0
5
10
15 T (°C)
71. (a) Strategy Heat is transferred from the hotter block of iron to the colder block of iron. During this process, the
temperatures of both blocks change continuously. For this estimate, use the average temperature. Use Eqs.
(14-4) and (15-20).
Solution Estimate the entropy change of the universe.
∆Suniverse = ∆S1 + ∆S2
Q
Q
m c∆T m c∆T2
≈ 1 + 2 = 1 1+ 2
T1av T2av
T1av
T2av
⎛
40.0°C − 20.0°C
40.0°C − 60.0°C
= (0.500 kg)[0.44 × 103 J (kg ⋅ K)] ⎜
+
+
20.0
K
40.0
K
⎜ 273.15 K +
273.15 K + 60.0 K +2 40.0 K
⎝
2
∆Suniverse ≈ 0.90 J K
⎞
⎟
⎟
⎠
(b) Strategy Heat is transferred from the colder block of iron to the hotter block of iron (which is impossible).
During this process, the temperatures of both blocks change continuously. For this estimate, use the average
temperature.
Solution Estimate the entropy change of the universe.
⎛
0.0°C − 20.0°C
80.0°C − 60.0°C
∆Suniverse = (0.500 kg)[0.44 × 103 J (kg ⋅ K)] ⎜
+
⎜ 273.15 K + 20.0 K+0.0 K 273.15 K + 60.0 K +80.0 K
⎝
2
2
= −2.7 J K
634
⎞
⎟
⎟
⎠
Physics
Chapter 15: Thermodynamics
72. Strategy Assume the freezer is reversible. Use Eqs. (14-4), (14-9), (15-16), and (15-19).
Solution Find the minimum work input required to freeze the ice.
Q
heat removed
1
Kr =
= C =
, so
net work input Wnet TH TC − 1
⎛T
⎞
⎛T
⎞
Wnet = QC ⎜ H − 1⎟ = (mLf − mcw ∆Tw − mcice∆Tice ) ⎜ H − 1⎟
⎝ TC
⎠
⎝ TC
⎠
⎛ 273.15 K + 20.0 K ⎞
= (1.20 kg){333,700 J kg − [4186 J (kg ⋅ K) ](−20.0 K) − [2100 J (kg ⋅ K) ](−20.0 K)} ⎜
− 1⎟
⎝ 273.15 K − 20.0 K ⎠
= 87.1 kJ
73. Strategy Use Eqs. (14-9) and (15-12), and Wnet = QH − QC .
Solution Find the time required to freeze the water.
W
⎛1 ⎞
Wnet = QH − QC , so QC = QH − Wnet = net − Wnet = Wnet ⎜ − 1⎟ .
e
⎝e ⎠
QC
1
heat removed
rate of heat removed ∆Q ∆t
=
=
=
, so
Therefore, − 1 =
e
Wnet net work input
power input
P
∆t =
∆Q
P
(
1
e
)
−1
=
mLf
P
(
1
e
)
−1
=
(1.0 kg)(333,700 J kg) ⎛ 1 min ⎞
⎜
⎟ = 15 min .
(186 W) 1 − 1 ⎝ 60 s ⎠
( 0.333 )
74. Strategy In (a) and (b), find the amount of heat that flows into or out of the reservoir; then use Eq. (15-20) to find
the change in entropy of the reservoir. In (c), add the changes in entropy found in (a) and (b) to find the change in
entropy of the total system.
Solution
(a) Stage B:
W = 0 and ∆U = Q for a constant volume process.
5
∆U = Q = nR∆T for an ideal diatomic gas, and Qcold = −Q.
2
Calculate the change in entropy.
− 5 nR∆T
Q
5(1.000 mol)[8.314 J (mol ⋅ K) ](273 K − 373 K)
∆S = cold = 2
=−
= 7.61 J K
T
T
2(273 K)
(b) Stage D:
W = 0 and ∆U = Q for a constant volume process.
5
∆U = Q = nR∆T for an ideal diatomic gas, and Qhot = −Q.
2
Calculate the change in entropy.
− 5 nR∆T
Q
5(1.000 mol)[8.314 J (mol ⋅ K) ](373 K − 273 K)
∆S = hot = 2
=−
= −5.57 J K
T
T
2(373 K)
(c) ∆Ssystem + reservoirs = 7.61 J K + (−5.57 J K) = 2.04 J K
635
Chapter 15: Thermodynamics
Physics
75. (a) Strategy For an isobaric process, W = P∆V is the work done by the gas, and P∆V = nR∆T according to the
ideal gas law.
Solution Find the work done by the air in the swim bladder.
PVi
(1.1 atm)(1.013 × 105 Pa atm)(8.16 × 10−3 L)(10−3 m3 L)
W = nR∆T =
R ∆T =
(2.0 K) = 6.2 mJ
Ti R
273.15 K + 20.0 K
(b) Strategy CP = 72 R for a diatomic ideal gas. Use Eq. (15-7).
Solution Find the heat gained by the air in the swim bladder.
7
7
7
Q = nCP ∆T = nR∆T = W = (6.2 mJ) = 22 mJ
2
2
2
(c) Strategy Use Eq. (14-4).
Solution Find the temperature change.
Q
−21.7 × 10−3 J
∆T =
=
= −1.2 mK
mc (5.00 g)[3.5 J (g ⋅ °C)]
The temperature will decrease 1.2 mK .
76. (a) Strategy and Solution Find the work done by the engine, the heat input, and the change in internal energy
for each step in the cycle.
Stage A:
V
2.50
W = nRT ln f = (1.000 mol)[8.314 J (mol ⋅ K) ](373 K) ln
= 692 J
Vi
2.00
∆U = 0 for an isothermal process.
To perform the work, Q = W = 692 J of heat was required.
State B:
W = 0 for a constant volume process.
∆U = Q for a constant volume process, and CV = 52 R for a diatomic ideal gas.
∆U = Q = nCV ∆T =
Stage C:
5
5
nR∆T = (1.000 mol)[8.314 J (mol ⋅ K) ](273 K − 373 K) = −2080 J
2
2
V
2.00
W = nRT ln f = (1.000 mol)[8.314 J (mol ⋅ K) ](273 K) ln
= −506 J
Vi
2.50
∆U = 0 and Q = W = −506 J.
Stage D:
5
5
W = 0 and ∆U = Q = nR∆T = (1.000 mol)[8.314 J (mol ⋅ K) ](373 K − 273 K) = 2080 J.
2
2
Stage
W (J)
Q (J)
∆U (J)
A
692
692 into the gas
0
B
0
2080 out of the gas
−2080
C
−506
506 out of the gas
0
D
0
2080 into the gas
2080
ABCD
186
186 into the gas
0
636
Physics
Chapter 15: Thermodynamics
(b) Strategy Use Eq. (15-12).
Solution Find the efficiency of the engine.
W
185.5 J
e = net =
= 0.0670
Qin
692 J + 2078.5 J
(c) Strategy Use Eq. (15-17).
Solution Compute the efficiency of a reversible engine and compare it to the efficiency of this engine.
T
273 K
e
0.268
er = 1 − C = 1 −
= 0.268 and r =
= 4.00, or er = 4.00e .
TH
373 K
e 0.0670
77. (a) Strategy Use Eq. (15-17).
Solution Compute the efficiency.
T
273.15 K + 22 K − 15 K
er = 1 − C = 1 −
= 0.051
TH
273.15 K + 22 K
(b) Strategy The power supplied to the town is equal to the efficiency times the rate at which heat is supplied by
the lake. Use Eq. (14-4) and the relationship between mass, density, and volume.
Solution Find the volume of water used each second.
∆Q emc∆T e ρVc∆T
P=e
, so
=
=
∆t
∆t
∆t
P∆t
(1.0 × 108 W)(1.0 s)
V=
=
= 31 m3 .
e ρ c∆T 0.051(1.00 × 103 kg m3 )[4186 J (kg ⋅ K) ](15 K)
(c) Strategy The incident power of the Sun must be greater than the power required to run the engine. Power is
equal to intensity times area.
Solution Compare the power supplied to the power required by the town.
PSun = IA = (200 W m 2 )(8.0 × 107 m 2 ) = 1.6 × 1010 W
P
1.0 × 108 W
Pengine = town =
= 2.0 × 109 W
e
0.051
1.6 × 1010 W > 2.0 × 109 W, so Psun > Pengine , and yes , the lake can supply enough heat to meet the town’s
needs.
78. (a) Strategy Use the ideal gas law, PV = nRT .
Solution First, n and T are constant, so P ∝ V −1. Next, the volume decreases while the pressure is constant.
Finally, the volume is constant as the pressure increases. Compute the unknown values of pressure and
volume.
nRT (3.00 mol)[8.314 J (mol ⋅ K) ](650.0 K)
V1 =
=
= 0.0400 m3
P1
(4.00 atm)(1.013 × 105 Pa atm)
V2 = 9.50V1 = 0.380 m3
P2 =
P
nRT
nRT
4.00 atm
=
= 1 =
= 0.421 atm
nRT
V2
9.50
9.50
9.50 P
1
637
Chapter 15: Thermodynamics
Physics
The P-V diagram is shown.
P (atm)
4.00
0.421
0 0.0400
0.380 V (m3)
(b) Strategy Refer to the diagram in part (a). Calculate the quantities for each step in the cycle.
Solution Step 1, isothermal process:
V
1
The work done on the gas is W = nRT ln i = (3.00 mol)[8.314 J (mol ⋅ K) ](650.0 K) ln
= −36.5 kJ .
Vf
9.50
The change in the internal energy of the gas is 0 for an isothermal process.
The heat transferred is Q = −W = 36.5 kJ .
Step 2, isobaric process:
The work done on the gas is
W = − P∆V = −(0.421 atm)(1.013 × 105 Pa atm)(0.0400 m3 − 0.380 m3 ) = 14.5 kJ .
The heat transferred is given by Q = nCP ∆T . Using the ideal gas law, ∆T = P∆V (nR ) and CP = 5 R 2.
5 P∆V 5
5
Q = nCP ∆T = n R
= P∆V = (0.421 atm)(1.013 × 105 Pa atm)(0.0400 m3 − 0.380 m3 )
2
nR
2
2
= −36.3 kJ
Applying the first law gives ∆U = Q + W = −36.3 kJ + 14.5 kJ = −21.8 kJ .
Step 3, isochoric process:
Without a displacement, work cannot be done, so W = 0 .
The change in the internal energy of the gas is equal to the heat that enters the system, so the change in
internal energy and the heat transferred are
3 ⎛
PV ⎞ 3
⎛3 ⎞
∆U = Q = nCV ∆T = n ⎜ R ⎟ ∆T = nR ⎜ T −
⎟ = (nRT − PV )
2 ⎝
nR ⎠ 2
⎝2 ⎠
3
= {(3.00 mol)[8.314 J (mol ⋅ K) ](650.0 K) − (0.421 atm)(1.013 × 105 Pa atm )(0.0400 m3 )} = 21.8 kJ .
2
(c) Strategy The efficiency is equal to the ratio of the net work done by the gas to the heat transferred into the
gas.
Solution The work done by the gas is negative the work done on the gas.
Wnet = −(−36.5 kJ + 14.5 kJ) = 22.0 kJ
The heat transferred into the gas is Qin = 36.5 kJ + 21.8 kJ = 58.3 kJ.
The efficiency of the engine is e =
Wnet
Qin
=
22.0 kJ
= 0.377 .
58.3 kJ
638
REVIEW AND SYNTHESIS: CHAPTERS 13–15
Review Exercises
1. Strategy Assume no heat is lost to the air. The potential energy of the water is converted into heating of the
water. The internal energy of the water increases by an amount equal to the initial potential energy.
Solution Find the change in internal energy.
∆U = mgh = (1.00 m3 )(1.00 × 103 kg m3 )(9.80 m s 2 )(11.0 m) = 108 kJ
2. Strategy Use Eq. (13-22). Form a proportion.
Solution Find the temperature of the nitrogen gas.
v
mHeTN
3kT
vrms =
, so N = 1 =
.
m
vHe
mNTHe
Therefore, TN =
mN
mHe
THe =
2 × 14.00674
(273.15 K + 20.0 K) = 2052 K = 1779°C .
4.00260
3. Strategy Set the increase of internal energy of the water equal to the kinetic energy of the rock and solve for the
final temperature, T.
Solution Find the final temperature of the water.
1
m v2
(1.0 × 10−3 kg)(8.4 × 103 m s) 2
mw c∆T = mw c(Tf − Ti ) = mr v 2 , so Tf = r + Ti =
+ 20.0°C = 28.4°C .
2
2mw c
2(1.0 kg)[4186 J (kg ⋅ K) ]
4. Strategy The change in internal energy of the water is equal to the work done on the water by the mixer plus the
heat that flows into the water: Q + W = ∆U = mc∆T .
Solution Find the quantity of heat that flowed into the water.
Q = mc∆T − W = (2.00 kg)[4.186 kJ (kg ⋅ K) ](4.00 K) − 6.0 kJ = 27.5 kJ .
639
Review and Synthesis: Chapters 13–15
Physics
5. Strategy Set the sum of the heat flows equal to zero. Use Eqs. (14-4) and (14-9).
Solution
(a) Find the mass of ice required.
0 = Qw + Qice = mw cw ∆Tw + mice Lf + mice cice ∆Tice = mw cw ∆Tw + mice ( Lf + cice ∆Tice ), so
m c ∆T
(0.250 kg)[4.186 kJ (kg ⋅ K)](−25.0 K)
mice = − w w w = −
= 74 g .
Lf + cice ∆Tice
333.7 kJ kg + [2.1 k J (kg ⋅ K)](10.0 K)
(b) Find the final temperature of the water, T, which includes the melted ice.
0 = Qw + Qice
0 = mw cw ∆Tw + mice Lf + mice cice ∆T1 + mice cw ∆T2
0 = mw cw (T − Tw ) + mice Lf + mice cice ∆Tice + mice cw (T − 273.15 K)
0 = (mw + mice )cw T + mice ( Lf + cice ∆Tice ) − cw [mw Tw + mice (273.15 K)]
c [m T + mice (273.15 K)] − mice ( Lf + cice ∆Tice )
T= w w w
(mw + mice )cw
[4.186 kJ (kg ⋅ K)][(0.250 kg)(273.15 K + 25.0 K) + (0.037 kg)(273.15 K)]
− (0.037 kg){333.7 kJ kg + [2.1 kJ (kg ⋅ K)](10.0 K)}
T=
− 273.15 K = 11°C
(0.250 kg + 0.037 kg)[4.186 kJ (kg ⋅ K)]
6. Strategy Determine how much larger the volume of the Pyrex container is than the volume of the water after the
temperature decrease. Use Eq. (13-7).
Solution Find the amount of water that can be added.
∆V
= β ∆T , so ∆VPyrex − ∆Vwater = V0 ( β Pyrex − β water )∆T
V0
= (40.0 L)(9.75 × 10−6 K −1 − 207 × 10−6 K −1 )(20.0°C − 90.0°C) = 0.552 L .
7. Strategy Find the heat required to melt the ice, and compare it to the available heat energy in the water. If the ice
does not melt completely, the final temperature will be 0°C. If the ice does melt completely, find the final
temperature of the system.
Solution The heat required to melt the ice is
Q = mice Lf + mice cice ∆T = (0.075 kg){333.7 kJ kg + [2.1 kJ (kg ⋅ K)](10.0 K)} = 27 kJ.
The heat available in the water is Q = mw cw ∆T = (0.500 kg)[4.186 kJ (kg ⋅ K)](50.0 K) = 105 kJ.
Since 105 kJ > 27 kJ, the ice will melt completely .
Set the sum of the heat flows equal to zero and solve for the final temperature, T.
0 = Qw + Qice
0 = mw cw ∆Tw + mice Lf + mice cice ∆T1 + mice cw ∆T2
0 = mw cw (T − Tw ) + mice Lf + mice cice (273.15 K − Tice ) + mice cw (T − 273.15 K)
0 = (mw cw + mice cw )T + mice Lf + mice (cice − cw )(273.15 K) − mw cw Tw − mice ciceTice
m c T + mice ciceTice − mice Lf − mice (cice − cw )(273.15 K)
T= w w w
cw (mw + mice )
(0.500 kg)[4.186 kJ (kg ⋅ K)](50.0 + 273.15) K
+ (0.075 kg){[2.1 kJ (kg ⋅ K)](−10.0 + 273.15) K − 333.7 kJ kg}
− (0.075 kg)[2.1 kJ (kg ⋅ K) − 4.186 kJ (kg ⋅ K)](273.15 K)
T=
= 305.6 K = 32°C
[4.186 kJ (kg ⋅ K)](0.500 kg + 0.075 kg)
640
Physics
Review and Synthesis: Chapters 13–15
8. Strategy Use the ideal gas law.
Solution Find the number of moles of air when the balloon is at 40.0°C.
PV (1.00 atm)(1.013 × 105 Pa atm)(12.0 m3 )
PV = nRT , so n2 =
=
= 467 mol .
RT2
[8.314 J (mol ⋅ K) ](273.15 K + 40.0 K)
9. (a) Strategy The maximum power emission is inversely proportional to the absolute temperature. Use Wien’s
law.
Solution Compute the surface temperature of the star.
2.898 × 10−3 m ⋅ K 2.898 × 10−3 m ⋅ K
=
= 4140 K
T=
λmax
700.0 × 10−9 m
(b) Strategy Use Stefan’s law of blackbody radiation.
Solution Compute the power radiated.
ᏼ = σ AT 4 = [5.670 × 10−8 W (m 2 ⋅ K 4 )][4π (7.20 × 108 m)2 ](4140 K) 4 = 1.09 × 1026 W
(c) Strategy Intensity is power radiated per unit area.
Solution Compute the intensity measured by the Earth-based observer.
ᏼ
1.085 × 1026 W
I= =
= 1.01× 10−9 W m 2
A 4π (9.78 ly)2 (9.461× 1015 m ly)2
10. (a) Strategy Use Eqs. (14-12) and (14-13).
Solution Find the thermal resistance for each material.
d
d
Rwood = Rw = w and Rinsulation = Ri = i .
κw A
κi A
Compute the rate of heat flow.
∆T
A∆T
(2.74 m)(3.66 m)[23.0°C − (−5.00°C)]
ᏼ=
=
=
= 320 W
0.0100 m +
0.0300 m
Rw + Ri d w + di
κw
κi
0.13 W (m⋅K)
0.038 W (m⋅K)
(b) Strategy The area of the insulated wall has been reduced by half, so the rate of heat flow through it is
reduced by half. The other half is glass. The total rate of heat flow is the sum of the two rates.
Solution Find the rate of heat flow.
∆T A∆T κ g A∆T
ᏼg =
=
=
, so
dg
Rg
dg
κg
ᏼ total =
κ g A∆T
dg
+
ᏼ [0.63 W (m ⋅ K)](1 2)(2.74 m)(3.66 m)[23.0°C − (−5.00°C)] 324 W
=
+
= 18 kW .
2
0.00500 m
2
641
Review and Synthesis: Chapters 13–15
Physics
11. (a) Strategy Use the ideal gas law.
Solution Find the pressure at point A, which is the same as the pressure at point D.
nRTA (2.00 mol)[8.314 J (mol ⋅ K) ](800.0 K)
PD = PA =
=
= 8.87 kPa
VA
1.50 m3
Find the temperature at point D.
P V
(8.87 × 103 Pa)(2.25 m3 )
TD = D D =
= 1200 K
(2.00 mol)[8.314 J (mol ⋅ K) ]
nR
(b) Strategy The net work done on the gas is equal to the area inside the graph.
Solution Find the net work done on the gas as it is taken though four cycles.
W = 4(8.87 kPa − 1.30 kPa)(2.25 m3 − 1.50 m3 ) = 23 kJ
(c) Strategy The internal energy of an ideal monatomic gas is given by U = 32 nRT .
Solution Compute the internal energy of the gas at point A.
3
3
U = nRT = (2.00 mol)[8.314 J (mol ⋅ K) ](800.0 K) = 20.0 kJ
2
2
(d) Strategy and Solution The total change in internal energy in four complete cycles is 0 , since the change
in temperature is zero.
12. Strategy Use the latent heat of vaporization for water to determine the rate of heat flow. Then, use Fourier’s law
of heat conduction to find the temperature at the base of the pan.
Solution The temperature of the water and the top side of the bottom of the pan is T = 100°C. Determine the rate
of heat flow.
mLv
Q mLv
Q
=
0.730 =
, so
∆t
∆t
∆t 0.730∆t
Find the temperature at the base of the pan, Tb .
T −T
mLv
Q
∆T
=κA
=κA b
=
, so
d
d
∆t
0.730∆t
dmLv
(0.00300 m)(10.0 g)(2256 J g)
+T =
+ 100.0°C = 112.0°C .
Tb =
0.730κ A∆t
0.730[237 W (m ⋅ K)](325 × 10−4 m 2 )(1.00 s)
13. Strategy The temperature is constant and the heat entering the system is Q = mLf .
Solution Find the change in entropy of the ice.
Q mLf (2.00 kg)(333.7 kJ kg )
∆S = =
=
= 2.44 kJ K
T
T
273.15 K + 0.0 K
642
Physics
Review and Synthesis: Chapters 13–15
14. Strategy Use Stefan’s law of radiation, Eq. (14-16). Intensity is power per unit area.
Solution Let r1 = 0.40 m be the radius of the sphere and r2 = 2.0 m be the distance from the center of the sphere
where the intensity is measured. Compute the emissivity.
ᏼ = eσ AT 4 = eσ (4π r12 )T 4 = 4π r22 I , so
e=
r22 I
σ r12T 4
=
(2.0 m) 2 (102 W m 2 )
[5.670 × 10−8 W (m 2 ⋅ K 4 )](0.40 m)2 (250 K + 273.15 K)4
= 0.60 .
15. Strategy The gravitational potential energy of the steel ball is converted into heat. Set the sum of the heat flows
equal to zero.
Solution Find the final temperature of the system, T.
0 = U s + Qs + Qw
0 = −ms gh + ms cs ∆Ts + mw cw ∆Tw
0 = −ms gh + ms cs (T − Ts ) + mw cw (T − Tw )
0 = −ms ( gh + csTs ) + (ms cs + mw cw )T − mw cw Tw
m c T + ms ( gh + csTs )
T= w w w
ms cs + mw cw
T=
(4.50 L)(1 kg L)[4186 J (kg ⋅ K)](10.1°C) + (7.30 kg){(9.80 m s 2 )(10.0 m) + [450 J (kg ⋅ K)](15.2°C)}
(7.30 kg)[450 J (kg ⋅ K)] + (4.50 L)(1 kg L)[4186 J (kg ⋅ K)]
= 10.9°C
16. (a) Strategy Use the definition of pressure.
Solution Find the weight of the car.
F
P = , so F = PA = (36.0 lb in 2 )(2 × 24.0 in 2 + 2 × 20.0 in 2 ) = 3170 lb .
A
(b) Strategy Since the contact areas of the front tires are greater than those of the back tires, the front half of the
car weighs more than the back half. Use the definition of center of mass.
Solution Compute the y-coordinate of the car’s center of mass. Since F = ma = PA,
the mass each tire supports is proportional to the contact area, so the contact areas can
be used instead of the masses when computing the center of mass. Due to symmetry,
the front tires can be combined, as well as the rear tires.
m
y
+ mrear yrear Afront yfront + Arear yrear
yCM = front front
=
mfront + mrear
Afront + Arear
=
(48.0 in 2 )(0) + (40.0 in 2 )(−7.00 ft)
48.0 in 2 + 40.0 in 2
= −3.18 ft
17. Strategy The heat loss is proportional to the temperature difference.
Solution Compute the reduction in heat loss.
∆T
81°C − 36°C
ᏼ 2 = 2 ᏼ1 =
ᏼ1 = 0.75ᏼ1
∆T1
81°C − 21°C
The heat loss was reduced to 75% of the original .
643
y (ft)
0
−7.00
x (ft)
Review and Synthesis: Chapters 13–15
Physics
18. (a) Strategy Assume the refrigerator is reversible. Use Eq. (15-17) and the definition of efficiency of an engine.
Solution Find the amount of heat exhausted.
T
W
273.15 K + 0.0 K
1.0 kJ
er = 1 − C = 1 −
= 0.1277, so QH = net =
= 7.8 kJ .
TH
273.15 K + 40.0 K
er
0.1277
(b) Strategy Use Wnet = QH − QC .
Solution Find the heat removed.
QC = QH − Wnet = 7.8 kJ − 1.0 kJ = 6.8 kJ
19. Strategy Use Eq. (15-17) and the definition of efficiency.
Solution
T
W
273.15 K − 4 K
1.0 kJ
er = 1 − C = 1 −
= 0.085, so QH = net =
= 12 kJ .
TH
273.15 K + 21 K
er
0.085
20. Strategy Use Fourier’s law of heat conduction for the copper rod and the heat of fusion for the ice.
Solution Find the rate of melting.
Q mLf
∆T
m κ A∆T [401 W (m ⋅ K)]π (0.0100 m)2 (100.0 K) ⎛ 3600 s ⎞
=
=κA
=
=
, so
⎜
⎟ = 136 g h .
∆t
∆t
∆t
d
Lf d
(333.7 J g)(1.00 m)
⎝ 1h ⎠
21. (a) Strategy and Solution
The boiling temperature of water varies with pressure. If the pressure is high, the water molecules are
pushed close together, making it harder for them to form a gas. (Gas molecules are farther apart from
each other than are liquid molecules.) A higher pressure raises the temperature at which the coolant
fluid will boil.
(b) Strategy and Solution
If you were to remove the cap on your radiator without first bringing the radiator pressure down to
atmospheric pressure, the fluid would suddenly boil, sending out a jet of hot steam that could burn you.
22. Strategy The work done per stroke (cycle) is equal to the average pressure times the change in volume. The
average power output is equal to the operating frequency times the work per stroke.
Solution Find the operating frequency f. Let pav be the average power output.
f Wstroke = f Pav ∆V = pav , so f =
pav
Pav ∆V
=
27.6 × 103 W
(1.3 × 105 Pa)π ( 0.150 m 2 )2 (0.200 m)
= 60 Hz .
23. Strategy and Solution
(a) Since the blocks are made of the same material and are at the same temperature, they will have the same
internal energy if they have the same mass .
(b)
Since they are at the same temperature, there is no net energy transfer between the two blocks .
(c)
The blocks need not touch each other in order to be in thermal contact. They can be in thermal contact
due to convection and radiation.
644
Physics
Review and Synthesis: Chapters 13–15
24. (a) Strategy The maximum possible efficiency is that of a reversible engine.
Solution Compute the maximum possible efficiency.
T
323 K
er = 1 − C = 1 −
= 0.396 or 39.6%
TH
535 K
(b) Strategy Use energy conservation and the definition of efficiency of an engine.
Solution Find the rate at which heat must be removed by means of a cooling tower.
Wnet QH QC Wnet e QC
, so
=
−
=
−
∆t
∆t ∆t
∆t
∆t
⎞
QC Wnet Wnet Wnet ⎛ 1
⎛
⎞
1
=
−
=
− 1⎟⎟ = (1.23 × 108 W) ⎜
− 1⎟ = 4.98 × 108 W .
⎜⎜
e∆t
∆t
∆t
∆t ⎝ 0.500er
⎝ 0.500(0.396) ⎠
⎠
25. (a) Strategy Use the ideal gas law, PV = nRT . Draw a qualitative diagram.
Solution First, the temperature is constant, so P ∝ V −1. Since the volume is
reduced to one-eighth of its initial size, the pressure increases by a factor of
eight. Next, the volume is constant, while the temperature and pressure
increases. Then, the temperature is again constant. Finally, the volume is
constant as the temperature and pressure decreases. The P-V diagram is shown.
P
P2
3
2
4
1
P1
V1
V2
V
(b) Strategy Refer to the diagram in part (a). Calculate the quantities for each step in the cycle. Note that the gas
is diatomic.
Solution Step 1, isothermal process:
V
The work done on the gas is W = nRT ln i = (2.00 mol)[8.314 J (mol ⋅ K) ](325 K) ln 8 = 11.2 kJ.
Vf
The change in the internal energy of the gas is 0 for an isothermal process.
The heat transferred is Q = −W = −11.2 kJ.
Step 2, isochoric process:
Without a displacement, work cannot be done, so W = 0.
The change in the internal energy of the gas is equal to the heat that enters the system, so the change in
internal energy and the heat transferred are
5
⎛5 ⎞
∆U = Q = nCv ∆T = n ⎜ R ⎟ ∆T = (2.00 mol)[8.314 J (mol ⋅ K) ](985 K − 325 K) = 27.4 kJ.
2
⎝2 ⎠
Step 3, isothermal process:
V
1
The work done on the gas is W = nRT ln i = (2.00 mol)[8.314 J (mol ⋅ K) ](985 K) ln = −34.1 kJ.
Vf
8
The change in the internal energy of the gas is 0 for an isothermal process.
The heat transferred is Q = −W = 34.1 kJ.
645
Review and Synthesis: Chapters 13–15
Physics
Step 4, isochoric process:
Without a displacement, work cannot be done, so W = 0.
The change in the internal energy of the gas is equal to the heat that enters the system, so the change in
internal energy and the heat transferred are
5
⎛5 ⎞
∆U = Q = nCv ∆T = n ⎜ R ⎟ ∆T = (2.00 mol)[8.314 J (mol ⋅ K) ](325 K − 985 K) = −27.4 kJ.
2
⎝2 ⎠
The results of the processes and the totals are shown in the table. (Note that the totals for work and heat differ
slightly from the sums of the values for each step due to round-off error.)
Process
W (kJ)
∆U (kJ)
Q (kJ)
Step 1
11.2
0
−11.2
Step 2
0
27.4
27.4
Step 3
−34.1
0
34.1
Step 4
0
−27.4
−27.4
Total
−22.8
0
22.8
(c) Strategy The efficiency is equal to the ratio of the net work done by the gas to the heat transferred into the
gas.
Solution The work done by the gas is negative the work done on the gas.
Wnet = −(−22.8 kJ) = 22.8 kJ and the heat transferred into the gas is Qin = 27.4 kJ + 34.1 kJ = 61.5 kJ.
The efficiency of the engine is e =
Wnet
Qin
=
22.8 kJ
= 0.371 or 37.1% .
61.5 kJ
(d) Strategy Use Eq. (15-17).
Solution Compute the efficiency of a Carnot engine operating at the same extreme temperatures.
T
325 K
er = 1 − C = 1 −
= 0.670 or 67.0%
TH
985 K
26. Strategy The potential energy of the ball is converted into kinetic energy as it falls. When it hits the ground, the
kinetic energy is converted into heat. Use Eq. (15-20) and U = mgh = Q.
Solution Find the increase in the entropy of the universe.
Q mgh (0.15 kg)(9.80 m s 2 )(24 m)
∆S = =
=
= 0.12 J K
T
T
19 K + 273.15 K
27. Strategy Use Eq. (13-4).
Solution Find the temperature at which the two strips have the same length.
L + ∆Lb = 1.00100( L + ∆Ls )
L + Lα b ∆T = 1.00100 L + 1.00100 Lα s ∆T
∆T (α b − 1.00100α s ) = 1.00100 − 1 = 0.00100
0.00100
∆T =
= TH − TL
α b − 1.00100α s
0.00100
0.00100
= 275°C −
= 132°C
TL = TH −
6
−
α b − 1.00100α s
19 × 10 − 1.00100(12 × 10−6 )
646
Physics
Review and Synthesis: Chapters 13–15
28. Strategy Use Eqs. (14-4) and (14-9).
Solution
(a) Compute the heat required to bring the temperature of the solid lead up to its melting point.
Q = mc∆T = (0.360 kg)[0.13 kJ (kg ⋅ K)](327°C − 20°C) = 14.4 kJ
Compute the heat required to bring the temperature of the liquid lead down to its melting point.
Q = mc∆T = (0.980 kg)[0.13 kJ (kg ⋅ K)](327°C − 420°C) = −11.8 kJ
Since more heat is required to melt the solid lead than is available, at least some of the liquid lead solidifies.
The answer is yes .
(b) The final mixture of solid and liquid lead is at its melting point, 327°C .
29. (a) Strategy Use conservation of energy. Ignore air resistance.
Solution Find the escape speed.
Ki + U i = K f + U f
1 2 GMm
mv −
= 0+0
RE
2
v=
2GM
=
RE
2(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg)
6.37 × 106 m
= 11.2 km s
(b) Strategy Use Eq. (13-22).
Solution Calculate the average speed.
v=
3kT
=
m
3(1.381× 10−23 J K)(273.15 K)
(2.00 u)(1.6605 × 10−27 kg u)]
= 1850 m s
(b) Strategy Use Eq. (13-22).
Solution Calculate the average speed.
v=
3kT
=
m
3(1.381× 10−23 J K)(273.15 K)
(32.0 u)(1.6605 × 10−27 kg u)]
= 461 m s
(d) Strategy and Solution
The atoms in the high end of the distribution are much faster than the average. Some of the
hydrogen atoms have speeds greater than the escape speed, thus they can escape. This is not
the case for oxygen, which is much more massive and, thus, much slower.
647
Review and Synthesis: Chapters 13–15
Physics
30. Strategy Use the ideal gas law and Hooke’s law.
Solution Find the final pressure in terms of the initial pressure.
1 π d 2h
P2V2 nRT
V1
h
4
=
= 1, so P2 =
P1 =
P1 =
P.
2
1
PV
nRT
V2
h + ∆x 1
π d (h + ∆x)
1 1
4
The magnitudes of the forces due to the spring, on the outside of the piston, and on the inside of the piston are
k ∆x, k ∆x + Patm A, and P2 A, respectively. Set the forces equal, substitute for P2 , and solve for ∆x.
k ∆x + Patm A = P2 A =
Thus, ∆x =
h
P A, so k (∆x) 2 + (kh + Patm A)∆x − Ah( P1 − Patm ) = 0.
h + ∆x 1
−(kh + Patm A) ± (kh + Patm A) 2 + 4kAh( P1 − Patm )
2k
.
1
1
Substituting k = 1.00 × 103 N m , h = 0.100 m, Patm = 1.013 × 105 Pa, A = π d 2 = π (0.0500 m)2 , and
4
4
P1 = 5.00 × 105 Pa, we find that ∆x = 0.168 m , where −0.467 m is extraneous, since the gas expands.
MCAT Review
1. Strategy and Solution According to the second law of thermodynamics, heat never flows spontaneously from a
colder body to a hotter body, therefore, heat will not flow from bar A to bar B. The correct answer is C .
2. Strategy Assume that the specific heat capacity of seawater is approximately the same at 0°C and 5°C.
Solution Find the approximate temperature T.
0 = Q0 + Q5 = mc(T − 0°C) + mc(T − 5°C), so 2T = 5°C or T = 2.50°C.
The correct answer is B .
3. Strategy Use the latent heat of fusion for water.
Solution The heat gained by the ice when melting is Q = mLf = (0.0180 kg)(333.7 kJ kg) = 6.01 kJ.
The correct answer is C .
4. Strategy and Solution Since e = 1 − QC QH = 1 − TC TH , decreasing the exhaust temperature will increase the
steam engine’s efficiency. The correct answer is B .
5. Strategy and Solution Since refrigerators remove heat by transferring it to a liquid that vaporizes, refrigerators
are primarily dependent upon the heat of vaporization of the refrigerant liquid. The correct answer is A .
6. Strategy and Solution Steam is generally at a higher temperature than water and the specific heat of steam is
lower than that of water, so water would be more effective than steam for changing steam to water. Circulating
water brings more mass of water in contact with the condenser than stationary water, so it can carry away heat at a
faster rate, therefore, it would be more effective for changing steam to water. The correct answer is D .
7. Strategy and Solution Since it is not possible to convert all of the input heat into output work, the amount of
useful work that can be generated from a source of heat can only be less than the amount of heat. The correct
answer is A .
648
Physics
Review and Synthesis: Chapters 13–15
8. Strategy and Solution The internal energy of the steam is converted into mechanical energy as it expands and
moves the piston of the steam engine to the right, therefore, the correct answer is C .
9. Strategy and Solution The refrigerant must be able to vaporize (boil) at temperatures lower than the freezing
point of water so that it can carry away heat (as a gas) from the contents of the refrigerator (which contain water)
to cool and possibly freeze the contents. The correct answer is B .
10. Strategy The heat transferred to the water by the heaters was Qw = mw cw ∆Tw . The heat required for the oil is
Qo = mo co ∆To .
Solution Form a proportion and use the temperature changes of the oil and water and the specific heat and the
specific gravity of the oil to obtain a ratio of heat required for the oil to that transferred to the water.
Qo
m c ∆T
(0.7 mw )(0.60cw )(60 − 20)
= o o o =
= 0.21
Qw mw cw ∆Tw
mw cw (100 − 20)
So, 21% of the amount of heat transferred to the water is required to heat the oil to 60°C. Assuming the heaters
work at the same rate for both the water and the oil, the time required to raise the temperature of the oil from
20°C to 60°C is 0.21(15 h) = 3.2 h. The correct answer is A .
11. Strategy and Solution The high pressure would increase the pressure on the plug, making it more difficult to lift.
The pressure difference between the air in the tank and the air outside of the tank would increase the fluid velocity
when the tank is drained, thus, decreasing the time required to drain the tank. The time required to heat the oil
would be the least likely affected, since the oil is fairly incompressible. The correct answer is A .
649
Chapter 16
ELECTRIC FORCES AND FIELDS
Conceptual Questions
1. This proposition would not work because even with small net charges some objects would be observed to repel
each other, which does not happen with gravity. To account for the weight of an object one may say, for example,
that the Earth is slightly positively charged and the object negatively charged. A slightly positively charged object
should then be repelled from the Earth and fall upward. Furthermore, increasing the charge on an object would
increase the force from the Earth, but the weight is not observed to change by increasing an object’s charge.
2. The clothes transfer some charge to one another as they rub together in the dryer causing static cling between parts
with opposite charges. A charged piece of clothing may also cling to something that is uncharged due to
polarization. When the clothes are slightly damp they do not hold a charge very long because the moisture allows
the charge to leak off more quickly. We would expect clothes to cling more when made of different materials,
since some materials more readily give up or take on electrons by rubbing than others. Rubbing of two such
different materials would therefore result in a greater transfer of charge than rubbing of two objects of the same
material (an affinity to give up or take on electrons).
3. In equilibrium, charges are not in motion. Thus, the electric field inside any conductor in equilibrium must be
zero, or else charges would move inside the conductor as a result. If the electric field inside the conductor is zero,
there cannot be any net charge distributed inside the conductor, or else there would be an electric field produced
by these charges. Therefore, any net charge on a conductor in equilibrium is found on the outside surface.
4. Electric field lines point in the direction of the electric field. Near a positive charge the electric field points
outward away from the charge, so the field lines emanate from a positive charge. Near a negative charge the
electric field points inward toward the charge, and so do the field lines. Thus, field lines must begin on positive
charges and end on negative charges.
5. (a) Since the sphere is positively charged, it must have lost electrons to the charged rod, so its mass will be
smaller. We know that electrons were transferred from the sphere to the rod, and that positive ions were not
transferred from the rod to the sphere, because electrons in a metal are much more mobile than ions in any
solid.
(b) The rod must have been positively charged since it attracted electrons off of the sphere. Also, after the two
objects are touched they must be at the same potential. Since the sphere ends up positively charged, the rod
must end up positively charged as well.
6. (a) Whenever there is an external electric field, the mobile charges in the conducting box move in response in
such a way as to make the electric field inside the box zero.
(b) The charges take some time to respond to an external field and redistribute themselves to a new equilibrium
position. As a result, the shielding works better for constant or slowly varying fields.
(c) Gravitational fields cannot be shielded because gravity is always attractive. The electrical shielding requires
that there be two types of charges. Electric field lines terminate on negative charges; gravitational field lines
never terminate but instead extend out to infinity.
7. First discharge the electroscope by touching it with your hand. Then bring the glass rod near the conducting bulb
and see if the leaves diverge. If they do, the rod is charged. It is not possible to tell whether the charges are
positive or negative. The rod keeps the same charge as long as you do not touch the end of the rod to the bulb.
650
Physics
Chapter 16: Electric Forces and Fields
8. The first plastic rod becomes negatively charged from rubbing it with the fur. When it is brought near the second
rod, the second rod becomes polarized and is attracted to first. When the rods touch some negative charge is
transferred from the first rod to the second. With the second rod now negatively charged, it is repelled by the first.
9. No, the net charge―sum of all the charges with signs―is the same before and after.
10. It is false that an electric field never exists within a conductor. For example, fields do exist for short periods of
time after a conductor enters an external field. These transient fields exist until the charges reach their equilibrium
positions. There is no electric field inside a conductor in electrostatic equilibrium.
11. If two electric fields lines crossed, the direction of the electric field at their intersection would be indeterminate
and a charge placed at this location would have to choose which direction to travel.
12. A buildup of charge on the truck could cause a catastrophic sparking as a result of the potential difference with the
ground. The dragging chain or conducting tires serve to maintain the truck’s grounding.
13. (a) The foils are positively charged and will repel each other even after the rod is removed.
(b) As another positively charged rod is brought close to the conducting sphere, the foils become further
positively charged by induction. They will move farther apart, but will return to their previous position if the
rod is removed.
(c) If a negatively charged rod is now brought near the sphere, the foils will become less positively charged and
will move closer together.
14. (a) Electrons in a conductor are much more mobile than ions in any kind of solid. Therefore, if the rod is a
conductor, electrons will quickly flow in such a way as to minimize any potential difference between the fur
and the rod. That is, it will not be possible to achieve a measurable transfer of charge by rubbing with a
conductor. The rod however receives a negative charge, from which we may surmise that it is an insulator.
(b) The charge of the second rod must also be negative because the force between the two rods is repulsive.
15. As the negatively charged rod is brought near the conducting sphere, electrons flow from the sphere through the
ground wire—the sphere is positively charged by induction. When the ground wire is removed, the sphere is left
with a net positive charge which distributes itself uniformly over the sphere’s surface after the rod is removed.
16. The electric flux through a surface is defined as the product of the magnitude of the perpendicular component of
the electric field through the surface and the area of the surface. The term “flux density” therefore refers to the
fact that the electric field is equal to the electric flux divided by an area.
17. Electric flux is the total “amount” of electric field “flowing” through a surface. It is analogous to the volume flow
rate of water through a pipe. For water, the product of the perpendicular velocity and the surface area gives the
volume of water flowing through the surface per unit time. Electric flux originates from positive charges just as
the flow of water originates from a faucet. Both are therefore given the name “sources”. Electric flux flows
toward negative charges just as water flows toward drains. Both are therefore given the name “sinks”.
18. Given that the flux through a closed surface is zero, we cannot conclude that the electric field in the region of the
surface is zero. If the surface were in a constant electric field, the positive flux into the surface would be negated
by the negative flux out of the surface. We can however conclude that the total charge inside the surface is zero.
This is a direct consequence of Gauss’s law.
19. (a) All four charges contribute to the electric field at point P.
(b) The electric flux through the surface from all four charges is identical to the electric flux obtained from the
interior charges only. The flux from the exterior charges must therefore equal zero.
651
Chapter 16: Electric Forces and Fields
Physics
Problems
1. Strategy There are 10 protons in each water molecule. Multiply the elementary charge by Avogadro’s number
and the number of protons per molecule.
Solution Find the total positive charge.
10(1.0 mol)(6.022 × 1023 mol−1)(1.602 × 10−19 C) = 9.6 × 105 C
2. Strategy There are 79 electrons per neutral gold atom. Use the elementary charge, Avogadro’s number, and the
molar mass of gold.
Solution Find the net charge after 1.0% of the electrons are removed.
(
0.010 79
electrons
atom
) (1.0 g)(6.022 ×1023 mol−1)(1.602 ×10−19 C) =
196.96654 g mol
390 C
3. (a) Strategy and Solution Since electrons have negative charge, and since the balloon acquired a negative net
charge, electrons were added to the balloon.
(b) Strategy Divide the net charge by the charge of an electron.
Solution Compute the number of electrons transferred.
− 0.60 × 10−9 C
= 3.7 × 109
−1.602 × 10−19 C
4. Strategy The electrons were transferred from the negatively charged rod to the positively charged sphere.
Solution Compute the charges on each object.
Sphere: qs = 4.0 × 10−9 C + (8.2 × 109 )(−1.602 × 10−19 C) = 2.7 nC
Rod: qr = −6.0 × 10−9 C − (8.2 × 109 )(−1.602 × 10−19 C) = − 4.7 nC
5. Strategy and Solution
(a) When the rod is brought near sphere A, negative charge flows from sphere B to sphere A. The spheres are
then moved apart and the rod is removed, so A is left with a net negative charge .
(b) Sphere B has an equal magnitude of positive charge , since the two spheres were initially uncharged.
6. Strategy Each time a pair of spheres makes contact, their net charge is shared equally.
Solution After A and B make contact and are separated, each sphere has a charge of Q 2. After B and C make
contact and are separated, each sphere has a charge of Q 4. After A and C make contact and are separated, each
sphere has a charge of (Q 2 + Q 4) 2 = 3Q 8. The charges on spheres A and C are 3Q 8 . The charge on
sphere B is Q 4 .
652
Physics
Chapter 16: Electric Forces and Fields
7. Strategy Each time a pair of spheres makes contact, their net charge is shared equally, with the exception of the
time when C is grounded.
Solution After A and B make contact and are separated, each sphere has a charge of Q 2. After B and C make
contact and are separated, each sphere has zero charge because sphere C was grounded. After A and C make
contact and are separated, each sphere has a charge of (Q 2 + 0) 2 = Q 4. The charges on spheres A and C are
Q 4 . The charge on sphere B is 0 .
8. Strategy Like charges repel one another; unlike charges attract one another.
Solution Spheres A and C have the same sign of charge, positive, so they repel each other, but are attracted to the
negatively-charged spheres B and D. Spheres B and D have the same sign of charge, negative, so they repel each
other. Sphere E has no charge, but when brought close to any one of the other spheres, it will become polarized by
induction and the spheres will attract each other.
Therefore, pairs AB, AD, AE, CB, CD, CE, BE, and DE attract and pairs AC and BD repel .
9. Strategy Use Coulomb’s law, Eq. (16-2).
Solution Find the distance between the charges.
F=
k q1 q2
r2
, so r =
k q1 q2
F
=
(8.988 × 109 N ⋅ m 2 C2 )(1 C) 2
= 30 km .
10 N
10. Strategy Use Coulomb’s law, Eq. (16-2).
Solution Find the charge on each sphere.
kq 2
= F , so q =
r2
Fr 2
(0.036 N)(0.250 m) 2
=
= 5.0 × 10−7 C.
k
8.988 × 109 N ⋅ m 2 C2
The charge is negative, so each sphere has −5.0 × 10−7 C of charge on it.
11. Strategy Divide the magnitude of the Coulomb force by the magnitude of the gravitational force.
Solution Compute the ratio.
Fq
Fg
=
kq 2
r2
Gmp me
r2
=
(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)2
kq 2
=
= 2.268 × 1039
Gmp me (6.674 × 10−11 N ⋅ m 2 kg 2 )(1.673 × 10−27 kg)(9.109 × 10−31 kg)
12. Strategy Set the magnitudes of the Coulomb force and the gravitational force equal and solve for the charge; then
divide the charge by the elementary charge to find the number of electrons.
Solution Set the magnitudes of the forces equal.
Fq =
kq 2
r2
= Fg =
Gm 2
r2
, so q =
number of electrons = N =
Gm 2
.
k
q
1
(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.0 kg) 2
=
= 2.7 × 109
e 1.602 × 10−19 C
8.988 × 109 N ⋅ m 2 C2
653
Chapter 16: Electric Forces and Fields
Physics
13. Strategy The force is attractive. Use Coulomb’s law, Eq. (16-2).
Solution
(a) Find the electric force on the positive charge.
kq q
(8.988 × 109 N ⋅ m 2 C2 )(2.0 × 10−9 C)(3.0 × 10−9 C)
F = − 12 2 = −
= − 6.0 × 10−5 N
r
(0.030 m)2
K
So, F = 6.0 × 10−5 N toward the − 3.0-nC charge .
(b) The force is equal in magnitude and opposite in direction to that found in part (a).
K
So, F = 6.0 × 10−5 N toward the 2.0-nC charge .
14. Strategy Set the magnitudes of the Coulomb force and the gravitational force equal and solve for the ratio of the
charge to mass.
Solution Find the ratio of charge to mass.
kq 2
r2
=
Gm 2
r2
, so
q
G
=
=
m
k
6.674 × 10−11 N ⋅ m 2 kg 2
8.988 × 109 N ⋅ m 2 C2
= 8.617 × 10−11 C kg .
15. Strategy Use Coulomb’s law, Eq. (16-2).
Solution Find the force on the negative charge.
F = Fq + F2q = −
kq 2
d2
+
2kq 2
(2d )2
=
K
kq 2 ⎛ 1 ⎞
kq 2
kq 2
F
−
=
−
=
1
,
so
to the left .
⎜
⎟
d2 ⎝ 2 ⎠
2d 2
2d 2
16. Strategy Use Coulomb’s law, Eq. (16-2).
Solution The original force magnitude is
F=
kq 2
r2
, so the new magnitude is F0.25 =
⎛ kq 2 ⎞
=
16
⎜ 2 ⎟ = 16 F .
⎜ r ⎟
(0.25r )2
⎝
⎠
kq 2
17. Strategy The force is attractive. Use Coulomb’s law, Eq. (16-2).
Solution Find the electric force on the potassium ion.
kq q
ke2
(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)2
F = − 12 2 = − 2 = −
= −2.8 × 10−12 N
−9
2
r
r
(9.0 × 10 m)
K
−12
−
So, F = 2.8 × 10
N toward the Cl ion .
654
Physics
Chapter 16: Electric Forces and Fields
18. Strategy Use Coulomb’s law, Eq. (16-2). The force due to the 0.80-µC charge is upward and that due to the
1.0-µC charge is to the right.
Solution Calculate the components of the force.
(8.988 × 109 N ⋅ m 2 C2 )(0.80 × 10−6 C)(0.60 × 10−6 C)
= 0.67 N
Fy =
(0.080 m)2
Fx =
(8.988 × 109 N ⋅ m 2 C2 )(1.0 × 10−6 C)(0.60 × 10−6 C)
⎛ (0.100 m) 2 − (0.080 m)2 ⎞
⎜
⎟
⎝
⎠
Calculate the magnitude of the force.
2
Fy
θ
Fx
= 1.5 N
F = Fx2 + Fy2 = (1.498 N) 2 + (0.6741 N) 2 = 1.6 N
Calculate the direction.
Fy
0.6741 N
θ = tan −1
= tan −1
= 24°
Fx
1.498 N
K
So, F = 1.6 N at 24° above the positive x-axis .
19. Strategy Use Coulomb’s law, Eq. (16-2). The force on the 1.0-µC charge due to the − 0.60-µC charge is to the
left and that due to the 0.80-µC charge is along the line between the charges and away from the 0.80-µC charge.
Solution Calculate the components of the force.
Fx = −
+
9
2
2
(8.988 × 10 N ⋅ m C )(0.60 × 10
−6
Fx
C)(1.0 × 10
⎛ (0.100 m)2 − (0.080 m) 2 ⎞
⎜
⎟
⎝
⎠
−6
(8.988 × 109 N ⋅ m 2 C2 )(0.80 × 10−6 C)(1.0 × 10−6 C) ⎛⎜ (0.100 m) 2 − (0.080 m) 2
⎜
0.100 m
(0.100 m)2
⎝
Fy = −
θ
C)
2
(8.988 × 109 N ⋅ m 2 C2 )(0.80 × 10−6 C)(1.0 × 10−6 C) ⎛ 0.080 m ⎞
⎜ 0.100 m ⎟ = − 0.58 N
(0.100 m)2
⎝
⎠
Calculate the magnitude of the force.
F = Fx2 + Fy2 = (−1.067 m)2 + (−0.575 m)2 = 1.2 N
Calculate the direction.
Fy
−0.575
θ = tan −1
= tan −1
= 28°
Fx
−1.067
K
So, F = 1.2 N at 28° below the negative x-axis .
655
⎞
⎟ = −1.1 N
⎟
⎠
Fy
Chapter 16: Electric Forces and Fields
Physics
20. Strategy Use Coulomb’s law and Hooke’s law. By symmetry, we know that the horizontal components of the
forces cancel. Draw a diagram.
Solution The magnitude of the force in the y-direction on the charge on the
spring due to one of the negative charges is
⎞
kQ q
kQ q⎛
b
F=
cos θ =
⎜
⎟.
2
2
⎜
⎟
r
r
⎝ a 2 + b2 ⎠
By symmetry, the net force is twice this. Set 2F equal to ks ∆y and to find the
7.0 µC
θθ
y
r
r
b
− 4.0 µC
− 4.0 µC
a
a
spring constant, ks .
⎞
2k Q q ⎛
b
⎜
⎟ , so
2
⎜ 2
⎟
r
⎝ a + b2 ⎠
2k Q q b
2(8.988 × 109 N ⋅ m 2 C2 )(7.0 × 10−6 C)(4.0 × 10−6 C)(0.040 m)
ks =
=
= 4500 N m .
∆y (a 2 + b 2 )3 2
(0.050 m)[(0.020 m)2 +(0.040 m)2 ]3 2
ks ∆y =
21. Strategy The force is repulsive, so the charges have the same sign. Since we are concerned only with the
magnitude of the charge on each sphere, we assume they are both positive for simplicity. Use Coulomb’s law,
Eq. (16-2).
Solution Let the total charge be Q. Then, Q = q1 + q2 . Find the charge on each sphere.
F=
kq1q2
r2
k (Q − q2 )q2
=
r2
q2 =
2
kQq2 − kq22
r2
, so q22 − Qq2 +
r2F
= 0. Solve for q2 .
k
7.50 × 10−6 C ± (7.50 × 10−6 C) 2 −
2
Q ± Q 2 − 4rk F
=
=
4(0.0600 m) 2 (20.0 N)
8.988×109 N⋅m 2 C2
2
= 6.21× 10−6 C or 1.29 × 10−6 C
Note that 7.50 × 10−6 C − 6.21× 10−6 C = 1.29 × 10−6 C.
Thus, the magnitudes of the charges are 6.21 µC and 1.29 µC .
22. Strategy Use Newton’s second law to find the force of one ball on the other, then use Coulomb’s law to find Q.
Solution Find the charge on the balls.
θ
Td
ΣFx = T sin − F = 0, so F = T sin = T =
.
L 2L
2
2
θ
mg
mg
ΣFy = T cos − mg = 0, so T =
=
=
2
2
2
cos θ2
L −( d )
2
L
Solve for Q.
kQ 2 kQ 2 Td
=
F= 2 = 2 =
2L
r
d
Q=
mgd 3
k 4 L2 − d 2
=
mgd
2L 1 −
( )
d
2L
θ/2
d
2
θ
2
=
θ/2
L
L
mg
1−
( )
d
2L
2
.
T
T
F
F
d
mg
mgd
4 L2 − d 2
, so
(9.0 × 10−8 kg)(9.80 m s 2 )(0.020 m)3
(8.988 × 109 N ⋅ m 2 C2 ) 4(0.98 m)2 − (0.020 m)2
656
= 0.020 nC .
mg
Physics
Chapter 16: Electric Forces and Fields
23. Strategy The force due to each charge is attractive. Use Coulomb’s law, Eq. (16-2).
G
K K
K
Solution The net force is F = F21 + F23. Separate F into components.
r21
k q2 q1 ⎛ r23 ⎞ k q2 q3
k q2 q1 ⎛ r13 ⎞
.
Fx = −
⎜
⎟−
⎜
⎟ and Fy = −
2
2
2
r21
r23
r21 ⎝ r21 ⎠
⎝ r21 ⎠
G
Calculate the magnitude of F.
2
F=
=
Fx2
+ Fy2
⎛ k q2 q1 r13 ⎞ ⎛ k q2 q1 r23 k
= ⎜
+
⎟ +⎜
3
3
⎜
⎟ ⎜
r21
r21
⎝
⎠ ⎝
r23 = 0.50 m
r13 = 1.20 m
2
q2 q3 ⎞
k q2 q1
⎟ =
2
3
r23 ⎟⎠
r21
⎛
2 ⎜
r13
+ r23 +
⎜
⎝
2
3 ⎞
q3 r21
⎟
2 ⎟
q1 r23
⎠
(8.988 × 109 N ⋅ m 2 C2 )(0.60 × 10−6 C)(1.2 × 10−6 C)
⎛ (1.20 m)2 + (0.50 m)2 ⎞
⎜
⎟
⎝
⎠
3
2
3⎤
⎡
(0.20 × 10−6 C) ⎛⎜ (1.20 m) 2 + (0.50 m)2 ⎞⎟ ⎥
⎢
⎝
⎠ ⎥ = 6.8 mN
× (1.20 m) 2 + ⎢0.50 m +
2
−6
⎢
⎥
(1.2 × 10 C)(0.50 m)
⎢
⎥
⎣
⎦
24. Strategy Since q is positive, Q must be negative. Let +y be up. Use Coulomb’s law, Eq. (16-2).
Solution By symmetry, Fx = 0. Set Fy = 0 at A and solve for Q.
Fy = 2 Fqy + FQy =
Therefore, Q = −
2kq 2
a2
kq Q
2q ⎛ 3a 2 ⎞
3a 2
×
−
= 0, so Q = 3 ⎜
⎟
2
a
a ⎜⎝ 4 ⎟⎠
( 3a 2)
3/ 2
⎛3⎞
= 2⎜ ⎟
⎝4⎠
3/ 2
a
3 3
q=
q.
4
a/2
3 3
q ≈ −1.3q .
4
25. Strategy Use Eq. (16-4b).
Solution Compute the force on the sphere.
K
K
F = qE = (− 6.0 × 10−7 C)(1.2 × 106 N C west) = 0.72 N to the east
E
N
− 0.60 µC
26. (a) Strategy and Solution Since positive charges move along the direction of electric field lines, the sodium
ions flow into the cell .
(b) Strategy Use Eq. (16-4b).
Solution Compute the magnitude of the electric force on the sodium ion.
F = eE = (1.602 × 10−19 C)(1.0 × 107 N C) = 1.6 × 10−12 N
27. Strategy Use Newton’s second law and Eq. (16-4b).
G
G
G
Solution F = qE = eE for a proton. Find the acceleration.
K
K
K
K eE (1.602 × 10−19 C)(33 × 103 N C up)
ma = eE, so a =
=
= 3.2 × 1012 m s 2 up .
−27
m
1.673 × 10
kg
657
E
+
p
a
3 a/2
Chapter 16: Electric Forces and Fields
Physics
28. Strategy Use Newton’s second law and Eq. (16-4b).
G
G
G
Solution F = qE = −eE for an electron. Find the acceleration.
K
K
eE
(1.602 × 10−19 C)(6100 N C north)
K
K
ma = − eE, so a = −
=−
= 1.1× 1015 m s 2 south .
−31
m
9.109 × 10
kg
E
−
e−
a
29. Strategy The electric field at the midpoint is directed away from the positive charge and toward the negative
charge. The magnitude of the field is the sum of the magnitudes of the fields due to each charge. Use Eq. (16-5).
Solution Find the electric field midway between the two charges.
k Q1
k Q2
4k
4(8.988 × 109 N ⋅ m 2 C2 )
E=
Q
Q
+
=
+
=
(15 × 10−6 C + 12 × 10−6 C) = 1.5 × 108 N C
(
)
1
2
(d 2) 2 (d 2) 2 d 2
(0.080 m)2
K
So, E = 1.5 × 108 N C directed toward the − 15-µC charge .
30. Strategy and Solution Since the electric field is directed upward and the charge of an electron is negative, the
direction of the electric force is downward .
31. Strategy Electric field lines begin on positive charges and end on negative charges. The same number of field
lines begins on the plate and ends on the negative point charge. Field lines never cross. Use the principles of
superposition and symmetry.
Solution The electric field lines for the system of the point charge and metal plate:
32. Strategy The electric field at x = d is the vector sum of the electric fields due to each positive charge. Use Eq.
(16-5).
Solution Find the electric field.
k q k 2q
kq kq
kq
Ex = 2 −
= 2 − 2 = 2
2
d
d
(2d )
2d
2d
The electric field is
kq
2d 2
to the right .
33. Strategy The electric field at x = 2d is the vector sum of the electric fields due to each positive charge. Use Eq.
(16-5).
Solution Find the electric field.
kq
k 2q k q 2k q
7k q
Ex =
− 2 = 2 − 2 =−
2
d
d
(2d )
4d
4d 2
The electric field is
7k q
4d 2
to the left .
34. Strategy and Solution Since both charges are positive, the electric field is nonzero at all locations not on the
x-axis. The answer is no.
658
Physics
Chapter 16: Electric Forces and Fields
35. Strategy and Solution The electric field due to each charge is directed to the left for x < 0, and to the right for
x > 3d; therefore, the electric field cannot be zero in these regions. In the region between the charges (on the
x-axis), the electric fields due to the charges are in opposite directions; thus, the electric field is zero in the region
0 < x < 3d .
36. Strategy The electric field at any point x is the vector sum of the electric fields due to each positive charge. Use
Eq. (16-5) and the quadratic formula.
Solution Find where the electric field is equal to zero.
kq
k 2q
Ex = 0 = 2 −
, so ( x − 3d ) 2 = 2 x 2. Solve for x.
2
x
(3d − x)
x 2 − 6dx + 9d 2 = 2 x 2
0 = x 2 + 6dx − 9d 2
x=
−6d ± (6d ) 2 − 4(1)(−9d 2 )
2(1)
−6d ± 72d 2
2
x = −3d ± 3d 2 = 3d (−1 ± 2) ≈ 1.24d or − 7.24d
The value −7.24d is extraneous, since the electric field is nonzero at x = −7.24d .
x=
The electric field is zero at x = 3d (−1 + 2) ≈ 1.24d .
37. Strategy Electric field lines begin on positive charges and end on negative charges. The magnitude of the
negative charge is twice that of the positive charges (which have equal magnitude). The same number of field
lines begins on each of the positive charges and all end on the negative charge. Field lines never cross. Use the
principles of superposition and symmetry.
Solution The electric field lines for the system of three charges:
+
–
+
659
Chapter 16: Electric Forces and Fields
Physics
38. Strategy Electric field lines begin on positive charges and end on negative charges. In each situation, the
magnitudes of the charges are equal, so the number of field lines beginning (for positive) or ending (for negative)
on each is the same. Field lines never cross. Use the principles of superposition and symmetry.
Solution
(a) Electric field lines due to two positive point charges:
+
+
(b) Electric field lines due to two negative point charges:
−
−
39. Strategy Let the x-direction be to the right and the y-direction be up. Due to symmetry, the x-components of the
fields due to the two charges add to zero at point B; the vector sum of the y-components of the two charges is
equal to twice that due to either one.
Solution Find the electric field at point B.
k q 2(8.988 × 109 N ⋅ m 2 C2 )(7.00 × 10−6 C)
E = 2E y = 2
=
= 1.12 × 106 N C
2
2
2
r
(0.300 2 m) + (0.300 m)
The electric field is 1.12 × 106 N C up .
40. Strategy Let the x-direction be to the right and the y-direction be up. Due to symmetry, the x-components of the
fields due to the two charges add to zero at point C; the vector sum of the y-components of the two charges is
equal to twice that due to either one.
Solution Find the electric field at point C.
k q 2(8.988 × 109 N ⋅ m 2 C2 )(7.00 × 10−6 C)
E = 2E y = 2
=
= 2.80 × 106 N C
2
2
2
r
(0.300 2 m) + (0.300 2 m)
The electric field is 2.80 × 106 N C up .
660
Physics
Chapter 16: Electric Forces and Fields
41. Strategy Let the x-direction be to the right and the y-direction be up. Label the charge on the left 1 and the charge
on the right 2. Let d equal the side length of the square. The electric field due to the charge on the left is directed
upward at point A, so it only has a y-component. Use Eq. (16-5).
Solution Find the electric field at point A.
kq
2k q
E x = E1x + E2 x = 0 −
cos 45° =
2
2d
4d 2
E y = E1 y + E2 y =
kq
d2
+
kq
2d 2
sin 45° =
kq
d2
2k q
+
4d 2
=
k q (4 + 2)
4d 2
Compute the magnitude.
⎛ 2k q
E= ⎜
⎜ 4d 2
⎝
=
2
2
⎞ ⎡ k q (4 + 2) ⎤
kq
⎟ +⎢
⎥ = 2
2
⎟ ⎢
4d
4d
⎠ ⎣
⎦⎥
2 + (4 + 2) 2
(8.988 × 109 N ⋅ m 2 C2 )(7.00 × 10−6 C) 2 + (4 + 2) 2
4(0.300 m) 2
= 9.78 × 105 N C
Compute the direction.
Ey
4+ 2
θ = tan −1
= tan −1
= 14.6° CCW from a vertical axis through the left side of the square
Ex
2
The electric field is 9.78 × 105 N C at 14.6° CCW from a vertical axis through the left side of the square .
42. Strategy Let the origin of the coordinate system be at A. The charges are positive, so the electric field due to the
left-hand charge (1) points upward at A and that due to the right-hand charge (2) is directed at an angle of 135° as
measured from the positive x-axis. Find the net electric field at A. Then, locate the third object such that its
electric field is equal in magnitude and opposite in direction to the field generated by the first two objects.
Solution Let d = 0.300 m. Find the net electric field due to the two objects.
kq
k q cos135°
E x = E1x + E2 x = 0 +
cos135° =
2
2d
2d 2
kq kq
k q ( 2 + sin135° )
E y = E1 y + E2 y =
+
sin135° =
2
2
d
2d
2d 2
Compute the magnitude.
2
y
E1
E2
2
x
E3
⎛ k q cos135° ⎞ ⎡ k q ( 2 + sin135° ) ⎤
kq
+⎢
=
E= ⎜
(cos135°)2 + (2 + sin135°)2
⎟
⎥
2
2
2
2d
2d
2d
⎝
⎠ ⎣
⎦
Set the magnitude of the electric field due to the third object equal to E and solve for the distance r that the third
object should be from A.
kq kq
(cos135°)2 + (2 + sin135°)2 , so
=
2d 2
r2
2
2
r=d
= (0.300 m)
= 0.254 m.
(cos135°)2 + (2 + sin135°)2
(cos135°) 2 + (2 + sin135°)2
Compute the direction.
Ey
2 + sin135°
θ = tan −1
= tan −1
= 75.4° above the − x-axis
cos135°
Ex
The charge should be placed on a line that makes an angle of 75.4° above the negative x -axis at
a distance of 0.254 m along that line from point A. Thus, the charge is above and to the left of A.
661
Chapter 16: Electric Forces and Fields
Physics
43. Strategy The net force on the charge at the origin is the vector sum of the forces due to each of the other two
charges. Use Coulomb’s law, Eq. (16-2).
Solution The force due to the negative charge is to the right (positive)
and that due to the positive charge is to the left (negative).
k q1 q2 k q1 q3
ΣF =
−
r122
r132
3.00 µC
−5.00 µC
8.00 µC
40.0 x (cm)
20.0
2
0
1
3
⎡ (3.00 × 10−6 C)(5.00 × 10−6 C) (3.00 × 10−6 C)(8.00 × 10−6 C) ⎤
= (8.988 × 109 N ⋅ m 2 C2 ) ⎢
−
⎥ = 1.61 N
(0.200 m)2
(0.350 m)2
⎥⎦
⎣⎢
So, the force on the charge at the origin is 1.61 N in the +x-direction .
44. Strategy Let s be the side of the square. Use Eq. (16-5) and the principles of superposition and symmetry.
Solution Find the magnitude of the electric field at point D.
2
EB
D
2
kQ 2 (8.988 × 109 N ⋅ m2 C2 )(1.00 × 10−9 C) 2
⎛ kQ ⎞ ⎛ kQ ⎞
E = EB2 + E A2 = ⎜ 2 ⎟ + ⎜ 2 ⎟ =
=
s2
(1.0 m) 2
⎝s ⎠ ⎝s ⎠
= 13 N C
E
EA
45. Strategy E x = 0 due to symmetry. If r is the distance to x = 4.0 m, y = 3.0 m from the two known charges, then
sin θ = y r . The field due to the known charges is upward, so that due to the unknown charge Q must be
downward, thus Q < 0. Use Eq. (16-5) and the principle of superposition.
Solution Find the unknown charge.
2kq ⎛ y ⎞ k Q
E y = 2 ⎜ ⎟ − 2 = 0, so
r ⎝r⎠ y
Q =
y (cm)
4.0
Eq
Eq
EQ
2qy 3
2q(3.0 m)3
=
= 0.43q.
3
r3
⎛ (4.0 m)2 + (3.0 m) 2 ⎞
⎜
⎟
⎝
⎠
q
Q
q
8.0 x (cm)
4.0
Q < 0, so Q = − 0.43q .
46. Strategy Use Eq. (16-5) and the principles of superposition and symmetry. Set the sum of the electric fields equal
to 0. x = d is the point at which the sum is zero.
Solution Find the point where the electric field is zero.
K
K
kq
kq2
0 = E1 + E2 , so 0 = 21 −
. Solve for d.
d
( x2 − d ) 2
( x2 − d )2
d2
20.0 nC
0
1
10.0 nC
0.50
1.00
2
x (m)
2
q
x2
1.00 m
⎛x
⎞
= ⎜ 2 − 1⎟ = 2 , so d =
=
= 0.586 m or 3.41 m.
q1
1 ± q2 q1 1 ± 10.0 20.0
⎝ d
⎠
3.41 m is extraneous (the field lines both point in the same direction at that point), so d = 0.586 m .
662
Physics
Chapter 16: Electric Forces and Fields
47. Strategy Use Eq. (16-5) and the principles of superposition and symmetry. If r is the distance to (x, y) =
(0.50 m, 0.50 m) from each charge, then cos θ = x r and sin θ = y r .
Solution Find the magnitude of the electric field.
y (m)
2
⎡ kq ⎛ x ⎞ kq ⎛ x ⎞ ⎤ ⎡ kq ⎛ y ⎞ kq ⎛ y ⎞ ⎤
E = E x2 + E y2 = ⎢ 21 ⎜ ⎟ − 22 ⎜ ⎟ ⎥ + ⎢ 21 ⎜ ⎟ + 22 ⎜ ⎟ ⎥
⎣ r ⎝ r ⎠ r ⎝ r ⎠⎦ ⎣ r ⎝ r ⎠ r ⎝ r ⎠⎦
k
= 3 [(q1 − q2 ) x]2 + [(q1 + q2 ) y]2
r
=
[(20.0 × 10−9 C − 10.0 × 10−9 C)(0.50 m)]2
8.988 × 109 N ⋅ m 2 C2
⎛ (0.50 m)2 + (0.50 m)2 ⎞
⎜
⎟
⎝
⎠
2
3
+ [(20.0 × 10−9 C + 10.0 × 10−9 C)(0.50 m)]2
E1
E2
0.50
10.0 nC
20.0 nC
0
1
0.50
1.00
2
x (m)
= 400 N C
48. Strategy Electrons have negative charge. The electric field points in the positive x-direction, so the electron will
be accelerated in the negative x-direction. Use the acceleration of a point charge in an electric field.
Solution The y-coordinate of the electron’s position is zero for all t. Find the x-coordinate of the electron.
1
1⎛ F ⎞
1 ⎛ qE ⎞
(1.602 × 10−19 C)(232 N C)(2.30 × 10−9 s)2
2
∆ x = a x ( ∆t ) 2 = ⎜ − ⎟ ( ∆ t ) 2 = ⎜ −
= − 0.108 mm
⎟ ( ∆t ) = −
2
2⎝ m⎠
2⎝ m ⎠
2(9.109 × 10−31 kg)
The coordinates of the electron’s position are (− 0.108 mm, 0) .
49. (a) Strategy Use Eq. (16-4b).
Solution Find the force on the electron.
K
K
F = − eE = − (1.602 × 10−19 C)(500.0 N C up) = 8.010 × 10−17 N down
(b) Strategy Use the work-kinetic energy theorem. The work done on the electron is equal to the force on the
electron times the deflection.
Solution Find the increase in the kinetic energy of the electron.
∆K = W = Fd = (8.010 × 10−17 N)(0.00300 m) = 2.40 × 10−19 J
663
Chapter 16: Electric Forces and Fields
Physics
50. (a) Strategy and Solution Since electrons travel opposite electric field lines and the electrons are deflected
upward, the electric field is directed from the top plate to the bottom plate .
(b) Strategy Use Eq. (16-6).
Solution Find the charge per unit area.
Q
= ⑀0 E = [8.854 × 10−12 C2 (N ⋅ m 2 )](2.00 × 104 N C) = 1.77 × 10−7 C m 2
A
(c) Strategy Use Newton’s second law.
Solution Find the deflection d in terms of the time.
1
∆y = a y (∆t )2 = d
2
Find the time ∆t.
∆x
∆t =
vi
Find a y .
ma y = eE, so a y =
eE
.
m
Substitute.
2
d=
1 ⎛ eE ⎞ ⎛ ∆x ⎞
(1.602 × 10−19 C)(2.00 × 104 N C)(0.020 m) 2
= 0.44 mm
⎜
⎟ =
⎜
⎟
2 ⎝ m ⎠ ⎝ vi ⎠
2(9.109 × 10−31 kg)(4.0 × 107 m s) 2
51. (a) Strategy Compare the electrical and gravitational forces.
Solution The gravitational force is mg = (0.00230 kg)(9.80 m s 2 ) = 2.25 × 10−2 N. The electrical force is
qE = (10.0 × 10−6 C)(6.50 × 103 N C) = 6.50 × 10−2 N.
The gravitational force is about 1 3 of the electrical force, so the gravitational force can’t be neglected.
(b) Strategy Add the forces and find the total acceleration using Newton’s
second law. Then, use the formula for the range of a projectile.
vi
55.0°
Solution Find the downward acceleration.
Fg + Fe
a=
m
Find ∆x.
∆ x = 1.78 m
v 2 sin 2θ mvi 2 sin 2θ (0.00230 kg)(8.50 m s)2 sin[2(55.0°)]
∆x = R = i
=
=
= 1.78 m
a
Fg + Fe
2.25 × 10−2 N + 6.50 × 10−2 N
664
Physics
Chapter 16: Electric Forces and Fields
52. (a) Strategy Compare the electrical and gravitational forces.
Solution The gravitational force is mg = (1.673 × 10−27 kg)(9.80 m s 2 ) = 1.64 × 10−26 N . The electrical
force is qE = (1.602 × 10−19 C)(6.50 × 103 N C) = 1.04 × 10−15 N .
Yes, gravity can be neglected. The electrical force is about 10 orders of magnitude larger than the
gravitational force.
(b) Strategy Find the acceleration using Newton’s second law. Then, use the
formula for the range of a projectile.
v
55.0°
Solution Find the downward acceleration.
F
a= E
m
Find ∆x.
∆ x = 1.09 m
v 2 sin 2θ mv 2 sin 2θ (1.673 × 10−27 kg)(8.50 × 105 m s) 2 sin[2(55.0°)]
∆x = R = i
=
=
= 1.09 m
a
FE
1.04 × 10−15 N
53. Strategy Find the necessary acceleration in terms of the average electric field.
Solution The acceleration is related to the average electric field by a = qE m .
vfx 2 − vix 2 = vfx 2 − 0 = 2a x ∆x =
mv 2 (1.673 × 10−27 kg)(1.0 × 107 m s) 2
2qE ∆x
= 1.3 × 105 N C .
, so E = fx =
−19
m
2q∆x
2(1.602 × 10
C)(4.0 m)
54. (a) Strategy and Solution Electrons have negative charge, so they are deflected toward the positive plate .
(b) Strategy Use Newton’s second law.
Solution Find the deflection d in terms of the time.
1
∆y = a y (∆t )2 = d
2
Find the time ∆t.
∆x
∆t =
vi
Find a y .
ma y = eE, so a y =
eE
.
m
Substitute.
2
1 ⎛ eE ⎞ ⎛ ∆x ⎞
(1.602 × 10−19 C)(1.0 ×103 N C)(0.0250 m)2
d= ⎜
= 0.78 mm
⎜
⎟ =
⎟
2 ⎝ m ⎠ ⎝ vi ⎠
2(9.109 × 10−31 kg)(8.4 × 106 m s) 2
665
Chapter 16: Electric Forces and Fields
Physics
55. (a) Strategy and Solution Electrons have negative charge, so the field must be oriented vertically downward
for them to be deflected upward.
(b) Strategy Use Newton’s second law.
Solution Find the deflection d in terms of the time.
1
∆y = a y (∆t )2 = d
2
Find the time ∆t.
∆x
∆t =
vi
Find a y .
2dvi2
=
(∆t )2 (∆x)2
ΣFy = eE = ma y , so E = ma y /e. Calculate E.
ay =
E=
2d
m ⎡ 2dvi2 ⎤ 2(9.109 × 10−31 kg)(0.0020 m)(8.4 × 106 m s)2
= 2600 N C
⎢
⎥=
e ⎢⎣ (∆x)2 ⎥⎦
(1.602 × 10−19 C)(0.0250 m)2
(c) Strategy Use d =
1
g (∆t )2 and ∆x = v∆t.
2
Solution Find the deflection of the electrons due to the gravitational force.
2
d=
1
1 ⎛ ∆x ⎞
(9.80 m s 2 )(0.0250 m)2
g (∆t )2 = g ⎜
= 4.3 × 10−17 m
⎟ =
6
2
2
2 ⎝ vi ⎠
2(8.4 × 10 m s)
56. Strategy and Solution The − 6 µC of charge on the conducting sphere induces a positive charge of 6 µC on the
inner surface of the conducting shell. This, in turn, induces a negative charge on the outer surface of the shell. The
conducting shell has a net 1-µC charge, so −6 µC + 1 µC = −5 µC is the charge on the outer surface.
57. Strategy The charge on the inner surface is induced by the net charge contained within the shell. The charge on
the outer surface is equal in magnitude and opposite in sign to the charge on the inner surface plus the net charge.
Solution
(a) The 6 µC of charge within the shell induces a − 6 µC charge on the inner surface of the shell.
(b) The shell has a net charge of 6 µC, so the charge on the outer surface is 6 µC + 6 µC = 12 µC .
666
Physics
Chapter 16: Electric Forces and Fields
58. Strategy The electric field is zero at any point inside a conductor in electrostatic equilibrium. The charge is on
the surface of the sphere. Use Eq. (16-5) and symmetry.
Solution
K
(a) Inside a conducting sphere (r < R), the electric field is zero, or E(r < R) = 0 . Outside a conducting sphere
with charge − q, the field points radially toward the sphere (symmetry, negative charge). So,
K
kq
E(r > R) = 2 radially toward the sphere .
r
(b) Sketch the graph using the results of part (a).
E(r)
kq
R2
0
R
r
59. (a) Strategy The electric field between the charged sphere and the spherical shell is the same as that due to a
point charge at the center of the sphere with the same charge as the sphere. Use Eq. (16-5).
Solution Compute the magnitude of the electric field.
kq (8.988 × 109 N ⋅ m 2 C2 )(230 × 10−9 C)
E=
=
= 6.8 × 106 N C
r2
(0.0175 m)2
(b) Strategy The electric field is zero at any point within a conducting material in electrostatic equilibrium.
Solution The spherical shell is in electrostatic equilibrium. All points 2.50 cm from the center of the sphere
are within the spherical shell. Therefore, the electric field there is zero .
(c) Strategy The electric field outside of the spherical shell is the same as that due to a point charge at the center
of the sphere with the same charge as the sphere. Use Eq. (16-5).
Solution Compute the magnitude of the electric field.
kq (8.988 × 109 N ⋅ m 2 C2 )(230 × 10−9 C)
E=
=
= 2.3 × 106 N C
2
2
r
(0.0300 m)
60. Strategy The charge on the inner surface is induced by the net charge contained within the cavity. The charge on
the outer surface is equal in magnitude and opposite in sign to the charge on the inner surface plus the net charge.
Solution
(a) q1 + q2 = 5 µC + (−12 µC) = −7 µC is the net charge contained within the cavity. So, the charge on the inner
surface is 7 µC .
(b) The conductor has a net charge of − 4 µC, so the outer surface has a charge of −7 µC − 4 µC = −11 µC .
667
Chapter 16: Electric Forces and Fields
Physics
61. (a) Strategy Since the electric field points toward Earth, the charge is negative. Just outside of a conducting
sphere, the field is nearly uniform since the curved surface is approximately flat for a small area A. Thus, the
expression for the electric field between two oppositely-charged plates can be used.
Solution Calculate the total charge.
Q
E=
, so Q = ⑀0 AE = [8.854 × 10−12 C2 (N ⋅ m 2 )]4π (6.371× 106 m) 2 (−150 N C) = − 6.8 × 105 C .
⑀0 A
Calculate the charge per unit area.
Q
= ⑀0 E = [8.854 × 10−12 C2 (N ⋅ m 2 )](−150 N C) = −1.3 nC m 2
A
(b) Strategy Since the Earth has negative charge, the charge density of the air must be positive.
Solution Calculate the charge density of the air.
∆Q ∆Q ⎛ h ⎞ 1 ∆Q
h, so
∆E =
=
=
⑀0 A ⑀0 ⎜⎝ V ⎟⎠ ⑀0 V
∆Q ⑀0∆E [8.854 × 10−12 C2 (N ⋅ m 2 )](150 N C − 120 N C)
=
=
= 1× 10−12 C m3 .
250 m
V
h
62. Strategy Use the definition of electric flux, Eq. (16-8).
Solution
(a) Φ E = E⊥ A = E⊥ a 2 , so Φ E& = 0, Φ E ⊥out = Ea 2 , and Φ E ⊥in = − Ea 2.
(b) Φ E = Ea 2 − Ea 2 = 0
G
63. Strategy E is at 60.0° with respect to A, so it is at 30.0° with respect to the normal of A. Use the definition of
electric flux, Eq. (16-8).
Solution Find the flux through the rectangle.
Φ E = EA cosθ = EA cos 30.0° = 0.866 EA
64. Strategy Since the charge is located at the center of the cube, the electric flux through one side of the cube is
one-sixth of the total flux. Use Gauss’s law, Eq. (16-9).
Solution Find the flux through one side of the cube.
4π (8.988 × 109 N ⋅ m 2 C2 )(0.890 × 10−6 C)
Φ E = 4π kq =
= 1.68 × 104 N ⋅ m 2 C
6
668
Physics
Chapter 16: Electric Forces and Fields
65. Strategy Use the definition of electric flux, Eq. (16-8), Gauss’s law, Eq. (16-9), and Coulomb’s law.
Solution
(a) The expression for the electric flux is Φ E = E⊥ A = E (4π r 2 ) = 4π r 2 E .
(b) Use Gauss’s law.
Φ E = cq = 4π r 2 E, so E =
cq
4π r
2
, and by Coulomb’s law, E =
q
4π ⑀0r 2
.
Solve for c.
cq
q
1
, so c = .
=
2
2
⑀0
4π r
4π ⑀0r
66. Strategy Use Gauss’s law, Eq. (16-9), and symmetry.
Solution
(a) The electric field outside a spherically symmetric charge distribution (with total charge q) is directed radially
away from its center, and is parallel at any point to the normal of a spherical Gaussian surface outside the
distribution and concentric with it. So, E⊥ = E.
kq
Φ E = E⊥ A = EA = E (4π r 2 ) = 4π kq, so E = 2 .
r
The result is the same as the electric field due to a point charge q.
(b) If a spherical Gaussian surface is placed within the charge distribution, q = 0 (no enclosed charge).
Φ E = E⊥ A = EA = 4π kq, so E = 4π kq A = 4π k (0) A = 0.
67. Strategy Use the results of Problem 66.
Solution
(a) The electric field magnitude due to a solid sphere of radius R with a uniform charge q spread throughout is
kq
E (r ≥ R ) = 2 .
r
(b) At some point r ≤ R, find the total charge enclosed within a spherical Gaussian surface.
4
q = π R3 ρ , where ρ is the uniform volume charge density.
3
3
Therefore, qenc =
r3
kq
The area is A = 4π r 2 , so E (4π r 2 ) = 4π kq 3 or E (r ≤ R) = 3 r . (E varies linearly with r.)
R
R
(c) The maximum magnitude of the electric field is Emax = E ( R) = kq R 2 . Sketch the graph.
E(r)
kq
R2
0
R
3
4 3
4
q
⎛r⎞
⎛r⎞
π r ρ = π r3 ×
= q ⎜ ⎟ . By Gauss’s law, Φ E = EA = 4π kqenc = 4π kq ⎜ ⎟ .
4 π R3
3
3
R
⎝ ⎠
⎝R⎠
3
2R
3R r
669
Chapter 16: Electric Forces and Fields
Physics
68. Strategy Since the electron has negative charge and is suspended above the line of charge, the electric force on
the electron must be upward to counteract the force due to gravity. For the electric force to be upward, the electric
field must be downward; therefore, the line of charge must be negative. Refer to Example 16.12. Use Newton’s
second law.
Solution According to Newton’s second law, ΣFy = (−e)(− E ) − mg = 0, so E = mg e .
From Example 16.12, the electric field due to a long, thin wire is given by Er = 2k λ r . Set
these two expressions for the electric field equal and solve for the magnitude of the linear
charge density.
mgr (9.109 × 10−31 kg)(9.80 m s 2 )(0.0120 m)
2k λ mg
=
, so λ =
=
= 3.72 × 10−23 C m.
r
e
2ke 2(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)
E
e−
λ
FE
y
mg
The linear charge density is −3.72 × 10−23 C m .
69. Strategy Use the properties of electric fields and the rules for sketching field lines, and Gauss’s law, Eq. (16-9).
Solution
(a) The electric field lines due to the (finite) sheet:
(b) The electric field lines due to an infinitely large sheet:
(c) The electric field lines in (b) are uniform, so
the field strength is independent of the distance from the sheet.
(d) The electric field lines in (a) are nearly uniform close to the sheet and far from the edges, so the answer is
yes .
(e) The Gaussian surface is a “pill box.” It is a cylinder with its top and bottom circular surfaces parallel to the
surface of the sheet, which bisects the cylinder. The electric field lines are approximately parallel to the side
of the cylinder, so Φ E side = E⊥ Aside = 0, or E⊥ = 0.
q
Φ E net = Etop Atop + Ebottom Abottom =
⑀0
K
K
E top = −Ebottom , and the outward normal of Atop is opposite to that for Abottom and the areas are equal.
Find the magnitude.
EA + (− E )(− A) = 2 EA =
q
⑀0
, so E =
1 ⎛q⎞ σ
.
=
2⑀0 ⎜⎝ A ⎟⎠ 2⑀0
670
Physics
Chapter 16: Electric Forces and Fields
70. (a) Strategy and Solution The electric field is zero inside a conductor in static equilibrium, so the electric field
inside the sheet is 0 .
(b) Strategy Use Gauss’s law, Eq. (16-9).
Solution The Gaussian surface is a “pill box.” It is a cylinder with its top and bottom circular surfaces
parallel to the surface of the sheet. The electric field lines are approximately parallel to the side of the
cylinder, so Φ E side = E⊥ Aside = 0, or E⊥ = 0.
q
Φ E net = Etop Atop + Ebottom Abottom = , where the top is outside the sheet and the bottom is inside the sheet.
⑀0
Ebottom = 0, and let Atop = Abottom = A and Etop = E, the field just outside the sheet.
Find E.
EA + (0) A = EA =
q
⑀0
, so E =
1 ⎛q⎞ σ
= .
⑀0 ⎜⎝ A ⎟⎠ ⑀0
(c) Strategy and Solution No, the results do not contradict each other. Applying the superposition principle to
two parallel sheets of charge gives the same result.
71. (a) Strategy Use the properties of electric fields and the rules for sketching field lines.
+++
+
++
+
+
+
+
++
+
++
+
Solution The electric field lines for the parallel-plate
capacitor are shown at right. The figure is drawn with
more field lines between the plates than outside due to
the fact that
the fields due to each plate have the same direction
(adding fields) between the plates and opposite
directions (canceling fields) outside. Thus, the field
is much stronger between the plates
––– – –– – – – – –– – –––
and the charge is mostly on the inner surfaces of the plates.
(b) Strategy Use Gauss’s law, Eq. (16-9).
Solution The Gaussian surface is a cylinder with one end (with area A) within the +q plate and the other
between the plates, and with the ends’ surfaces parallel to the plates. Φ E through the curved surface is 0
( EA cos 90° = 0). Φ E ends = EA + 0 (E inside the plate is 0.)
σA σ
q
q
q = σ A and Φ E = EA = , so E =
=
= .
⑀0
⑀0 A ⑀0 A ⑀0
(c) Strategy and Solution The result agrees with that of Problem 69. The positive charge on each side of the
plate (q) creates the same magnitude electric field outside the plate as the positive and negative charges (±q)
do between plates.
(d) Strategy Use the result of Problem 69 and the principle of superposition.
Solution From Problem 61, E = σ (2⑀0 ) for a single charged plate.
Etotal = E+ + E− =
σ
+
σ
=
σ
2⑀0 2⑀0 ⑀0
The fields due to each plate have the same direction, so the fields add.
671
Chapter 16: Electric Forces and Fields
Physics
72. Strategy Use the properties of electric fields and the rules for sketching field lines, and Gauss’s law, Eq. (16-9).
Solution
(a) The electric field lines for the cable:
(b)
K
E (r ≤ a) = 0 , since E is zero inside a conductor.
a < r < b : Let the Gaussian surface be a cylinder coaxial with the wire and shell.
Φ E = 0 at the ends of the cylinder, since E⊥ = E cos θ = E cos 90° = 0. At the side
of the cylinder, E⊥ = E, and Φ E = EA = 4π kq. So,
4π kq 4π k (λ L)
2k λ
or E (a < r < b) =
.
=
2π rL
A
r
r > b: Use the same Gaussian surface with r > b. The same arguments hold, but the
enclosed charge is different.
4π kq 4π k
2k
E=
(qwire + qshell ) =
[λ L + (−λ ) L], so E (r ≥ b) = 0 .
=
A
2π rL
rL
E=
73. Strategy Use the definition of electric flux, Eq. (16-8), and Gauss’s law, Eq. (16-9). The appropriate closed
surface for the shell is a sphere. Due to symmetry, the electric field lines must be normal to the closed surface.
Since the charge is positive, the electric field is directed radially away from the center of the spherical shell.
Solution Find the electric field outside of the shell.
Φ E = 4π kq = EA cos θ = EA = E (4π r 2 ), so kq = Er 2 and E =
The electric field is
kq
r2
kq
r2
.
(r > R) directed radially away from the center of the shell .
74. Strategy Set the magnitude of the weight equal to that of the electric force. Use Coulomb’s law, Eq. (16-2).
Solution Find the mass.
ke2
ke2 (8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)2
mg = 2 , so m = 2 =
= 5.9 kg .
r
gr
(9.80 m s 2 )(2.0 × 10−15 m)2
75. Strategy Use Eq. (16-5).
Solution Find the distance from the point charge where the magnitude of the electric field is 5.0 kN C.
E=
kq
r
, so r =
2
kq
=
E
(8.988 × 109 N ⋅ m 2 C2 )(1.0 × 10−6 C)
5.0 × 103 N C
= 1.3 m .
76. Strategy Since the charge is negative and the electric field is directed upward, the force on the charge is directed
downward. Use Eq. (16-4b).
Solution Compute the magnitude of the force on the raindrop.
F = 8eE = 8(1.602 × 10−19 C)(2.0 × 106 N C) = 2.6 pN
The force on the raindrop is 2.6 pN down.
672
Physics
Chapter 16: Electric Forces and Fields
77. Strategy Use Eq. (16-4b).
Solution Find the force on each electron.
K
K
K
F = qE = −eE = −(1.602 × 10−19 C)(2.00 × 105 N C downward) = 3.20 × 10−14 N upward
78. Strategy Use Coulomb’s law, Eq. (16-2). Set the sum of the forces due to q2 and q3 equal to zero.
Solution Find the location of the third point charge.
kq q
k q1 q3
q
q
0 = − F12 + F13 = − 1 2 2 +
, so 22 = 32 and
2
x2
x3
x2
x3
q3
x3 = ±
q2
x2 = ±
−3.0 µC
5.0 µC
−20.0
2
0
1
−10.0
x (cm)
8.0 µC
(−20.0 cm) = ± 33 cm.
3.0 µC
−33 cm is extraneous (the force on q1 ≠ 0), so q3 must be placed at x = 33 cm .
79. Strategy Use Newton’s second law and Coulomb’s law, Eq. (16-2).
Solution Find the speed of the electron.
ΣFr =
ke2
r
2
= mar =
mv 2
, so v =
r
ke2
=
mr
(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)2
(9.109 × 10
−31
kg)(5.3 × 10
−11
m)
= 2.2 × 106 m s .
80. Strategy Use Coulomb’s law and the principle of superposition.
Solution
(a) Find the electric field at point P.
⎛ q
K
kq
kq
q ⎞
E( P) = 2t down + 2b up = k ⎜ − 2t + 2b ⎟ up
⎜
rt
rb
rb ⎟⎠
⎝ rt
⎡
50 C
20 C ⎤
= (8.988 × 109 N ⋅ m 2 C2 ) ⎢ −
+
up = 4.0 × 104 N C up
2
2⎥
(10,000
m)
(2000
m)
⎣⎢
⎦⎥
(b) Since the electric field is directed upward at the surface, positive charge would accumulate there.
81. Strategy Use Coulomb’s law and the principle of superposition.
Solution Find q.
0 = E x (1.0 m, 0) =
kq0
d2
+
kq
(d 2) 2
= q0 + 4q, so
1
1
q = − q0 = − (6.0 nC) = −1.5 nC .
4
4
673
6.0 nC
q
E=0
0
0.50
1.0
x (m)
Chapter 16: Electric Forces and Fields
Physics
82. Strategy Use Coulomb’s law, Eq. (16-2). Form a ratio.
Solution According to Coulomb’s law,
kQ Q
kQcQa
kQ Q
Fba = b2 a and Fca =
= c 2a .
2
d
(d 2)
2d
Find the ratio of Fca to Fba .
kQcQa
Fca
2d 2 = Qc = 1 , since Q = Q .
= kQ
c
b
bQa
Fba
2Qb
2
d2
83. Strategy Qnet = 5.0 µC − 1.0 µC = 4.0 µC. After the spheres are brought into contact, each will have a charge
Qnet / 2 = 2.0 µC = Q. Use Coulomb’s law, Eq. (16-2). Form a ratio.
Solution Find the ratio of the magnitudes.
kQ 2
2
Fafter
Q2
(2.0 µC)2
= L
=
=
= 0.80
Fbefore k Q1 Q2
Q1 Q2 (5.0 µC)(1.0 µC)
L2
84. Strategy Use Coulomb’s law, Eq. (16-2).
Solution
(a) Approximate the force.
k q1 q2 (8.988 × 109 N ⋅ m 2 C2 )(1.0 × 10−6 C)(0.2 × 10−6 C)
F=
=
= 2 mN
r2
(1.00 m)2
(b)
Coulomb’s law is only valid for point charges or when the sizes of the charge distributions are much
smaller than their separation.
(c) Since the positive charges on each sphere will move toward the outer (opposite) sides of the spheres due to
the repulsive force between them, the average distance separating charges will be larger than 12 cm, so the
actual force would be smaller .
674
Physics
Chapter 16: Electric Forces and Fields
85. Strategy The net electric field at P is the vector sum of the electric fields at that location due to both of the
charges. Let the left-hand charge by 1 and the right-hand charge be 2. Also, let a = 0.0340 m and b = 0.0140 m.
Use Eq. (16-5) and the principle of superposition.
Solution Find the components of the electric field.
⎞
k q2 ⎛
k q2 a
a
=
E x = E1x + E2 x = 0 +
⎜
⎟
a 2 + b 2 ⎜⎝ a 2 + b 2 ⎟⎠ (a 2 + b 2 )3 2
(8.988 × 109 N ⋅ m 2 C2 )(47.0 × 10−9 C)(0.0340 m)
=
[(0.0340 m)2 +(0.0140 m)2 ]3 2
E1
y
x
P
E2
= 2.89 × 105 N C
⎞
⎡q
⎤
k q2 ⎛
q b
b
⎜
⎟ = k ⎢ 12 − 2 2 2 3 2 ⎥
2
2
2
⎜
⎟
(a + b ) ⎥⎦
b
a + b ⎝ a 2 + b2 ⎠
⎢⎣ b
9
−
⎧⎪ 63.0 × 10 C
(47.0 × 10−9 C)(0.0140 m) ⎫⎪
= (8.988 × 109 N ⋅ m 2 C2 ) ⎨
−
⎬
2
[(0.0340 m)2 +(0.0140 m)2 ]3 2 ⎭⎪
⎩⎪ (0.0140 m)
E y = E1 y + E2 y =
k q1
−
a
b
b
a
= 2.77 × 106 N C
86. Strategy Find the electric field at the location of charge D due to charges A, B, and C. Then, use Newton’s
second law to compute the charge on D. The direction of the electric field must be in the direction of charge D’s
acceleration or anti-parallel to it.
Solution Let s = 0.0250 m. Compute the components of the electric field.
⎞
k qB ⎛ 1 ⎞ k qC
k ⎛ q
−
= ⎜ B − qC ⎟
E x = EAx + EBx + ECx = 0 +
⎜
⎟
⎟
2s 2 ⎝ 2 ⎠
s2
s 2 ⎜⎝ 2 2
⎠
⎞
k qA k qB ⎛ 1 ⎞
k ⎛ q
E y = EAy + EBy + ECy = −
+
− 0 = ⎜ B − qA ⎟
2
2 ⎜ 2⎟
2 ⎜2 2
⎟
s
s ⎝
2s ⎝
⎠
⎠
Compute the magnitude.
2
2
E = Ex + E y =
k
s2
2
⎛ qB
⎞ ⎛ q
⎞
− qC ⎟ + ⎜ B − qA ⎟
⎜⎜
⎟
⎜
⎟
⎝2 2
⎠ ⎝2 2
⎠
2
2
2
⎞ ⎛ 0.150 × 10−6 C
⎞
8.988 × 109 N ⋅ m 2 C2 ⎛ 0.150 × 10−6 C
=
− 0.300 × 10−6 C ⎟ + ⎜
− 0.200 × 10−6 C ⎟
⎜
2
⎜
⎟
⎜
⎟
2 2
2 2
(0.0250 m)
⎝
⎠ ⎝
⎠
= 4.13 × 106 N C
Because both components are negative, the direction of the electric field at D is in the direction of the
acceleration, thus, the sign of the charge on D is positive. According to Newton’s second law, ΣF = qE = ma, so
q=
ma (0.00200 kg)(248 m s 2 )
=
= 0.120 µC .
E
4.13 × 106 N C
675
Chapter 16: Electric Forces and Fields
Physics
87. (a) Strategy Use the acceleration of a charged particle in an electric field.
Solution Find the speed of the electrons.
⎛ eE ⎞
vf2 − vi2 = vf2 − 0 = 2a∆x = 2 ⎜ ⎟ ∆x, so
⎝m⎠
2(1.602 × 10−19 C)(4.0 × 105 N C)(0.050 m)
⎛ eE ⎞
vf = 2 ⎜ ⎟ ∆x =
⎝m⎠
(b) Strategy Use ∆x =
9.109 × 10
−31
kg
= 8.4 × 107 m s .
1
a x (∆t )2 and ∆x = v∆t.
2
Solution Find the time intervals during the acceleration of a the electrons and their motion at constant speed.
1
2∆x1
2m∆x1
∆x
∆x2
=
. ∆x2 = v∆t2 , so ∆t2 = 2 =
.
∆x1 = a(∆t1)2 , so ∆t1 =
2eE ∆x1
2
a
eE
v
m
Find the total time to travel the length of the tube.
⎛
m
2m∆x1
∆x2 ⎞ 2m∆x1
∆t = ∆t1 + ∆t2 =
+ ∆x2
= ⎜1 +
⎟
eE
eE
2eE ∆x1 ⎝ 2∆x1 ⎠
⎡
45 cm ⎤ 2(9.109 × 10−31 kg)(5.0 × 10−2 m)
= ⎢1 +
= 6.6 ns
⎥
−19
C)(4.0 × 105 N C)
⎣ 2(5.0 cm) ⎦ (1.602 × 10
88. Strategy The electric field due to the 2-µC charge is directed away from it, and that due to the −4-µC charge is
directed toward it. In region B, the electric field is directed to the right for both charges. In regions A and C, the
electric fields due to the charges have opposite directions. So, only regions A and C can have zero field. Let
QR = −4 µC and QL = 2 µC. Use Coulomb’s law, Eq. (16-2), and the principle of superposition.
Solution Region A:
Let +x be to the left with x = 0 at QR and the distance between QL and QR be d. Set E = 0.
0=
QL
QR
k QL
(x − d )
2
⎛ d⎞
= ⎜1 − ⎟
x⎠
⎝
d
= 1±
x
1±
k QR
x2
2
QL
QR
d
x=
−
QL
QR
=
d
1±
2
4
= 0.6d or 3d
0.6d is extraneous, since it lies in region B. 3d is in region A, and E = 0 there.
676
Physics
Chapter 16: Electric Forces and Fields
Region C:
Let +x be to the right with x = 0 at QL and the distance between QL and QR be d.
k QL
0=
x
2
−
k QR
( x − d )2
2
Q
⎛ d⎞
1
−
= R
⎜
⎟
x⎠
QL
⎝
1±
QR
d
=
QL
x
d
x=
1±
QR
QL
=
d
1±
4
2
= −2d or 0.4d
Neither result is in region C. So, the only region where E = 0 is A .
89. Strategy Use the properties of electric fields and the rules for sketching field lines.
Solution Since the semicircle is positively charged, the field lines point toward the center of curvature. Let
the semicircle be oriented
such that its ends are on the x-axis and its midpoint is on the negative y-axis. The
K
x-components of E all cancel due to symmetry, and the y-components all add and point in the positive
y-direction. So, the electric field at the center points away from the midpoint of the semicircle.
E
r
90. Strategy Since the electric field points down the incline and the force on the block is up the incline, the sign of
the net charge on the block must be negative. The block does not slide because the electrical force on it is equal in
magnitude and opposite in direction to the component along the incline of the gravitational force on the block.
Use Newton’s second law.
Solution Find the magnitude of the charge on the block.
ΣFx = qE − mg sin17.0° = 0, so
mg sin17.0° (0.00235 kg)(9.80 m s 2 ) sin17.0°
q=
=
= 1.45 × 10−5 C.
E
465 N C
FE
x
E
17.0°
mg
mg sin17.0°
Thus, the charge on the block is −1.45 × 10−5 C .
677
17.0°
Chapter 16: Electric Forces and Fields
Physics
91. (a) Strategy Set the magnitude of the electric force equal to the magnitude of the gravitational force.
Solution Find the required magnitudes of the net charges on the Earth and the Sun.
k qS qE GmSmE
G
, so qS qE = mSmE .
=
2
2
k
r
r
q
q
m
Let E = S . Then, qE = E qS . Find qS .
mS
mE mS
⎛ mS
⎛ 1 ⎞ G
G
mSmE ⎜
⎟⎟ = mSmE ⎜⎜
⎜
k
⎝ qE ⎠ k
⎝ mE qS
2
GmS
qS2 =
k
qS =
qS = mS
⎞
⎟⎟
⎠
G
6.674 × 10−11 N ⋅ m 2 kg 2
= (1.987 × 1030 kg)
= 1.712 × 1020 C
k
8.988 × 109 N ⋅ m 2 C2
Find qE .
⎛ 1 ⎞ G
⎛ mE
G
mSmE ⎜
⎟⎟ = mSmE ⎜⎜
⎜
k
⎝ mS qE
⎝ qS ⎠ k
2
GmE
qE2 =
k
qE =
qE = mE
⎞
⎟⎟
⎠
6.674 × 10−11 N ⋅ m 2 kg 2
G
= (5.974 × 1024 kg)
= 5.148 × 1014 C
9
2
2
k
8.988 × 10 N ⋅ m C
(b) Strategy and Solution If the magnitude of the charges of the proton and electron were not exactly equal,
astronomical bodies would have net charges with the same sign, so the force between them would be
repulsive. The force responsible for the Earth’s orbit is attractive, so this charge imbalance could not possibly
be the explanation for Earth’s orbit. The answer is no .
92. Strategy Using Coulomb’s law, sum the components of the forces due to the other three ions on the chloride ion.
Solution Calculate the components of the force.
kq 2
kq 2
kq 2
Fx = −
+
cos
45°
−
cos 75°
(0.5 nm)2 (0.8 nm)2
(0.3 nm)2
(8.988 × 109 N ⋅ m 2 C2 )(2 × 10−21 C) 2 ⎛
1
1
cos 75° ⎞
=
−
= −2 × 10−13 N ( − 2.07 × 10−13 N)
⎜− 2 +
2
2 ⎟
−18 2
10
m
0.5
0.8
2
0.3
⎝
⎠
Fy = −
kq 2
(0.8 nm)
sin 45° +
2
kq 2
(0.3 nm)
sin 75° =
2
= 3 × 10−13 N (3.46 × 10−13 N)
Calculate the magnitude of the force.
(8.988 × 109 N ⋅ m 2 C2 )(2 × 10−21 C)2 ⎛
1
sin 75° ⎞
+
⎜− 2
⎟
−18 2
10
m
0.32 ⎠
⎝ 0.8 2
F = Fx2 + Fy2 = (−2.07 × 10−13 N)2 + (3.46 × 10−13 N) 2 = 4 × 10−13 N
Calculate the direction.
Fy
3.46
θ = tan −1
= tan −1
= 60° above the negative x-axis
Fx
−2.07
K
So, F = 4 × 10−13 N at 60° above the negative x-axis .
678
Physics
Chapter 16: Electric Forces and Fields
93. Strategy Let θ + / θ − be the angle between r+ / r− and the x-axis. By symmetry, E x, net = 0 on the x-axis, and the
G
y-components of E due to each charge are directed downward. Use the binomial approximation (1 ± x)n ≈ 1 ± nx
for x << 1, and Coulomb’s law.
Solution
(a) Write an expression for the magnitude of the electric field of the dipole on the positive x-axis.
⎛
⎞
d
d
d
kq
kq
kq ⎛ 2 ⎞ kq ⎛ 2 ⎞
kq ⎜
kq
⎟
2
⎜
⎟
⎜
⎟
E y ( x > 0, 0) = − 2 sin θ + − 2 sin θ − = − 2
− 2
=−
−
2 ⎜
2
⎟
2
2
2
d
⎜
⎟
⎜
⎟
r
r
r+
r−
r+ ⎝ + ⎠ r− ⎝ − ⎠
x + 4 ⎜ x 2 + d ⎟ x + d4
4 ⎠
⎝
kqd
kqd
=−
=−
3/ 2
2 3/ 2
2 d2
x + 4
x3 1 + d 2
)
(
(
⎛
d
⎜
2
⎜ 2
2
⎜ x +d
4
⎝
)
4x
If x >> d, then
E y ( x >> d , 0) ≈ −
So, E y = E =
kqd ⎛ 3d 2 ⎞
kqd
⎜1 − 2 ⎟ ≈ − 3
3 ⎜
⎟
x ⎝ 8x ⎠
x
kqd
[3d 2 /(8 x 2 ) << 1]
.
x3
K
(b) E+ y and E− y are both directed downward, so E( x, 0) is directed in the negative y -direction for all x.
94. (a) Strategy Use Coulomb’s law and the principle of superposition. The direction of the electric field is up if
E > 0 and down if E < 0.
Solution Write expressions for the electric field, specify directions in each of the four regions.
1 ⎞
kq
⎛
E ⎜ 0, y > d ⎟ =
2 ⎠
⎝
y − d2
(
kq
−
) ( y + d2 )
2
1 ⎞
kq
⎛
E ⎜ 0, 0 < y < d ⎟ = −
2 ⎠
⎝
y − d2
(
1
kq
⎛
⎞
E ⎜ 0, − d < y < 0 ⎟ = −
2
⎝
⎠
y − d2
(
1 ⎞
kq
⎛
E ⎜ 0, y < − d ⎟ = −
2 ⎠
⎝
y − d2
(
y + d2
−
)
2
; down
kq
) ( y + d2 )
+
2
) (
2
kq
−
) (
2
; up
2
kq
y + d2
)
2
2
; down
; up
679
⎞
⎟
⎟
⎟
⎠
Chapter 16: Electric Forces and Fields
Physics
(b) Strategy Use the binomial approximation (1 ± x)n ≈ 1 ± nx for x << 1.
Solution Approximate the expression found in part (a) at distant points from the dipole ( ( y >> d ).
1 ⎞
kq
kq
kq ⎡ d ⎛ d ⎞ ⎤ 2kqd
⎛
E ⎜ 0, y > d ⎟ =
( y > d/2)
−
≈ 2 ⎢1 + − ⎜ 1 − ⎟ ⎥ = 3
2
2
2
y ⎠ ⎦⎥
y ⎢⎣ y ⎝
y
⎝
⎠ y2 1 − d
2
d
y
1
+
2y
2y
and
1 ⎞
kq
kq
kq ⎡ ⎛ d ⎞ ⎛ d ⎞ ⎤
2kqd
⎛
E ⎜ 0, y < − d ⎟ = −
( y < −d/2)
+
≈ 2 ⎢− ⎜1 + ⎟ + ⎜1 − ⎟ ⎥ = − 3
2
2
2
y
y
y
y
⎝
⎠
2
2
⎢
⎥
⎝
⎠
⎝
⎠
d
d
⎣
⎦
y 1− 2y
y 1+ 2y
)
(
(
)
(
)
(
)
G
2kqd
up .
So, E ( y >> d ) =
3
y
The field is proportional to 1 y 3 . The result does not conflict with Coulomb’s law, which applies to
individual point charges. E is due to the superposition of the fields due to two point charges. The field due to
each charge obeys Coulomb’s law.
95. Strategy Use the conditions for equilibrium, Eqs. (8-8).
Solution
(a) Calculate the net electric force acting on the dipole.
G
ΣFx = qE x − qE x = 0 and ΣFy = qE y − qE y = q(0) − q(0) = 0, so Fnet = 0 .
(b) Calculate the net torque on the dipole.
⎛d⎞
⎛d⎞
Στ = − F−q r−q sin θ − Fq rq sin θ = −qE ⎜ ⎟ sin θ − qE ⎜ ⎟ sin θ = −qEd sin θ
⎝2⎠
⎝2⎠
(c) Evaluate the net torque for each angle.
Στ (θ ) = − (3.0 × 10−6 C)(2.0 × 104 N C)(0.070 m) sin θ
θ (°)
Torque (N ⋅ m)
0
0
36.9
− 0.0025
90.0
− 0.0042
680
Chapter 17
ELECTRIC POTENTIAL
Conceptual Questions
1. (a) The electric field does positive work on –q as it moves closer to +Q.
(b) The potential increases as –q moves closer to +Q.
(c) The potential energy of –q decreases.
(d) If the fixed charge instead has a value –Q, the electric field does negative work, the potential decreases, and
the potential energy increases.
2. Such a capacitor can be built by replacing the air between the capacitor plates with a dielectric material. This
change not only increases the maximum possible voltage across the capacitor but also increases the amount of
charge on the capacitor plates for a given potential difference.
3. While standing on a high voltage wire, the magnitude of a bird’s electric potential varies between –100 kV and
+100 kV. Important for the bird is the fact that although its body is at a non-zero potential, the potential difference
across its body is small. If a large potential difference existed across its body, the bird would be electrocuted.
4. A positive charge in an electric field moves toward a position of lower potential. A negative charge in this
situation moves toward a position of higher potential.
5. Zero work is required to move a charge between two points at the same potential. An external force may need to
be applied to move the charge but the work done to start the charge in motion will be negated by the work done to
stop it.
6. If the charge of a point particle is negative, its electric potential energy decreases as it is moved towards a region
of higher electric potential.
7. If all parts of a conductor in electrostatic equilibrium were not at the same potential, electric fields would exist
within the conductor and charges would not remain stationary. The assumption of electrostatic equilibrium would
therefore be invalid.
8. There is no physical significance to zero potential—only potential differences have physical consequences. The
potential of the earth is often taken to be zero and therefore an object that is grounded has zero potential. This is
only a reference value however and the potential of the earth could be taken to be any other quantity as long as
other values were appropriately offset by the same potential.
9. If the electric field is zero throughout a region of space, the electric potential must be constant throughout that
region.
10. The woman’s head has acquired a net charge. The microscopic charges (electrons or ions) distribute themselves so
as to maximize their separation from one another, as a result of their mutual repulsion. This is why the charges
move out onto the woman’s hair, which then spreads out in response to the repulsive electrical forces. The
charged strands of hair orient themselves parallel to the electric field lines, which emanate radially outward from
the woman’s head, as though it were a charged conducting sphere.
11. If the potential is constant throughout a region of space, the electric field must be zero throughout that region.
681
Chapter 17: Electric Potential
Physics
12. If a uniform electric field exists throughout a region of space, the potential must be linearly increasing in the
direction parallel to the field and unchanging in the other two directions.
13. It doesn’t matter which points we choose because the potential on each plate is constant over the whole plate.
14. The factor of 1/2 appears because the average height of the water in the pool is (1/2)h. The work required to fill
the pool is exactly equal to the potential energy of the water in the pool, (1/2)Mgh. This is analogous to the
charging of a capacitor. The charge Q on the capacitor is like the mass M of water in the pool, and the electrical
potential energy Q∆V of the charge is like the gravitational potential energy Mgh. Thus, by analogy, the total
potential energy of a charged capacitor is (1/2)Q∆V, which is correct.
15. Nothing happens to the capacitance, which depends only on the geometry and electrical properties of the materials
in the capacitor. Since C = Q/V, if the charge doubles then the voltage doubles as well.
16. Cow A is more likely to be killed because the potential difference between its front and hind legs will be greater
than that for cow B.
17. We can’t say anything about the electric field if all we know is the potential at a single point. The electric field
tells us how the potential changes if we move from one point to another.
18. As long as the person touching the dome is isolated from the ground, there is no complete circuit for current to
flow through, so she is safe.
19. The electric field points from regions of higher potential to regions of lower potential. Therefore, the upper
atmosphere is at a higher potential than the Earth.
20. Since the capacitor is connected to a battery the whole time, its voltage ∆V remains constant. The capacitance is
proportional to κ, so when the dielectric is removed the capacitance decreases by a factor of 3. Since Q = C∆V,
and ∆V remains constant, the charge on the capacitor Q decreases by a factor of 3 as well. The electric field
remains constant, since ∆V is constant. The energy stored decreases by a factor of 3.
21. In this case the capacitor plates are isolated, so it is the charge Q that remains constant. Again the capacitance
decreases by a factor of 3, so ∆V must increase by a factor of 3 to keep Q = C∆V the same. Thus, the electric field
increases by a factor of 3 and the energy stored in the capacitor increases by a factor of 3 as well.
22. The plates are isolated so the charge remains constant. The capacitance is given by C = κ ⑀0 A / d . As the plates are
moved closer, all these quantities remain constant except d, which decreases. Therefore, the capacitance increases.
Since Q = C∆V, and Q is constant, ∆V must decrease. The electric field remains constant. The energy stored in the
capacitor decreases as well.
23. The potential close to the positive charge must be positive, while close to the negative charge it must be negative.
Therefore, there is a point in region B where the potential is zero. If we move very far away from the two charges,
they will look like a single point of charge –3 µC, so the potential very far away must be negative. Thus, there
must be a point in region A with a potential of zero as well. The electric field can only be zero in region A, and
this does not occur at the same point where the potential is zero.
682
Physics
Chapter 17: Electric Potential
Problems
1. Strategy Use Eq. (17-1).
Solution Compute the electric potential energy.
qq
(8.988 × 109 N ⋅ m 2 C2 )(5.0 × 10−6 C)( − 2.0 × 10−6 C)
UE = k 1 2 =
= −18 mJ
r
5.0 m
2. (a) Strategy Use Eq. (17-1).
Solution Compute the electric potential energy.
qq
(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)( − 1.602 × 10−19 C)
UE = k 1 2 =
= − 4.36 × 10−18 J
r
0.0529 × 10−9 m
(b) Strategy and Solution The negative sign signifies that the force between the two charges is attractive; the
potential energy is lower than if the two were separated by a larger distance.
3. Strategy The work done by the applied force is positive, since the direction of the applied force was in the
direction of motion. (The force between the two charges is repulsive.) The potential energy of the charges is
positive, so the work done on the charges is equal to their potential energy. Use Eq. (17-1).
Solution Compute the work done on the charges.
qq
(8.988 × 109 N ⋅ m 2 C2 )(6.5 × 10−6 C)2
W = UE = k 1 2 =
= 8.4 J
r
0.045 m
4. Strategy The work done by the external agent is positive since the potential energy increases. Use Eq. (17-1).
Solution Find the work done by the external agent.
ke2
(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)2
W = ∆U = U f − U i =
− U∞ =
− 0 = 2.3 × 10−13 J
rf
1.0 × 10−15 m
5. Strategy The work done on the charges is equal to their potential energy. Let the upper charge by 1, the lower
left-hand charge be 2, and the right-hand charge be 3. Also, let a = 0.16 m and b = 0.12 m. Use Eq. (17-2).
Solution Compute the work done on the charges.
⎛q q
q1q3
q q ⎞
+ 2 3⎟
W = UE = k ⎜ 1 2 +
⎜ b
a ⎟⎠
a 2 + b2
⎝
⎡ (5.5 × 10−6 C)(−6.5 × 10−6 C) (5.5 × 10−6 C)(2.5 × 10−6 C)
= (8.988 × 109 N ⋅ m 2 C2 ) ⎢
+
0.12 m
⎢
(0.16 m) 2 + (0.12 m) 2
⎣
(−6.5 × 10−6 C)(2.5 × 10−6 C) ⎤
+
⎥ = −3.0 J
0.16 m
⎥⎦
6. Strategy Let q1 = −q2 = q = 10.0 nC and d = 4.00 cm. Use Eq. (17-2).
Solution Find the total electric potential energy for the two charges.
kq q
kq 2
(8.988 × 109 N ⋅ m 2 C2 )(10.0 × 10−9 C)2
UE = 1 2 = −
=−
= −11.2 µJ
r
2d
2(0.0400 m)
683
Chapter 17: Electric Potential
Physics
7. Strategy Let q1 = −q2 = q = 10.0 nC, d = 4.00 cm, and q3 = − 4.2 nC. Use Eq. (17-2).
Solution Find the total electric potential energy of the three charges at point a.
⎡ q 2 qq3 qq3 ⎤ kq ⎡ q
⎛qq
qq
q q ⎞
⎛ 1 ⎞⎤
+
−
U E = k ⎜ 1 2 + 1 3 + 2 3 ⎟ = k ⎢−
⎥=
⎢ − + q3 ⎜ 1 − ⎟ ⎥
2
3
2
r
r
r
d
d
d
d
⎝ 3 ⎠⎦
⎣
13
23 ⎠
⎝ 12
⎣⎢
⎦⎥
(8.988 × 109 N ⋅ m 2 C2 )(10.0 × 10−9 C) ⎡ 10.0 × 10−9 C 2(−4.2 × 10−9 C) ⎤
=
+
⎢−
⎥ = −17.5 µJ
0.0400 m
2
3
⎢⎣
⎥⎦
8. Strategy Let q1 = −q2 = q = 10.0 nC, d = 4.00 cm, and q3 = − 4.2 nC. Use Eq. (17-2).
Solution Find the total electric potential energy of the three charges at point b.
⎛ q 2 qq3 qq3 ⎞
⎛qq
qq
q q ⎞
kq 2
UE = k ⎜ 1 2 + 1 3 + 2 3 ⎟ = k ⎜ −
+
−
⎟=−
⎜ 2d
r13
r23 ⎠
d
d ⎟⎠
2d
⎝ r12
⎝
(8.988 × 109 N ⋅ m 2 C2 )(10.0 × 10−9 C) 2
=−
= −11.2 µJ
2(0.0400 m)
9. Strategy Let q1 = −q2 = q = 10.0 nC, d = 4.00 cm, and q3 = − 4.2 nC. Use Eq. (17-2).
Solution Find the total electric potential energy of the three charges at point c.
⎛ q 2 qq3 qq3 ⎞
⎛qq
qq
q q ⎞
kq 2
UE = k ⎜ 1 2 + 1 3 + 2 3 ⎟ = k ⎜ −
+
−
⎟=−
⎜ 2d 2d
r13
r23 ⎠
2d ⎟⎠
2d
⎝ r12
⎝
(8.988 × 109 N ⋅ m 2 C2 )(10.0 × 10−9 C) 2
=−
= −11.2 µJ
2(0.0400 m)
10. Strategy Use Eq. (17-2).
Solution Find the electric potential energy.
⎛qq
qq
q q ⎞
UE = k ⎜ 1 2 + 1 3 + 2 3 ⎟
r13
r23 ⎠
⎝ r12
⎡
⎛
3.0 × 10−6 C
−1.0 × 10−6 C ⎞⎟
= (8.988 × 109 N ⋅ m 2 C2 ) ⎢ (4.0 × 10−6 C) ⎜
+
⎢
⎜ (4.0 m)2 + (3.0 m)2
⎟
3.0 m
⎝
⎠
⎣
−6
−6
⎤
(3.0 × 10 C)( − 1.0 × 10 C)
+
⎥ = 2.8 mJ
4.0 m
⎥⎦
y (m)
4.0
3
2
2.0
0.0
1
0.0
2.0
11. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
Wfield = −∆U = U i − U f = U12 − (U12 + U13 + U 23 ) = −U13 − U 23 = −
kq1q3 kq2q3
−
r13
r23
⎛ 8.00 × 10−9 C
⎞
−8.00 × 10−9 C
= − (8.988 × 109 N ⋅ m 2 C2 )(2.00 × 10−9 C) ⎜
+
⎟ = −2.70 µJ
⎜ 0.0400 m
0.0400 m + 0.1200 m ⎟⎠
⎝
684
4.0
x (m)
Physics
Chapter 17: Electric Potential
12. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
kq1q3 kq2q3
−
r13
r23
1
1
⎛
⎞
C) ⎜
−
⎟ = 1.80 µJ
⎝ 0.0800 m 0.0400 m ⎠
Wfield = −∆U = U i − U f = U12 − (U12 + U13 + U 23 ) = −U13 − U 23 = −
= − (8.988 × 109 N ⋅ m 2 C2 )(2.00 × 10−9 C)(8.00 × 10−9
13. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
Wfield = −∆U = U i − U f = U12 + U13i + U 23i − (U12 + U13f + U 23f ) = U13i − U13f + U 23i − U 23f
⎡
⎛ 1
⎛ 1
1 ⎞
1 ⎞⎤
= k ⎢ q1q3 ⎜
−
−
⎟ + q2q3 ⎜
⎟⎥
⎝ r13i r13f ⎠
⎝ r23i r23f ⎠ ⎦⎥
⎣⎢
⎡
1
1
⎛
⎞
= (8.988 × 109 N ⋅ m 2 C2 ) ⎢(8.00 × 10−9 C)(2.00 × 10−9 C) ⎜
−
⎟
⎝ 0.0400 m 0.0800 m ⎠
⎣
1
1
⎛
⎞⎤
+ (−8.00 × 10−9 C)(2.00 × 10−9 C) ⎜
−
⎟ ⎥ = 4.49 µJ
⎝ 0.0400 m + 0.1200 m 0.0400 m ⎠ ⎦
14. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
Wfield = −∆U = U i − U f = U12 + U13i + U 23i − (U12 + U13f + U 23f ) = U13i − U13f + U 23i − U 23f
⎡
⎛ 1
⎛ 1
1 ⎞
1 ⎞⎤
= k ⎢ q1q3 ⎜
−
−
⎟ + q2q3 ⎜
⎟⎥
⎢⎣
⎝ r13i r13f ⎠
⎝ r23i r23f ⎠ ⎥⎦
1
1
1
1
⎛
⎞
= (8.988 × 109 N ⋅ m 2 C2 )(8.00 × 10−9 C)(2.00 × 10−9 C) ⎜
−
−
+
⎟
0.0800
m
0.120
m
0.0400
m
0.120
m
⎝
⎠
= −1.80 µJ
15. Strategy The potential difference is the change in electric potential energy per unit charge. Use ∆U E = q∆V .
Solution Find the change in the electric potential energy.
∆U E = q∆V = (3.0 nC)(25 V) = 75 nJ
16. Strategy The potential difference is the change in electric potential energy per unit charge. Use ∆U E = q∆V .
Solution Find the change in the electric potential energy.
∆U E = q∆V = q(Vf − Vi ) = e(VB − VA ) = (−1.602 × 10−19 C)[−360 V − (−240 V)] = 1.92 × 10−17 J
685
Chapter 17: Electric Potential
Physics
17. Strategy Use the principle of superposition and Eq. (17-9).
Solution Sum the electric fields at the center due to each charge.
K K
K
K
K
K
K
K
K
E = Ea + Eb + Ec + Ed = Ea + Eb − Ea − Eb = 0
Do the same for the potential at the center.
kQ
4kQ 4(8.988 × 109 N ⋅ m 2 C2 )(9.0 × 10−6 C)
V =Σ i =
=
= 2.3 × 107 V
ri
r
(0.020 m) 2 + (0.020 m) 2
2
18. Strategy Use the principle of superposition and Eq. (17-9).
Solution Sum the electric fields at the center due to each charge.
K K
K
K
K
K
K
K
K
K
K
⎛k q
kq ⎞
E = Ea + Eb + Ec + Ed = Ea + Eb + Ec − Eb = Ea + Ec = ⎜ 2a + 2c ⎟ toward c
⎜ r
r ⎟⎠
⎝
8.988 × 109 N ⋅ m 2 C2
(9.0 × 10−6 C + 3.0 × 10−6 C) toward c = 5.4 × 108 N C toward c
=
2
⎛ (0.020 m)2 + (0.020 m)2 ⎞
⎜⎜
⎟⎟
2
⎝
⎠
Do the same for the potential at the center.
kQ
k
V = Σ i = (3q − q − q − q) = 0
ri
r
19. Strategy Use Eqs. (17-7), (17-8), (17-9), and (6-8).
Solution Find the potentials, potential difference, change in electric potential energy, and work done by the
electric field.
(a) V =
kQ (8.988 × 109 N ⋅ m 2 C2 )(−50.0 × 10−9 C)
=
= −1.5 kV
r
0.30 m
(b) V =
(8.988 × 109 N ⋅ m 2 C2 )(−50.0 × 10−9 C)
= −900 V
0.50 m
⎛ 1 1 ⎞
1 ⎞
⎛ 1
(c) ∆V = kQ ⎜ − ⎟ = (8.988 × 109 N ⋅ m 2 C2 )(−50.0 × 10−9 C) ⎜
−
= 600 V
r
r
0.50
m
0.30
m ⎟⎠
⎝
A⎠
⎝ B
∆V > 0, so the potential increases .
(d) ∆U E = q∆V = (−1.0 × 10−9 C)(6.0 × 102 V) = −6.0 × 10−7 J
∆U E < 0, so the potential energy decreases .
(e) Wfield = −∆U E = 6.0 × 10−7 J
686
Physics
Chapter 17: Electric Potential
20. Strategy Use Eq. (17-9).
Solution Find the electric potential at point P due to the charges.
⎛ 2.0 × 10−3 C
kQ
− 4.0 × 10−3 C
+
V = Σ i = (8.988 × 109 N ⋅ m 2 C2 ) ⎜
⎜ 4.0 m
ri
(4.0 m) 2 + (3.0 m)2
⎝
= −2.7 MV
y (m)
⎞
⎟
⎟
⎠
4.0
− 4.0 mC
2.0
0.0
2.0 mC
0.0
2.0
P
4.0
x (m)
21. Strategy and Solution
(a) Since V is positive, q is positive .
1
(b) V ∝ , so since the potential is doubled, the distance is halved or 10.0 cm .
r
22. Strategy Just outside the surface of the sphere, the electric potential is given by V = Er , where r is the radius of
the sphere.
Solution Find the electric potential.
V = Er = (8.40 × 105 V m)(0.750 m) = 6.30 × 105 V
kqi
, the minimum or most negative value of the potential is the case
ri
where the two negative charges are closer to x = 0 than the two positive charges.
23. (a) Strategy and Solution Since V = Σ
y
+
_
_
+
x
(b) Strategy Let d = 1.0 m and q = 1.0 µC. Use Eq. (17-9).
Solution Find the potential at the origin.
⎛ q
kQ
q q
q ⎞ kq ⎛ 2
2 ⎞ 4(8.988 × 109 N ⋅ m 2 C2 )(1.0 × 10−6 C)
V =Σ i = k⎜ 3 + d + d − 3 ⎟ =
2
2
+
+
−
=
⎜
⎜ d
ri
3 ⎟⎠
1.0 m
d⎟ d ⎝3
2
2
2 ⎠
⎝2
= 36 kV
24. (a) Strategy Use Eq. (16-5).
Solution Find RE .
E0 =
kQ0
R02
=E=
k (3Q0 )
RE2
, so RE =
3 R0 .
(b) Strategy Use Eq. (17-8).
Solution Find RV .
V0 =
kQ0
k (3Q0 )
=V =
, so RV = 3R0 .
R0
RV
687
Chapter 17: Electric Potential
Physics
25. Strategy Use Eq. (17-9).
Solution Find the electric potential at the third corner, B.
kQ
k
8.988 × 109 N ⋅ m 2 C2
V = Σ i = (QA + QB ) =
(2.0 × 10−9 C − 1.0 × 10−9 C) = 9.0 V
ri
r
1.0 m
26. (a) Strategy Use Eq. (17-9).
Solution Find the electric potential at each point.
⎛ 4.2 × 10−9 C − 6.4 × 10−9 C ⎞
kq
kq
+
Va = + + − = (8.988 × 109 N ⋅ m 2 C2 ) ⎜
⎟ = 270 V
⎜ 0.060 m
r+
r−
0.159 m ⎟⎠
⎝
⎛ 4.2 × 10−9 C − 6.4 × 10−9 C ⎞
+
Vb = (8.988 × 109 N ⋅ m 2 C2 ) ⎜
⎟ = −160 V
⎜ 0.120 m
0.120 m ⎟⎠
⎝
(b) Strategy Use Eq. (17-6).
Solution Compute the potential difference for the trip from a to b.
∆V = Vb − Va = −160 V − 270 V = − 430 V
(c) Strategy The work done by an external agent is equal to the change in electric potential energy of the point
charge when moved from a to b. Use Eq. (17-7).
Solution Compute the work done.
W = q∆V = (1.50 × 10−9 C)( − 430 V) = − 6.5 × 10−7 J
27. (a) Strategy Use Eq. (17-9).
Solution Find the potential at the points.
⎛ 2.50 × 10−9 C −2.50 × 10−9 C ⎞
kQ1 kQ2
+
= (8.988 × 109 N ⋅ m 2 C2 ) ⎜
+
⎟ = 300 V
⎜ 0.050 m
⎟
r1
r2
0.150 m
⎝
⎠
kQ
kQ
kQ
kQ
If Q = 2.50 × 10−9 C, Vb = 1 + 2 =
−
= 0 .
r1
r2
r
r
Va =
(b) Strategy The work done by an external agent is equal to the change in electric potential energy of the point
charge when moved from infinity to b. Use Eq. (17-7).
Solution Compute the work done.
W = q∆V = q(Vb − V∞ ) = q(0 − 0) = 0
688
Physics
Chapter 17: Electric Potential
28. (a) Strategy Let d = 4.00 cm, r = 12.0 cm, and q = 8.00 nC. Use Eq. (17-9). Let V = 0 at infinity.
Solution Find the potentials at points a and b.
kq kq
kq k (−q) 3kq 3(8.988 × 109 N ⋅ m 2 C2 )(8.00 × 10−9 C)
Va = 1 + 2 =
+
=
=
= 1350 V
r1
r2
d
4d
4d
4(0.0400 m)
Vb =
kq k (−q)
kq
(8.988 × 109 N ⋅ m 2 C2 )(8.00 × 10−9 C)
+
=−
=−
= −899 V
2d
d
2d
2(0.0400 m)
(b) Strategy The potential difference is the change in electric potential energy per unit charge. Use ∆U E = q∆V .
Solution Compute the change in electric potential energy.
∆U E = q∆V = (2.00 × 10−9 C)( − 899 V − 1348 V) = − 4.49 µJ
29. (a) Strategy Let d = 4.00 cm, r = 12.0 cm, and q = 8.00 nC. Use Eq. (17-9). Let V = 0 at infinity.
Solution Find the potentials at points b and c.
kq kq
kq k (−q)
kq
(8.988 × 109 N ⋅ m 2 C2 )(8.00 × 10−9 C)
Vb = 1 + 2 =
+
=−
=−
= −899 V
r1
r2
2d
d
2d
2(0.0400 m)
kq k (−q)
Vc =
+
= 0
r
r
(b) Strategy The potential difference is the change in electric potential energy per unit charge. Use ∆U E = q∆V .
Solution Compute the change in electric potential energy.
∆U E = q∆V = (2.00 × 10−9 C)(0 + 899 V) = 1.80 µJ
30. Strategy Rewrite each unit in terms of kg, m, s, and C.
Solution Show that 1 N C = 1 V m.
1 N C=
kg ⋅ m s 2
J
kg ⋅ m 2 s 2 kg ⋅ m s 2
and 1 V m =
=
=
, therefore 1 N C = 1 V m.
C
m⋅C
m⋅C
C
31. (a) Strategy Use Eq. (16-4b).
Solution Find the electric force that acts on the particle.
G
G
F = qE = (4.2 × 10−9 C)(240 N C to the right) = 1.0 µΝ to the right
(b) Strategy The work done on the particle is equal to the electric force times the displacement of the particle.
Solution Compute the work done on the particle.
W = Fd = (1.0 µN)(0.25 m) = 0.25 µJ
(c) Strategy The electric field points in the direction of decreasing potential, so Va > Vb and Va − Vb > 0. Use
Eq. (17-10).
Solution Compute the potential difference.
Va − Vb = Ed = (240 N C)(0.25 m) = 60 V
689
Chapter 17: Electric Potential
Physics
32. (a) Strategy and Solution Positive work is required to move an electron (negative charge) from low potential to
high potential, so Y is at the higher potential.
(b) Strategy Use Eqs. (6-8) and (17-5).
Solution Find the potential difference.
q∆V = −e∆V = ∆U E = −Wfield , so ∆V = VY − VX =
Wfield
8.0 × 10−19 J
=
= 5.0 V .
e
1.602 × 10−19 J
33. Strategy ∆V = Ed for a uniform electric field.
Solution Find the distance between the equipotential surfaces.
∆V
1.0 V
d=
=
= 1.0 cm
E
100.0 N C
34. Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces
drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together
where the field is stronger. The electric field always points in the direction of maximum potential decrease.
K
Solution Outside the sphere, E is radially directed (toward the sphere), and V ∝ r −1. The equipotential surfaces
G
K
are perpendicular to E at any point, so they are spheres . Inside the sphere, E = 0 and V is constant.
35. Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces
drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together
where the field is stronger. The electric field always points in the direction of maximum potential decrease.
K
Solution Outside the cylinder, E is radially directed away from the axis of the cylinder. The equipotential
K
surfaces are perpendicular to E at any point, so they are cylinders .
690
Physics
Chapter 17: Electric Potential
36. Strategy The rate at which work is done by the electric organs is equal to the rate of change of the electric
potential energy. Use Eq. (17-7). The total amount of work done in one pulse is equal to the rate times the
duration of the pulse.
Solution
(a) Compute the rate at which work is done.
∆W q∆V
⎛ q ⎞
=
= ∆V ⎜ ⎟ = (0.20 × 103 V)(18 C s) = 3.6 kW
∆t
∆t
⎝ ∆t ⎠
(b) Compute the total amount of work done.
∆W
W=
∆t = (3.6 × 103 W)(0.0015 s) = 5.4 J
∆t
37. (a) Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential
surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer
together where the field is stronger. The electric field always points in the direction of maximum potential
decrease.
Solution The electric field lines are radial. They begin on the point charge and end on the inner surface of
the shell. Then they begin again on the surface and extend to infinity.
+
(b) Strategy Use Eqs. (16-5) and (17-8), and the principle of superposition.
Solution For r < r1, E is that due to the point charge, E = kq r 2 . For r1 < r < r2 , E = 0, since this is inside a
conductor. For r > r2 , E once again is that due to the point charge, kq r 2 . For r < r1, V = kq r (point
charge). For r1 < r < r2 , V = kq r1 , since V is continuous, and it is constant in a conductor. For r > r2 ,
V=
kq ⎛ kq kq ⎞
+ ⎜ − ⎟ (to preserve continuity). The graphs of the electric field magnitude and potential:
r1 ⎝ r
r2 ⎠
E
V
kq
r 12
kq
r1
kq
r22
0
r1
r2
r
0
691
r1
r2
r
Chapter 17: Electric Potential
Physics
38. Strategy Since the electric field is uniform, we can use Eq. (17-10).
Solution Find the magnitude of the charge on the drop in terms of the elementary charge e.
F
Fe
Fe
dFe
(0.16 m)(9.6 × 10−16 N)
F = qE and ∆V = − Ed , so q =
e = 2e .
=
=
=
=
E
Ee
(−∆V d )e e∆V (1.602 × 10−19 C)(480 V)
39. Strategy Since the electric field is uniform, we can use Eq. (17-10). Use Newton’s second law.
Solution Find the magnitude of the charge on the drop.
ΣF = qE − mg = 0, so
q =
mg
mg
mgd (1.0 × 10−15 kg)(9.80 m s 2 )(0.16 m)
=
=
=
= 1.6 × 10−19 C = e .
E
−∆V d
∆V
9.76 × 103 V
qE
mg
40. Strategy Use conservation of energy and Eq. (17-7).
Solution Find the change in kinetic energy.
∆K = −∆U = −q∆V = −2(1.602 × 10−19 C)(200.0 × 103 V − 500.0 × 103 V) = 9.612 × 10−14 J
41. Strategy Use conservation of energy and Eq. (17-7).
Solution Find the potential difference.
1
mv 2 (9.109 × 10−31 kg)(7.26 × 106 m s)2
∆U = −e∆V = −∆K = − mv 2 , so ∆V =
=
= 150 V .
2
2e
2(1.602 × 10−19 C)
42. (a) Strategy The electric field always points in the direction of maximum potential decrease. Electrons, being
negatively charged, move in the direction opposite the direction of the electric field; that is, in the direction of
potential increase.
Solution Since the speed of the electron decreased, it must have traveled in the direction of the electric field,
so it moved in the direction of potential decrease; that is, to a lower potential .
(b) Strategy The kinetic energy of the electron decreased, so its potential energy increased. Use conservation of
energy and Eq. (17-7).
Solution Compute the potential difference the electron moved through.
∆V =
∆U −∆K m(vf 2 − vi 2 ) (9.109 × 10−31 kg)[(2.50 × 106 m s)2 − (8.50 × 106 m s) 2 ]
=
=
=
= −188 V
q
−e
2e
2(1.602 × 10−19 C)
43. Strategy According to Example 17.8, the speed of the electrons at the anode is proportional to the square root of
the potential difference. Use a proportion.
Solution Find the speed of the electrons.
v
∆V2
∆V2
6.0 kV
v ∝ ∆V , so 2 =
and v2 = v1
= (6.5 × 107 m s)
= 4.6 × 107 m s .
v1
∆V1
∆V1
12 kV
692
Physics
Chapter 17: Electric Potential
44. Strategy According to Example 17.8, the speed of the electrons at the anode is proportional to the square root of
the potential difference. Use a proportion.
Solution Find the potential difference.
2
2
2
⎛ 3.0 × 107 m s ⎞
⎛ v2 ⎞
∆V2 ⎛ v2 ⎞
v ∝ ∆V , so
= ⎜ ⎟ and ∆V2 = ∆V1 ⎜ ⎟ = (12 kV) ⎜
⎟ = 2.6 kV .
⎜ 6.5 × 107 m s ⎟
∆V1 ⎝ v1 ⎠
⎝ v1 ⎠
⎝
⎠
45. Strategy and Solution
K
(a) Electrons travel opposite the direction of the electric field, so E is directed upward .
(b) For a uniform electric field, ΣFy = eE =
vy
v y md
e∆V
e ∆V
.
=
= ma y , so a y =
. Thus, ∆t =
ay
e∆V
md
d
(c) Since the electron gains kinetic energy, its potential energy decreases .
46. Strategy The field is uniform. Use conservation of energy and Eq. (17-10).
Solution Find the kinetic energy increase.
∆K = −∆U = −q∆V = eEd = (1.602 × 10−19 C)(500.0 N C)(0.0030 m) = 2.4 × 10−19 J .
47. Strategy Use conservation of energy and Eq. (17-7).
Solution Find the final kinetic energy.
∆K = K f − Ki = −∆U = −q∆V = −2e∆V , so
K f = Ki − 2e∆V = 1.20 × 10−16 J − 2(1.602 × 10−19 C)( − 0.50 × 103 V) = 2.8 × 10−16 J .
48. Strategy The force between the nuclei is repulsive, since they both have positive charge. Use conservation of
energy and Eq. (17-1).
Solution Find the closest distance that a helium nucleus approaches the gold nucleus.
kq q
1
U f = 1 2 = Ki = mHe vi 2 , so
2
r
2kqAu qHe 2k (79e)(2e) 316(8.988 × 109 N ⋅ m 2 C2 )(1.602 × 10−19 C)2
=
=
= 4.85 × 10−14 m .
r=
(6.68 × 10−27 kg)(1.50 × 107 m s)2
mHe vi 2
mHe vi 2
Note that the radius of the gold nucleus is about 7 × 10−15 m and the radius of the gold atom is about 1× 10−10 m.
49. Strategy The electron must have enough kinetic energy at point A to overcome the potential decrease between A
and C. Use conservation of energy and Eq. (17-7).
Solution Find the required kinetic energy.
K A = ∆U = −e∆V = −(1.602 × 10−19 C)( − 60.0 V − 100.0 V) = 2.56 × 10−17 J
50. Strategy and Solution Since positive charges move through decreases in potential, and since the potential and
potential energy are greatest at A, the proton will spontaneously travel from point A to point E. So, K A = 0 .
693
Chapter 17: Electric Potential
Physics
51. Strategy Use the definition of capacitance, Eq. (17-14).
Solution Find the magnitude of the charge on each plate.
Q = C ∆V = (2.0 µF)(9.0 V) = 18 µC
52. (a) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Find the potential difference between the plates.
Q 0.75 µC
= 50 mV .
Q = C ∆V , so ∆V = =
C 15.0 µF
(b) Strategy and Solution The plate with the positive charge is at the higher potential, so the +0.75-µC plate.
53. Strategy Use the definition of capacitance, Eq. (17-14).
Solution
Q = C ∆V = (10.2 × 10−6 F)( − 60.0 V) = − 6.12 × 10−4 C
612 µC of charge must be removed from each plate.
54. (a) Strategy Use Eq. (17-10).
Solution Compute the maximum potential difference across the capacitor.
∆Vmax = Emax d = (3 × 106 V m)(0.0010 m) = 3 kV
(b) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the magnitude of the greatest charge.
Q = C ∆V = (2.0 × 10−6 F)(3 × 103 V) = 6 mC
55. Strategy and Solution
K
(a) Since E does not depend upon the separation of the plates ( E = σ ⑀0 ), it stays the same .
(b) Since ∆V ∝ d , ∆V increases if d increases.
56. Strategy Use the definition of capacitance, Eq. (17-14), and the fact that the potential difference depends upon
the plate separation.
Solution
(a) Compute the potential difference between the plates.
Q 0.800 × 10−6 C
∆V = =
= 667 V
C
1.20 × 10−9 F
(b) Since ∆V ∝ d , ∆V doubles if d doubles.
694
Physics
Chapter 17: Electric Potential
57. Strategy and Solution
(a) Since ∆V ∝ d , ∆V decreases if d decreases.
K
(b) Since E does not depend upon the separation of the plates ( E = σ ⑀0 ), it stays the same .
(c) Since ∆V ∝ d , C ∝ d −1, and Q = C ∆V , the charge on the plates stays the same .
58. Strategy Use the definition of capacitance, Eq. (17-14), and the fact that the potential difference depends upon
the plate separation.
Solution
(a) Find the magnitude of the charge on each plate.
Q = C ∆V = (1.20 nF)(12 V) = 14 nC
K
(b) Since ∆V ∝ d , C ∝ d −1, and Q = C ∆V , the charge on the plates stays the same . Since E depends upon the
magnitude of the charge on each plate ( E = σ ⑀0 ), the electric field stays the same .
59. Strategy The capacitance of a parallel plate capacitor is directly proportional to its area. Form a proportion.
Solution Find the capacitance for each situation.
(a)
1A
C2 A2
A
1
1
, so C2 = 2 C1 = 2 C1 = (0.694 pF) = 0.347 pF .
=
C1 A1
A1
A1
2
2A
C2 A2
A2
2
1
, so C2 =
=
C1 = 3 C1 = (0.694 pF) = 0.463 pF .
(b)
C1 A1
A1
A1
3
60. Strategy Use Eq. (17-10).
Solution Compute the plate separation.
∆V
1.5 V
=
= 1500 km
d=
E
1.0 × 10−6 V m
61. Strategy Use the definition of capacitance, Eq. (17-14).
Solution Find the capacitance of the spheres.
Q
3.2 × 10−14 C
=
= 8.0 pF .
Q = C ∆V , so C =
∆V
0.0040 V
62. Strategy Use the definition of capacitance and the definition of potential for a spherical conductor.
Solution Find the capacitance of the Moon if wrapped in aluminum foil.
Q
Q r
1.737 × 106 m
Q = C ∆V , so C =
=
= =
= 1.933 × 10−4 F .
∆V kQ k 8.988 × 109 N ⋅ m 2 C2
r
695
Chapter 17: Electric Potential
Physics
63. Strategy The electrons are accelerated by the electric field between the plates of the capacitor. When they emerge
from the positive plate, their speed will be greater than their speed when they entered the capacitor.
Solution Find the acceleration of the electrons while they are inside the capacitor.
F
ma
e∆V
∆V = Ed = d =
d , so a =
.
e
e
md
Find the speed of the electrons as they emerge from the capacitor.
2e∆V
⎛ e∆ V ⎞
vfx 2 − vix 2 = 2a x ∆x = 2ad = 2 ⎜
, so
⎟d =
m
⎝ md ⎠
vfx = vix 2 +
2e∆V
2(1.602 × 10−19 C)(40.0 V)
= (2.50 × 106 m s)2 +
= 4.51× 106 m s .
m
9.109 × 10−31 kg
64. Strategy Use the definition of electric flux and Gauss’s law.
Solution
(a) The Gaussian surface is a cylinder whose axis is parallel to a radius vector (of the sphere) through it. One end
is just within the conductor and the other is just outside it. The ends have area A, with a radius much smaller
than that of the sphere, so the electric field is approximately uniform. Find E just outside the conductor.
Q
Φ E = EA cos θ =
⑀0
Cylindrical surface: Φ E = EA cos 90° = EA(0) = 0
End inside conductor: Φ E = EA cos θ = (0) A cos θ = 0, since the electric field is zero inside a conductor.
σ
Q
Q
End just outside the conductor: Φ E = EA cos 0° = EA = , so E =
= .
A⑀0 ⑀0
⑀0
(b) Consider an area A of the surface of an arbitrary conductor. If A is small enough such that its surface is
approximately flat, then the electric field will be nearly uniform just outside the surface. Comparing an area
of the same size on a spherical conductor with the same charge density to that of the arbitrary conductor, we
see that the electric field just outside either conductor should be σ ⑀0 ; as long as A is small enough that it is
approximately flat, then this holds for any conductor.
65. (a) Strategy Use Eq. (16-6).
Solution The electric field between the plates is
Q
4.0 × 10−11 C
E=
=
= 3.3 × 103 V m .
⑀0 A [8.854 × 10−12 C2 (N ⋅ m 2 ) ](0.062 m)(0.022 m)
(b) Strategy Use the definition of the dielectric constant, Eq. (17-17).
Solution Find the electric field between the plates of the capacitor with the dielectric.
E
E
3.3 × 103 V m
= 6.0 × 102 V m .
κ = 0 , so E = 0 =
κ
E
5.5
66. Strategy Assume the field is uniform. Use Eq. (17-10).
Solution Find the maximum possible height for the bottom of the thundercloud.
∆V
1.00 × 108 V
d=
=
= 300 m .
E
3.33 × 105 V m
696
Physics
Chapter 17: Electric Potential
67. (a) Strategy Use Eq. (17-10).
Solution Compute the magnitude of the average electric field between the cow’s front and hind legs.
∆V (400 − 200) × 103 V
E=
=
= 1.1× 105 V m
d
1.8 m
Since the electric field always points in the direction of decreasing potential, the average electric field is
1.1× 105 V m toward the hind legs .
(b) Strategy and Solution The front and hind legs of Cow B are nearly at the same potential, whereas those for
Cow A span a potential difference of approximately 200 kV, thus Cow A is more likely to be killed.
68. Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the capacitance of the capacitor.
Q
0.020 × 10−6 C
C=
=
= 83 pF .
∆V
240 V
69. Strategy The spark flies between the spheres when the electric field between them exceeds the dielectric
strength. The magnitude of the electric field is given by ∆V d , where d is the distance between the spheres.
Solution Find d.
∆V
∆V
900 V
E=
, so d =
=
= 0.30 mm .
d
E
3.0 × 106 V m
70. Strategy Use Eq. (17-16).
Solution
(a) Find the greatest κ d , since A and ⑀0 are constant.
3.5
7.0
2.0
= 35,
= 3.5, and
= 0.2. Since 35 > 3.5 > 0.2, the paper is the best choice.
0.10
2.0
10.0
C =κ
⑀0 A
d
=
3.5[8.854 × 10−12 C2 (N ⋅ m 2 )](120 × 10− 4 m 2 )
0.10 × 10−3 m
= 3.7 nF
(b) Compute the smallest capacitance.
2.0[8.854 × 10−12 C 2 (N ⋅ m 2 )](120 × 10−4 m 2 )
C=
= 21 pF
10.0 × 10−3 m
71. Strategy Use Eq. (17-16).
Solution Compute the capacitance of the capacitor.
⑀ A 2.5[8.854 ×10−12 C2 (N ⋅ m 2 )](0.30 m)(0.40 m)
C =κ 0 =
= 89 nF
d
0.030 × 10−3 m
72. Strategy Use Eq. (17-16).
Solution Find the average dielectric constant of the tissue in the limb.
⑀ A
dC
(0.030 m)(0.59 × 10−12 F)
C = κ 0 , so κ =
=
= 5.0 .
d
⑀0 A [8.854 ×10−12 C2 (N ⋅ m 2 )](4.0 × 10−4 m 2 )
697
Chapter 17: Electric Potential
Physics
73. (a) Strategy Use Eq. (17-18c).
Solution Compute the capacitance.
Q2
Q 2 (8.0 × 10−2 C)2
U=
, so C =
=
= 7.1 µF .
2C
2U
2(450 J)
(b) Strategy Use Eq. (17-18a).
Solution Compute the potential difference.
1
2U
2(450 J)
U = Q∆V , so ∆V =
=
= 1.1× 104 V .
Q 8.0 × 10−2 C
2
74. Strategy Use Eq. (17-19). The dielectric strength of air is 3 kV mm , which is equal to the maximum electric
field.
Solution Compute the maximum electric energy density in dry air.
1
1
u = κ ⑀0 E 2 = (1.00054)[8.854 × 10−12 C2 (N ⋅ m 2 )](3 × 106 V m) 2 = 40 J m3
2
2
75. Strategy The capacitance of a capacitor is inversely proportional to the distance between the plates and
U = Q 2 (2C ).
Solution Form a proportion to find the ratio of the new capacitance to the old.
C2 d1
d1
1
=
=
=
C1 d 2 1.50d1 1.50
Form a proportion to find the energy stored in the capacitor in terms of the old.
U 2 Q 2 (2C2 ) C1
=
=
= 1.50, so U 2 = 1.50U1.
U1 Q 2 (2C1 ) C2
Thus, the energy increases by 50% .
76. Strategy The energy stored in the capacitor is given by U = 12 C (∆V )2 . When the plate separation is increased,
the capacitance changes but the potential difference stays the same, so the energy in the capacitor changes as well.
The work done on the capacitor in separating the plates is negative the change in energy.
Solution Form a proportion. The capacitance is inversely proportional to the plate separation.
U f Cf (∆V )2 Cf
d
1.00 cm
=
=
= i =
= 0.500. So, the energy is reduced by half.
2
U i Ci (∆V )
Ci d f 2.00 cm
Find the work done on the capacitor.
1
0.500 ⑀0 A
W = U i − U f = U i − 0.500U i = 0.500U i = 0.500 Ci (∆V )2 =
( ∆V ) 2
di
2
2
=
0.500[8.854 × 10−12 C2 (N ⋅ m 2 )](314 × 10− 4 m 2 )
(20.0 V) 2 = 2.78 nJ
2(0.0100 m)
698
Physics
Chapter 17: Electric Potential
77. (a) Strategy Use Eq. (17-15).
Solution Find the capacitance for the thundercloud.
⑀ A [8.854 × 10−12 C2 (N ⋅ m 2 )](4500 m)(2500 m)
C= 0 =
= 0.18 µF
d
550 m
(b) Strategy Use Eq. (17-18c).
Solution Find the energy stored in the capacitor.
Q2
(18 C)2
U=
=
= 8.9× 108 J
2C 2(0.1811× 10−6 F)
78. (a) Strategy The capacitance after the slab is removed is equal to the capacitance with the slab divided by the
dielectric constant.
Solution Compute the capacitance.
C 6.0 µF
C0 = =
= 2.0 µF
3.0
κ
(b) Strategy Use Eqs. (17-10) and (17-17).
Solution Find the potential difference across the capacitor.
E0 = κ E
E0d = κ Ed
∆V0 = κ ∆V = 3.0(1.5 V) = 4.5 V
(c) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the charge on the plates.
Q = C ∆V = (2.0 µF)(4.5 V) = 9.0 µC
(d) Strategy Use Eq. (17-18c).
Solution Compute the energy stored in the capacitor.
Q 2 (9.0 × 10−6 C)2
U=
=
= 20 µJ
2C 2(2.0 × 10−6 F)
79. (a) Strategy Use the definition of capacitance, Eq. (17-14), and Eq. (17-15).
Solution Find the charge on the capacitor.
⑀ A
[8.854 × 10−12 C2 (N ⋅ m 2 )](0.100 m)2 (150 V)
Q = C ∆V = 0 ∆V =
= 18 nC
d
0.75 × 10−3 m
(b) Strategy Use Eq. (17-18a).
Solution Compute the energy stored in the capacitor.
1
1
U = Q∆V = (17.7 × 10−9 C)(150 V) = 1.3 µJ
2
2
699
Chapter 17: Electric Potential
Physics
80. (a) Strategy Use Eq. (17-15).
Solution Compute the capacitance.
⑀ A [8.854 × 10−12 C2 (N ⋅ m 2 )](0.100 m)2
C= 0 =
= 59.0 pF
d
0.75 × 10−3 m + 0.750 × 10−3 m
(b) Strategy From Problem 63, Q = 18 nC. Use Eq. (17-18c) and conservation of energy.
Solution Compute the new energy stored in the capacitor.
Q2
(18 × 10−9 C)2
U=
=
= 2.7 µJ
2C 2(59.0 × 10−12 F)
Work was done on the capacitor when the plates were separated; that work has been stored in the capacitor
as potential energy.
81. (a) Strategy U = P∆t where P = 10.0 kW and ∆t = 2.0 ms. Use Eq. (17-18b).
Solution Find the initial potential difference.
1
2U
2(10.0 kW)(2.0 ms)
U = C (∆V ) 2 , so ∆V =
=
= 630 V .
2
C
100.0 × 10−6 F
(b) Strategy Use Eq. (17-18c).
Solution Find the initial charge.
Q2
U=
, so Q = 2CU = 2(100.0 × 10−6 F)(10.0 kW)(2.0 ms) = 0.063 C .
2C
82. Strategy Use Eq. (17-18a).
Solution Compute the energy stored in the capacitor.
1
1
U = Q∆V = (0.020 × 10−6 C)(240 V) = 2.4 µJ
2
2
83. Strategy The work done by the external agent is equal to the change in potential energy of the capacitor. Use
Eq. (17-18c) and the fact that the capacitance is inversely proportional to the plate separation.
Solution Find the work required to double the plate separation.
⎞ Q 2 ⎛ df
⎞ (0.80 × 10−6 C)2
Q 2 Q 2 Q 2 ⎛ Ci
W = ∆U =
−
=
− 1⎟ =
(2 − 1) = 0.27 mJ
⎜
⎜ − 1⎟ =
−9
2Cf 2Ci 2Ci ⎝ Cf
⎠ 2Ci ⎝ di
⎠ 2(1.20 × 10 F)
84. Strategy Use Eq. (17-18b).
Solution Find the required potential difference.
1
2U
2(300 J)
U = C (∆V ) 2 , so ∆V =
=
= 8 kV .
2
C
9 × 10−6 F
700
Physics
Chapter 17: Electric Potential
85. (a) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the charge that passes through the body tissues.
Q = C ∆V = (15 × 10−6 F)(9.0 × 103 V) = 0.14 C
(b) Strategy Use Eq. (17-18b) and the definition of average power.
Solution Find the average power delivered to the tissues.
∆E U C (∆V )2 (15 × 10−6 F)(9.0 × 103 V) 2
Pav =
=
=
=
= 0.30 MW
∆t ∆t
2∆t
2(2.0 × 10−3 s)
86. Strategy Assume the that thundercloud and Earth system behaves like a capacitor. Use Eq. (17-18a).
Solution Find the electric potential energy released by the lightning strike.
1
1
U = Q∆V = (25.0 C)(1.00 × 108 V) = 1.25 GJ .
2
2
87. (a) Strategy Assume that the thundercloud and Earth system acts like a capacitor. Use Eq. (17-18a).
Solution Find the electric potential energy released by the lightning strike.
1
1
U = Q∆V = (20.0 C)(1.00 × 109 V) = 10.0 GJ
2
2
(b) Strategy Use the definition of latent heat.
Solution Find the mass of sap that is vaporized.
Q = energy absorbed = mLV = 0.100(10.0 × 109 J), so m =
1.00 × 109 J
= 443 kg .
2,256,000 J kg
(c) Strategy Divide 10.0% of the total energy released from the lightning strike by the homeowner’s monthly
energy use.
Solution
∆t =
0.100(10.0 × 109 J)
(400.0 × 103 W ⋅ h month)(3600 s h)
= 0.694 month
88. Strategy Use Eq. (17-9).
Solution Find the potential midway between the charges.
kq kq
k
8.988 × 109 N ⋅ m 2 C2
V = 1 + 2 = (q1 + q2 ) =
(−12.0 × 10−9 C − 22.0 × 10−9 C) = −873 V
0.700 m
r
r
r
2
701
Chapter 17: Electric Potential
Physics
89. (a) Strategy Let qL = −qR = 10.0 nC. Use Eqs. (17-5) and (17-9).
Solution Find the potential energy of the point charge at each location.
⎛ kq
⎛ 1
kq ⎞
1 ⎞
U a = qVa = q ⎜ L + R ⎟ = kqqL ⎜ − ⎟
rR ⎠
⎝ rL
⎝ rL rR ⎠
1
1
⎛
⎞
= (8.988 × 109 N ⋅ m 2 C2 )(− 4.2 × 10−9 C)(10.0 × 10−9 C) ⎜
−
⎟ = − 6.3 µJ
0.0400
m
0.1200
m
⎝
⎠
1
1
⎛
⎞
9
2
2
−9
−9
U b = qVb = (8.988 × 10 N ⋅ m C )(− 4.2 × 10 C)(10.0 × 10 C) ⎜
−
⎟= 0
⎝ 0.0400 m 0.0400 m ⎠
1
1
⎛
⎞
U c = qVc = (8.988 × 109 N ⋅ m 2 C2 )(− 4.2 × 10−9 C)(10.0 × 10−9 C) ⎜
−
⎟= 0
⎝ 0.0800 m 0.0800 m ⎠
(b) Strategy The work done by the external force is negative the work done by the field. Use Eq. (6-8).
Solution Find the work required to move the point charge.
W = −Wfield = ∆U = U a − U b = − 6.3 µJ − 0 = − 6.3 µJ
90. Strategy Use Eqs. (6-8) and (17-7).
Solution Compute the work done by the electric field.
Wfield = −∆U = − (− e)∆V = (1.602 × 10−19 C)[100.0 V − (−100.0 V)] = 3.204 × 10−17 J
91. Strategy The potential at the surface of a conducting sphere is equal to the magnitude of the electric field times
the radius of the sphere.
Solution Compute the potential.
V = Er = (3.0 × 106 N C)(0.15 m) = 450 kV
92. Strategy Let q = 2.0 × 10−21 C and r = 1.0 nm. Use Eq. (17-9).
Solution Find the potential at the sodium ions due to the other three ions.
kq
kq kq
⎛ q 2q q ⎞ kq (8.988 × 109 N ⋅ m 2 C2 )(2.0 × 10−21 C)
V = 1 + 2 + 3 = k ⎜− +
− ⎟=
=
= 9.0 mV
r1
r2
r3
2(1.0 × 10−9 m)
⎝ r r 2r ⎠ 2r
93. (a) Strategy Electric field lines begin on positive charges and end on negative charges. The same number of
field lines begins on the plate and ends on the negative point charge. Field lines never cross. Use the
principles of superposition and symmetry.
Solution The electric field lines for the cylinder and sheet:
702
Physics
Chapter 17: Electric Potential
(b) Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential
surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer
together where the field is stronger. The electric field always points in the direction of maximum potential
decrease.
Solution The equipotential surfaces for the cylinder and sheet:
94. Strategy Use Newton’s second law.
Solution Find the acceleration.
eE
ΣFy = eE = ma y , so a y =
.
m
Find the time to reach the lower plate.
∆y =
1
a y (∆t )2 , so ∆t =
2
2(9.109 × 10−31 kg)(0.040 m)
2m∆y
=
eE
(1.602 × 10−19 C)(5.0 × 104 N C)
= 3.0 ns .
95. Strategy Assume the field is uniform. Use Eq. (17-10).
Solution Compute the magnitude of the electric field in the membrane.
∆V 90 × 10−3 V
E=
=
= 9 × 106 V m .
d
10 × 10−9 m
96. Strategy Use Newton’s second law.
Solution
K
(a) Since E points downward, the negatively charged electron’s change in velocity is directed upward.
Find the acceleration.
eE
ΣFy = eE = ma y , so a y =
.
m
Find the time interval.
∆x
∆t =
vx
Find the change in velocity.
⎛ eE ⎞ ⎛ ∆x ⎞ (1.602 × 10−19 C)(2.0 × 104 N C)(0.060 m)
∆v y = a y ∆t = ⎜
= 7.0 × 106 m s
⎟⎜ v ⎟ =
7
−31
m
(9.109 × 10
kg)(3.0 × 10 m s)
⎝
⎠⎝ x ⎠
K
So, ∆v = 7.0 × 106 m s upward .
(b) Find the deflection of the electrons.
2
∆y =
1
1 ⎛ eE ⎞ ⎛ ∆x ⎞
(1.602 × 10−19 C)(2.0 × 104 N C)(0.060 m)2
a y (∆t )2 = ⎜
=
= 7.0 mm
⎜
⎟
2
2 ⎝ m ⎠⎟ ⎝ vx ⎠
2(9.109 × 10−31 kg)(3.0 × 107 m s) 2
703
Chapter 17: Electric Potential
Physics
97. Strategy The negatively charged particle will accelerate toward the positively charged plate while it is between
the plates of the capacitor.
Solution The particle is between the plates for a time given by ∆t = ∆x vx . During this time, the particle travels a
vertical distance ∆y = 0.00100 m. Find the acceleration of the particle.
2
2v 2 ∆y
1
1 ⎛ ∆x ⎞
a ( ∆x ) 2
∆y = a (∆t )2 = a ⎜⎜ ⎟⎟ =
, so a = x
.
2
2 ⎝ vx ⎠
2v x 2
( ∆x ) 2
∆V
, where d is the plate separation and N is the
d
number of excess electrons on the particle. According to Newton’s second law, the acceleration of the particle is
Ne ∆dV Ne∆V
. We set the two expressions for the acceleration of the particle equal and solve for N.
=
a=
m
md
The magnitude of the electrical force on the particle is NeE = Ne
2md vx 2 ∆y 2(5.00 × 10−19 kg)(0.00200 m)(35.0 m s)2 (0.00100 m)
Ne∆V 2vx 2 ∆y
=
, so N =
=
= 51 .
md
(∆x)2
(1.602 × 10−19 C)(3.00 V)(0.0100 m)2
e∆V (∆x)2
98. (a) Strategy Compute the electrical and gravitational forces on the particle and compare. Refer to Problem 83.
Solution The gravitational force on the particle is mg = (5.00 × 10−19 kg)(9.80 m s 2 ) = 4.90 × 10−18 N .
The electrical force on the particle is
Ne∆V 51(1.602 × 10−19 C)(3.00 V)
=
= 1.23 × 10−14 N .
0.00200 m
d
Compare the forces.
1.226 × 10−14 N
= 2.50 × 103
4.90 × 10−18 N
The electrical force is 2.50 × 103 times larger than the gravitational force.
(b) Strategy Use the results from Problem 83. The horizontal component of the velocity doesn’t change.
Solution The horizontal component of the velocity is vx = 35.0 m s .
Compute the y-component of the particle’s velocity.
2v 2 ∆y ⎛ ∆x ⎞ 2vx ∆y 2(35.0 m s)(0.00100 m)
v y = a∆t = x
=
= 7.00 m s
⎜ ⎟=
0.0100 m
∆x
(∆x) 2 ⎜⎝ vx ⎟⎠
99. Strategy Find the charge on the capacitor. Use Eqs. (17-14) and (17-15).
Solution The charge on the capacitor is Q = Ne, where N is the number of excess electrons.
Q = Ne = C ∆V =
⑀0 A
d
∆V , so N =
⑀0 A∆V
de
=
[8.854 × 10−12 C2 (N ⋅ m 2 )](0.0100 m)2 (3.00 V)
(0.00200 m)(1.602 × 10−19 C)
704
= 8.29 × 106 .
Physics
Chapter 17: Electric Potential
100. (a) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the capacitance.
Q
0.020 × 10−6 C
=
= 83 pF
C=
∆V
240 V
(b) Strategy Use Eq. (17-15).
Solution Find the area of a single plate.
⑀ A
dC (0.40 × 10−3 m)(8.33 ×10−11 F)
=
= 3.8 × 10−3 m 2 .
C = 0 , so A =
−12 2
2
⑀0
d
8.854 × 10
C (N ⋅ m )
(c) Strategy Use Eq. (17-10).
Solution Compute the voltage required to ionize the air between the plates.
∆V = Ed = (3.0 × 103 V mm)(0.40 mm) = 1.2 kV
101. Strategy The energy in the capacitor is converted into heat in the water. Use Eqs. (14-4) and (17-18b).
Solution Find the temperature change of the water.
1
C (∆V )2
(200.0 × 10−6 F)(12.0 V)2
=
= 3.44 mK .
Q = mc∆T = U = C (∆V )2 , so ∆T =
2
2mc
2(1.00 cm3 )(1.00 g cm3 )[4.186 J (g ⋅ K)]
102. (a) Strategy Use Eqs. (17-16) and (17-18b).
Solution Find the energy stored in the capacitor.
1
1 ⎛ κ⑀ A ⎞
5.2[8.854 ×10−12 C 2 (N ⋅ m 2 )](1.0 × 10−7 m 2 )(90.0 ×10−3 V) 2
U = C (∆V )2 = ⎜ 0 ⎟ (∆V ) 2 =
2
2⎝ d ⎠
2(7.5 × 10−9 m)
= 2.5 × 10−12 J
(b) Strategy Divide the total charge by the charge of one ion. Use the definition of capacitance, Eq. (17-14), and
Eq. (17-16).
Solution Find the number of ions outside of the membrane.
Q C ∆V κ ⑀0 A∆V 5.2[8.854 × 10−12 C2 (N ⋅ m 2 )](1.0 × 10−7 m 2 )(90.0 × 10−3 V)
=
=
=
= 3.4 × 108 ions
e
e
ed
(1.602 × 10−19 C ion)(7.5 × 10−9 m)
103. Strategy Use Eq. (17-16).
Solution Find the capacitance of the axon.
⑀ A 5[8.854 ×10−12 C2 (N ⋅ m 2 )](5 × 10−12 m 2 )
C =κ 0 =
= 5 × 10−14 F
d
4.4 × 10−9 m
705
Chapter 17: Electric Potential
Physics
104. Strategy Use conservation of energy, Wfield = −∆U , and the fact that the field is uniform.
Solution Find the kinetic energy of each electron when it leaves the space between the plates.
⎛ ∆V ⎞
∆K = K f − Ki = −∆U = Wfield = eE ∆y = e ⎜
⎟ ∆y, so
⎝ d ⎠
(1.602 × 10−19 C)(100.0 × 103 V)(0.0030 m)
⎛ ∆V ⎞
−15
∆
=
×
+
= 6.0 × 10−15 J .
2.0
10
J
K f = Ki + e ⎜
y
⎟
0.0120 m
⎝ d ⎠
105. (a) Strategy For a parallel plate capacitor, E = σ ⑀0 and ∆V = Ed .
Solution Find the potential difference between the plates.
σ d (4.0 × 10−6 C m 2 )(0.0060 m)
∆V = Ed =
=
= 2.7 kV
⑀0
8.854 × 10−12 C2 (N ⋅ m 2 )
(b) Strategy Use conservation of energy and the fact that ∆U = q∆V .
Solution Find the kinetic energy of each point charge just before it hits the positive plate.
∆K = K f − Ki = −∆U = −q∆V , so K f = Ki − q∆V = 0 − (−2.5 × 10−9 C)(2711 V) = 6.8 µJ .
106. Strategy Use conservation of energy, Wfield = −∆U , and the fact that the field is uniform.
Solution Find the final kinetic energy of the alpha particle.
∆K = K f − Ki = −∆U = Wfield = 2eEd , so
K f = Ki + 2eEd = 0 + 2(1.602 × 10−19 C)(10.0 × 103 V m)(0.010 m) = 3.2 × 10−17 J .
107. Strategy Use Eqs. (6-8) and (17-7).
Solution Find the work done by the electric field.
Wfield = −∆U = − q∆V = − e∆V = −(1.602 × 10−19 C)( − 90.0 × 10−3 V) = 1.44 × 10−20 J
108. Strategy Electric field lines begin on positive charges and end on negative charges. The same number of field
lines begins on the plate and ends on the negative point charge. Field lines never cross. Use the principles of
superposition and symmetry. The electric field always points in the direction of maximum potential decrease.
Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such
that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field
is stronger.
Solution The electric field lines and equipotential surfaces for the cube:
706
Physics
Chapter 17: Electric Potential
109. Strategy The energy stored in the capacitor is directly proportional to the capacitance. Use Eq. (17-16) and the
fact that ∆V is constant. Form a proportion.
Solution Determine what happens to the energy stored in the capacitor.
(
)
∆U ∆C κ ⑀0 A df − di
d
1
=
=
= i −1 =
− 1 = − 0.200, so the energy is reduced by 20.0%.
κ ⑀0 A
U0
C0
df
1.25
1
1
di
110. Strategy Use Eq. (17-18b) and form a proportion. ∆V is constant.
Solution Find the energy stored in the capacitor after the dielectric is inserted.
1 C (∆V ) 2
U
C κ C0
= 2
=
=
= κ , so U = κ U 0 = 3.0U 0 .
2
1
U0
C0
C0
C0 (∆V )
2
111. (a) Strategy Treat the axon as a parallel plate capacitor. Use Eq. (17-16) and the fact that the area of the curved
surface of a cylinder is (2π r ) L, where r and L are the radius and length of a cylinder, respectively.
Solution Calculate the capacitance per unit length of the axon.
κ ⑀ A κ ⑀ [(2π r ) L]
, so
C= 0 = 0
d
d
C 2πκ ⑀0r 2π (7.0)[8.854 × 10−12 C2 (N ⋅ m 2 )](5.0 × 10−6 m)
=
=
= 3.2 × 10−7 F m .
−9
L
d
6.0 × 10 m
(b) Strategy Use Eq. (17-10) and the fact that the magnitude of the electric field inside a parallel plate capacitor
is given by σ ⑀0 .
Solution The outside of the membrane has the positive charge, since the potential is higher outside than
inside.
κ ⑀ ∆V 7.0[8.854 × 10−12 C2 (N ⋅ m 2 )](0.085 V)
E
σ
∆V = Ed = 0 d =
=
= 8.8 × 10−4 C m 2 .
d , so σ = 0
κ
κ ⑀0
d
6.0 × 10−9 m
112. (a) Strategy Use Eq. (17-18b), the definition of capacitance, Eq. (17-14), and the relationships between the
quantities (energy, potential difference, capacitance) before and after the dielectric is inserted.
Solution Calculate the initial energy stored in the capacitor (without the dielectric).
1
1
U i = Ci (∆Vi ) 2 = (4.00 × 10−6 F)(100.0 V)2 = 20.0 mJ
2
2
C
∆Vi
Cf = κ Ci and Q = Ci ∆Vi = Cf ∆Vf . So, ∆Vf = i ∆Vi =
.
κ
Cf
Calculate the final energy.
2
Uf =
1
1
1 ⎡1
20.0 mJ
⎛ ∆V ⎞
⎤ 1
= 3.3 mJ
Cf (∆Vf ) 2 = κ Ci ⎜ i ⎟ = ⎢ Ci (∆Vi ) 2 ⎥ = U i =
κ
κ
κ
2
2
2
6.0
⎝
⎠
⎣
⎦
(b) Strategy and Solution Since the energy of the capacitor increases when the dielectric is removed, an
external agent has to do positive work to remove the dielectric .
707
Chapter 17: Electric Potential
Physics
113. (a) Strategy Use Eq. (17-10).
Solution Compute the minimum thickness of the titanium dioxide.
5.00 V
∆V
d=
=
= 1.25 µm
E
4.00 × 106 V m
(b) Strategy Use Eq. (17-16).
Solution Find the area of the plates.
⑀ A
(1.25 × 10−6 m)(1.0 F)
dC
=
= 1600 m 2 .
C = κ 0 , so A =
κ ⑀0 90.0[8.854 × 10−12 C2 (N ⋅ m 2 )]
d
114. Strategy Use the definition of capacitance, Eq. (17-14), to find the charge that moves through the membrane.
Then divide the charge by e.
Solution
⎛C⎞
(a) Q = C ∆V = ⎜ ⎟ A∆V = (1× 10−6 F cm 2 )(0.05 cm 2 )[20 × 10−3 V − (−90 × 10−3 V)] = 6 nC
⎝ A⎠
(b)
5.5 × 10−9 C
Q
=
= 3 × 1010 ions
e 1.602 × 10−19 C ion
115. Strategy Let E0 = 20.0 V m , E1 be the field outside of the dielectric after it is inserted, and E2 be the field
inside the dielectric. Use the principle of superposition for the potential after the dielectric is inserted and the fact
that E2 = E1 κ .
Solution Find the electric field inside the dielectric.
Initially: ∆V = E0d
d
d
d ⎛E ⎞d Ed
+ E2 = E1 + ⎜ 1 ⎟ = 1
2
2
2 ⎝κ ⎠2
2
Solve for E1 in terms of E0.
Finally: ∆V = E1
E1d
2
1⎞
⎛
⎜1 + κ ⎟
⎝
⎠
2 E0
⎛ 1⎞
.
⎜ 1 + κ ⎟ = E0d , so E1 =
1 + κ1
⎝
⎠
Calculate E2.
2 E0 2(20.0 V m)
E
=
= 8.0 V m
E2 = 1 =
κ κ +1
4.0 + 1
708
Chapter 18
ELECTRIC CURRENT AND CIRCUITS
Conceptual Questions
1.
S
Ᏹ
R
R
2. Voltmeters have very large internal resistances, so there is usually no danger of a large current. Ammeters on the
other hand have very small internal resistances, so they may draw very large currents in a circuit if they are not
connected in series with any other significant resistance.
3. The resistance of a light-bulb filament increases significantly as it heats up. The largest current therefore flows
through the bulb when it is cold. Thus, it is most likely to burn out just after being switched on.
4. If he connects three 300 Ω resistors in parallel, the equivalent resistance will be the desired 100 Ω.
5. This statement is not exactly true. The current flowing through a branch in a circuit is inversely proportional to the
resistance of the branch. Thus, more current follows the path of least resistance than follows any other path, but
every path has some current.
6. An ideal ammeter has zero resistance so as to have no effect on the current it is supposed to measure when
connected in series in a circuit. An ideal voltmeter has an infinite resistance so that it does not perturb the voltage
it is supposed to measure when connected in parallel in a circuit.
7. As the temperature increases, the atoms in the metal begin to vibrate with greater amplitudes. The chance of an
electron colliding with one of the atoms is therefore increased. This effectively reduces the mean free path of the
electrons and increases the resistance of the metal.
8. Some of the energy is dissipated as heat by the resistors and some of it gets stored in the electric fields of the
capacitors as they are charged up.
9. Electric stoves and clothes dryers require relatively large amounts of power to operate. Supplying them with
240 V instead of 120 V decreases the magnitude of required current to supply the power. This reduces the rate
(P = I2R) at which energy is dissipated in the wiring.
10. Elements connected in series in a circuit have the identical current flowing through them, while elements in
parallel have identical potential differences. Therefore, ammeters are connected in series and voltmeters in
parallel. It is true that if the element under consideration is a resistor, a measurement of the current allows one to
calculate the voltage via Ohm’s law and vice versa. So, for a resistor, an ammeter could be used to measure the
voltage in a sense. This doesn’t work for other elements though. A fully charged capacitor for example has no
current flowing through it and an ammeter connected in series would tell us the current but give us no information
about the voltage. Similarly, a voltmeter connected in parallel to a capacitor would allow us to measure the
voltage but give us no information about the current flowing through it.
11. Most of the electrical resistance of the body is due to the skin. The resistance of wet skin is much lower than the
resistance of dry skin. Thus, it is more dangerous to touch a “live” electrical wire with wet hands because more
current will flow through the body.
709
Chapter 18: Electric Current and Circuits
Physics
12. An electric field is required to start a current flowing within a conductor. The electric field inside a conductor is
therefore not equal to zero if a current is flowing through it.
13. Batteries convert chemical energy into electrical energy. As a battery is used, its supply of chemical energy is
depleted. Recharging the battery does not actually put additional charges back into the battery, but instead
converts electrical energy into chemical energy.
14. Current flows from B to C through the clock. Current flows from D to A through the battery. Terminal A of the
battery is at the higher potential. Side B of the clock is at the higher potential. Current can be made to flow across
a circuit element from a lower to a higher potential if work is done by an external agent such as a battery.
15. The total resistance of two resistors connected in series is equal to the sum of their resistances. If resistance is
proportional to length, then the above statement tells us that the total resistance of two resistors of length L/2
connected in series is proportional to the length of the combined resistor L. If resistance had any other relationship
to length, the total resistance would not agree with that given by the first statement.
16. The inverse of the total resistance of two resistors connected in parallel is equal to the sum of the inverse
resistances of each resistor. If the resistance of a wire is inversely proportional to its cross-sectional area, then the
above statement tells us that the total resistance of two resistors of cross-sectional area A connected in parallel is
proportional to 0.5A. If resistance had any other relationship to cross-sectional area, the total resistance would not
agree with that given by the first statement.
17. An electrician working on live wiring wears insulated shoes to avoid being grounded and therefore to reduce the
chance of starting a flow of current through the body if a live wire is touched. Similarly, an electrician using two
hands would risk completing a circuit so that current flows from one hand to the other through the body and near
the heart.
18. A 20-A circuit breaker would allow more current into the circuit than it was designed to handle. The additional
current may allow additional appliances to operate, but resistive heating in the wires due to the increased current
may be hazardous.
19. The bird perched on the power line is at the same electric potential as the line but it is isolated from the ground so
that no current flows through its body. When a person standing on the ground touches a power line with a metal
pole, a potential difference exists between the line and the person’s grounded feet—a current will therefore flow
through their body.
20. The total emf produced by several batteries in series is the sum of the emf of each battery—each battery may have
any emf. If batteries connected in parallel have different emfs, significantly larger currents will flow through the
batteries than if they had the same emf. One way to see this is that in parallel the batteries must have the same
potential drop across them, and usually the internal resistance of a battery is very small. For example, consider
two batteries connected in parallel, one with an emf of 9 V and one with 10 V, and each with an internal
resistance of 0.1 Ω. Since the potential drops across the batteries must be equal, there must be a difference of 10 A
of current flowing through the 9-V battery as compared to the current flowing through the 10-V battery, to make
up for the 1-V difference. Even if the batteries are disconnected from the rest of the circuit, there will be
significant currents flowing around the loop composed of just the batteries in parallel. The result in either case is a
significant loss of power to resistive heating in the batteries—a waste of the energy stored in the batteries.
Furthermore, these large currents could result in overheating of the batteries, which could be dangerous. To avoid
this problem, batteries connected in parallel should have the same emf.
21. (a) It increases.
(b) It decreases.
(c) It increases.
710
Physics
Chapter 18: Electric Current and Circuits
22. (a) Bulbs C and D are equally bright and they are brighter than bulbs A or B. Bulbs A and B are also equally
bright, but less so than bulbs C and D.
(b) The brightness of bulb B increases.
(c) The brightness of bulb C remains the same.
23. (a) Bulbs B and C get brighter.
(b) Bulb A gets dimmer and bulb C gets brighter. With all three resistances in the circuit, the currents are
IA = 2V/(3R), IB = IC = V/(3R). When B is removed, the current in both A and C is V/(2R).
(c) Bulb A gets brighter, bulb C stops glowing entirely.
Problems
1. Strategy Use the definition of electric current.
Solution Compute the total charge.
∆q
I=
, so ∆q = I ∆t = (3.0 A)(4.0 h)(3600 s h) = 4.3 × 104 C .
∆t
2. (a) Strategy Use the definition of electric current.
Solution Compute the charge.
∆q = I ∆t = (0.500 A)(10.0 s) = 5.00 C .
(b) Strategy Divide the charge found in part (a) by the magnitude of the charge of an electron.
Solution Compute the number N of electrons.
∆q
5.00 C
N=
=
= 3.12 × 1019 electrons
C
e 1.602 × 10−19
electron
3. (a) Strategy and Solution The electrons flow from the filament to the anode; since they are negatively charged,
the current flows from the anode to the filament.
(b) Strategy Use the definition of electric current.
Solution Compute the current in the tube.
∆q
= ∆q × f = (1.602 × 10−19 C)(6.0 × 1012 s −1 ) = 0.96 µA
I=
∆t
4. Strategy Use the definition of electric current.
Solution Compute the beam current.
∆q
= ∆q × f = [2(1.602 × 10−19 C)](3.0 × 1013 s −1 ) = 9.6 µA
I=
∆t
5. Strategy Use the definition of electric current and the elementary charge of an electron.
Solution Find the number of electrons per second that hit the screen.
N I
∆q Ne
320 × 10−6 A
I=
=
, so
= =
= 2.0 × 1015 electrons/s .
C
∆t ∆t
∆t e 1.602 × 10−19
electron
711
Chapter 18: Electric Current and Circuits
Physics
6. Strategy Since the negatively charged electrons and positive ions move in opposite directions, they both
contribute to the current in the same direction. Use the definition of electric current.
Solution Compute the current in the tube.
∆q (3.8 × 1016 + 1.2 × 1016 )(1.602 × 10−19 C)
I=
=
= 8.0 mA
∆t
1.0 s
7. Strategy Since the oppositely charged ions move in opposite directions, they both contribute to the current in the
same direction. Use the definition of electric current.
Solution Compute the current in the solution.
∆q Ne [2(3.8 × 1016 ) + 6.2 × 1016 ](1.602 × 10−19 C)
I=
=
=
= 22.1 mA
∆t ∆t
1.0 s
8. Strategy The energy delivered by each battery is equal to the total work done by each battery. Use Eq. (18-2).
Solution Compute the energy delivered by each battery, assuming they are ideal.
Scooter:
W = q = (12 V)(4.0 kC) = 48 kJ
Automobile:
W = (12 V)(30.0 kC) = 360 kJ
9. Strategy The total energy stored in a battery is equal to the total work the battery is able to do. Use Eq. (18-2).
Solution Compute the energy stored in the battery.
W = q = (1.20 V)(675 C) = 810 J
10. (a) Strategy A coulomb is an A ⋅ s. Convert A ⋅ h to A ⋅ s to find the amount of charge that can be pumped by
the battery.
Solution Compute the charge.
(180.0 A ⋅ h)(3600 s h ) = 6.480 × 105 C
(b) Strategy The total energy stored in a battery is equal to the total work the battery is able to do. Use Eq.
(18-2).
Solution Compute the electrical energy that the battery can supply.
W = q = (12.0 V)(6.480 × 105 ) = 7.78 MJ
(c) Strategy Use the definition of electric current.
Solution Find the time required to drain the battery.
∆q 6.480 × 105 C ⎛ 1 h ⎞
∆t =
=
⎜
⎟ = 54.5 h
I
3.30 A
⎝ 3600 s ⎠
712
Physics
Chapter 18: Electric Current and Circuits
11. (a) Strategy Use the definition of electric current.
Solution Compute the amount of charge pumped by the battery.
∆q = I ∆t = (220.0 A)(1.20 s) = 264 C
(b) Strategy The electrical energy supplied is equal to the work done by the battery. Use Eq. (18-2).
Solution Compute the amount of electrical energy supplied by the battery.
W = q = (12.0 V)(264 C) = 3.17 kJ
12. (a) Strategy Calculate the work done using Eq. (18-2) and the definition of electric current.
Solution Compute the amount of electrical energy supplied by the solar cell.
W = ∆q = I ∆t = (0.45 V)(18.0 × 10−3 A)(9.0 h)(3600 s h ) = 260 J
(b) Strategy The power is equal to the rate at which the solar cell supplies electrical energy.
Solution Find the average power by dividing the energy supplied by the time.
W I ∆t
P=
=
= I = (0.45 V)(18.0 × 10−3 A) = 8.1 mW
∆t
∆t
13. Strategy Use Eq. (18-3).
Solution Form a proportion.
(
(
)
)
2
2
ne 14 π d12 v1
⎛d ⎞
I1
neA1v1
d 2v
⎛2⎞
=1=
=
= 1 1 , so v1 = ⎜⎜ 2 ⎟⎟ v2 = ⎜ ⎟ v2 = 4v2 .
I2
neA2 v2 ne 1 π d 2 v
⎝1⎠
d 22 v2
⎝ d1 ⎠
2
2
4
The relationship between the drift speeds is v1 = 4v2 .
14. Strategy Use Eq. (18-3).
Solution Find the drift speed of the conduction electrons in the wire.
I
I
2.50 A
I = neAvD , so vD =
=
=
= 5.86 × 10−5 m s .
2
28
−
3
neA ne(π r ) (8.47 × 10 m )(1.602 × 10−19 C)π (0.00100 m)2
15. Strategy Use Eq. (18-3) and ∆x = vD ∆t.
Solution Find the drift speed of the conduction electrons in the wire.
I
I
I
I = neAvD , so vD =
=
=
.
2
neA ne(π r ) π ner 2
Find the time to travel 1.00 m along the wire.
∆t =
∆x
∆x
π ner 2 ∆x π (8.47 × 1028 m −3 )(1.602 × 10−19 C)( 0.00100 m 2 )2 (1.00 m) ⎛ 1 min ⎞
=
=
=
⎜
⎟
I
vD
I
10.0 A
⎝ 60 s ⎠
2
π ner
= 17.8 min
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Chapter 18: Electric Current and Circuits
Physics
16. Strategy Find the time of travel ∆t using ∆x = vD ∆t and Eq. (18-3).
Solution Find the time of travel.
28
−3
−19 C) 1 π (0.0010 m)2 (0.010 m)
∆x neAd (5.8 × 10 m )(1.602 × 10
⎛ 1 min ⎞
4
∆t =
=
=
⎜
⎟ = 8.1 min
vD
I
0.15 A
⎝ 60 s ⎠
17. Strategy Let h be the thickness of the strip so that the cross-sectional area is A = hw, where w is the width. Use
Eq. (18-3).
Solution Find the thickness of the strip.
I
130 × 10−6 A
I = neAvD = ne(hw)vD , so h =
=
= 81 µm .
newvD (8.8 × 1022 m −3 )(1.602 × 10−19 C)(260 × 10−6 m)(0.44 m s)
18. Strategy The cross-sectional area of the wire is 14 π d 2 . Use Eq. (18-3).
Solution Find the current in the wire.
π
I = neAvD = ne 14 π d 2 vD = ned 2 vD
4
=
π
4
(
)
(5.90 × 1028 m −3 )(1.602 × 10−19 C)(0.00050 m)2 (6.5 × 10−6 m s) = 12 mA
19. Strategy Use Eq. (18-3) and n = 1.3ρ N A M , the number of electrons per unit volume.
Solution Find the drift speed of the conduction electrons.
I
IM
(2.0 A)(64 g mol)
=
=
vD =
6
3
neA 1.3ρ N A eA 1.3(9.0 × 10 g m )(6.022 × 1023 mol−1 )(1.602 × 10−19 C)(1.00 × 10−6 m 2 )
= 0.11 mm s
20. Strategy Find the average time ∆t it takes for an electron to move 12 m along the wire using ∆x = vD ∆t , Eq.
(18-3), and n = 3.5 ρ N A M , the number of electrons per unit volume. The cross-sectional area of the wire is
1 π d 2.
4
Solution Find ∆t.
∆x neA∆x 3.5 ρ N A eA∆x
∆t =
=
=
vD
I
MI
=
3.5(2.7 × 106 g m3 )(6.022 × 1023 mol−1 )(1.602 × 10−19 C) 14 π (0.0026 m)2 (12 m)
= 50 h
(27 g mol)(12 A)(3600 s h )
21. Strategy Use the definition of resistance.
Solution Compute the current through the resistor.
∆V
∆V 16 V
R=
, so I =
=
= 1.3 A .
I
R 12 Ω
714
Physics
Chapter 18: Electric Current and Circuits
22. (a) Strategy Use the definition of resistance.
Solution Compute the resistance.
∆V
4.50 V
R=
=
= 54 Ω
I
0.083 A
(b) Strategy and Solution The current flows from right to left through the battery (from low to high potential).
Thus, the current flows from left to right through the resistor.
23. Strategy Use Eq. (18-8). The cross-sectional areas of the wires are given by 14 π d 2 .
Solution Form a proportion to find the ratio of diameters.
ρ Al
RAl
=1=
RCu
ρCu
L
AAl
L
ACu
ρ A
ρ d 2
d
= Al Cu = Al Cu , so Cu =
ρCu AAl ρCu d Al 2
d Al
ρCu
1.67
=
= 0.794 .
2.65
ρ Al
24. Strategy Use the definition of resistance and Eq. (18-8). The cross-sectional area of the wire is 14 π d 2 .
Solution Find ∆V , the potential difference between the bird’s feet.
∆V = IR =
I ρ L (150 A)(2.65 × 10−8 Ω ⋅ m)(0.020 m)
=
= 0.25 mV
1 π (0.020 m) 2
A
4
25. (a) Strategy Use the definition of resistance.
Solution Compute the required potential difference between the electrician’s hands.
∆V = IR = (50 mA)(1 kΩ) = 50 V
(b) Strategy and Solution An electrician working on a “live” circuit keeps one hand behind his or her back
to avoid becoming part of the circuit.
26. Strategy Use the definition of resistance.
Solution
(a) Compute the resistance given the data at point 1.
∆V
0.30 V
R=
=
= 15 Ω
I
0.020 A
(b) Compute the resistance given the data at point 2.
0.40 V
R=
= 10 Ω
0.040 A
27. Strategy Use Eq. (18-8). The cross-sectional area of the wire is 14 π d 2 .
Solution Find the diameter of the nichrome wire.
R=ρ
L
L
ρL
(108 × 10−8 Ω ⋅ m)(46 m)
=ρ
, so d = 2
=2
= 2.5 mm .
1πd2
A
πR
π (10.0 Ω)
4
715
Chapter 18: Electric Current and Circuits
Physics
28. Strategy As found in Example 18.4, R R0 = 1 + α∆T .
Solution Find α .
1 + α∆T =
⎞
R
1 ⎛ R
1
⎛ 25.0 Ω ⎞
− 1⎟ =
− 1⎟ = 3.8 × 10−3 °C−1 .
, so α =
⎜
⎜
∆T ⎝⎜ R0 ⎠⎟ 85.0° C − 15.0° C ⎝ 19.8 Ω ⎠
R0
29. Strategy As found in Example 18.4, R R0 = 1 + α∆T . Find T using this and the definition of resistance.
Solution Estimate the temperature of the tungsten filament.
⎛ 2.90 V
⎞
⎞
R
1⎛ R
1
⎜ 0.300 A − 1⎟ + 20.0°C = 1750°C .
− 1⎟⎟ + T0 =
1 + α∆T =
, so T = ⎜⎜
R0
α ⎝ R0 ⎠
4.50 × 10−3 °C−1 ⎜⎝ 1.10 Ω ⎟⎠
30. Strategy The terminal voltage is V = − Ir.
Solution Solve for I, the maximum current.
− V 1.5 V − 1.0 V
I=
=
= 5A
r
0.10 Ω
31. Strategy The current is equal to the terminal voltage divided by the resistance of the resistor connected across the
battery terminals. Use Eq. (18-10).
Solution Find the terminal voltage, V.
V
12.0 V
=
= 4.0 V .
V = IR and V = − Ir , so V = − r , or V =
r
2.0 Ω
R
1 + R 1 + 1.0
Ω
Compute the current through the 1.0-Ω resistor.
V 4.0 V
I= =
= 4.0 A
R 1.0 Ω
32. Strategy For wires of the same length and diameter, R ∝ ρ.
Solution Find the ratios; then compare.
(a)
(b)
RAg
RCu
RAl
RCu
=
=
ρ Ag
ρCu
=
1.59 × 10−8 Ω ⋅ m
1.67 × 10−8 Ω ⋅ m
2.65 × 10−8 Ω ⋅ m
1.67 × 10−8 Ω ⋅ m
= 0.952
= 1.59
(c) The material with the lowest resistivity is the best conductor. That material is silver.
33. Strategy Use the definition of resistance, the relationship between voltage and uniform electric field, and Eq.
(18-8).
Solution V = IR = EL and R = ρ L A . Find E.
V = EL = IR = I ρ
L
I
, so E = ρ , where ρ is the resistivity .
A
A
716
Physics
Chapter 18: Electric Current and Circuits
34. Strategy As found in Example 18.4, R R0 = 1 + α∆T . Use Eq. (18-8) and the definition of resistance.
Solution LAl = 3LCu , rAl = 2rCu , ρCu = 0.6 ρ Al , and α Cu = α Al .
(a) Form a proportion.
RAl
RCu
=
LAl
AAl
L
ρCu ACu
Cu
ρ Al
=
2
ρ Al LAl ACu
ρ Al (3LCu )π rCu
3
, so RAl =
RCu =
(24 Ω) = 30 Ω .
2
ρCu LCu AAl
(0.6 ρ Al ) LCu π (2rCu )
0.6(2)2
(b) At I = 10 A, V = 300 V, so R =
V 300 V
=
= 30 Ω .
I
10 A
(c) Find T.
1 + α∆T =
⎞
R
1⎛ R
1
⎛ 30 Ω ⎞
− 1⎟⎟ + T0 =
− 1⎟ + 20°C = 80°C .
, so T = ⎜⎜
−1 ⎜⎝ 24 Ω
R0
α ⎝ R0 ⎠
⎠
0.004°C
35. Strategy and Solution E does not depend upon T, but ρ is directly proportional to T, and vD ∝ I . So,
the electric field stays the same, the resistivity decreases, and the drift speed increases.
36. (a) Strategy Sum the individual emfs with those with their left terminal at the higher potential being positive.
Solution Compute the equivalent emf.
eq = 3.0 V + 4.5 V − 1.5 V + 2.0 V − 5.0 V = 3.0 V
(b) Strategy Use the definition of resistance.
Solution Compute the current through the resistor.
eq 3.0 V
∆V eq
R=
=
, so I =
=
= 0.94 A .
I
I
R
3.2 Ω
37. (a) Strategy Sum the individual emfs with those with their left terminal at the higher potential being positive.
Solution Compute the equivalent emf.
eq = 3.0 V + 3.0 V + 2.5 V − 1.5 V = 7.0 V
(b) Strategy Use the definition of resistance.
Solution Find the value of the resistor.
∆V eq
7.0 V
R=
=
=
= 18 Ω
I
I
0.40 A
717
Chapter 18: Electric Current and Circuits
Physics
38. (a) Strategy Ceq = ΣCi for the capacitors, which are in parallel.
Solution Compute the equivalent capacitance.
Ceq = 2.0 µF + 6.0 µF + 3.0 µF = 11.0 µF
(b) Strategy Use Eq. (17-14).
Solution Compute the charge on the capacitor.
Q = C ∆V = C = (6.0 × 10−6 F)(44.0 V) = 260 µC .
39. (a) Strategy Ceq = ΣCi for the capacitors, which are in parallel.
Solution Compute the equivalent capacitance.
Ceq = 4.0 µF + 2.0 µF + 3.0 µF + 9.0 µF + 5.0 µF = 23.0 µF
(b) Strategy Use Eq. (17-14).
Solution Compute the charge on the equivalent capacitor.
Q = C ∆V = Ceq = (23.0 × 10−6 F)(16.0 V) = 368 µC .
(c) Strategy Use Eq. (17-14).
Solution Compute the charge on the capacitor.
Q = C ∆V = C = (3.0 × 10−6 F)(16.0 V) = 48 µC .
40. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the equivalent resistance between points A and B.
1 ⎞
⎛ 1
Req = 15 Ω + ⎜
+
⎟
12
24
Ω
Ω⎠
⎝
−1
= 23 Ω
(b) Strategy Label the currents on a diagram. Use Kirchhoff’s rules.
Solution The current through the emf is I = Req = I1 + I 2 , where the
currents labeled 1 and 2 are shown in the diagram. Applying the loop rule,
R
we have I 2 R2 − I1R1 = 0, so I 2 = 1 I1. Solve for the current through the
R2
I
I1
12-Ω resistor, I1.
I = I1 + I 2 = I1 +
⎛
⎛
R1
R ⎞
R ⎞
I1 = ⎜⎜1 + 1 ⎟⎟ I1 , so I1 = ⎜⎜ 1 + 1 ⎟⎟
R2
⎝ R2 ⎠
⎝ R2 ⎠
718
−1
⎛
R ⎞
I = ⎜⎜ 1 + 1 ⎟⎟
⎝ R2 ⎠
−1
I2
R1 = 12 Ω
⎛ 12 ⎞
= ⎜1 + ⎟
Req ⎝ 24 ⎠
R2 = 24 Ω
I
−1
276 V
= 8.0 A .
23 Ω
Physics
Chapter 18: Electric Current and Circuits
41. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the resistance between points A and B.
1
⎛ 1
⎞
Req = ⎜
+
⎟
2.0
1.0
1.0
Ω
Ω
+
Ω
⎝
⎠
−1
+ 4.0 Ω = 5.0 Ω
(b) Strategy Label the currents on a diagram. Use Kirchhoff’s rules.
Solution The current through the emf is I = Req = I1 + I 2 , where the
currents labeled 1 and 2 are shown in the diagram. Applying the loop rule,
R
2.0 Ω
I = 1.0 I1.
we have I 2 (2 R2 ) − I1R1 = 0, so I 2 = 1 I1 =
2 R2
2(1.0 Ω) 1
Solve for the current through the 2.0-Ω resistor, I1.
I = I1 + I 2 = I1 + 1.0 I1 = 2.0 I1 , so I1 =
I2
R2 = 1.0 Ω
I
I1
R2 = 1.0 Ω
R1 = 2.0 Ω
I
I
20 V
=
=
= 2.0 A .
2.0 2.0 Req 2.0(5.0 Ω)
42. Strategy Use Eqs. (18-13) and (18-17). The current through the 4.0-Ω resistor is the same as that through the
emf, I = Req .
Solution Find Req .
Req =
1
1
2.0 Ω
1
+ R +1.0
Ω
Find R.
, so Req =
I=
Req
1
1
2.0 Ω
1
+ R +1.0
Ω
+ 4.0 Ω
1
1
2.0 Ω
+
1
R +1.0 Ω
+ 4.0 Ω =
93.5 V
=
= 5.5 Ω .
17 A
I
= 1.5 Ω
1
1
1
+
=
2.0 Ω R + 1.0 Ω 1.5 Ω
1
1
1
=
−
R + 1.0 Ω 1.5 Ω 2.0 Ω
−1
1 ⎞
⎛ 1
R=⎜
−
⎟ − 1.0 Ω = 5.0 Ω
⎝ 1.5 Ω 2.0 Ω ⎠
43. (a) Strategy Use Eqs. (18-15) and (18-18).
Solution Find the equivalent capacitance of the circuit.
1
= 1.5 µF
Ceq =
1 +
1
−1
4.0 µF
(
1 + 1
1.0 µF 1.0 µF
)
+ 2.0 µF
(b) Strategy The charge on the 4.0-µF capacitor is the same charge as on the equivalent capacitor. Use
Eq. (17-14).
Solution Find the charge.
Q = C ∆V = Ceq = (1.54 µF)(24 V) = 37 µC
719
Chapter 18: Electric Current and Circuits
Physics
44. (a) Strategy Ceq = 1.63 µF. Use Eqs. (18-15) and (18-18).
Solution Find the unknown capacitance.
1
1
1
=
+
−1
1.63 µF 4.0 µF
1 + 1
+ 2.0 µF
1.0 µF C
⎛ 1
1⎞
+ ⎟
⎜
C
1.0
F
µ
⎝
⎠
−1
(
⎛ 1
1 ⎞
=⎜
−
⎟
1.63
F
4.0
µ
µF ⎠
⎝
)
−1
− 2.0 µF
−1
⎤
1 ⎡⎛ 1
1 ⎞
= ⎢⎜
−
⎟ − 2.0 µF⎥
C ⎢⎝ 1.63 µF 4.0 µF ⎠
⎥
⎣
⎦
−1
−
1
1.0 µF
−1
⎧⎡
⎫
−1
⎤
1 ⎞
1 ⎪
⎪ ⎢⎛ 1
⎥
C=⎨ ⎜
−
−
µ
−
2.0
F
⎬
⎟
1.0 µF ⎪
⎥
⎪ ⎢⎣⎝ 1.63 µF 4.0 µF ⎠
⎦
⎩
⎭
−1
= 3.0 µF
(b) Strategy The charge on the 4.0-µF capacitor is the same charge as on the equivalent capacitor. Use Eq.
(17-14).
Solution Find the charge.
Q = C ∆V = Ceq = (1.63 µF)(24 V) = 39 µC
45. Strategy Use the concept of equivalent resistance. The equivalent resistance of two identical resistances R in
parallel is half or R 2.
Solution
(a) The two 2.0-Ω resistors are in series, so their equivalent resistance is 2.0 Ω + 2.0 Ω = 4.0 Ω . These two
resistors are in parallel with the rightmost 4.0-Ω resistor. Because the resistances of each branch of this
parallel circuit are equal, the current is split evenly. Let the current through each branch be called I3 . We
must determine I3 . Now, the equivalent resistance of this parallel circuit is 2.0 Ω , and this is in series with
the rightmost 3.0-Ω resistor and the rightmost 1.0-Ω , so the equivalent series resistance is 6.0 Ω . This
resistance is in parallel with the 6.0-Ω resistor, so the current is again split evenly. Let it be called I 2 ; then,
I 3 = I 2 2. The equivalent resistance of this parallel circuit is 3.0-Ω, and this is in series with the middle
1.0-Ω resistor, so the equivalent series resistance is 4.0 Ω . This equivalent resistance is in parallel with the
leftmost 4.0-Ω resistor, so the current is again split evenly. Let it be called I1; then, I 3 = I 2 2 = I1 4. The
equivalent resistance of this parallel circuit is 2.0-Ω , and this is in series with the leftmost 1.0-Ω and 3.0-Ω
resistors, so the equivalent resistance of the entire circuit is 6.0 Ω . If the current through the emf is I; then,
I 3 = I 2 2 = I1 4 = I 8. The current though the emf is given by I = Req . Compute the current through one
of the 2.0-Ω resistors.
I
24 V
I3 = =
=
= 0.50 A
8 8 Req 8(6.0 Ω)
(b) The current through the 6.0-Ω resistor is I 2 , which is one-fourth of the current through the emf.
I
24 V
I2 = =
=
= 1.0 A
4 4 Req 4(6.0 Ω)
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Physics
Chapter 18: Electric Current and Circuits
(c) The current through the leftmost 4.0-Ω resistor is I1 , which is half of the current through the emf.
I1 =
I
24 V
=
=
= 2.0 A
2 2 Req 2(6.0 Ω)
46. (a) Strategy Use Eqs. (8-13) and (8-17).
Solution Compute the resistance between points A and B.
1
= 3.0 Ω
Req = 1.0 Ω +
1
1
+
−1
2.0 Ω+1.0 Ω
3.3 Ω+
( 4.01 Ω + 8.01 Ω )
(b) Strategy The current through the 1.0-Ω resistor connected directly to point A is the same as the current
through the emf.
Solution Find the current.
18 V
I=
=
= 6.0 A
Req 3.0 Ω
(c) Strategy Use the concept of equivalent resistance and Kirchhoff’s rules. The equivalent resistance of
resistances R and 2R in parallel is 2 R 3.
Solution The 4.0-Ω and 8.0-Ω resistances are in parallel and eight is twice four, so the equivalent resistance
of this parallel circuit is 2.67 Ω . Let the current through the 4.0-Ω resistor (directed to the right) be I 3 and
that through the 8.0-Ω resistor (directed to the right) be I 4 . Let the current entering this parallel circuit from
the left be I 2 ; then, I 2 = I 3 + I 4 . According to the loop rule, I 3 (4.0 Ω) − I 4 (8.0 Ω) = 0, so I 3 = 2.0 I 4 .
Find I 2 in terms of I 4 .
I 2 = I 3 + I 4 = 2.0 I 4 + I 4 = 3.0 I 4
The 3.3-Ω resistor and the 2.67-Ω equivalent resistance are in series, so the equivalent resistance of these
two is 6.0 Ω . This resistance is in parallel with the series resistance of the 2.0-Ω resistor and the rightmost
1.0-Ω resistor with an equivalent resistance of 3.0 Ω . Since six is twice three, the equivalent resistance of
the 6.0-Ω and 3.0-Ω resistances is 2.0 Ω . This resistance is in series with the leftmost 1.0-Ω resistor, so
the equivalent resistance of the entire circuit is 1.0 Ω + 2.0 Ω = 3.0 Ω . Let the current through the 2.0-Ω
resistor and the rightmost 1.0-Ω resistor be I1 and the current through the emf be I; then, I = I1 + I 2 .
According to the loop rule, I1 (3.0 Ω) − I 2 (6.0 Ω) = 0, so I1 = 2.0 I 2 . Find I 4 in terms of I.
I = I1 + I 2 = 2.0 I 2 + I 2 = 3.0 I 2 = 3.0(3.0 I 4 ) = 9.0 I 4 , so I 4 =
I=
I
18 V
, so I 4 =
=
=
= 0.67 A .
Req
9.0 9.0 Req 9.0(3.0 Ω)
721
I
. The current through the emf is
9.0
Chapter 18: Electric Current and Circuits
Physics
47. (a) Strategy The resistors are in parallel; they all begin at A and end at B, or vice versa.
A
Solution Find the equivalent resistance.
⎛1 1 1 1 1 1 1 1⎞
Req = ⎜ + + + + + + + ⎟
⎝R R R R R R R R⎠
−1
=
R
8
B
(b) Strategy and Solution There is a “short” circuit between points B and C, so R = 0 .
(c) Strategy The potential across each resistor is 32 V.
Solution Compute the current in one of the resistors.
V 32 V
I= =
= 16 A
R 2.0 Ω
48. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Compute the resistance between points A and B.
1
Req =
= 3.0 Ω
1 +
1
1
6.0 Ω
2.0 Ω+
1
+
1.0 Ω+1.0 Ω
+ 3.0 Ω
1
( 4.01 Ω + 4.01 Ω )
−1
(b) Strategy Use the concept of equivalent resistance and Kirchhoff’s rules. The equivalent resistance of two
identical resistances R in parallel is half or R 2.
Solution The 4.0-Ω resistors are in parallel, along with the series combination of the two 1.0-Ω resistors, so
the potential difference across them is the same. Let this potential difference be V. The equivalent resistance
of the two 4.0-Ω resistors is 2.0 Ω . The equivalent resistance of the two 1.0-Ω resistors is 2.0 Ω , as well.
Thus, the equivalent resistance of all four resistors is 1.0 Ω . This resistance is in series with the 2.0-Ω and
3.0-Ω resistors, so their equivalent resistance is 6.0 Ω . This resistance is in parallel with the 6.0-Ω
resistance, so their equivalent resistance is 3.0 Ω , which agrees with the result obtained in part (a). The
current through the emf is I =
12 V
=
= 4.0 A. This current is split evenly between the 6.0-Ω
Req 3.0 Ω
equivalent resistance and the actual 6.0-Ω resistor. So, 2.0 A flows through the 2.0-Ω and 3.0-Ω resistors
and 1.0-Ω equivalent resistance (of the 1.0-Ω and 4.0-Ω resistors). According to the loop rule (taking the
outer loop that excludes the 6.0-Ω resistor), 12 V − (2.0 A)(2.0 Ω) − V − (2.0 A)(3.0 Ω) = 0, so the potential
difference across the 4.0-Ω resistors is V = 12 V − (2.0 A)(2.0 Ω) − (2.0 A)(3.0 Ω) = 2.0 V .
(c) Strategy and Solution As found in part (b), the current through the 3.0-Ω resistor is 2.0 A.
722
Physics
Chapter 18: Electric Current and Circuits
49. (a) Strategy Use Eqs. (18-15) and (18-18).
Solution Find the equivalent capacitance.
Ceq
⎛ 1
⎞
1
=⎜
+
⎟
⎝ 12 µF 12 µF + 12 µF ⎠
−1
= 8.0 µF
(b) Strategy Since the capacitor at the left side of the diagram (1) is in series with the parallel combination of
the other two capacitors (2), the charge Q on the capacitor 1 is the same as that on capacitor 2. (Think of the
parallel combination as one capacitor with capacitance C2 = 12 µF + 12 µF = 24 µF.) Use the definition of
capacitance.
Solution Find the potential difference across C1. Let this potential difference be V1 and the potential
difference across C2 be V2 . Then, = V1 + V2 . Form a proportion.
V2 − V1 Q C2 C1
25 V
, so V1 =
=
= −1 =
=
=
= 17 V .
C
12 µF
V1
V1
V1
Q C1 C2
1 + 1 1 + 24 µF
C2
(c) Strategy The charge on the capacitor at the far right of the circuit (1) is half of the charge on the capacitor at
the left of the circuit (2).
Solution Find the charge on the capacitor.
1
1
Q2 = C2V2 = 2Q1 , so Q1 = C2V2 = (12 × 10−6 F)(17 V) = 1.0 × 10−4 C .
2
2
50. Strategy Use Eqs. (18-15) and (18-18). Draw a circuit diagram.
Solution The circuit diagram:
A
B
Ceq =
1
1
9.0 pF
+ 9.0 pF+1 9.0 pF
= 6.0 pF
51. (a) Strategy Use Eqs. (18-13) and (18-17).
Solution Find the equivalent resistance.
1
Req =
1 +
1
4.00 Ω
1
2.00 Ω+
1 +
4.00 Ω
= 2.00 Ω
1
1 ⎞ −1
⎛ 1
2.00 Ω+ ⎜
+
⎟
⎝ 4.00 Ω 4.00 Ω ⎠
(b) Strategy Use the definition of resistance and the equivalent resistance found in part (a).
Solution Compute the current that flows through the emf.
6.00 V
I=
=
= 3.00 A
Req 2.00 Ω
723
Chapter 18: Electric Current and Circuits
Physics
(c) Strategy Use Eqs. (18-13) and (18-17) and the definition of resistance. Redraw the circuit.
Solution Let R = 2.00 Ω; then Rc = 4.00 Ω = 2 R is the 4.00-Ω resistor at the bottom. The emf has been
left out for convenience.
A
I1
R
I2
R
2R
2R
Rc = 2R
2R
B
⎛ 1
1 ⎞
From part (a), Req = ⎜⎜
+
⎟⎟
⎝ 4.00 Ω RL ⎠
the circuit. Find RL .
−1
= 2.00 Ω, where RL is the equivalent resistance of the left branch of
−1
1
1
1
1 ⎞
⎛ 1
, so RL = ⎜
=
+
−
⎟ = 4.00 Ω.
2.00 Ω 4.00 Ω RL
⎝ 2.00 Ω 4.00 Ω ⎠
So, the resistances are the same for the left and right branches. Since 3.00 A flows through the emf, 1.50 A
flows through each branch. This current is equal to I1 + I 2. Find I1/I 2 considering that ∆V is the same
across each branch.
−1
⎡
I1
1 ⎞ ⎤
2R
1.50 A
⎛ 1
∆V = I1 ⎢ R + ⎜
+
=
= 1, or I1 = I 2 =
= 0.750 A.
⎟ ⎥ = I 2 (2 R), so
I2 R + R
2
⎝ 2 R 2 R ⎠ ⎥⎦
⎢⎣
This current is split evenly again since the parallel circuit containing Rc has equal resistances.
So, Rc =
0.750 A
= 0.375 A .
2
52. Strategy Use Kirchhoff’s rules. Let I1 be the top branch, I 2 be the middle branch, and I3 be the bottom branch.
Assume that each current flows right to left.
Solution Find the current in each branch of the circuit.
(1) I1 = − I 2 − I3 (2) 0 = 5.00 V + (56 Ω) I 2 − (22 Ω) I1 (3) 0 = 1.00 V + (56 Ω)I 2 − (75 Ω)I3
Substitute (1) into (2).
0 = 5.00 V + (56 Ω)I 2 − (22 Ω)( − I 2 − I 3 ) = 5.00 V + (78 Ω)I 2 + (22 Ω)I 3 (4)
Solve (3) for I 2 and substitute into (4).
(56 Ω)I 2 = (75 Ω)I3 − 1.00 V, so I 2 =
75
1
I −
A.
56 3 56
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Physics
Chapter 18: Electric Current and Circuits
Solve for I3 .
1
5
39
⎛ 75
⎞
0 = 5.00 V + (78 Ω) ⎜ I 3 −
A ⎟ + (22 Ω) I 3 =
A+
(75 I 3 − 1 A) + I3 = 0.164 A + 5.75 I 3 , so
56
56
22
616
⎝
⎠
I3 = −0.0285 A.
Calculate I 2 .
75
1
(−0.0285 A) −
A = −0.0560 A
56
56
Calculate I1.
I2 =
I1 = − I 2 − I3 = 0.0560 A + 0.0285 A = 0.0845 A
To two significant figures, the currents are:
Branch
I (mA)
Direction
Top
85
right to left
Middle
56
left to right
Bottom
29
left to right
53. Strategy Use Kirchhoff’s rules. Let I1 be the top branch, I 2 be the middle branch, and I3 be the bottom branch.
Assume that each current flows right to left.
Solution Find the current in each branch of the circuit.
(1) I1 = − I 2 − I3 (2) 0 = 25.00 V + (5.6 Ω)I 2 − (122 Ω)I1
(3) 0 = 25.00 V + 5.00 V + (75 Ω)I 3 − (122 Ω) I1
Substitute (1) into (2).
(5) 0 = 25.00 V + (122 Ω + 5.6 Ω)I 2 + (122 Ω) I 3
Multiply (4) by 5 and subtract from (5).
(4) 0 = 5.00 V + (75 Ω)I3 − (5.6 Ω) I 2
0 = [122 Ω + 5.6 Ω + 5(5.6 Ω)]I 2 + [122 Ω − 5(75 Ω)]I3 , so I 2 =
5(75 Ω) − 122 Ω
I = 1.6 I3 (1.626 I 3 ).
122 Ω + 5.6 Ω + 5(5.6 Ω) 3
Substitute the result above into (4).
0 = 5.00 V + (75 Ω)I3 − (5.6 Ω)(1.626I3 ), so I3 =
5.00 V
= − 0.076 A.
1.626(5.6 Ω) − 75 Ω
So, I 2 = 1.626(−0.076 A) = −0.12 A and I1 = −(−0.12 A) − (−0.076 A) = 0.20 A.
Branch
I (A)
Direction
AB
0.20
right to left
FC
0.12
left to right
ED
0.076
left to right
725
Chapter 18: Electric Current and Circuits
Physics
54. Strategy Use Kirchhoff’s rules. Let I be the current flowing up through the 5.00-Ω resistor and I1 be the current
flowing to the left through the 4.00-Ω resistor.
Solution Find the unknown emf and the unknown currents.
(1) I = I1 − 0.0500 A
0 = − (1.00 Ω)I1 + 1.20 V − (4.00 Ω) I1 − (2.00 Ω)(0.0500 A)
(2) 0 = + 1.10 V − (5.00 Ω)I1
0 = + (5.00 Ω)I − 1.00 V − (2.00 Ω)(0.0500 A)
(3) 0 = − 1.10 V + (5.00 Ω)I
Subtract (3) from (2).
2.20 V
0 = 2.20 V − (5.00 Ω)( I + I1 ), so I + I1 =
, or I = 0.440 A − I1.
5.00 Ω
Set this result equal to (1).
0.490 A
0.440 A − I1 = I1 − 0.0500 A, so I1 =
= 0.245 A.
2
Find I.
I = I1 − 0.0500 A = 0.245 A − 0.0500 A = 0.195 A
Solve for .
0 = − 1.10 V + (5.00 Ω)I , so = 1.10 V − (5.00 Ω)