1.1.
Solve:
1.2.
Solve:
1.3.
Solve:
1.4.
Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is
concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable
approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object,
and (ii) rotations and internal motions are not significant features of the object’s motion. The particle model is
important in that it allows us to simplify a problem. Complete reality—which would have to include the motion
of every single atom in the object—is too complicated to analyze. By treating an object as a particle, we can
focus on the most important aspects of its motion while neglecting minor and unobservable details.
(b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance.
(c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird
flies, or how water flows through a pipe.
1.5.
Solve: (a) An operational definition defines a concept or an idea in terms of a procedure, or a set of
operations, that is used to identify or measure the concept.
G
(b) The displacement Δr of an object is a vector found by drawing an arrow from the object’s initial location to
G G G
G
its final location. Mathematically, Δr = rf − ri . The average velocity v of an object is a vector that points in the
G
G
same direction as the displacement Δr and has length, or magnitude, Δr / Δt , where Δt = tf − ti is the time
interval during which the object moves from its initial location to its final location.
1.6.
Solve: The player starts from rest and moves faster and faster (accelerates).
1.7.
Solve: The particle starts with an initial velocity but as it slides it moves slower and slower till coming
to rest. This is a case of negative acceleration because it is an acceleration opposite to the positive direction of
motion.
G
Solve: The acceleration of an object is a vector formed by finding the ratio of Δv , the change in the
G
object’s velocity, to Δt , the time in which the change occurs. The acceleration vector a points in the direction
G
of Δv , which is found by vector subtraction.
1.8.
G
(a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector
G
between points 0 and 1 and v1 be the velocity vector between points 1 and 2.
1.9.
Solve:
(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
G
Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector
G
between points 0 and 1 and v1 be the velocity vector between points 1 and 2.
1.10.
(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.
1.11.
(a)
Solve:
(b)
1.12.
(a)
Solve:
(b)
1.13.
Model:
Represent the car as a particle.
Visualize: The dots are equally spaced until brakes are applied to the car. Equidistant dots indicate constant
average speed. On braking, the dots get closer as the average speed decreases.
1.14.
Model:
Represent the (child + sled) system as a particle.
Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dots
indicate constant average speed. On encountering a rocky patch, the average speed decreases and the sled comes
to a stop. This part of the motion is indicated by a separation between the dots that becomes smaller and smaller.
1.15.
Model: Represent the tile as a particle.
Visualize: The tile falls from the roof with an acceleration equal to a = g = 9.8 m/s2. Starting from rest, its
velocity increases until the tile hits the water surface. This part of the motion is represented by dots with
increasing separation, indicating increasing average velocity. After the tile enters the water, it settles to the
bottom at roughly constant speed.
1.16.
Model: Represent the tennis ball as a particle.
Visualize: The particle falls freely for the three stories under the acceleration of gravity. It strikes the ground and
very quickly decelerates to zero (while decompresses) and finally travels upward with negative acceleration under
gravity to zero velocity at a height of two stories. The downward and upward motions of the ball are shown in the
figure. The increasing length between the dots during downward motion indicates increasing average velocity or
downward acceleration. On the other hand, the decreasing length between the dots during upward motion indicates
acceleration in a direction opposite to its motion; that is, in the downward direction.
Assess: For a free-fall motion, acceleration due to gravity is always vertically downward.
1.17.
Model: Represent the toy car as a particle.
Visualize: As the toy car rolls down the ramp, its average speed increases. This is indicated by the increasing
G
length of the velocity arrows. That is, motion down the ramp is under an acceleration a. At the bottom of the
ramp, the toy car continues with the speed obtained with no change in velocity.
1.18. Solve:
(a)
Dot
1
2
3
4
5
6
7
8
9
Time (s)
0
2
4
6
8
10
12
14
16
x (m)
0
30
95
215
400
510
600
670
720
(b)
1.19. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant
speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was
getting dark, he walks faster back to his house covering the same distance in 200 s.
1.20. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English
professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking
that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one
hour of driving at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back
very slowly, confused and tired as he was, and reached El Dorado in two hours.
Visualize: The bicycle is moving with an acceleration of 1.5 m/s2. Thus, the velocity will increase by
1.5 m/s each second of motion.
1.21.
G
Visualize: The particle moves upward with a constant acceleration a . The final velocity is 200 m/s
and is reached at a height of 1000 m.
1.22.
⎛ 10−6 s ⎞
−6
Solve: (a) 9.12 μ s = (9.12 μs) ⎜
⎟ = 9.12 × 10 s
⎝ 1 μs ⎠
⎛ 103 m ⎞
3
(b) 3.42 km = (3.42 km) ⎜
⎟ = 3.42 × 10 m
⎝ 1 km ⎠
1.23.
−2
⎛ cm ⎞ ⎛ 10 m ⎞ ⎛ 1 ms ⎞
2
(c) 44 cm/ms = 44 ⎜
⎟ ⎜ −3 ⎟ = 4.4 × 10 m/s
⎟⎜
⎝ ms ⎠ ⎝ 1 cm ⎠ ⎝ 10 s ⎠
⎛ km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞
(d) 80 km/hour = 80 ⎜
⎟⎜
⎟⎜
⎟ = 22 m/s
⎝ hour ⎠ ⎝ 1 km ⎠ ⎝ 3600 s ⎠
3
1.24.
−2
⎛ 2.54 cm ⎞ ⎛ 10 m ⎞
Solve: (a) 8.0 inches = 8.0 (inch) ⎜
⎟ = 0.20 m
⎟⎜
⎝ 1 inch ⎠ ⎝ 1 cm ⎠
1m
⎛ feet ⎞⎛ 12 inch ⎞⎛
⎞
(b) 66 feet /s = 66 ⎜
⎟⎜
⎟⎜
⎟ = 20 m/s
⎝ s ⎠⎝ 1 foot ⎠⎝ 39.37 inch ⎠
⎛ miles ⎞⎛ 1.609 km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞
(c) 60 mph = 60 ⎜
⎟⎜
⎟⎜
⎟⎜
⎟ = 27 m/s
⎝ hour ⎠⎝ 1 mile ⎠ ⎝ 1 km ⎠ ⎝ 3600 s ⎠
3
2
1m
⎛
⎞
−3
(d) 14 square inches = 14 (inches) 2 ⎜
⎟ = 9.0 × 10 square meter
⎝ 39.37 inches ⎠
1.25.
Solve: (a)
⎛ 3600 s ⎞
3
1 hour = 1(hour) ⎜
⎟ = 3600 s = 3.60 × 10 s
1
hour
⎝
⎠
⎛ 24 hours ⎞ ⎛ 3600 s ⎞
4
(b) 1 day = 1 (day) ⎜
⎟⎜
⎟ = 8.64 × 10 s
⎝ 1 day ⎠ ⎝ 1 hour ⎠
⎛ 365.25 days ⎞ ⎛ 8.64 × 104 s ⎞
7
(c) 1 year = 1 (year) ⎜
⎟ = 3.16 × 10 s
⎟⎜
⎝ 1 year ⎠⎝ 1 day ⎠
1m
⎛ ft ⎞⎛ 12 inch ⎞⎛
⎞
2
(d) 32 ft /s 2 = 32 ⎜ 2 ⎟⎜
⎟⎜
⎟ = 9.75 m/s
⎝ s ⎠⎝ 1 ft ⎠⎝ 39.37 inch ⎠
1.26.
Solve: (a)
⎛1 m ⎞
20 ft = 20 ( ft ) ⎜
⎟ = 7.0 m
⎝ 3 ft ⎠
⎛ 1 km ⎞⎛ 1000 m ⎞
5
(b) 60 miles = 60(miles) ⎜
⎟⎜
⎟ = 1.0 × 10 m
⎝ 0.6 miles ⎠⎝ 1 km ⎠
⎛ 1 m/s ⎞
(c) 60 mph = 60(mph) ⎜
⎟ = 30 m/s
⎝ 2 mph ⎠
⎛ 1 cm ⎞ ⎛ 10−2 m ⎞
(d) 8 in = 8(in) ⎜
⎟⎜
⎟ = 0.16 m
⎝ 1/2 in ⎠⎝ 1 cm ⎠
1.27.
Solve:
⎛ 4 in ⎞
(a) (30 cm ) ⎜
⎟ = 12 in
⎝ 10 cm ⎠
⎛ 2 mph ⎞
(b) ( 25 m/s ) ⎜
⎟ = 50 mph
⎝ 1 m/s ⎠
⎛ 0.6 mi ⎞
(c) (5 km ) ⎜
⎟ = 3 mi
⎝ 1 km ⎠
⎛1 ⎞
in
⎛1
⎞⎜ 2 ⎟ 1
(d) ⎜ cm ⎟ ⎜
⎟ = in
⎝2
⎠ ⎝ 1 cm ⎠ 4
Solve: (a) 33.3 × 25.4 = 846
(b) 33.3 − 25.4 = 7.9
1.28.
(c) 33.3 = 5.77
(d) 333.3 ÷ 25.4 = 13.1
1.29.
Solve: (a) (33.3) 2 = 1.109 × 103. For numbers starting with “1” an extra digit is kept.
(b) 33.3 × 45.1 = 1.50 × 103
Scientific notation is an easy way to establish significance.
(c) 22.2 − 1.2 = 3.5
(d) 1/ 44.4 = 0.0225
1.30.
Solve: The length of a typical car is 15 ft. Or
1m
⎛ 12 inch ⎞⎛
⎞
15(ft) ⎜
⎟⎜
⎟ = 4.6 m
1
ft
39.37
inch
⎝
⎠⎝
⎠
This length of 15 ft is approximately two-and-a-half times my height.
1.31.
Solve: The height of a telephone pole is estimated to be around 50 ft or 15 m. This height is
approximately
8 times my height.
1.32.
Solve: I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home to
campus. My average speed is
⎛ 0.447 m/s ⎞
6 miles ⎛ 60 min ⎞
⎟ = 11 m/s
⎜
⎟ = 24 mph = 24(mph) ⎜
15 min ⎝ 1 hour ⎠
⎝ 1 mph ⎠
1.33.
Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair
growth is
1(inch) ⎛ 2.54 cm ⎞ ⎛ 10−2 m ⎞⎛ 1 month ⎞ ⎛ 1 day ⎞⎛ 1 h ⎞
−9
⎟⎜
⎟⎜
⎜
⎟⎜
⎟⎜
⎟ = 9.8 × 10 m/s
(month) ⎝ 1 inch ⎠ ⎝ 1 cm ⎠ ⎝ 30 days ⎠ ⎝ 24 h ⎠⎝ 3600 s ⎠
6
⎛ m ⎞ ⎛ 10 μ m ⎞ ⎛ 3600 s ⎞
= 9.8 × 10−9 ⎜ ⎟ ⎜
⎟⎜
⎟ = 35 μ m/h
⎝ s ⎠⎝ 1 m ⎠⎝ 1 h ⎠
1.34.
Model:
Visualize:
Represent the Porsche as a particle for the motion diagram.
1.35.
Model:
Visualize:
Represent the watermelon as a particle for the motion diagram.
1.36.
Model:
Visualize:
Represent (Sam + car) as a particle for the motion diagram.
1.37.
Model:
Visualize:
Represent the speed skater as a particle for the motion diagram.
1.38.
Model:
Visualize:
Represent the wad as a particle for the motion diagram.
1.39.
Model:
Visualize:
Represent the ball as a particle for the motion diagram.
1.40.
Model:
Visualize:
Represent the ball as a particle for the motion diagram.
1.41.
Model:
Visualize:
Represent the motorist as a particle for the motion diagram.
1.42.
Model:
Visualize:
Represent Bruce and the puck as particles for the motion diagram.
1.43.
Model:
Visualize:
Represent Fred and yourself as particles for the motion diagram.
1.44.
Solve: Rahul was coasting on interstate highway I-35 from Wichita to Kansas City at 65 mph. Seeing
an accident at a distance of 200 feet in front of him, he braked his car to a stop with steady deceleration.
1.45.
Solve: A car starts coasting at an initial speed of 30.0 m/s up a 10° incline. 230 m up the incline the
road levels out to a flat road and the car continues coasting at a reduced speed along the road.
1.46.
Solve: A skier starts from rest down a 25° slope with very little friction. At the bottom of the 100 m
slope the skier moves to a flat area and continues at constant velocity.
1.47.
Solve: A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its
original height.
1.48.
Solve: Two boards lean against each other at equal angles to the vertical direction. A ball rolls up the
incline, over the peak, and down the other side.
1.49.
Solve:
(a)
(b) Sue passes 3rd Street doing 40 mph, slows steadily to the stop sign at 4th Street, stops for 1 s, then speeds up
and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Sue to
drive from 3rd Street to 5th Street?
(c)
1.50.
Solve:
(a)
(b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The driver
maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find
the length of the tunnel.
(c)
1.51. Solve:
(a)
(b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his
car to rest in 10 seconds with a constant deceleration. Then he turns into an adjoining street. Starting from rest,
Jeremy accelerates with exactly the same magnitude as his earlier deceleration and reaches the same speed of 60
mph over the same distance in exactly the same time. Find the car’s acceleration or deceleration.
(c)
1.52. Solve:
(a)
(b) A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.0 m/s2. At the same instant,
the rabbit (B) begins to run away from the coyote with an acceleration of 2.0 m/s2. The coyote catches the rabbit
after running 40 m. How far away was the rabbit when the coyote first saw it?
(c)
Solve: Since area equals length × width, the smallest area will correspond to the smaller length and the
smaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, the
smallest area is (64 m)(100 m) = 6.4 × 103 m2 and the largest area is (75 m)(110 m) = 8.3 × 103 m2.
1.53.
1.54.
Solve: (a) We need kg/m3. There are 100 cm in 1 m. If we multiply by
3
⎛ 100 cm ⎞
3
⎜
⎟ = (1)
⎝ 1m ⎠
we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density of
aluminum is
3
⎛ kg ⎞⎛ 100 cm ⎞
3 kg
2.7 × 10−3 ⎜ 3 ⎟⎜
⎟ = 2.7 × 10 3
m
⎝ cm ⎠⎝ 1 m ⎠
(b) Likewise, the mass density of alcohol is
3
kg
⎛ g ⎞⎛ 100 cm ⎞ ⎛ 1 kg ⎞
0.81⎜ 3 ⎟⎜
⎟ = 810 3
⎟ ⎜
cm
1
m
1000
g
m
⎝
⎠⎝
⎠ ⎝
⎠
1.55.
Model:
The car is represented by the particle model as a dot.
Solve:
(a)
Time t (s)
0
10
20
30
40
50
60
70
80
90
Position x (m)
1200
975
825
750
700
650
600
500
300
0
(b)
1.56. Solve: Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a
constant leisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward
the same seat, so Susan walks more quickly to cover the remaining 30 m in another 20 seconds, beating Ella to
the seat. Susan stands next to the seat for 10 seconds to remove her backpack.
1.57. Solve: A crane operator holds a ton of bricks 30 m above the ground. Four seconds after he is told to lower the
bricks, he takes four seconds to lower them 15 m at a constant rate before stopping the bricks to make an eight-second
safety check. He then continues lowering the bricks the remaining 15 m, taking four more seconds.
1-1
2.1. Model:
Visualize:
Solve:
We will consider the car to be a particle that occupies a single point in space.
Since the velocity is constant, we have xf = xi + vx Δt. Using the above values, we get
x1 = 0 m + (10 m/s)(45 s) = 450 m
Assess: 10 m/s ≈ 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is
reasonable and expected.
2.2.
Model:
Visualize:
Solve:
We will consider Larry to be a particle.
Since Larry’s speed is constant, we can use the following equation to calculate the velocities:
vs =
sf − si
tf − ti
(a) For the interval from the house to the lamppost:
v1 =
200 yd − 600 yd
= −200 yd/min
9:07 − 9:05
For the interval from the lamppost to the tree:
1200 yd − 200 yd
v2 =
= +333 yd/min
9:10 − 9:07
(b) For the average velocity for the entire run:
1200 yd − 600 yd
vavg =
= +120 yd/min
9:10 − 9:05
2.3.
Model:
Visualize:
Solve:
Cars will be treated by the particle model.
Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows:
Δx x1 − x0
x −x
v=
=
⇒ t1 = t0 + 1 0
Δt t1 − t0
v
Using the known values identified in the pictorial representation, we find:
x
−x
400 mile
tAlan 1 = tAlan 0 + Alan 1 Alan 0 = 8:00 AM +
= 8:00 AM + 8 hr = 4:00 PM
v
50 miles/hour
x
−x
400 mile
tBeth 1 = tBeth 0 + Beth 1 Beth 0 = 9:00 AM +
= 9:00 AM + 6.67 hr = 3:40 PM
v
60 miles/hour
(a) Beth arrives first.
(b) Beth has to wait tAlan 1 − tBeth 1 = 20 minutes for Alan.
Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.
Solve: (a) The time for each segment is Δt1 = 50 mi/40 mph = 5/4 hr and Δt2 = 50 mi/60 mph = 5/6hr.
The average speed to the house is
100 mi
= 48 mph
5/6 h + 5/4 h
2.4.
(b) Julie drives the distance Δx1 in time Δt1 at 40 mph. She then drives the distance Δx2 in time Δt2 at 60 mph.
She spends the same amount of time at each speed, thus
Δt1 = Δt2 ⇒ Δx1 /40 mph = Δx2 /60 mph ⇒ Δx1 = (2/3)Δx2
But Δx1 + Δx2 = 100 miles, so (2 / 3)Δx2 + Δx2 = 100 miles. This means Δx2 = 60 miles and Δx1 = 40 miles. Thus,
the times spent at each speed are Δt1 = 40 mi/40 mph = 1.00 h and Δt2 = 60 mi/60 mph = 1.00 h. The total time
for her return trip is Δt1 + Δt2 = 2.00 h. So, her average speed is 100 mi/2 h = 50 mph.
2.5.
Model: The bicyclist is a particle.
Visualize: Please refer to Figure EX2.5.
Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t
= 10 s is
Δs 100 m − 50 m
v=
=
= 2.5 m/s
20 s
Δt
The slope at t = 25 s is
v=
100 m − 100 m
= 0 m/s
10 s
v=
0 m − 100 m
= −10 m/s
10 s
The slope at t = 35 s is
2.6.
Solve:
Visualize: Please refer to Figure EX2.6.
(a) We can obtain the values for the velocity-versus-time graph from the equation v = Δs/Δt.
(b) There is only one turning point. At t = 3 s the velocity changes from +5 m/s to −20 m/s, thus reversing the
direction of motion. At t = 1 s, there is an abrupt change in motion from rest to +5 m/s, but there is no reversal
in motion.
Visualize: Please refer to Figure EX2.7. The particle starts at x0 = 10 m at t0 = 0. Its velocity is
initially
in
the
–x direction. The speed decreases as time increases during the first second, is zero at t = 1 s, and then increases
after the particle reverses direction.
Solve: (a) The particle reverses direction at t = 1 s, when vx changes sign.
(b) Using the equation xf = x0 + area of the velocity graph between t1 and tf ,
2.7.
x2 s = 10 m − (area of triangle between 0 s and 1 s) + (area of triangle between 1 s and 2 s)
1
1
(4 m/s)(1 s) + (4 m/s)(1 s) = 10 m
2
2
x3 s = 10 m + area of trapazoid between 2 s and 3 s
= 10 m −
1
= 10 m + (4 m/s + 8 m/s)(3 s − 2 s) = 16 m
2
x4 s = x3 s + area between 3 s and 4 s
1
= 16 m + (8 m/s + 12 m/s)(1 s) = 26 m
2
2.8.
Visualize: Please refer to Figure EX2.8.
Solve: A constant velocity from t = 0 s to t = 2 s means zero acceleration. On the other hand, a linear increase
in velocity between t = 2 s and t = 4 s implies a constant positive acceleration.
2.9.
Visualize: Please refer to Figure EX2.9.
Solve: (a) The acceleration of the train at t = 3.0 s is the slope of the v vs t graph at t = 3 s. Thus
a = ( 2 m/s − ( − 2 m/s) ) (8 s ) = 0.5 m/s 2 .
(b)
2.10.
Solve:
Visualize: Please refer to Figure EX2.10.
(a) At t = 2.0 s, the position of the particle is
x2 s = 2.0 m + area under velocity graph from t = 0 s to t = 2.0 s
1
= 2.0 m + (4 m/s)(2.0 s) = 6 m
2
(b) From the graph itself at t = 2.0 s, v = 4 m/s.
(c) The acceleration is
ax =
Δvx vfx − vix 6 m/s − 0 m/s
=
=
= 2 m/s 2
3s
Δt
Δt
2.11.
Solve:
Visualize: Please refer to Figure EX2.11.
(a) Using the equation
xf = xi + area under the velocity-versus-time graph between ti and tf
we have
x (at t = 1 s) = x (at t = 0 s) + area between t = 0 s and t = 1 s
= 2.0 m + (4 m/s)(1 s) = 6 m
Reading from the velocity-versus-time graph, vx (at t = 1 s) = 4 m/s. Also, ax = slope = Δv/Δt = 0 m/s 2 .
(b) x (at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s
= 2.0 m + 4 m/s × 2 s + 2 m/s × 1 s + (1/2) × 2 m/s × 1 s = 13.0 m
Reading from the graph, vx (t = 3 s) = 2 m/s. The acceleration is
ax (t = 3 s) = slope =
vx (at t = 4 s) − vx (at t = 2 s)
= −2 m/s 2
2s
2.12.
Model:
Visualize:
Solve:
Represent the jet plane as a particle.
(a) Since we don’t know the time of acceleration, we will use
v12 = v02 + 2a ( x1 − x0 )
v 2 − v02 (400 m/s) 2 − (300 m/s) 2
⇒a= 1
=
= 8.75 m/s 2
2 x1
2(4000 m)
(b) The acceleration of the jet is approximately equal to g, the acceleration due to gravity.
2.13.
Model: We are using the particle model for the skater and the kinematics model of motion under
constant acceleration.
Solve: Since we don’t know the time of acceleration we will use
vf2 = vi2 + 2a ( xf − xi )
⇒a=
Assess:
v −v
(6.0 m/s) 2 − (8.0 m/s) 2
=
= −2.8 m/s 2
2( xf − xi )
2(5.0 m)
2
f
2
i
A deceleration of 2.8 m/s2 is reasonable.
2.14.
Model:
Visualize:
Solve:
We are assuming both cars are particles.
The Porsche’s time to finish the race is determined from the position equation
1
xP1 = xP0 + vP0 (tP1 − tP0 ) + aP (tP1 − tP0 ) 2
2
1
⇒ 400 m = 0 m + 0 m + (3.5 m/s 2 )(tP1 − 0 s)2 ⇒ tP1 = 15.1 s
2
The Honda’s time to finish the race is obtained from Honda’s position equation as
1
xH1 = xH0 + vH0 (tH1 − tH0 ) + aH0 (tH1 − tH0 ) 2
2
1
400 m = 50 m + 0 m + (3.0 m/s 2 )(tH1 − 0 s)2 ⇒ tH1 = 15.3 s
2
So, the Porsche wins.
2.15.
Model:
Visualize:
Solve:
Represent the spherical drop of molten metal as a particle.
(a) The shot is in free fall, so we can use free fall kinematics with a = − g . The height must be such that
the shot takes 4 s to fall, so we choose t1 = 4 s. Then,
1
1
1
y1 = y0 + v0 (t1 − t0 ) − g (t1 − t0 ) 2 ⇒ y0 = gt12 = (9.8 m/s 2 )(4 s) 2 = 78.4 m
2
2
2
(b) The impact velocity is v1 = v0 − g (t1 − t0 ) = − gt1 = −39.2 m/s.
G
Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v is negative
because the vector points downward.
2.16.
Model:
Visualize:
Solve:
Assume the ball undergoes free-fall acceleration and that the ball is a particle.
(a) We will use the kinematic equations
v = v0 + a (t − t0 ) and
1
y = y0 + v0 (t − t0 ) + a(t − t0 ) 2
2
as follows:
v(at t = 1.0 s) = 19.6 m/s + (−9.8 m/s 2 )(1.0 s − 0 s) = 9.8 m/s
y (at t = 1.0 s) = 0 m + (19.6 m/s)(1.0 s − 0 s) + 1/2(−9.8 m/s 2 )(1.0 s − 0 s) 2 = 14.7 m
v(at t = 2.0 s) = 19.6 m/s + (−9.8 m/s 2 )(2.0 s − 0 s) = 0 m/s
y (at t = 2.0 s) = 0 m + (19.6 m/s)(2.0 s − 0 s) + 1/2(−9.8 m/s 2 )(2.0 s − 0 s) 2 = 19.6 m
v(at t = 3.0 s) = 19.6 m/s + (−9.8 m/s 2 )(3 s − 0 s) = 9.8 m/s
y (at t = 3.0 s) = 0 m + (19.6 m/s)(3.0 s − 0 s) + 1/2(−9.8 m/s 2 )(3.0 s − 0 s)2 = 14.7 m
v(at t = 4.0 s) = 19.6 m/s + (−9.8 m/s 2 )(4.0 s − 0 s) = −19.6 m/s
y (at t = 4.0 s) = 0 m + (19.6 m/s)(4.0 s − 0 s) + 1/2(−9.8 m/s 2 )(4.0 s − 0 s) 2 = 0 m
(b)
Assess: (a) A downward acceleration of 9.8 m/s 2 on a particle that has been given an initial upward velocity of
+19.6 m/s will reduce its speed to 9.8 m/s after 1 s and then to zero after 2 s. The answers obtained in this
solution are consistent with the above logic.
(b) Velocity changes linearly with a negative uniform acceleration of 9.8 m/s 2 . The position is symmetrical in
time around the highest point which occurs at t = 2 s.
2.17.
Model:
Visualize:
We represent the ball as a particle.
Solve: Once the ball leaves the student’s hand, the ball undergoes free fall and its acceleration is equal to the
acceleration due to gravity that always acts vertically downward toward the center of the earth. According to the
constant-acceleration kinematic equations of motion
1
y1 = y0 + v0 Δt + a Δt 2
2
Substituting the known values
−2 m = 0 m + (15 m/s)t1 + (1/2)(−9.8 m/s 2 )t12
The solution of this quadratic equation gives t1 = 3.2 s. The other root of this equation yields a negative value for
t1 , which is not valid for this problem.
Assess: A time of 3.2 s is reasonable.
2.18.
Model:
Visualize:
Solve:
We will use the particle model and the constant-acceleration kinematic equations.
(a) Substituting the known values into y1 = y0 + v0 Δt + 12 a Δt 2 , we get
1
−10 m = 0 m + 20 (m/s)t1 + (−9.8 m/s 2 )t12
2
One of the roots of this equation is negative and is not relevant physically. The other root is t1 = 4.53 s, which is
the answer to part (b). Using v1 = v0 + aΔt , we obtain
v1 = 20(m/s) + (−9.8 m/s 2 )(4.53 s) = −24 m/s
(b) The time is 4.5 s.
Assess: A time of 4.5 s is a reasonable value. The rock’s velocity as it hits the bottom of the hole has a negative
sign because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed
up, is consistent with the fact that the rock travels an additional distance of 10 m into the hole.
2.19.
Model:
Visualize:
We will represent the skier as a particle.
Note that the skier’s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the
acceleration parallel to the incline is equal to g sin 10°.
Solve: Using the following constant-acceleration kinematic equations,
vf2x = vi2x + 2ax ( xf − xi )
⇒ (15 m / s) 2 = (3.0 m / s) 2 + 2(9.8 m / s 2 )sin10°( x1 − 0 m) ⇒ x1 = 64 m
vfx = vix + ax (tf − ti )
⇒ (15 m / s) = (3.0 m / s) + (9.8 m / s 2 )(sin10°)t ⇒ t = 7.1 s
Assess:
A time of 7.1 s to cover 64 m is a reasonable value.
2.20.
Model:
Visualize:
Represent the car as a particle.
Solve: Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s
before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without
friction, then we have motion on a frictionless inclined plane with an accleration
a = − g sin θ = − g sin 20° = −3.35 m/s 2 . Constant acceleration kinematics gives
v2
(30 m/s) 2
v12 = v02 + 2a ( x1 − x0 ) ⇒ 0 m 2 /s 2 = v02 + 2ax1 ⇒ x1 = − 0 = −
= 134 m
2a
2( −3.35 m/s 2 )
Notice how the two negatives canceled to give a positive value for x1.
G
Assess: We must include the minus sign because the a vector points down the slope, which is in the negative
x-direction.
Solve: (a) The position t = 2 s is x2 s = [2(2) 2 − 2 + 1] m = 7 m.
(b) The velocity is the derivative v = dx / dt and the velocity at t = 2 s is calculated as follows:
2.21.
v = (4t 2 − 1) m/s ⇒ v2 s = [4(2) − 1] m/s = 7 m/s
(c) The acceleration is the derivative a = dv / dt and the acceleration at t = 2 s is calculated as follows:
a = (4) m/s 2 ⇒ a2 s = 4 m/s 2
2.22.
Solve: The formula for the particle’s position along the x-axis is given by
tf
xf = xi + ∫ vx dt
ti
Using the expression for vx we get
2
xf = xi + ⎡⎣tf3 − ti3 ⎤⎦
3
ax =
dvx d
= (2t 2 m/s) = 4t m/s 2
dt dt
(a) The particle’s position at t = 1 s is x1 s = 1 m + 23 m = 53 m.
(b) The particle’s speed at t = 1 s is v1 s = 2 m/s.
(c) The particle’s acceleration at t = 1 s is a1 s = 4 m/s 2 .
2.23.
Solve: The formula for the particle’s velocity is given by
vf = vi + area under the acceleration curve between ti and tf
For t = 4 s, we get
1
v4 s = 8 m/s + (4 m/s 2 )4 s = 16 m/s
2
Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will
therefore increase. A value of 16 m/s is reasonable.
2.24.
Solve: (a)
Time (s)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Position (m)
−4
−2
0
1.75
3
4
5
6
7
(b)
(c) Δs = s (at t = 1 s) − s (at t = 0 s) = 0 m − (−4 m) = 4 m.
(d) Δs = s (at t = 4 s) − s (at t = 2 s) = 7 m − 3 m = 4 m.
(e) From t = 0 s to t = 1 s, vs = Δs / Δt = 4 m/s.
(f) From t = 2 s to t = 4 s, vs = Δs / Δt = 2 m/s.
(g) The average acceleration is
a=
Δv 2 m/s − 4 m/s
=
= −2 m/s 2
2 s −1 s
Δt
2.25.
Solve: (a)
(b) To be completed by student.
dx
(c)
= vx = 2t − 4 ⇒ vx (at t = 1 s) = [2 m/s 2 (1 s) − 4 m/s] = −2 m/s
dt
(d) There is a turning point at t = 2 s.
(e) Using the equation in part (c),
vx = 4 m/s = (2t − 4) m/s ⇒ t = 4
Since x = (t 2 − 4t + 2) m, x = 2 m.
(f)
2.26.
Visualize: Please refer to Figure P2.26.
Solve: The graph for particle A is a straight line from t = 2 s to t = 8 s. The slope of this line is −10 m/s,
which is the velocity at t = 7.0 s. The negative sign indicates motion toward lower values on the x-axis. The
velocity of particle B at t = 7.0 s can be read directly from its graph. It is −20 m/s, The velocity of particle C
can be obtained from the equation
vf = vi + area under the acceleration curve between ti and tf
This area can be calculated by adding up three sections. The area between t = 0 s and t = 2 s is 40 m/s, the area
between t = 2 s and t = 5 s is 45 m/s, and the area between t = 5 s and t = 7 s is −20 m/s. We get
(10 m/s) + (40 m/s) + (45 m/s) − (20 m/s) = 75 m/s.
2.27.
Solve:
Visualize: Please refer to Figure P2.27.
(a) We can calculate the position of the particle at every instant with the equation
xf = xi + area under the velocity-versus-time graph between ti and tf
The particle starts from the origin at t = 0 s, so xi = 0 m. Notice that the each square of the grid in Figure P2.27
has “area” (5 m/s) × (2 s) = 10 m. We can find the area under the curve, and thus determine x, by counting
squares. You can see that x = 35 m at t = 4 s because there are 3.5 squares under the curve. In addition, x = 35 m
at t = 8 s because the 5 m represented by the half square between 4 and 6 s is cancelled by the –5 m represented
by the half square between 6 and 8 s. Areas beneath the axis are negative areas. The particle passes through x =
35 m at t = 4 s and again at t = 8 s.
(b) The particle moves to the right for 0 s ≥ t ≥ 6 s, where the velocity is positive. It reaches a turning point at x =
40 m at t = 6 s. The motion is to the left for t > 6 s. This is shown in the motion diagram below.
2.28. Visualize:
Solve: We will determine the object’s velocity using graphical methods first and then using calculus.
Graphically, v(t ) = v0 + area under the acceleration curve from 0 to t. In this case, v0 = 0 m/s. The area at each time t
requested is a triangle.
t =0 s
v(t = 0 s) = v0 = 0 m/s
t=2s
1
v(t = 2 s) = (2 s)(5 m/s) = 5 m/s
2
t=4s
1
v(t = 4 s) = (4 s)(10 m/s) = 20 m/s
2
t =6 s
1
v(t = 6 s) = (6 s)(10 m/s) = 30 m/s
2
t =8 s
v(t = 8 s) = v (t = 6 s) = 30 m/s
The last result arises because there is no additional area after t = 6 s. Let us now use calculus. The acceleration
function a(t) consists of three pieces and can be written:
0 s≤t ≤4 s
⎧ 2.5t
⎪
a (t ) = ⎨ −5t + 30 4 s ≤ t ≤ 6 s
⎪ 0
6 s≤t ≤8 s
⎩
These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity
function is found by integration as follows: For 0 ≤ t ≤ 4 s,
t
t
v(t ) = v(t = 0 s) + ∫ a (t )dt = 0 + 2.5
0
This gives
t2
= 1.25t 2
20
t =0 s
v(t = 0 s) = 0 m / s
t=2s
v(t = 2 s) = 5 m / s
t=4s
v(t = 4 s) = 20 m / s
For 4 s ≤ t ≤ 6 s,
t
t
⎡ −5t 2
⎤
+ 30t ⎥ = −2.5t 2 + 30t − 60
v(t ) = v(t = 4 s) + ∫ a(t )dt = 20 m/s + ⎢
4
⎣ 2
⎦4
This gives:
t=6s
v(t = 6 s) = 30 m/s
For 6 s ≤ t ≤ 8 s,
t
v(t ) = v(t = 6 s) + ∫ a(t ) dt = 30 m/s + 0 m/s = 30 m/s
6
This gives:
t=8s
v(t = 8 s) = 30 m/s
Assess:
graphs.
The same velocities are found using calculus and graphs, but the graphical method is easier for simple
2.29.
Visualize: Please refer to Figure P2.29.
Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time
graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive
when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero
at t = 0, 1 s, 2 s, 3 s, . . . ; the slope is most positive at t = 0.5 s, 2.5 s, . . ; and the slope is most negative at t = 1.5 s,
3.5 s, . . .
(b)
2.30.
Solve:
Visualize: Please refer to Figure P2.30.
(a) We can determine the velocity as follows:
vx = vx (at t = ti ) + area under the acceleration-versus-time graph from t = ti to tf
vx (at t = 4 s) = 0 m/s + (−1 m/s 2 )(4 s) = −4 m/s
vx (at t = 8 s) = −4 m/s + (3 m/s 2 )(4 s) m/s = 8 m/s
vx (at t = 10 s) = 8 m/s + (−2 m/s 2 )(4 s) = 4 m/s
(b) If vx (at t = 0 s) = 2.0 m/s, the entire velocity-versus-time graph will be displaced upward by 2.0 m/s.
2.31.
Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position
function:
vx =
dx
= (6t 2 − 18t )m/s
dt
For vx = 0 m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s.
(b) At the first of these solutions,
x(at t = 0 s) = 2(0 s)3 − 9(0 s) 2 + 12 = 12 m
The acceleration is the derivative of the velocity function:
ax =
dvx
= (12t − 18) m/s 2 ⇒ a (at t = 0 s) = −18 m/s 2
dt
At the second solution,
x(at t = 3 s) = 2(3 s)3 − 9(3 s) 2 + 12 = −15 m
ax (at t = 3 s) = 12(3 s) − 18 = +18 m/s 2
2.32.
Model: Represent the object as a particle.
Solve: (a) Known information: x0 = 0 m, v0 = 0 m/s, x1 = 40 m, v1 = 11 m/s, t1 = 5 s. If the acceleration is
uniform (constant a), then the motion must satisfy the three equations
1
x1 = at12 ⇒ a = 3.20 m/s 2
2
v1 = at1 ⇒ a = 2.20 m/s 2
v12 = 2ax1 ⇒ a = 1.51 m/s 2
But each equation gives a different value of a. Thus the motion is not uniform acceleration.
(b) We know two points on the velocity-versus-time graph, namely at t0 = 0 and t1 = 5 s. What shape does the
function have between these two points? If the acceleration was uniform, which it’s not, then the graph would be
a straight line. The area under the graph is the displacement Δx. From the figure you can see that Δx = 27.5 m
for a straight-line graph. But we know that, in reality, Δx = 40 m. To get a larger Δx, the graph must bulge
upward above the straight line. Thus the graph is curved, and it is concave downward.
2.33.
Solve: The position is the integral of the velocity.
t1
t1
t1
t0
0
0
x1 = x0 + ∫ vx dt = x0 + ∫ kt 2 dt = x0 + 13 kt 3 = x0 + 13 kt13
We’re given that x0 = −9.0 m and that the particle is at x1 = 9.0 m at t1 = 3.0 s. Thus
9.0 m = ( −9.0 m ) + 13 k (3.0 s)3 = ( −9.0 m ) + k ( 9.0 s3 )
Solving for k gives k = 2.0 m/s3 .
2.34.
Solve: (a) The velocity is the integral of the acceleration.
v1x = v0 x + ∫ ax dt = 0 m/s + ∫ (10 − t ) dt = (10t − 12 t 2 ) = 10t1 − 12 t12
t1
t1
t1
t0
0
0
The velocity is zero when
v1x = 0 m/s = (10t1 − 12 t12 ) = (10 − 12 t1 ) × t1
⇒ t1 = 0 s or t1 = 20 s
The first solution is the initial condition. Thus the particle’s velocity is again 0 m/s at t1 = 20 s.
(b) Position is the integral of the velocity. At t1 = 20 s, and using x0 = 0 m at t0 = 0 s, the position is
t1
20
t0
0
x1 = x0 + ∫ vx dt = 0 m + ∫
(10t − t ) dt = 5t
1 2
2
2 20
0
− 16 t 3
20
0
= 667 m
2.35.
Model: Represent the ball as a particle.
Visualize: Please refer to Figure P2.35.
Solve: In the first and third segments the acceleration as is zero. In the second segment the acceleration is
negative and constant. This means the velocity vs will be constant in the first two segments and will decrease
linearly in the third segment. Because the velocity is constant in the first and third segments, the position s will
increase linearly. In the second segment, the position will increase parabolically rather than linearly because the
velocity decreases linearly with time.
2.36.
Model: Represent the ball as a particle.
Visualize: Please refer to Figure P2.36. The ball rolls down the first short track, then up the second short track,
and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t
= 0 at the beginning, that is, when the ball starts to roll down the first short track.
Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks.
Assess: Note that the derivative of the s versus t graph yields the vs versus t graph. And the derivative of the vs
versus t graph gives rise to the as versus t graph.
2.37.
Model: Represent the ball as a particle.
Visualize: Please refer to Figure P2.37. The ball moves to the right along the first track until it strikes the wall,
which causes it to move to the left on a second track. The ball then descends on a third track until it reaches the
fourth track, which is horizontal.
Solve:
Assess: Note that the time derivative of the position graph yields the velocity graph, and the derivative of the
velocity graph gives the acceleration graph.
2.38.
Solve:
Visualize:
Please refer to Figure P2.38.
2.39.
Solve:
Visualize:
Please refer to Figure P2.39.
2.40.
Visualize: Please refer to Figure P2.40. There are four frictionless tracks.
Solve: For the first track, as is negative and constant and vs is decreasing linearly. This is consistent with a ball
rolling up a straight track but not so far that vs goes to zero. For the second track, as is zero and vs is constant but
greater than zero. This is consistent with a ball rolling on a horizontal track. For the third track, as is positive and
constant and vs is increasing linearly. This is consistent with a ball rolling down a straight track. For the fourth
track, as is zero and vs is constant. This is again consistent with a ball rolling on a horizontal track. The as on the
first track has the same absolute value as the as on the third track. This means the slope of the first track up is the
same as the slope of the third track down.
2.41.
Model: The plane is a particle and the constant-acceleration kinematic equations hold.
Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s × 1 mi/1609 m × 3600 s/h = 179 mph.
(b) Using as = Δv / Δt , we have,
as (t = 0 to t = 10 s) =
23 m/s − 0 m/s
= 2.3 m/s 2
10 s − 0 s
as (t = 20 s to t = 30 s) =
69 m/s − 46 m/s
= 2.3 m/s 2
30 s − 20 s
For all time intervals a is 2.3 m/s2.
(c) Using kinematics as follows:
vfs = vis + a (tf − ti ) ⇒ 80 m/s = 0 m/s + (2.3 m/s 2 )(tf − 0 s) ⇒ tf = 35 s
(d) Using the above values, we calculate the takeoff distance as follows:
1
1
sf = si + vis (tf − ti ) + as (tf − ti ) 2 = 0 m + (0 m/s)(35 s) + (2.3 m/s 2 )(35 s) 2 = 1410 m
2
2
For safety, the runway should be 3 ×1410 m = 4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so
the takeoff is not safe.
2.42.
Solve:
Model:
(a)
The automobile is a particle.
The acceleration is not constant because the velocity-versus-time graph is not a straight line.
(b) Acceleration is the slope of the velocity graph. You can use a straightedge to estimate the slope of the graph
at t = 2 s and at t = 8 s. Alternatively, we can estimate the slope using the two data points on either side of 2 s and
8 s. Either way, you need the conversion factor 1 mph = 0.447 m/s from Table 1.4.
ax ( at 2 s ) ≈
vx ( at 4 s ) − vx ( at 0 s )
mph 0.447 m/s
= 11.5
×
= 5.1 m/s 2
4 s−0 s
s
1 mph
ax ( at 8 s ) ≈
vx ( at 10 s ) − vx ( at 6 s )
mph 0.447 m/s
= 4.5
×
= 2.0 m/s 2
10 s − 6 s
s
1 mph
(c) The displacement, or distance traveled, is
Δx = x ( at 10 s ) − x ( at 0 s ) = ∫
10 s
0s
vx dx = area under the velocity curve from 0 s to 10 s
We can approximate the area under the curve as the area of five rectangular steps, each with width Δt = 2 s and
height equal to the average of the velocities at the beginning and end of each step.
0 s to 2 s
vavg = 14 mph = 6.26 m/s
area = 12.5 m
2 s to 4 s
vavg = 37 mph = 16.5 m/s
area = 33.0 m
4 s to 6 s
6 s to 8 s
vavg = 53 mph = 23.7 m/s
area = 47.4 m
vavg = 65 mph = 29.1 m/s
area = 58.2 m
8 s to 10 s
vavg = 74 mph = 33.1 m/s
area = 66.2 m
The total area under the curve is ≈ 217 m , so the distance traveled in 10 s is ≈ 217 m .
2.43.
Solve:
Model: Represent the car as a particle.
(a) First, we will convert units:
60
miles 1 hour 1610 m
×
×
= 27 m/s
hour 3600 s 1 mile
The motion is constant acceleration, so
v −v
(27 m/s − 0 m/s)
v1 = v0 + aΔt ⇒ a = 1 0 =
= 2.7 m/s 2
Δt
10 s
(b) The fraction is a/g = 2.7 / 9.8 = 0.28. So a is 28% of g.
(c) The distance is calculated as follows:
1
1
x1 = x0 + v0 Δt + a (Δt ) 2 = a (Δt ) 2 = 1.3 × 102 m = 4.3 × 102 feet
2
2
2.44.
Model: Represent the spaceship as a particle.
Solve: (a) The known information is: x0 = 0 m, v0 = 0 m/s, t0 = 0 s, a = g = 9.8 m/s 2 , and v1 = 3.0 × 108 m/s.
Constant acceleration kinematics gives
v −v
v1 = v0 + aΔt ⇒ Δt = t1 = 1 0 = 3.1 × 107 s
a
The problem asks for the answer in days, so we need a conversion:
t1 = (3.1× 107 s) ×
1 hour 1 day
×
= 3.6 × 102 days
3600 s 24 hour
(b) The distance traveled is
1
1
x1 − x0 = v0 Δt + a (Δt ) 2 = at12 = 4.6 × 1015 m
2
2
(c) The number of seconds in a year is
1 year = 365 days ×
24 hours 3600 s
×
= 3.15 × 107 s
1 day
1 hour
In one year light travels a distance
1 light year = (3.0 × 108m/s)(3.15 × 107 s) = 9.46 × 1015 m
The distance traveled by the spaceship is 4.6 × 1015m/9.46 × 1015m = 0.49 of a light year.
Assess:
Note that x1 gives “Where is it?” rather than “How far has it traveled?” “How far” is represented by
x1 − x0 . They happen to be the same number in this problem, but that isn’t always the case.
2.45.
Model:
Visualize:
Solve:
The car is a particle, and constant-acceleration kinematic equations hold.
This is a two-part problem. During the reaction time, we can use
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2
2
= 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m
During deceleration,
v22 = v12 + 2a1 ( x2 − x1 )
0 = (20 m/s) 2 + 2(−6.0 m/s 2 )( x2 − 10 m) ⇒ x2 = 43 m
She has 50 m to stop, so she can stop in time.
2.46.
Model:
Visualize:
Solve:
The car is a particle and constant-acceleration kinematic equations hold.
(a) This is a two-part problem. During the reaction time,
x1 = x0 + v0 (t1 − t0 ) + 1/2a0 (t1 − t0 ) 2
= 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m
After reacting, x2 − x1 = 110 m − 10 m = 100 m, that is, you are 100 m away from the intersection.
(b) To stop successfully,
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ (0 m/s) 2 = (20 m/s) 2 + 2a1 (100 m) ⇒ a1 = −2 m/s 2
(c) The time it takes to stop once the brakes are applied can be obtained as follows:
v2 = v1 + a1 (t2 − t1 ) ⇒ 0 m/s = 20 m/s + (−2 m/s 2 )(t2 − 0.50 s) ⇒ t2 = 11 s
The total time to stop since the light turned red is 11.5 s.
2.47.
Model:
Visualize:
Solve:
(a)
To
We will use the particle model and the constant-acceleration kinematic equations.
find
x2 ,
we
first
need
to
determine
x1.
Using
x1 = x0 + v0 (t1 − t0 ),
we
get
x1 = 0 m + (20 m/s)(0.50 s − 0 s) = 10 m. Now,
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (20 m/s) 2 + 2(−10 m/s 2 )( x2 − 10 m) ⇒ x2 = 30 m
The distance between you and the deer is ( x3 − x2 ) or (35 m − 30 m) = 5 m.
(b) Let us find v0 max such that v2 = 0 m/s at x2 = x3 = 35 m. Using the following equation,
v22 − v02 max = 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 − v02 max = 2(−10 m/s 2 )(35 m − x1 )
Also, x1 = x0 + v0 max (t1 − t0 ) = v0 max (0.50 s − 0 s) = (0.50 s)v0 max . Substituting this expression for x1 in the above
equation yields
−v02 max = (−20 m/s 2 )[35 m − (0.50 s) v0 max ] ⇒ v02 max + (10 m/s)v0 max − 700 m 2 /s 2 = 0
The solution of this quadratic equation yields v0 max = 22 m/s. (The other root is negative and unphysical for the
present situation.)
Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional
distance of 5 m with a reaction time of 0.50 s and a deceleration of 10 m/s2.
2.48.
Model:
Visualize:
The car is represented as a particle.
Solve: (a) This is a two-part problem. First, we need to use the information given to determine the acceleration
during braking. Second, we need to use that acceleration to find the stopping distance for a different initial
velocity. First, the car coasts at constant speed before braking:
x1 = x0 + v0 (t1 − t0 ) = v0t1 = (30 m/s)(0.5 s) = 15 m
Then, the car brakes to a halt. Because we don’t know the time interval, use
v22 = 0 = v12 + 2a1 ( x2 − x1 )
v12
(30 m/s) 2
=−
= −10 m/s 2
2( x2 − x1 )
2(60 m − 15 m)
G
We used v1 = v0 = 30 m/s. Note the minus sign, because a1 points to the left.
⇒ a1 = −
We can repeat these steps now with v0 = 40 m/s. The coasting distance before braking is
x1 = v0t1 = (40 m/s)(0.5 s) = 20 m
The position x2 after braking is found using
v22 = 0 = v12 + 2a1 ( x2 − x1 )
⇒ x2 = x1 −
v12
(40 m/s) 2
= 20 m −
= 100 m
2a1
2(−10 m/s 2 )
(b) The car coasts at a constant speed for 0.5 s, traveling 20 m. The graph will be a straight line with a slope of
40 m/s. For t ≥ 0.5 the graph will be a parabola until the car stops at t2.We can find t2 from
v
v2 = 0 = v1 + a1 (t2 − t1 ) ⇒ t2 = t1 − 1 = 4.5 s
a1
The parabola will reach zero slope (v = 0 m/s) at t = 4.5 s. This is enough information to draw the graph shown
in the figure.
2.49.
Model:
Visualize:
The rocket is represented as a particle.
Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with
a = − g . The maximum altitude is y2.. In the acceleration phase:
1
1
1
y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2 = at12 = (30 m/s 2 )(30 s) 2 = 13,500 m
2
2
2
v1 = v0 + a (t1 − t0 ) = at1 = (30 m/s 2 )(30 s) = 900 m/s
In the coasting phase,
v22 = 0 = v12 − 2 g ( y2 − y1 ) ⇒ y2 = y1 +
v12
(900 m/s) 2
= 13,500 m +
= 54,800 m = 54.8 km
2g
2(9.8 m/s 2 )
The maximum altitude is 54.8 km (≈ 33 miles).
(b) The rocket is in the air until time t3 . We already know t1 = 30 s. We can find t2 as follows:
v
v2 = 0 m/s = v1 − g (t2 − t1 ) ⇒ t2 = t1 + 1 = 122 s
g
Then t3 is found by considering the time needed to fall 54,800 m:
1
1
2 y2
y3 = 0 m = y2 + v2 (t3 − t2 ) − g (t3 − t2 ) 2 = y2 − g (t3 − t2 ) 2 ⇒ t3 = t2 +
= 228 s
g
2
2
(c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then
begins to decrease linearly with a slope of −9.8(m/s)/s. The velocity passes through zero (the turning point at
y2 ) at t2 = 122 s. The impact velocity at t3 = 228 s is calculated to be v3 = v2 − g (t3 − t2 ) = −1040 m/s.
Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it
falls, and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that
is a more complicated problem.
2.50.
Model:
Visualize:
Solve:
We will model the rocket as a particle. Air resistance will be neglected.
(a) Using the constant-acceleration kinematic equations,
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (16 s − 0 s) = a0 (16 s)
1
1
y1 = y0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = a0 (16 s − 0 s) 2 = a0 (128 s 2 )
2
2
1
y2 = y1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2
2
1
2
⇒ 5100 m = 128 s a0 + 16 s a0 (20 s − 16 s) + ( − 9.8 m/s 2 )(20 s − 16 s) 2 ⇒ a0 = 27 m/s 2
2
(b) The rocket’s speed as it passes through a cloud 5100 m above the ground can be determined using the kinematic
equation:
v2 = v1 + a1 (t2 − t1 ) = (16 s) a0 + (−9.8 m/s 2 )(4 s) = 4.0 × 102 m/s
Assess: 400 m/s ≈ 900 mph, which would be the final speed of a rocket that has been accelerating for 20 s at a
rate of approximately 20 m/s2 or 66 ft/s2.
2.51.
Model:
equations.
Visualize:
We will model the lead ball as a particle and use the constant-acceleration kinematic
Note that the particle undergoes free fall until it hits the water surface.
Solve: The kinematics equation y1 = y0 + v0 (t1 − t0 ) + 12 a0 (t1 − t0 ) 2 becomes
1
−5.0 m = 0 m + 0 m + (−9.8 m/s 2 )(t1 − 0) 2 ⇒ t1 = 1.01 s
2
Now, once again,
1
y2 = y1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2
2
⇒ y2 − y1 = v1 (3.0 s − 1.01 s) + 0 m/s = 1.99 v1
v1 is easy to determine since the time t1 has been found. Using v1 = v0 + a0 (t1 − t0 ) , we get
v1 = 0 m/s − (9.8 m/s 2 )(1.01 s − 0 s) = − 9.898 m/s
With this value for v1 , we go back to:
y2 − y1 = 1.99v1 = (1.99)(−9.898 m/s) = −19.7 m
y2 − y1 is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative
sign shows the direction of the displacement vector.
Assess: A depth of about 60 ft for a lake is not unusual.
2.52.
Model:
Visualize:
Solve:
The elevator is a particle moving under constant-acceleration kinematic equations.
(a) To calculate the distance to accelerate up:
(v1 ) 2 = v02 + 2a0 ( y0 − y0 ) ⇒ (5 m/s) 2 = (0 m/s) 2 + 2(1 m/s 2 )( y1 − 0 m) ⇒ y1 = 12.5 m
(b) To calculate the time to accelerate up:
v1 = v0 + a0 (t1 − t0 ) ⇒ 5 m/s = 0 m/s + (1 m/s 2 )(t1 − 0 s) ⇒ t1 = 5 s
To calculate the distance to decelerate at the top:
v32 = v22 + 2a2 ( y3 − y2 ) ⇒ (0 m/s) 2 = (5 m/s) 2 + 2(−1 m/s 2 )( y3 − y2 ) ⇒ y3 − y2 = 12.5 m
To calculate the time to decelerate at the top:
v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 5 m/s + (−1 m/s 2 )(t3 − t2 ) ⇒ t3 − t2 = 5 s
The distance moved up at 5 m/s is
y2 − y1 = ( y3 − y0 ) − ( y3 − y2 ) − ( y1 − y0 ) = 200 m − 12.5 m − 12.5 m = 175 m
The time to move up 175 m is given by
1
y2 − y1 = v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ 175 m = (5 m/s)(t2 − t1 ) ⇒ (t2 − t1 ) = 35 s
2
To total time to move to the top is
(t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) = 5 s + 35 s + 5 s = 45 s
Assess: To cover a distance of 200 m at 5 m/s (ignoring acceleration and deceleration times) will require a time
of 40 s. This is comparable to the time of 45 s for the entire trip as obtained above.
2.53.
Model:
Visualize:
The car is a particle moving under constant-acceleration kinematic equations.
Solve: This is a three-part problem. First the car accelerates, then it moves with a constant speed, and then it
decelerates.
First, the car accelerates:
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (4.0 m/s 2 )(6 s − 0 s) = 24 m/s
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + (4.0 m/s 2 )(6 s − 0 s) 2 = 72 m
2
2
Second, the car moves at v1:
1
x2 − x1 = v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 = (24 m/s)(8 s − 6 s) + 0 m = 48 m
2
Third, the car decelerates:
v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 24 m/s + (−3.0 m/s 2 )(t3 − t2 ) ⇒ (t3 − t2 ) = 8 s
1
1
x3 = x2 + v2 (t3 − t2 ) + a2 (t3 − t2 ) 2 ⇒ x3 − x2 = (24 m/s)(8 s) + (−3.0 m/s 2 )(8 s) 2 = 96 m
2
2
Thus, the total distance between stop signs is:
x3 − x0 = ( x3 − x2 ) + ( x2 − x1 ) + ( x1 − x0 ) = 96 m + 48m + 72 m = 216 m
Assess: A distance of approximately 600 ft in a time of around 10 s with an acceleration/deceleration of the
order of 7 mph/s is reasonable.
2.54.
Model:
Visualize:
Solve:
The car is a particle moving under constant linear acceleration.
Using the kinematic equation for position:
1
1
x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ x1 + 30 m = x1 + v1 (5.0 s − 4.0 s) + (2 m/s 2 )(5.0 s − 4.0 s) 2
2
2
1
⇒ 30 m = v1 (1.0 s) + (2 m/s 2 )(1.0 s) 2 ⇒ v1 = 29 m/s
2
And 4.0 seconds before:
v1 = v0 + a0 (t1 − t0 ) ⇒ 29 m/s = v0 + (2 m/s 2 )(4.0 s − 0 s) ⇒ v0 = 21 m/s
Assess:
21 m/s ≈ 47 mph and is a reasonable value.
2.55.
Model: Santa is a particle moving under constant-acceleration kinematic equations.
Visualize: Note that our x-axis is positioned along the incline.
Solve:
Using the following kinematic equation,
v12 = v02 + 2a|| ( x1 − x2 ) = (0 m/s) 2 + 2(4.9 m/s 2 )(10 m − 0 m) ⇒ v1 = 9.9 m/s
Assess:
Santa’s speed of 20 mph as he reaches the edge is reasonable.
2.56.
Model:
Visualize:
The cars are represented as particles.
Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at
time t1 they have the same position. It is important in a problem such as this to express information in terms of
positions (that is, coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity
kinematics gives
xA1 = xA0 + vA (t1 − tA0 ) = vA (t1 − tA0 )
xC1 = xC0 + vC (t1 − tC0 ) = xC0 + vCt1
The critical piece of information is that Ann and Carol have the same position at t1 , so xA1 = xC1. Equating these
two expressions, we can solve for the time t1 when Ann passes Carol:
vA (t1 − tA0 ) = xC0 + vCt1
⇒ (vA − vC )t1 = xC0 + vAtA0
⇒ t1 =
xC0 + vAtA0 2.4 mi + (50 mph)(0.5 h)
=
= 2.0 h
vA − vC
50 mph − 36 mph
(b) Their position is x1 = xA1 = xC1 = xC0 + vCt1 = 74 miles.
(c) Note that Ann’s graph doesn’t start until t = 0.5 hours, but her graph has a steeper slope so it intersects
Carol’s graph at t ≈ 2.0 hours.
2.57.
Model:
Visualize:
We will use the particle model for the puck.
We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the
motion because the speed does not change. So, the problem “starts” at the bottom of the uphill ramp and “ends”
at the bottom of the downhill ramp. At the top of the ramp the speed does not change along the horizontal
section. The final speed from the uphill roll (first problem) becomes the initial speed of the downhill roll (second
problem). Because the axes point in different directions, we can avoid possible confusion by calling the downhill
axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill
velocities as v. Note that the height information, h = 1 m, has to be transformed into information about positions
along the two axes.
Solve: (a) The uphill roll has a0 = − g sin 30D = −4.90 m/s 2 . The speed at the top is found from
v12 = v02 + 2a0 ( x1 − x0 )
⇒ v1 = v02 + 2a0 x1 = (5 m/s) 2 + 2(−4.90 m/s 2 )(2.00 m) = 2.32 m/s
(b) The downward roll starts with velocity u1 = v1 = 2.32 m/s and a1 = + g sin 20D = 3.35 m/s 2 . Then,
u22 = u12 + 2a1 ( z2 − z1 ) = (2.32 m/s) 2 + 2(3.35 m/s 2 )(2.92 m − 0 m) ⇒ u2 = 5.00 m/s
(c) The final speed is equal to the initial speed, so the percentage change is zero!
Assess: This result may seem surprising, but can be more easily understood after we introduce the concept of
energy. For now, imagine this is a one dimensional vertical problem. The total vertical change in height of the
puck is zero. We have already seen how an object with an initial velocity upward has the same velocity in the
opposite direction as it passes through that height going down.
2.58.
Model:
Visualize:
Solve:
We will model the toy train as a particle.
Using kinematics,
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 2 m + (2.0 m/s)(2.0 s − 0 s) + 0 m = 6.0 m
2
The acceleration can now be obtained as follows:
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (2.0 m/s)2 + 2a1 (8.0 m − 6.0 m) ⇒ a1 = −1.0 m/s 2
Assess:
A deceleration of 1 m/s 2 in bringing the toy train to a halt over a distance of 2.0 m is reasonable.
2.59.
Model:
Visualize:
Solve:
We will use the particle model and the kinematic equations at constant-acceleration.
To find x2 , let us use the kinematic equation
v22 = v12 + 2a1 ( x2 − x1 ) = (0 m/s) 2 = (50 m/s) 2 + 2(−10 m/s 2 )( x2 − x1 ) ⇒ x2 = x1 + 125 m
Since the nail strip is at a distance of 150 m from the origin, we need to determine x1 :
x1 = x0 + v0 (t1 − t0 ) = 0 m + (50 m/s)(0.60 s − 0.0 s) = 30 m
Therefore, we can see that x2 = (30 + 125) m = 155 m. That is, he can’t stop within a distance of 150 m. He is in
jail.
Assess: Bob is driving at approximately 100 mph and the stopping distance is of the correct order of
magnitude.
2.60.
Model:
Visualize:
Solve:
We will use the particle model with constant-acceleration kinematic equations.
The acceleration, being the same along the incline, can be found as
v12 = v02 + 2a ( x1 − x0 ) ⇒ (4.0 m/s) 2 = (5.0 m/s)2 + 2a (3.0 m − 0 m) ⇒ a = −1.5 m/s 2
We can also find the total time the puck takes to come to a halt as
v2 = v0 + a (t2 − t0 ) ⇒ 0 m/s = (5.0 m/s) + (−1.5 m/s 2 )t2 ⇒ t2 = 3.3 s
Using the above obtained values of a and t2 , we can find x2 as follows:
1
1
x2 = x0 + v0 (t2 − t0 ) + a (t2 − t0 ) 2 = 0 m + (5.0 m/s)(3.3 s) + (−1.5 m/s 2 )(3.3 s) 2 = 8.3 m
2
2
That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting
position x0 and the puck stops 0.2 m or 20 cm before the ramp ends, you are not a winner.
2.61.
Model:
Visualize:
We will use the particle model for the skier’s motion ignoring air resistance.
Solve: (a) As discussed in the text, acceleration along a frictionless incline is a = g sin 25°, where g is the
acceleration due to gravity. The acceleration of the skier on snow therefore is a = (0.90) g sin 25°. Also since
h/x1 = sin 25°, x = 200 m/ sin 25°. The final velocity can now be determined using kinematics
⎛ 200 m
⎞
v12 = v02 + 2a( x1 − x0 ) = (0 m/s) 2 + 2(0.90)(9.8 m/s 2 )sin 25° ⎜
− 0 m⎟
sin
25
°
⎝
⎠
⎛ 1 km ⎞⎛ 3600 s ⎞
⇒ v1 = 59.397 m/s = (59.397 m/s) ⎜
⎟⎜
⎟ = 214 km/h
⎝ 1000 m ⎠⎝ 1 h ⎠
(b) The speed lost to air resistance is (214 − 180)/214 × 100% = 16%.
Assess: A record of 180 km/h on such a slope in the presence of air resistance makes the obtained speed of 214
km/h (without air resistance) physical and reasonable.
2.62.
Model: We will use the particle model for the motion of the rocks, which move according to constantacceleration kinematic equations.
Visualize:
Solve:
(a) For Heather,
1
yH1 = yH0 + vH0 (tH1 − tH0 ) + a0 (tH1 − tH0 ) 2
2
1
⇒ 0 m = (50 m) + (−20 m/s)(tH1 − 0 s) + (−9.8 m/s 2 )(tH1 − 0 s) 2
2
2 2
⇒ 4.9 m/s tH1 + 20 m/s tH1 − 50 m = 0
The two mathematical solutions of this equation are −5.83 s and +1.75 s. The first value is not physically
acceptable since it represents a rock hitting the water before it was thrown, therefore, tH1 = 1.75 s. For Jerry,
1
yJ1 = yJ0 + vJ0 (tJ1 − tJ0 ) + a0 (tJ1 − tJ0 ) 2
2
1
⇒ 0 m = (50 m) + (+20 m/s)(tJ1 − 0 s) + (−9.8 m/s 2 )(tJ1 − 0 s) 2
2
Solving this quadratic equation will yield tJ1 = −1.75 s and +5.83 s. Again only the positive root is physically
meaningful. The elapsed time between the two splashes is tJ1 − tH1 = 5.83 s − 1.75 s = 4.08 s ≈ 4.1 s.
(b) Knowing the times, it is easy to find the impact velocities:
vH1 = vH0 + a0 (tH1 − tH0 ) = (−20 m/s) + (−9.8 m/s)(1.75 s − 0 s) = −37.1 m/s
vJ1 = vJ0 + a0 (tJ1 − tJ0 ) = (+20 m/s) + (−9.8 m/s 2 )(5.83 s − 0 s) = −37.1 m/s
Assess: The two rocks hit water with equal speeds. This is because Jerry’s rock has the same downward speed
as Heather’s rock when it reaches Heather’s starting position during its downward motion.
2.63.
Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration
kinematic equations.
Visualize:
Solve: Using the known values, we have
v12 = v02 + 2a0 ( y1 − y0 ) ⇒ (−10 m/s) 2 = v02 + 2(−9.8 m/s 2 )(5.0 m − 0 m) ⇒ v0 = 14 m/s
2.64.
Model:
Visualize:
Solve:
The car is a particle that moves with constant linear acceleration.
The reaction time is 1.0 s, and the motion during this time is
x1 = x0 + v0 (t1 − t0 ) = 0 m + (20 m/s)(1.0 s) = 20 m
During slowing down,
1
x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 = 200 m
2
1
= 20 m + (20 m/s)(15 s − 1.0 s) + a1 (15 s − 1.0 s)2 ⇒ a1 = −1.02 m/s 2
2
The final speed v2 can now be obtained as
v2 = v1 + a1 (t2 − t1 ) = (20 m/s) + (−1.02 m/s 2 )(15 s − 1 s) = 5.7 m/s
2.65. Solve: (a) The quantity
2 P 2(3.6 × 104 W)
=
= 60 m 2 /s3 . Thus
m
1200 kg
vx =
At t = 10 s, vx =
(b) With vx =
(60 m s )t
2
3
(60 m s ) (10 s) = 24 m/s ( ≈ 50 mph), and at t = 20 s, v = 35 m/s (≈ 75 mph).
2
3
x
2 P 12
t , we have
m
ax =
(c) At t = 1 s, ax =
dvx
2 P 1 −12
P
=
× t =
dt
m 2
2mt
P
(3.6 × 104 W)
=
= 3.9 m/s 2 . Similarly, at t = 10 s, ax = 1.2 m/s 2 .
2mt
2(1200 kg)(1 s)
(d) Consider the limiting case of very short times. Note that ax → ∞ as t → 0. This is physically impossible for
the Alfa Romeo.
dx
2 P 12
(e) We can use the relationship that vx =
and integrate to find x(t ). We have vx =
t and the initial
dt
m
condition xi = 0 at ti = 0. Thus
x
∫ dx =
0
and x =
2 P t 1/ 2
t dt
m ∫0
2P t 3 / 2 2 2P 3 / 2
=
t
m 3/ 2 3 m
(f) Time to travel a distance x is found by solving the above equation for t.
⎡3 m ⎤
t=⎢
x⎥
⎢⎣ 2 2 P ⎥⎦
For x = 402 m, t = 18.2 s.
2/3
2.66.
Model:
Visualize:
Solve:
Both cars are particles that move according to the constant-acceleration kinematic equations.
(a) David’s and Tina’s motions are given by the following equations:
1
xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2 = vD0tD1
2
1
1 2
xT1 = xT0 + vT0 (tT1 − tT0 ) + aT (tT1 − tT0 ) 2 = 0 m + 0 m + aTtT1
2
2
When Tina passes David the distances are equal and tD1 = tT1 , so we get
1 2
1
2v
2(30 m/s)
xD1 = xT1 ⇒ vD0tD1 = aTtT1
⇒ vD0 = aTtT1 ⇒ tT1 = D0 =
= 30 s
2
2
aT
2.0 m/s 2
Using Tina’s position equation,
1 2 1
xT1 = aTtT1
= (2.0 m/s 2 )(30 s) 2 = 900 m
2
2
(b) Tina’s speed vT1 can be obtained from
vT1 = vT0 + aT (tT1 − tT0 ) = (0 m/s) + (2.0 m/s 2 )(30 s − 0 s) = 60 m/s
Assess: This is a high speed for Tina (~134 mph) and so is David’s velocity (~67 mph). Thus the large distance
for Tina to catch up with David (~0.6 miles) is reasonable.
2.67.
Model:
Visualize:
We will represent the dog and the cat in the particle model.
Solve: We will first calculate the time tC1 the cat takes to reach the window. The dog has exactly the same time
to reach the cat (or the window). Let us therefore first calculate tC1 as follows:
1
xC1 = xC0 + vC0 (tC1 − tC0 ) + aC (tC1 − tC0 ) 2
2
1
2
⇒ 3.0 m = 1.5 m + 0 m + (0.85 m/s 2 )tC1
⇒ tC1 = 1.879 s
2
In the time tD1 = 1.879 s, the dog’s position can be found as follows:
1
xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2
2
1
= 0 m + (1.50 m/s)(1.879 s) + (−0.10 m/s 2 )(1.879 s) 2 = 2.6 m
2
That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe.
2.68.
Model:
Visualize:
Solve:
We use the particle model for the large train and the constant-acceleration equations of motion.
Your position after time tY1 is
1
xY1 = xY0 + vY0 (tY1 − tY0 ) + aY (tY1 − tY0 ) 2
2
= 0 m + (8.0 m/s)(tY1 − 0 s) + 0 m = 8.0 tY1
The position of the train, on the other hand, after time tT1 is
1
xT1 = xT0 + vT0 (tT1 − tT0 ) 2 + aT (tT1 − tT0 ) 2
2
1
2
= 30 m + 0 m + (1.0 m/s 2 )(tT1 ) 2 = 30 + 0.5tT1
2
The two positions xY1 and xT1 will be equal at time tY1 ( = tT1 ) if you are able to jump on the back step of the
train. That is,
2
2
30 + 0.5tY1
= 8.0tY1 ⇒ tY1
− (16 s)tY1 + 60 s 2 = 0 ⇒ tY1 = 6 s and 10 s
The first time, 6 s, is when you will overtake the train. If you continue to run alongside, the accelerating train will
then pass you at 10 s. Let us now see if the first time tY1 = 6.0 s corresponds to a distance before the barrier. From
the position equation for you, xY1 = (8.0 m/s)(6.0 s) = 48.0 m. The position equation for the train will yield the same
number. Since the barrier is at a distance of 50 m from your initial position, you can just catch the train before
crashing into the barrier.
2.69.
Model: Jill and the grocery cart will be treated as particles that move according to the constantacceleration kinematic equations.
Visualize:
Solve:
The final position of Jill when the cart is caught is given by
1
1
1
xJ1 = xJ0 + vJ0 (tJ1 − tJ0 ) + aJ0 (tJ1 − tJ0 ) 2 = 0 m + 0 m + aJ0 (tJ1 − 0 s) 2 = (2.0 m/s 2 )tJ12
2
2
2
The cart’s position when it is caught is
1
1
xC1 = xC0 + vC0 (tC1 − tC0 ) + aC0 (tC1 − tC0 ) 2 = 50 m + 0 m + (0.5 m/s 2 )(tC1 − 0 s) 2
2
2
2
= 50 m + (0.25 m/s 2 )tC1
Since xJ1 = xC1 and tJ1 = tC1 , we get
1
2
2
(2.0)tJ12 = 50 s 2 + 0.25tC1
⇒ 0.75tC1
= 50 s 2 ⇒ tC1 = 8.20 s
2
2
⇒ xC1 = 50 m + (0.25 m/s 2 )tC1
= 50 m + (0.25 m/s 2 )(8.2 s) 2 = 67.2 m
So, the cart has moved 17.2 m.
2.70.
Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations.
Visualize:
Solve:
The watermelon’s and Superman’s position as they meet each other are
1
yW1 = yW0 + vW0 (tW1 − tW0 ) + aW0 (tW1 − tW0 ) 2
2
1
yS1 = yS0 + vS0 (tS1 − tS0 ) + aS0 (tS1 − tS0 ) 2
2
1
⇒ yW1 = 320 m + 0 m + (−9.8 m/s 2 )(tW1 − 0 s) 2
2
⇒ yS1 = 320 m + (−35 m/s)(tS1 − 0 s) + 0 m
Because tS1 = t W1 ,
2
yW1 = 320 m − (4.9 m/s 2 ) t W1
yS1 = 320 m − (35 m/s) t W1
Since yW1 = yS1 ,
2
320 m − (4.9 m/s 2 )tW1
= 320 m − (35 m/s)tW1 ⇒ tW1 = 0 s and 7.1 s
Indeed, t W1 = 0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the
Empire State Building. The other value, tW1 = 7.1 s, is the time when the watermelon will catch up with
Superman. The speed of the watermelon as it passes Superman is
vW1 = vW0 + aW0 (tW1 − t W0 ) = 0 m/s + (−9.8 m/s 2 )(7.1 s − 0 s) = −70 m/s
Note that the negative sign implies a downward velocity.
Assess: A speed of 140 mph for the watermelon is understandable in view of the significant distance (250 m)
involved in the free fall.
2.71. Model:
equations.
Visualize:
Treat the car and train in the particle model and use the constant acceleration kinematics
Solve: In the particle model the car and train have no physical size, so the car has to reach the crossing at an
infinitesimally sooner time than the train. Crossing at the same time corresponds to the minimum a1 necessary to
avoid a collision. So the problem is to find a1 such that x2 = 45 m when y2 = 60 m.
The time it takes the train to reach the intersection can be found by considering its known constant velocity.
v0 y = v2 y = 30 m/s =
y2 − y0 60 m
=
⇒ t2 = 2.0 s
t 2 − t0
t2
Now find the distance traveled by the car during the reaction time of the driver.
x1 = x0 + v0 x ( t1 − t0 ) = 0 + ( 20 m/s )( 0.50 s ) = 10 m
The kinematic equation for the final position at the intersection can be solved for the minimum acceleration a1.
1
2
x2 = 45 m = x1 + v1x ( t2 − t1 ) + a1 ( t2 − t1 )
2
1
2
= 10 m + ( 20 m/s )(1.5 s ) + a1 (1.5 s )
2
⇒ a1 = 4.4 m/s 2
Assess: The acceleration of 4.4 m/s2 = 2.0 miles/h/s is reasonable for an automobile to achieve. However, you
should not try this yourself! Always pay attention when you drive! Train crossings are dangerous locations, and
many people lose their lives at one each year.
2.72.
Solve: A comparison of the given equation with the constant-acceleration kinematics equation
1
x1 = x0 + v0 (t1 − t0 ) + ax (t1 − t0 ) 2
2
yields the following information: x0 = 0 m, x1 = 64 m, t0 = 0, t1 = 4 s, and v0 = 32 m/s.
(a) After landing on the deck of a ship at sea with a velocity of 32 m/s, a fighter plane is observed to come to a
complete stop in 4.0 seconds over a distance of 64 m. Find the plane’s deceleration.
(b)
1
(c) 64 m = 0 m + (32 m/s)(4 s − 0 s) + ax (4 s − 0 s) 2
2
64 m = 128 m + (8 s 2 ) ax ⇒ ax = −8 m/s 2
2.73.
Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation
2
(v1y ) 2 = v0y
+ 2(ay )( y1 − y0 )
yields the following information: y0 = 0 m, y1 = 10 m, a y = −9.8 m/s 2 , and v1 y = 10 m/s. It is clearly a problem of
free fall. On a romantic Valentine’s Day, John decided to surprise his girlfriend, Judy, in a special way. As he
reached her apartment building, he found her sitting in the balcony of her second floor apartment 10 m above the
first floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her
attention. Judy noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he
tells her the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this
Valentine’s Day?
(b)
2
(c) (10 m/s) 2 = v0y
− 2(9.8 m/s 2 )(10 m − 0 m) ⇒ v0y = 17.2 m/s
Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very
reasonable.
2.74.
Solve: A comparison with the constant-acceleration kinematics equation
(v1x ) 2 = (v0x ) 2 + 2ax ( x1 − x0 )
yields the following quantities: x0 = 0 m, v0 x = 5 m/s, v1x = 0 m/s, and ax = −(9.8 m/s 2 )sin10D.
(a) A wagon at the bottom of a frictionless 10° incline is moving up at 5 m/s. How far up the incline does it move
before reversing direction and then rolling back down?
(b)
(c) (0 m/s) 2 = (5 m/s) 2 − 2(9.8 m/s 2 )sin10°( x1 − 0 m)
⇒ 25(m/s) 2 = 2(9.8 m/s 2 )(0.174) x1 ⇒ x1 = 7.3 m
2.75.
Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second
equation gives a position consistent with the first equation. The third equation gives a subsequent position
following the second equation with zero acceleration. A rocket sled accelerates from rest at 20 m/s2 for 5 sec and
then coasts at constant speed for an additional 5 sec. Draw a graph showing the velocity of the sled as a function
of time up to t = 10 s. Also, how far does the sled move in 10 s?
(b)
1
(c) x1 = (20 m/s 2 )(5 s)2 = 250 m
2
v1 = 20 m/s 2 (5 s) = 100 m/s
x2 = 250 m + (100 m/s)(5 s) = 750 m
2.76. Model:
Visualize:
Solve:
The masses are particles.
The rigid rod forms the hypotenuse of a right triangle, which defines a relationship between x2 and y1 :
x22 + y12 = L2 .
Taking the time derivative of both sides yields
2 x2
dx2
dx
+ 2 y1 1 = 0
dt
dt
dx2
dy
and v1y = 1 to write x2v2 x + y1v1 y = 0 .
dt
dt
⎛ y1 ⎞
y
Thus v2 x = − ⎜ ⎟ v1 y . But from the figure, 1 = tan θ ⇒ v2 x = −v1 y tan θ .
x
2
⎝ x2 ⎠
We can now use v2 x =
Assess:
As x2 decreases (v2 x < 0), y1 increases (v1 y > 0), and vice versa.
2.77.
Model:
Visualize:
The rocket and the bolt will be represented as particles to investigate their motion.
The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is,
vB0 = vR1 and it is positive since the rocket is moving upward. The bolt continues to move upward with a
deceleration equal to g = 9.8 m/s 2 before it comes to rest and begins its downward journey.
Solve:
To find aR we look first at the motion of the rocket:
1
yR1 = yR0 + vR0 (tR1 − tR0 ) + aR (tR1 − tR0 ) 2
2
1
= 0 m + 0 m/s + aR (4.0 s − 0 s)2 = 8aR
2
To find aR we must determine the magnitude of yR1 or yB0 . Let us now look at the bolt’s motion:
1
yB1 = yB0 + vB0 (tB1 − tB0 ) + aB (tB1 − tB0 ) 2
2
1
0 = yR1 + vR1 (6.0 s − 0 s) + (−9.8 m/s 2 )(6.0 s − 0 s) 2
2
⇒ yR1 = 176.4 m − (6.0 s) vR1
Since vR1 = vR0 + aR (tR1 − tR0 ) = 0 m/s + 4 aR = 4 aR the above equation for yR1 yields yR1 = 176.4 − 6.0(4aR ). We
know from the first part of the solution that yR1 = 8aR . Therefore, 8aR = 176.4 − 24.0aR and hence
aR = 5.5 m/s 2 .
2.78.
Model:
motion.
Visualize:
Solve:
The rocket car is a particle that moves according to the constant-acceleration equations of
This is a two-part problem. For the first part,
1
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (9.0 s − 0 s) 2 = (81 s 2 ) a0
2
2
2
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (9.0 s − 0 s) = (9.0 s) a0
During the second part of the problem,
1
x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2
2
1
1
⇒ 990 m = (81 s 2 )a0 + (9.0 s)a0 (12 s − 9.0 s) + (−5.0 m/s 2 )(12 s − 9.0 s)2
2
2
⇒ a0 = 15 m/s 2
This leads to:
v1 = (9.0 s)a0 = (9.0 s)(15 m/s 2 ) = 135 m/s
Using this value of v1 , we can now calculate v2 as follows:
v2 = v1 + a1 (t2 − t1 ) = (135 m/s) + (−5.0 m/s 2 )(12 s − 9.0 s) = 120 m/s
That is, the car’s speed as it passes the judges is 120 m/s.
Assess: This is a very fast motion (~250 mph), but the acceleration is large and the long burn time of 9 s yields
a high velocity.
2.79.
Model:
Visualize:
Solve:
Use the particle model.
(a) Substituting into the constant-acceleration kinematic equation
1
10 ⎞
⎛
x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ 100 m = x1 + v1 ⎜ t2 − ⎟ + 0 m
2
3⎠
⎝
100 − x1 10
t2 =
+
v1
3
Let us now find v1 and x1 as follows:
⎛ 10
⎞
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (3.6 m/s 2 ) ⎜
s − 0 s ⎟ = 12 m/s
3
⎝
⎠
2
1
1
⎛ 10
⎞
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + (3.6 m/s 2 ) ⎜
s − 0 s ⎟ = 20 m
2
2
⎝ 3
⎠
The expression for t2 can now be solved as
t2 =
100 m − 20 m 10 s
+
= 10 s
12 m/s
3
(b) The top speed = 12 m/s which means v1 = 12 m/s. To find the acceleration so that the sprinter can run the
100-meter dash in 9.9 s, we use
v1 = v0 + a0 (t1 − t0 ) ⇒ 12 m/s = 0 m/s + a0t1 ⇒ t1 =
12 m/s
a0
1
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0t12 = a0t12
2
2
2
Since x2 = x1 + v1 (t2 − t1 ) + 12 a1 (t2 − t1 ) 2 , we get
1
100 m = a0t12 + (12 m/s) (9.9 s − t1 ) + 0 m
2
Substituting the above equation for t1 in this equation,
2
⎛
12 m/s ⎞
⎛ 1 ⎞ ⎛ 12 m/s ⎞
2
100 m = ⎜ ⎟ a0 ⎜
⎟ + (12 m/s ) ⎜ 9.9 s −
⎟ ⇒ a0 = 3.8 m/s
a
⎝ 2 ⎠ ⎝ a0 ⎠
0
⎝
⎠
(c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s2 to 3.8 m/s2 for the sprint
time to be reduced from 10 s to 9.9 s, that is, by 1%. This decrease of time by 1% corresponds to an increase of
acceleration by
3.8 − 3.6
× 100% = 5.6%
3.6
2.80.
Solve:
(a) The acceleration is the time derivative of the velocity.
dv
d
ax = x = ⎡⎣ a (1 − e − bt ) ⎤⎦ = abe − bt
dt dt
With a = 11.81 m/s and b = 0.6887 s −1 , ax = 8.134e −0.6887 t m/s 2 . At the times t = 0 s, 2 s, and 4 s, this has the values
8.134 m/s2, 2.052 m/s2, and 0.5175 m/s2.
dx
dx
(b) Since vx =
= a − ae −bt and the initial condition
, the position x is the integral of the velocity. With vx =
dt
dt
that xi = 0 m at ti = 0 s ,
x
t
t
o
o
o
∫ dx = ∫ a dt − ∫ ae dt
− bt
Thus
t
a
a
a
t
x = at o + e − bt = at + e − bt −
b
b
b
o
This can be written a little more neatly as
a
x = ( bt + e− bt − 1)
b
= 17.15 ( 0.6887t + e−0.6887 t − 1) m
(c) By trial and error, t = 9.92 s yields x = 100.0 m.
Assess: Lewis’s actual time was 9.93 s.
2.81.
Model:
Visualize:
Solve:
We will use the particle-model to represent the sprinter and the equations of kinematics.
Substituting into the constant-acceleration kinematic equations,
1
1
1
1
x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (4 s − 0 s) 2 = a0t12 = a0 (4.0 s) 2
2
2
2
2
⇒ x1 = (8 s 2 ) a0
v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (4.0 s − 0 s) ⇒ v1 = (4.0 s) a0
From these two results, we find that x1 = (2 s)v1. Now,
1
x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2
2
⇒ 100 m = (2 s)v1 + v1 (10 s − 4 s) + 0 m ⇒ v1 = 12.5 m/s
Assess: Using the conversion 2.24 mph = 1 m/s, v1 = 12.5 m/s = 28 mph. This speed as the sprinter reaches the
finish line is physically reasonable.
2.82. Model:
Visualize:
Solve:
The balls are particles undergoing constant acceleration.
(a) The positions of each of the balls at t1 is found from kinematics.
1 2
1
gt1 = v0t1 − gt12
2
2
1
1
( y1 ) B = ( y0 )B + ( v0 y )B t1 − gt12 = h − gt12
2
2
In the particle model the balls have no physical extent, so they meet when ( y1 ) A = ( y1 ) B . This means
( y1 ) A = ( y0 ) A + ( v0 y ) A t1 −
1
1
h
v0t1 − gt12 = h − gt12 ⇒ t1 =
2
2
v0
1
gh 2
Thus the collision height is ycoll = h − gt12 = h −
.
2
2v0 2
(b) We need the collision to occur while ycoll ≥ 0 . Thus
h−
So hmax =
gh
gh 2
2v 2
⇒ h≤ 0
≥ 0 ⇒ 1≥
2
2
2v0
g
2v0
2v0 2
.
g
(c) Ball A is at its highest point when its velocity ( v1 y ) = 0 .
A
( v ) = ( v ) − gt ⇒ 0 = v − gt ⇒ t = vg
1y A
0y A
0
1
1
0
h
h v0
v2
. Equating these,
= ⇒h= 0 .
v0
v0 g
g
Interestingly, the height at which a collision occurs while Ball A is at its highest point is exactly half of
In (a) we found that the collision occurs at t1 =
Assess:
hmax .
1
2.83.
Model:
Visualize:
The space ships are represented as particles.
Solve: The difficulty with this problem is how to describe “barely avoid.” The Klingon ship is moving with
constant speed, so its position-versus-time graph is a straight line from xK0 = 100 km. The Enterprise will be
decelerating, so its graph is a parabola with decreasing slope. The Enterprise doesn’t have to stop; it merely has
to slow quickly enough to match the Klingon ship speed at the point where it has caught up with the Klingon
ship. (You do the same thing in your car when you are coming up on a slower car; you decelerate to match its
speed just as you come up on its rear bumper.) Thus the parabola of the Enterprise will be tangent to the straight
line of the Klingon ship, showing that the two ships have the same speed (same slopes) when they are at the same
position. Mathematically, we can say that at time t1 the two ships will have the same position ( xE1 = xK1 ) and the
same velocity (vE1 = vK1 ). Note that we are using the particle model, so the ships have zero length. At time t1,
xK1 = xK0 + vK0t1
1
xE1 = vE0t1 + at12
2
vK1 = vK0
vE1 = vE0 + at1
Equating positions and velocities at t1:
1
xK0 + vK0t1 = vE0t1 + at12
2
vK0 = vE0 + at1
We have two simultaneous equations in the two unknowns a and t1. From the velocity equation,
t1 = (vK0 − vE0 )/ a
Substituting into the position equation gives
2
xK0 = −(vK0 − vE0 ) ⋅
(vK0 − vE0 ) 1 ⎛ (vK0 − vE0 ) ⎞
(vK0 − vE0 ) 2
+ a⋅⎜
⎟ =−
a
a
2 ⎝
2a
⎠
⇒a=−
(vK0 − vE0 ) 2
(20,000 m/s − 50,000 m/s) 2
=−
= − 4500 m/s 2
2 xK0
2(100,000 m)
Assess: The magnitude of the deceleration is 4500 m/s2, which is a rather extreme ≈460g. Fortunately, the
Enterprise has other methods to keep the crew from being killed.
3.1.
Visualize:
G G
G
G
G
Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and connect the tail of vector A
G
with the tip of vector B.
G G G
G
G
G
(b) Since A − B = A + (− B) , we place the tail of the vector (− B ) on the tip of vector A and then connect the tail
G
G
of vector A with the tip of vector (− B ) .
3.2.
Visualize:
G G
G
G
G
Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and then connect vector A’s
G
tail with vector B’s tip.
G G G
G G
G
G
G
(b) To find A − B , we note that A − B = A + (− B) . We place the tail of vector − B on the tip of vector A and
G
G
then connect vector A’s tail with the tip of vector − B.
3.3.
Solve:
Visualize:
G
Vector E points to the left and up, so the components Ex and E y are negative and positive,
respectively, according to the Tactics Box 3.1.
(a) Ex = − E cosθ
and
E y = E sin θ .
(b) Ex = − E sin φ
and
E y = E cos φ .
Assess: Note that the role of sine and cosine are reversed because we are using a different angle. θ and φ are
complementary angles.
G
Visualize: The position vector r whose magnitude r is 10 m has an x-component of 6 m. It makes an
angle θ with the + x -axis in the first quadrant.
3.4.
Using trigonometry, rx = r cosθ , or 6 m = (10 m)cosθ . This gives θ = 53.1°. Thus the y-component of
G
the position vector r is ry = r sin θ = (10 m)sin 53.1° = 8 m.
Solve:
Assess: The y-component is positive since the position vector is in the first quadrant.
3.5.
Solve:
Visualize:
The figure shows the components vx and vy, and the angle θ.
We have, v y = −v sin 40°, or −10 m/s = −v sin 40°, or v = 15.56 m/s.
Thus the x-component is vx = v cos 40° = (15.56 m/s ) cos 40° = 12 m/s.
Assess: The x-component is positive since the position vector is in the fourth quadrant.
3.6.
Visualize:
We will follow rules in Tactics Box 3.1.
G
Solve: (a) Vector r points to the right and down, so the components rx and ry are positive and negative,
respectively:
rx = r cosθ = (100 m)cos 45° = 70.7 m
ry = − r sin θ = −(100 m)sin 45° = −70.7 m
G
(b) Vector v points to the right and up, so the components vx and v y are both positive:
vx = v cosθ = (300 m /s) cos 20° = 282 m/s
G
(c) Vector a has the following components:
ax = −a cosθ = −(5.0 m/s 2 )cos90° = 0 m/s 2
v y = v sin θ = (300 m/s)sin 20° = 103 m/s
a y = − a sin θ = −(5.0 m/s 2 )sin 90° = −5.0 m/s 2
Assess: The components have same units as the vectors. Note the minus signs we have manually inserted
according to Tactics Box 3.1.
3.7.
Visualize:
We will follow the rules given in Tactics Box 3.1.
Solve:
v y = (5 cm/s)cos90° = 0 cm/s
(a) vx = −(5 cm/s)sin 90° = −5 cm/s
(b) ax = −(10 m/s 2 )sin 40° = −6.4 m/s 2
(c) Fx = (50 N)sin 36.9° = 30 N
a y = −(10 m/s 2 )cos 40° = −7.7 m/s 2
Fy = (50 N)cos36.9° = 40 N
Assess: The components have the same units as the vectors. Note the minus signs we
have manually inserted according to Tactics Box 3.1.
3.8.
Visualize:
G
G
The components of the vector C and D, and the angles θ are shown.
K
K
Solve: For C we have C x = −(3.15 m)cos15° = −3.04 m and C y = (3.15 m)sin15° = 0.815 m. For D we
have Dx = 25.6sin 30° = 12.8 and Dy = −25.67cos30° = −22.2.
G
G
Assess: The components of the vector C have the same units as C itself. Dx and Dy
G
are unitless because D is without units. Note the minus signs we have manually inserted
following rules of Tactics Box 3.1.
3.9.
Visualize:
Solve: The magnitude of the vector is E = ( Ex ) 2 + ( E y ) 2 = (125 V/m) 2 + ( −250 V/m) 2 = 280 V/m. In the
G
K
expression for E , the − ĵ and +iˆ means that E is in quadrant IV. The angle θ is below the positive x-axis. We
have:
θ = tan −1
Assess:
| Ey |
Ex
⎛ 250 V/m ⎞
−1
= tan −1 ⎜
⎟ = tan 2 = 63.4°
125
V/m
⎝
⎠
Since | E y | > | Ex | , the angle θ made with the +x-axis is larger than 45°. θ = 45° for | E y | = | Ex | .
3.10.
Visualize:
Solve:
(a) Using the formulas for the magnitude and direction of a vector, we have:
4
4
θ = tan −1 = tan −1 1 = 45°
B = ( − 4) 2 + (4) 2 = 5.7
(b) r = (−2 cm) 2 + (−1 cm) 2 = 2.2 cm
1
2
θ = tan −1 = tan −1 0.5 = 26.6°
100
= tan −1 10 = 84.3°
10
10
(d) a = (10 m/s 2 ) 2 + (20 m/s 2 ) 2 = 22.4 m/s 2
θ = tan −1 = tan −1 0.5 = 26.6°
20
Assess: Note that θ ≤ 45° when | E y | ≤ | Ex |, where θ is the angle made with the x-axis. On the other hand,
(c) v = (−10 m/s) 2 + (−100 m/s) 2 = 100.5 m/s
θ > 45° when | E y | > | Ex | .
θ = tan −1
3.11.
Visualize:
Solve:
(a) Using the formulas for the magnitude and direction of a vector, we have:
6
4
θ = tan −1 = tan −1 1.5 = 56.3°
A = (4) 2 + ( −6) 2 = 7.21
(b) r = (50 m) 2 + (80 m) 2 = 94.3 m
⎛r ⎞
⎝ rx ⎠
⎛ 80 m ⎞
⎟ = 58.0°
⎝ 50 m ⎠
θ = tan −1 ⎜ y ⎟ = tan −1 ⎜
40
= tan −1 2 = 63.4°
20
2
(d) a = (2 m/s 2 ) 2 + (−6 m/s 2 ) 2 = 6.3 m/s 2
θ = tan −1 = tan −1 0.33 = 18.4°
6
Assess: Note that the angle θ made with the x-axis is smaller than 45° whenever | E y | < | Ex | , θ = 45° for
(c) v = (−20 m/s) 2 + (40 m/s) 2 = 44.7 m/s
θ = tan −1
| E y | = | Ex |, and θ > 45° for | E y | > | Ex | . In part (d), θ is with the y-axis, where the opposite of this rule applies.
3.12.
Visualize:
G G
G G G
G
G
K
K
We have C = A − B or C = A + (− B), where − B = ( B, direction opposite B ). Look back at Tactics Box 1.2,
which shows how to perform vector subtraction graphically.
G
G
G
G
G
Solve: To obtain vector C from A and B, we place the tail of − B on the tip of A, and then use the tip-totail rule of graphical addition.
3.13.
Visualize:
G G
G G G
The vectors A, B, and C = A + B are shown.
G
G G G
G
(a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. Thus, C = A + B = (5iˆ + 2 ˆj ) + (−3iˆ − 5 ˆj ) = 2iˆ − 3 ˆj.
G G
G
(b) Vectors A, B, and C are shown with their tails together.
G
K
(c) Since C = 2iˆ − 3 ˆj = C xiˆ + C y ˆj , Cx = 2, and C y = −3. Therefore, the magnitude and direction of C are
Solve:
C = (2) 2 + (−3) 2 = 3.6
θ = tan −1
| Cy |
⎛3⎞
= tan −1 ⎜ ⎟ = 56° below the + x-axis
Cx
⎝2⎠
G
Assess: The vector C is to the right and down, thus implying a negative y-component and positive xcomponent, as obtained above. Also θ > 45° since | C y | > | Cx | .
3.14.
Visualize:
G
G G
G
G
K
(a) We have A = 5iˆ + 2 ˆj , B = −3iˆ − 5 ˆj, and − B = +3iˆ + 5 ˆj. Thus, D = A + ( − B) = 8iˆ + 7 ˆj.
G G
G
(b) Vectors A, B and D are shown in the above figure.
G
G
(c) Since D = 8iˆ + 7 ˆj = Dxiˆ + Dy ˆj , Dx = 8 and Dy = 7. Therefore, the magnitude and direction of D are
Solve:
D = (8) 2 + (7) 2 = 10.6
Assess:
⎛ Dy ⎞
−1 ⎛ 7 ⎞
⎟ = tan ⎜ ⎟ = 41°
⎝8⎠
⎝ Dx ⎠
θ = tan −1 ⎜
Since | Dy | < | Dx | , the angle θ is less than 45°, as it should be.
3.15.
Visualize:
G
G
G
G
Solve: (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. This means 2 A = 10iˆ + 4 ˆj and 3B = −9iˆ − 15 ˆj. Hence,
G
G
G
E = 2 A + 3B = 1iˆ − 11 ˆj.
G G
G
(b) Vectors A , B, and E are shown in the above figure.
G
G
(c) From the E vector, Ex = 1 and E y = −11 . Therefore, the magnitude and direction of E are
E = (1) 2 + (−11) 2 = 11.05
⎛ Ex ⎞
⎛1⎞
= tan −1 ⎜ ⎟ = 5.19° right of the − y -axis
⎜ | E | ⎟⎟
⎝ 11 ⎠
⎝ y ⎠
φ = tan −1 ⎜
Assess: Note that φ is the angle made with the y-axis, and that is why φ = tan −1 ( Ex /| E y |) rather than
tan −1 (| E y |/ Ex ), which would be the case if φ were the angle made with the x-axis.
3.16.
Visualize:
G
G
Solve: (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj.
G G
G
F = A − 4 B = 17iˆ + 22 ˆj = Fxiˆ + Fy ˆj with Fx = 17 and Fy = 22.
G G
G
(b) The vectors A, B, and F are shown in the above figure.
G
(c) The magnitude and direction of F are
This
means
F = Fx2 + Fy2 = (17) 2 + (22) 2 = 27.8
⎛ Fy ⎞
−1 ⎛ 22 ⎞
⎟ = tan ⎜ ⎟ = 52.3°
⎝ 17 ⎠
⎝ Fx ⎠
θ = tan −1 ⎜
Assess:
Fy > Fx implies θ > 45°, as is observed.
G
− 4 B = +12iˆ + 20 ˆj.
Hence,
3.17.
Solve: A different coordinate system can only mean a different orientation of the grid and a different
origin of the grid.
(a) False, because the size of a vector is fixed.
(b) False, because the direction of a vector in space is independent of any coordinate system.
(c) True, because the orientation of the vector relative to the axes can be different.
3.18.
Visualize:
G
G
In coordinate system I, A = −(4 m) ˆj , so Ax = 0 m and Ay = −4 m. The vector B makes an angle of
G
60° counterclockwise from vertical, which makes it have an angle of θ = 30° with the –x-axis. Since B points
to the left and up, it has a negative x-component and a positive y-component. That is,
G
Bx = −(5.0 m)cos30° = −4.3 m and By = + (5.0 m)sin 30° = 2.5 m. Thus, B = −(4.3 m)iˆ + (2.5 m) ˆj.
G
In coordinate system II, A points to the left and down, and makes an angle of 30° with the –y-axis.
Ax = −(4.0 m)sin 30° = −2.0 m
and
Ay = −(4.0)cos30° = −3.5 m.
This
implies
Therefore,
G
ˆ
ˆ
A = −(2.0 m)i − (3.5 m) j.
The
vector
G
B makes an angle of 30° with the +y-axis and is to the left and up. This means we have to manually insert a
x-component.
Bx = − B sin 30° = −(5.0 m)sin 30° = −2.5 m,
and
minus
sign
with
the
G
ˆ
ˆ
By = + B cos30° = (5.0 m)cos30° = 4.3 m. Thus B = −(2.5 m)i + (4.3 m) j.
Solve:
3.19.
Visualize:
G
Refer to Figure EX3.19. The velocity vector v points west and makes an angle of 30° with
the
G
–x-axis. v points to the left and up, implying that vx is negative and v y is positive.
Solve:
We
have
vx = −v cos30°
v y = + v sin 30° = (100 m/s)sin 30° = 50.0 m/s.
G
Assess: vx and v y have the same units as v .
= −(100 m/s)cos30°
= −86.6 m/s
and
3.20.
Visualize:
(a)
G G
G
Solve: (b) The components of the vectors A, B, and C are
Ax = ( 3.0 m ) cos 20° = 2.8 m and Ay = − ( 3.0 m ) sin 20°= −1.0 m ; Bx = 0 m and By = 2 m;
Cx = − ( 5.0 m ) cos70° = −1.71 m and C y = − ( 5.0 m ) sin 70°= −4.7 m. This means the vectors can be written,
G
G
G
A = (2.8 m)iˆ − (1.0 m) ˆj
B = (2.0 m) ˆj
C = (−1.71 m)iˆ − (4.7 m) ˆj
G G G G
(c) We have D = A + B + C = (1.09 m)iˆ − (3.7 m) ˆj. This means
D = (1.09 m) 2 + (3.7 m) 2 = 3.9 m
θ = tan −1
G
The direction of D is south of east, 7 4 ° b elow the positive x-axis.
3.9
= tan −1 3.58 = 74°
1.09
3.21.
Visualize:
G G G
Solve: Using the method of tail-to-tip graphical addition, the diagram shows the resultant for D + E + F in (a),
G
G G
G
G
the resultant for D + 2 E in (b), and the resultant for D − 2 E + F in (c).
G
E = Exiˆ + E y ˆj = 2iˆ + 3 ˆj ,
which means
Ex = 2
and
E y = 3. Also,
G
F = Fxiˆ + Fy ˆj = 2iˆ − 2 ˆj , which means Fx = 2 and Fy = −2.
G
G
(a) The magnitude of E is given by E = Ex2 + E y2 = (2) 2 + (3) 2 = 3.6 and the magnitude of F is given by
3.22.
Solve: We
have
F = Fx2 + Fy2 = (2) 2 + (2) 2 = 2.8.
G G
G G
(b) Since E + F = 4iˆ + 1 ˆj , the magnitude of E + F is (4) 2 + (1) 2 = 4.1.
G
G
G
G
(c) Since − E − 2 F = −(2iˆ + 3 ˆj ) − 2(2iˆ − 2 ˆj ) = −6iˆ + 1 ˆj , the magnitude of − E − 2 F is
(−6) 2 + (1) 2 = 6.1.
G
G
Solve: We have r = (5iˆ + 4 ˆj )t 2 m. This means that r does not change the ratio of its components as t
G
G
increases, that is, the direction of r is constant. The magnitude of r is given by
3.23.
r = (5t 2 ) 2 + (4t 2 )2 m = (6.40t 2 ) m.
(a) The particle’s distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 25.6 m, and 160 m.
G
dt 2
G dr
= (5iˆ + 4 ˆj )
m/s = (5iˆ + 4 ˆj )2t m/s = (10iˆ + 8 ˆj )t m/s
(b) The particle’s velocity is v =
dt
dt
(c) The magnitude of the particle’s velocity is given by v = (10 t ) 2 + (8t ) 2 = 12.8t m/s. The particle’s speed at
t = 0 s, t = 2 s, and t = 5 s is 0 m/s, 25.6 m/s , and 64.0 m/s .
3.24.
Visualize:
G
G
G
Solve: (a) Vector C is the sum of vectors A and B, which is obtained using the tip-to-tip rule of graphical
addition. Its magnitude is measured to be 4.7 and its angle made with the +x-axis is measured to be 33°.
(b) Using the law of cosines, C 2 = A2 + B 2 − 2 AB cos φ , and the geometry of parallelograms, which shows that
φ = 180° − (θ B − θ A ) = 180° − (60° − 20°) = 140°, we obtain
C = (3) 2 + (2) 2 − 2(3)(2)cos(140°) = 4.71
Using the law of sines:
sin α sin140°
=
⇒ α = 15.8°
2
4.71
Thus, θC = α + 20° = 35.8°.
(c) We have:
Ax = A cos θ A = 3cos 20° = 2.82
Ay = A sin θ A = 3sin 20° = 1.03
Bx = B cos θ B = 2cos60° = 1.00
By = B sin θ B = 2sin 60° = 1.73
This means: Cx = Ax + Bx = 3.82 and C y = Ay + By = 2.76. The magnitude and direction of C are given by
C = Cx2 + C y2 = (3.82)2 + (2.76) 2 = 4.71
⎛ Cy ⎞
−1 ⎛ 2.76 ⎞
⎟ = tan ⎜
⎟ = 35.8°
C
⎝ 3.82 ⎠
⎝ x⎠
θC = tan −1 ⎜
Assess: Using the method of vector components and their algebraic addition to find the resultant vector yields
the same results as using the graphical addition of vectors.
3.25. Visualize: Refer to Figure
P3.25 in your textbook.
G
G G
G G G
G
Solve: (a) We are given that A + B + C = −2iˆ with A = 4iˆ, and C = −2 ˆj. This means A + C = 4iˆ − 2 ˆj. Thus,
G G G
G G
G
B = ( A + B + C ) − ( A + C ) = (−2iˆ) − (4iˆ − 2 ˆj ) = −6iˆ + 2 ˆj.
G
(b) We have B = Bxiˆ + By ˆj with Bx = −6 and By = 2. Hence, B = ( −6) 2 + (2) 2 = 6.3
θ = tan −1
By
| Bx |
= tan −1
2
= 18°
6
G
G
Since B has a negative x-component and a positive y-component, the angle θ made by B is with the –x-axis and
it is above the –x-axis.
Assess: Since | By | < | Bx |, θ < 45° as is obtained above.
3.26.
Visualize:
Solve:
(a) θ E = tan −1 ⎛⎜⎝ 11 ⎞⎟⎠ = 45°
θ F = tan −1 ⎛⎜⎝ 12 ⎞⎟⎠ = 63.4°
Thus φ = 180° − θ E − θ F = 71.6°
G
G
(b) From the figure, E = 2 and F = 5. Using
G 2 = E 2 + F 2 − 2 EF cos φ = ( 2) 2 + ( 5) 2 − 2( 2)( 5)cos(180° − 71.6°)
⇒ G = 3.00.
sin α sin(180° − 71.6°)
=
⇒ α = 45°
2.975
5
G
Since θ E = 45° , the angle made by the vector G with the +x-axis is θG = (α + θ E ) = 45° + 45° = 90°.
(c) We have
Furthermore, using
Ex = +1.0,
and E y = +1.0
Fx = −1.0,
and Fy = +2.0
Gx = 0.0,
and
⇒G =
G y = 3.0
( 0.0 ) + ( 3.0 ) = 3.0,
2
2
and θ = tan −1
|G y |
|Gx |
⎛ 3.0 ⎞
= tan −1 ⎜
⎟ = 90°
⎝ 0.0 ⎠
G
That is, the vector G makes an angle of 90° with the x-axis.
Assess: The graphical solution and the vector solution give the same answer within the given significance of
figures.
3.27.
Solve:
Visualize: Refer to Figure P3.27.
From the rules of trigonometry,
we
have
Ay = 4sin 40° = 2.6. Also,
G G G G
Since
A + B + C = 0,
Ax = 4cos 40° = 3.1 and
Bx = −2cos10° = −1.97
and
By = +2sin10° = 0.35.
G
G G
G
G
ˆ
ˆ
C = − A − B = (− A) + (− B) = (−3.1i − 2.6 j ) + (+1.97iˆ − 0.35 ˆj ) = −1.1iˆ − 3.0 ˆj.
3.28.
Visualize:
D u
G
G
In the tilted coordinate system, the vectors A and B are expressed as:
G
G
A = (2sin15° m)iˆ + (2cos15° m) ˆj and B = (4cos15° m)iˆ − (4sin15° m) ˆj.
G G
G
Therefore, D = 2 A + B = (4 m)[(sin15° + cos15°)iˆ + (cos15° − sin15°) ˆj ] = (4.9 m)iˆ + (2.9 m) ˆj. The magnitude of
Solve:
this vector is D = 5.7 m, and it makes an angle of θ = tan −1 (2.9 m/4.9 m) = 31° with the +x-axis.
Assess: The resultant vector can be obtained graphically by using the rule of tail-to-tip addition.
3.29.
Visualize:
G
The magnitude of the unknown vector is 1 and its direction is along iˆ + ˆj . Let A = iˆ + ˆj as shown in the
G
G
diagram. That is, A = 1iˆ + 1 ˆj and the x- and y-components of A are both unity. Since θ = tan −1 ( Ay / Ax ) = 45°,
the unknown vector must make an angle of 45° with the +x-axis and have unit magnitude.
G
Solve: Let the unknown vector be B = Bxiˆ + By ˆj where
Bx = B cos 45° =
1
B
2
and By = B sin 45° =
G
We want the magnitude of B to be 1, so we have
1
B
2
B = Bx2 + By2 = 1
2
2
⎛ 1 ⎞ ⎛ 1 ⎞
B⎟ + ⎜
B ⎟ = 1 ⇒ B2 = 1 ⇒ B = 1
⇒ ⎜
⎝ 2 ⎠ ⎝ 2 ⎠
Hence,
Bx = By =
Finally,
1
2
G
1 ˆ 1 ˆ
i+
j
B = Bxiˆ + By ˆj =
2
2
3.30.
Model: Carlos will be represented as a particle and the particle model will be used for motion under
constant acceleration kinetic equations.
Visualize:
Solve: Carlos runs at constant speed without changing direction. The total distance he travels is found from
kinematics:
r1 = r0 + v0 Δt = 0 m + ( 5 m/s )( 600 s ) = 3000 m
G
This displacement is north of east, or θ = 25° from the +x-axis. Thus the position r1 becomes
G
r1 = (3000 m)(cos 25°iˆ + sin 25° ˆj ) = 2.7 km iˆ + 1.27 km ˆj
That is, Carlos ends up 1.27 km north of his starting position.
Assess: The choice of our coordinate system is such that the x-component of the displacement is along the east
and the y-component is along the north. The displacement of 3.0 km is reasonable for Carlos to run in 10 minutes
if he is an athlete.
Visualize: The coordinate system (x,y,z) is shown here. While +x denotes east and +y denotes north,
G
G
G
G
the +z-direction is vertically up. The vectors S morning (shortened as S m ), Safternoon (shortened as Sa ), and the total
G
G G
displacement vector S total = Sa + S m are also shown.
3.31.
G
G
Solve: S m = (2000iˆ + 3000 ˆj + 200kˆ) m, and Sa = (−1500iˆ + 2000 ˆj − 300kˆ) m. The total displacement is the
sum of the individual displacements.
(a) The sum of the z-components of the afternoon and morning displacements is
Sza + Szm = −300 m + 200 m = −100 m, that is, 100 m lower.
G
G G
(b) S total = Sa + S m = (500iˆ + 5000 ˆj − 100kˆ) m, that is, (500 m east) + (5000 m north) – (100 m vertical). The
magnitude of your total displacement is
S total =
( 500 ) + ( 5000 ) + ( −100 ) m = 5.03 km
2
2
2
3.32.
Visualize:
Only the minute hand is shown in the figure.
G
G
Solve: (a) We have S8:00 = (2.0 cm) ˆj and S8:20 = (2.0 cm)cos30°iˆ − (2.0 cm)sin 30° ˆj. The displacement vector is
G
G G
Δr = S8:20 − S8:00
= (2.0 cm)[cos30°iˆ − (sin 30° + 1) ˆj ]
= (2.0 cm)[0.87iˆ − 1.50 ˆj ]
= (1.74 cm)iˆ − (3.00 cm) ˆj
G
G
G
G G
(b) We have S8:00 = (2.0 cm) ˆj and S9:00 = (2.0 cm) ˆj. The displacement vector is Δr = S9:00 − S8:00 = 0.
Assess: The displacement vector in part (a) has a positive x-component and a negative y-component. The
vector thus is to the right and points down, in quadrant IV. This is where the vector drawn from the tip of the
8:00 a.m. arm to the tip of the 8:20 a.m. arm will point.
3.33.
Visualize:
(a)
Note that +x is along the east and +y is along the north.
G
G
Solve: (b) We are given A = −(200 m) ˆj , and can use trigonometry to obtain B = −(283 m)iˆ − (283 m) ˆj and
G
G G G G
C = (100 m)iˆ + (173 m) ˆj. We want A + B + C + D = 0. This means
G G G
G
D = −A − B − C
= (200 m ˆj ) + (283 m iˆ + 283 m ˆj ) + (−100 m iˆ − 173 m ˆj ) = 183 m iˆ + 310 m ˆj
K
The magnitude and direction of D are
D = (183 m) 2 + (310 m) 2 = 360 m and θ = tan −1
Dy
Dx
⎛ 310 m ⎞
= tan −1 ⎜
⎟ = 59.4°
⎝ 183 m ⎠
G
This means D = (360 m, 59.4° north of Geast).
(c) The measuredGlength of the vector D on the graph (with a ruler) is approximately 1.75 times the measured
length of vector A . Since A = 200 m, this gives D = 1.75 × 200 m = 350 m. Similarly, the angle θ measured
with the protractor is close to 60°. These answers are in close agreement to part (b).
3.34.
Visualize:
(a) The figure shows Sparky’s individual displacements and his net displacement.
G
G
G
G
(b) Dnet = D1 + D2 + D3 , where individual displacements are
G
D1 = (50cos 45°iˆ + 50sin 45° ˆj ) m = (35.4iˆ + 35.4 ˆj ) m
G
D2 = −70iˆ m
G
D3 = −20 ˆj m
G
Thus Sparky’s displacement is Dnet = ( −35iˆ + 15.4 ˆj ) m.
(c) As a magnitude and angle,
Solve:
Dnet = ( Dnet ) 2x + ( Dnet ) 2y = 38 m,
⎛ ( Dnet ) y ⎞
⎟ = 24°
⎝ | ( Dnet ) x | ⎠
θ net = tan −1 ⎜
Sparky’s net displacement is 38 m in a direction 24° north of west.
3.35.
Visualize:
Solve:
G
G
G
We are given A = 5 m iˆ and C = (−1 m)kˆ. Using trigonometry, B = (3cos 45° m)iˆ − (3sin 45° m) ˆj. The
G
G G G G
is
displace-ment is r = A + B + C = (7.12 m)iˆ − (2.12 m) ˆj − (1 m) kˆ. The magnitude of r
total
r = (7.12)2 + (2.12) 2 + (1)2 m = 7.5 m.
Assess: A displacement of 7.5 m is a reasonable displacement.
3.36.
Visualize:
Solve:
We
have
G
v = vxiˆ + v y ˆj
= v||iˆ + v⊥ ˆj
= v cosθ iˆ + v sin θ ˆj.
Thus,
v|| = v cosθ = (100 m/s)cos30° = 86.6 m/s.
Assess:
For the small angle of 30°, the obtained value of 86.6 m/s for the horizontal component is reasonable.
3.37.
Visualize:
⎛ 2.5 m/s ⎞
(a) Since vx = v cosθ , we have 2.5 m/s = (3.0 m/s)cosθ ⇒ θ = cos −1 ⎜
⎟ = 34°.
⎝ 3.0 m/s ⎠
(b) The vertical component is v y = v sin θ = (3.0 m/s) sin 34° = 1.7 m/s.
Solve:
3.38.
Visualize:
The coordinate system used here is tilted with x-axis along the slope.
Solve: The component of the velocity parallel to the x-axis is
v|| = −v cos70° = −v sin 20° = −10 m/s (0.34) = −3.4 m/s. This is the speed down the slope. The component of the
velocity perpendicular v⊥ = −v sin 70° = −v cos 20° = −10 m/s (0.94) = −9.4 m/s. This is the speed toward the ground.
Assess: A speed of approximately 10 m/s implies a fall time of approximately 1 second under free fall. Note
that g =
–9.8 m/s2. This time is reasonable for a drop of approximately 5 m, or 16 feet.
3.39.
Visualize:
Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of 2.0 m/s, it will take her 50 s
to row across. But while she’s doing so, the current sweeps her boat sideways a distance 1 m/s × 50 s = 50 m.
Mary’s net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the
river’s current. She lands 50 m east of the point that was directly across the river from her when she started.
(b) Mary’s net displacement is shown on the figure.
3.40.
G
Visualize: Establish a coordinate system with origin at the tree and with the x-axis pointing east. Let
G
G
A be a displacement vector directly from the tree to the treasure. Vector A is A = (100iˆ + 500 ˆj ) paces.
This describes the displacement you would undergo by walking north 500 paces, then east 100 paces. Instead,
you follow the road for 300 paces and undergo displacement
G
B = (300sin 60°iˆ + 300cos60° ˆj ) paces = (260iˆ + 150 ˆj ) paces
G
G G G
Solve: Now let C be the displacement vector from your position to the treasure. From the figure A = B + C.
G G G
So the displacement you need to reach the treasure is C = A − B = (−160iˆ + 350 ˆj ) paces.
G
If θ is the angle measured between C and the y-axis,
⎛ 160 ⎞
⎟ = 24.6°
⎝ 350 ⎠
θ = tan −1 ⎜
You should head 24.6° west of north. You need to walk distance C = Cx2 + C y2 = 385 paces to get to the
treasure.
3.41. Visualize: A 3% grade rises 3 m for every 100 m horizontal distance. The angle of the ground is
thus α = tan −1 (3/100) = tan −1 (0.03) = 1.72°.
Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the
ground.
G
Solve: From the figure, the magnitude of the component vector of v perpendicular to the ground
is v⊥ = v sin α = 15.0 m/s.
But this is only the size. We also have to note that the direction of vG⊥ is down, so the
component is v⊥ = −15.0 m/s.
3.42.
Visualize:
G G
G
G
G
The resulting velocity is given by v = vfly + vwind , where vwind = 6 m/s iˆ and vfly = −v sin θ iˆ − v cosθ ˆj.
G
Substituting the known values we get v = −8 m/s sin θ iˆ − 8 m/s cosθ ˆj + 6 m/s iˆ. We need to have vx = 0. This
Solve:
means 0 = −8 m/s sin θ + 6 m/s, so sin θ =
6
and θ = 48.6°. Thus the ducks should head 48.6° west of south.
8
3.43.
Model:
Visualize:
Solve:
The car is treated as a particle in this problem.
(a) The tangential component is a|| = a sin 30° = (2.0 m/s 2 )(0.5) = 1.0 m/s 2 .
(b) The perpendicular component is a⊥ = a cos30° = (2.0 m/s 2 )(0.866) = 1.7 m/s 2 .
Assess: Magnitudes of the tangential and perpendicular components of acceleration are reasonable.
3.44.
Model:
Visualize:
We will treat the knot in the rope as a particle in static equilibrium.
G
G
Solve: Expressing the vectors using unit vectors, we have F1 = 3.0iˆ and F2 = −5.0sin 30°iˆ + 5.0cos30° ˆj. Since
G
G G
G
G G G G
F1 + F2 + F3 = 0, we can write F3 = − F1 − F2 = −0.5iˆ − 4.33 ˆj . The magnitude of F3 is given by
G
F3 = (−0.5) 2 + (−4.33) 2 = 4.4 units. The angle F3 makes is θ = tan −1 (4.33/0.5) = 83° and is below the negative xaxis.
Assess: The resultant vector has both components negative, and is therefore in quadrant III. Its magnitude and
G
direction are reasonable. Note the minus sign that we have manually inserted with the force F2 .
3.45.
Visualize:
Use a tilted coordinate system such that x-axis is down the slope.
G
G
Solve: Expressing all three forces in terms of unit vectors, we have F1 = −(3.0 N)iˆ, F2 = + (6.0 N) ˆj, and
G
F3 = (5.0 N)sin θ iˆ − (5.0 N)cosθ ˆj.
G
(a) The component of Fnet parallel to the floor is ( Fnet ) x = −(3.0 N) + 0 N + (5.0 N)sin 30° = −0.50 N, or 0.50 N
up the slope.
G
(b) The component of Fnet perpendicular to the floor is ( Fnet ) y = 0 N + (6.0 N) − (5.0 N)cos30° = 1.67 N.
G
G
(c) The magnitude of Fnet is Fnet = ( Fnet ) x + ( Fnet ) y = (−0.50 N) 2 + (1.67 N) 2 = 1.74 N. The angle Fnet makes
is
(F )
⎛
φ = tan −1 Gnet y = tan −1 ⎜
|( Fnet ) x |
G
G
Fnet is 73° above the floor on the left side of F2 . 00000
1.67 N ⎞
⎟ = 73°
⎝ 0.50 N ⎠
3.46.
Visualize:
Using trigonometry to calculate θ, we get θ = tan −1 (100 cm/141 cm) = 35.3°. Expressing the three forces in unit
G
G
G
vectors, FB = −(3.0 N)iˆ, FC = −(6.0 N) ˆj , and FD = + (2.0 N)cos35.3iˆ − (2.0 N)sin 35.3 ˆj = (1.63 N)iˆ − (1.16 N) ˆj. The total
G
G
G
G
G
force is Fnet = FB + FC + FD = −1.37 N iˆ − 7.2 N ˆj. The magnitude of Fnet is Fnet = (1.37 N) 2 + (7.2 N) 2 = 7.3 N.
Solve:
θ net = tan −1
G
Fnet = (7.3 N,79° below −x in quadrant III).
|( Fnet ) y |
⎛ 7.2 N ⎞
= tan −1 ⎜
⎟ = 79°
|( Fnet ) x |
⎝ 1.37 N ⎠
3-1
4.1.
Solve:
(a)
(b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making
the 90° turn the car accelerates back up to 100 mph in the same time it took to slow down.
4.2.
Solve:
(a)
(b) A car drives up a hill, over the top, and down the other side at constant speed.
4.3.
Solve:
(a)
(b) A ball rolls along a level table at 3 m/s. It rolls over the edge and falls 1 m to the floor. How far away from
the edge of the table does it land?
4.4.
Solve: (a) The figure shows a motion diagram of a pendulum as it swings from one side to the other. It’s
clear that the velocity at the lowest point is not zero. The velocity vector at this point is tangent to the circle. We
can use the method of Tactics Box 1.3 to find the acceleration at the lowest point. The acceleration is not zero.
Instead, you can see that the acceleration vector points toward the center of the circle.
(b) The end of the arc is like the highest point of a ball tossed straight up. The velocity is zero for an instant as
the vector changes from pointing outward to pointing inward. However, the acceleration is not zero at this point.
G
The velocity is changing at the end point, and this requires an acceleration. The motion diagram shows that Δv ,
G
and thus a , is tangent to the circle at the end of the arc.
4.5.
Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic
equations in a plane.
Visualize:
Solve:
Resolving the acceleration into its x and y components, we obtain
G
a = ( 0.80 m/s 2 ) cos 40°iˆ + ( 0.80 m/s 2 ) sin 40° ˆj = ( 0.613 m/s 2 ) iˆ + ( 0.514 m/s 2 ) ˆj
G G G
From the velocity equation v1 = v0 + a ( t1 − t0 ) ,
G
v1 = ( 5.0 m/s ) iˆ + ⎡⎣( 0.613 m/s 2 ) iˆ + ( 0.514 m/s 2 ) ˆj ⎤⎦ ( 6 s − 0 s ) = ( 8.68 m/s ) iˆ + ( 3.09 m/s ) ˆj
G
The magnitude and direction of v are
v=
( 8.68 m/s ) + ( 3.09 m/s ) = 9.21 m/s
2
⎛v ⎞
2
⎛ 3.09 m/s ⎞
⎟ = 20° north of east
⎝ 8.68 m/s ⎠
θ = tan −1 ⎜ 1 y ⎟ = tan −1 ⎜
⎝ v1x ⎠
Assess:
An increase of speed from 5.0 m/s to 9.21 m/s is reasonable.
G
Solve: (a) At t = 0 s, x = 0 m and y = 0 m, or r = (0iˆ + 0ˆj ) m. At t = 4 s, x = 0 m and y = 0 m, or
G
r = (0iˆ + 0ˆj ) m. In other words, the particle is at the origin at both t = 0 s and at t = 4 s. From the expressions for
4.6.
x and y,
⎤
dy ˆ ⎡⎛ 3 2
G dx
⎞
v = iˆ +
j = ⎢⎜ t − 4t ⎟ iˆ + ( t − 2 ) ˆj ⎥ m/s
dt
dt
⎠
⎣⎝ 2
⎦
G
G
At t = 0 s, v = −2ˆj m/s, v = 2 m/s. At t = 4 s, v = 8iˆ + 2ˆj m/s, v = 8.3 m/s.
G
(b) At t = 0 s, v is along − ĵ , or 90° south of + x. At t = 4 s,
(
)
⎛ 2 m/s ⎞
⎟ = 14° north of +x
⎝ 8 m/s ⎠
θ = tan −1 ⎜
4.7.
Solve:
Visualize: Refer to Figure EX4.7.
From the figure, identify the following:
x1 = 0 m
y1 = 0 m
x2 = 2000 m
y2 = 1000 m
v1x = 0 m/s
v1 y = 200 m/s
v2 x = 200 m/s
v2 y = −100 m/s
The components of the acceleration can be found by applying v22 = v12 + 2aΔs for the x and y directions. Thus
v22x − v12x ( 200 m/s ) − ( 0 m/s )
=
= 10.00 m/s 2
2Δx
2 ( 2000 m − 0 m )
2
ax =
( −100 m/s ) − ( 200 m/s ) = −15.00 m/s 2
2 (1000 m − 0 m )
2
ay =
2
2
G
So a = (10.00iˆ − 15.00 ˆj ) m/s 2 .
Assess: A time of 20 s is needed to change v1x = 0 m/s to v2 x = 200 m/s at ax = 10 m/s 2 . This is the same time
needed to change v1 y to v2 y at a y = −15 m/s 2 .
4.8.
Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion.
Visualize: Please refer to Figure EX4.8.
G
Solve: (a) At t = 2 s, the graphs give vx = 16 cm/s and v y = 30 cm/s. The angle made by the vector v with the
x-axis can thus be found as
⎛v ⎞
⎛ 30 cm/s ⎞
⎟ = 62° above the x-axis
⎝ 16 cm/s ⎠
θ = tan −1 ⎜ y ⎟ = tan −1 ⎜
⎝ vx ⎠
(b) After t = 5 s, the puck has traveled a distance given by:
5s
x1 = x0 + ∫ vx dt = 0 m + area under vx -t curve = 12 (40 cm/s)(5 s) = 100 cm
0
5s
y1 = y0 + ∫ v y dt = 0 m + area under v y -t curve = (30 cm/s)(5 s) = 150 cm
0
⇒ r1 = x12 + y12 =
(100 cm ) + (150 cm ) = 180 cm
2
2
4.9.
Model: Use the particle model for the puck.
Visualize: Please refer to Figure EX4.9
Solve: (a) Since the vx vs t and v y vs t graphs are straight lines, the puck is undergoing constant acceleration
along the x- and y- axes. The components of the puck’s acceleration are
ax =
dvx Δvx ( −10 m/s − 10 m/s)
=
=
= −2.0 m/s 2
dt
10 s − 0 s
Δt
ay =
(10 m/s − 0 m/s)
= 1.0 m/s 2
(10 s − 0 s)
The magnitude of the acceleration is a = a x2 + a y2 = 2.2 m/s 2 .
(b) The puck is undergoing constant acceleration in both the x and y directions. Identify from the graphs vix = 10 m/s,
viy = 0 m/s. Since the puck starts at the origin, xi = yi = 0 m, and set ti = 0 s. Using kinematics,
x = 0 m + (10 m/s)t + 12 ( −2.0 m/s 2 )t 2
y = 0 m + 0 m/s + 12 (1.0 m/s 2 )t 2
The distance from the origin at time t is r = x 2 + y 2 . The table below shows the values of x, y, and r at the times
t = 0, 5, 10 s.
t =0 s
5s
10 s
x
0m
25 m
0m
y
0m
12.5 m
50 m
R
0m
28 m
50 m
Assess: The puck turns around at t = 5 s in the x direction, and constantly accelerates in the y direction.
Traveling 50 m from the starting point in 10 s is reasonable.
4.10.
Model: Assume the particle model for the ball, and apply the constant-acceleration kinematic
equations of motion in a plane.
Visualize:
Solve:
G
(a) We know the velocity v1 = (2.0iˆ + 2.0jˆ) m/s at t = 1 s. The ball is at its highest point at t = 2 s, so
v y = 0 m/s. The horizontal velocity is constant in projectile motion, so vx = 2.0 m/s at all times. Thus
G
v2 = 2.0iˆ m/s at t = 2 s. We can see that the y-component of velocity changed by Δv y = −2.0 m/s.
between t = 1 s and t = 2 s. Because a y is constant, v y changes by –2.0 m/s in any 1-s interval. At t = 3 s, v y is
2.0 m/s less than its value of 0 at t = 2 s. At t = 0 s, v y must have been 2.0 m/s more than its value of 2.0 m/s at
t = 1 s. Consequently, at t = 0 s,
G
v0 = (2.0iˆ + 4.0 jˆ ) m/s
At t = 1 s,
G
v0 = (2.0iˆ + 2.0 jˆ ) m/s
At t = 2 s,
G
v0 = (2.0iˆ + 0.0 jˆ ) m/s
At t = 3 s,
G
v0 = (2.0iˆ − 2.0 jˆ ) m/s
(b) Because v y is changing at the rate –2.0 m/s per s, the y-component of acceleration is a y = −2.0 m/s 2 . But
a y = − g for projectile motion, so the value of g on Exidor is g = 2.0 m/s 2 .
G
(c) From part (a) the components of v0 are v0 x = 2.0 m/s and v0 y = 4.0 m/s. This means
⎛v ⎞
⎝ v0 x ⎠
⎛ 4.0 m/s ⎞
⎟ = 63.4° above +x
⎝ 2.0 m/s ⎠
θ = tan −1 ⎜ 0 y ⎟ = tan −1 ⎜
Assess: The y-component of the velocity vector decreases from 2.0 m/s at t = 1 s to 0 m/s at t = 2 s. This gives
an acceleration of −2 m/s 2 . All the other values obtained above are also reasonable.
4.11.
Model:
Visualize:
Solve:
The ball is treated as a particle and the effect of air resistance is ignored.
Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ,
2
50 m = 0 m + (25 m/s)(t1 − 0 s) + 0 m ⇒ t1 = 2.0 s
Now, using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ,
2
y1 = 0 m + 0 m + 12 (−9.8 m/s 2 )(2.0 s − 0 s) 2 = −19.6 m
Assess: The minus sign with y1 indicates that the ball’s displacement is in the negative y direction or
downward. A magnitude of 19.6 m for the height is reasonable.
4.12.
Model:
neglected.
Visualize:
Solve:
The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , we obtain
2
(−2.0 × 10−2 m) = 0 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 0.0639 s
(b) Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ,
2
(50 m) = 0 m + v0 x (0.0639 s − 0 s) + 0 m ⇒ v0 x = 782 m/s
Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity,
and a value of 782 m/s is understandable.
4.13.
Model: We will use the particle model for the food package and the constant-acceleration kinematic
equations of motion.
Visualize:
Solve: For the horizontal motion,
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (150 m/s)(t1 − 0 s) + 0 m = (150 m/s)t1
We will determine t1 from the vertical y-motion as follows:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2
⇒ 0 m = 100 m + 0 m + 12 (−9.8 m/s 2 )t12 ⇒ t1 =
200 m
= 4.518 s
9.8 m/s 2
From the above x-equation, the displacement is x1 = (150 m/s)(4.518 s) = 678 m.
Assess: The horizontal distance of 678 m covered by a freely falling object from a height of 100 m and with an
initial horizontal velocity of 150 m/s (≈ 335 mph) is reasonable.
4.14.
Model:
Visualize:
Assume the particle model for the spyglass and use the projectile motion equations.
Solve: (a) The spyglass has an initial horizontal velocity equal to that of the ship. As the spyglass falls, it and
the ship move forward together at the same velocity, so the spyglass lands at the bottom of the mast directly
vertically below on the ship where the sailor dropped it.
(b) The time the spyglass takes to fall can be found by considering the vertical motion:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 15 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 1.749 s
Therefore, x1 = 0 m/s + (4.0 m/s)(1.749 s) = 7.0 m. The spyglass lands 7.0 m to the right of the fisherman.
4.15.
Model:
G G G
G
G
The position vectors r and r ′ in frames S and S′ are related by the equation r = r′ + R, where
G
R is the position vector of the origin of frame S′ as measured in frame S. S is Ted’s frame and S′ is Stella’s
frame.
Visualize:
Solve:
G
G
The relation between R and the velocity of Stella V is
G
G
R
= V = (100 m/s ) cos 45°iˆ − (100 m/s ) sin 45° ˆj
5s
G
⎡100 ˆ 100 ˆ ⎤
⇒ R = ( 5.0 m ) ⎢
i−
j = 353.6iˆ − 353.6 ˆj m
2 ⎥⎦
⎣ 2
(
G G G
Because r = r′ + R,
)
G G G
r ′ = r − R = ( 200 m ) iˆ − ⎡⎣( 353.6 m ) iˆ − ( 353.6 m ) ˆj ⎤⎦ = − (154 m ) iˆ + ( 354 m ) ˆj
G
The vector r ′ determines the position of the exploding firecracker as seen by Stella.
4.16.
Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the
water be S′. Frame S′ moves relative to S with a velocity Vx .
Visualize:
Solve: Let vx be the velocity of the boat in frame S, and v′x be the velocity of the boat in S′. Then for travel
down the river,
30 km
vx = v′x + Vx =
= 10.0 km/h
3.0 hr
For travel up the river,
⎛ 30 km ⎞
−v′x + Vx = − ⎜
⎟ = −6.0 km/h
⎝ 5.0 hr ⎠
Adding these two equations yields Vx = 2.0 km/h. That is, the velocity of the flowing river relative to the earth is
2.0 km/h.
4.17. Motion: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the
moving sidewalk be S′. Frame S′ moves relative to S with velocity Vx .
Solve:
Let vx be your velocity in frame S and v′x be your velocity in S′. In the first case, when the moving
sidewalk is broken, Vx = 0 m/s and
vx W =
( x1 − x0 )
50 s
In the second case, when you stand on the moving sidewalk, v′x = 0 m/s. Therefore, using vx = v′x + Vx , we get
vx S = Vx =
x1 − x0
75 s
In the third case, when you walk while riding, v′x = vx W . Using vx = v′x + Vx , we get
x1 − x0 x1 − x0 x1 − x0
=
+
⇒ t = 30 s
t
50 s
75 s
Assess:
A time smaller than 50 s was expected.
4.18. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S
with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for
both frames. The frames S and S′ have their origins coincident at t = 0 s.
G
G G G G
Solve: (a) According to the Galilean transformation of position: r = r ′ + R = r′ + tV . We need to find Mary’s
G
position vector r from the earth frame S. The observer in frame S′ will observe the boat move straight north
G
G
and will find its position as r ′ = ( 2.0t ) ˆj m/s. We also know that V = ( 3.0 ) iˆ m/s. Since
r ′ = 100 m = ( 2.0 m/s ) t and t = 50 s, we have
G
r = ( 2.0 m/s )( 50 s ) ˆj + ( 50 s )( 3.0 m/s ) t iˆ = (150 m ) iˆ + (100 m ) ˆj
Thus she lands 150 m east of the point that was straight across the river from her when she started.
(b)
G
G
G
Note that r ′ is the displacement due to rowing, R is the displacement due to the river’s motion, and r is the net
displacement.
4.19. Model: Let Susan’s frame be S and Shawn’s frame be S′. S′ moves relative to S with velocity V.
Both Susan and Shawn are observing the intersection point from their frames.
G G G
G
Solve: The Galilean transformation of velocity is v = v′ + V , where v is the velocity of the intersection point
G
G
from Susan’s reference frame, v′ is the velocity of the intersection point from Shawn’s frame S′, and V is the
G
velocity of S′ relative to S or Shawn’s velocity relative to Susan. Because v = − ( 60 mph ) ˆj and
G
G
G
v′ = − ( 45 mph ) iˆ, we have V = v − v′ = ( 45 mph ) iˆ − ( 60 mph ) ˆj. This means that Shawn’s speed relative to
Susan is
V=
( 45 mph ) + ( −60 mph ) = 75 mph
2
2
4.20. Solve: (a) From t = 0 s to t = 1 s the particle does not rotate. From t = 1 s to t = 3 s, the particle
rotates clockwise from the angular position 0 rad to −2π rad. Therefore, Δθ = −2π rad in two seconds, or
ω = −π rad s. From t = 3 s to t = 4 s the particle rotates counterclockwise from the angular position −2π rad
to +4π rad. Thus Δθ = 4π − ( −2π ) = 6π rad and ω = +6π rad s.
(b)
4.21.
Solve: Since ω = ( dθ dt ) we have
θ f = θ i + area under the ω -versus-t graph between ti and tf
From t = 0 s to t = 2 s, the area is 12 ( 20 rad/s )( 2 s ) = 20 rad. From t = 2 s to t = 4 s, the area is
( 20 rad/s )( 2 s ) = 40 rad. Thus, the area under the ω -versus-t graph during the total time interval of 4 s is 60 rad or
(60 rad) × (1 rev/2π rad) =
9.55 revolutions.
4.22.
Solve: Since ω = ( dθ dt ) we have
θ f = θ i + area under the ω versus t graph between ti and tf
From t = 0 s to s to t = 4 s, the area is 12 ( 20 rad/s )( 4 s ) = 40 rad. From t = 4 s to t = 8 s, the area is
1
2
( −10 rad/s )( 4 s ) = −20 rad. Thus, the area under the ω versus t graph during the total time interval of 8 s is 20
rad or (20 rad) × (1 rev/2π rad) = 3.2 revolutions.
4.23.
Model: Treat the record on a turntable as a particle rotating at 45 rpm.
Solve: (a) The angular velocity is
ω = 45 rpm ×
1min 2π rad
×
= 1.5π rad/s
60 s 1 rev
(b) The period is
T=
2π rad
ω
=
2π rad
= 1.33 s
1.5π rad/s
4.24.
Model:
Visualize:
Solve:
The airplane is to be treated as a particle.
(a) The angle you turn through is
θ 2 − θ1 =
180°
s 5000 miles
=
= 1.2500 rad = 1.2500 rad ×
= 71.62°
π rad
r 4000 miles
(b) The plane’s angular velocity is
ω=
θ 2 − θ1
t2 − t1
=
1.2500 rad
rad
1h
= 0.13889 rad/h = 0.13889
×
= 3.858 × 10−5 rad/s
9 hr
h 3600 s
Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the
separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.
4.25. Solve: Let RE be the radius of the earth at the equator. This means RE + 300 m is the radius to the top
of the tower. Letting T be the period of rotation, we have
vtop − vbottom =
2π ( RE + 300 m) 2π RE 2π (300 m)
600π m
−
=
=
= 2.18 × 10−2 m/s
T
T
24 h
24(3600) s
4.26.
Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the
plane must fly from east to west. The speed is
km
km
1 mile
⎛ 2π rad ⎞
3
v = ωr = ⎜
= 1680
×
= 1040 mph
⎟ (6.4 × 10 km) = 1680
h
h 1.609 km
⎝ 24 h ⎠
4.27.
Solve:
Model: The rider is assumed to be a particle.
Since ar = v 2 / r , we have
v 2 = ar r = (98 m/s 2 )(12 m) ⇒ v = 34 m/s
Assess:
34 m/s ≈ 70 mph is a large yet understandable speed.
4.28.
Solve:
Model: The earth is a particle orbiting around the sun.
(a) The magnitude of the earth’s velocity is displacement divided by time:
v=
2π r
2π (1.5 × 1011 m)
=
= 3.0 × 104 m/s
24 h 3600 s
T
365 days ×
×
1 day
1h
(b) Since v = rω , the angular acceleration is
v
r
ω= =
3.0 × 104 m/s
= 2.0 × 10−7 rad/s
1.5 × 1011 m
(c) The centripetal acceleration is
ar =
v 2 (3.0 × 104 m/s) 2
=
= 6.0 × 10−3 m/s 2
r
1.5 × 1011 m
Assess: A tangential velocity of 3.0 × 104 m/s or 30 km/s is large, but needed for the earth to go through a
displacement of 2π (1.5 × 1011 m) ≈ 9.4 × 108 km in 1 year.
Solve: The pebble’s angular velocity ω = (3.0 rev/s)(2π rad/rev) = 18.9 rad/s. The speed of the pebble
as it moves around a circle of radius r = 30 cm = 0.30 m is
4.29.
v = ω r = (18.9 rad/s)(0.30 m) = 5.7 m/s
The radial acceleration is
v 2 ( 5.7 m/s )
=
= 108 m/s 2
0.30 m
r
2
ar =
4.30.
Model: The crankshaft is a rotating rigid body.
Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s,
maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to
t = 7 s. The angular acceleration (α ) graph is based on the fact that α is the slope of the ω -versus-t graph.
4.31. Model: The turntable is a rotating rigid body.
The angular velocity is the area under the α -versus-t graph:
Solve:
α=
dω
⇒ ω = ∫ α ( x ) dt = ω 0 + area under the α graph
dt
The values of ω at selected values of time (t) are:
t (s)
ω (rad/s)
0
0.5
0
(5 + 3.75)(0.5)/2 = 2.18
1.0
(5 + 2.5)(1)/2 = 3.75
1.5
(5 + 1.25)(1.5)/2 = 4.68
2.0
(5 + 0)(2)/2 = 5.0
2.5
5.0
3.0
5.0
4.32. Model: The wheel is a rotating rigid body.
Solve:
(a)
The angular acceleration (α ) is the slope of the ω -versus-t graph.
(b) The car is at rest at t = 0 s. It gradually speeds up for 4 s and then slows down for 4 s. The car is at rest from
t = 8 s to t = 12 s, and then speeds up again for 4 s.
4.33. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body.
Solve:
(a) The α-versus-t graph has a positive slope of 5 rad/s2 from t = 0 s to t = 2 s and a negative slope of
−5 rad/s 2 from t = 2 s to t = 4 s.
(b) The angular velocity is the area under the α -versus-t graph:
α=
dω
⇒ ω = ∫ α ( x ) dt = ω 0 + area under α graph.
dt
4.34.
Model:
Visualize:
Model the car as a particle in nonuniform circular motion.
Note that halfway around the curve, the tangent is 45° south of east. The perpendicular component of the
acceleration is 45° north of east.
Solve: The radial and tangential components of the acceleration are
ar = a cos 25° = ( 3.0 m s 2 ) cos 25° = 2.7 m s 2
at = a sin 25° = ( 3.0 m s 2 ) sin 25° = 1.27 m s 2
4.35.
Model:
Visualize:
Model the child on the merry-go-round as a particle in nonuniform circular motion.
Solve: (a) The speed of the child is v0 = rω = (2.5 m)(1.57 rad/s) = 3.9 m/s.
(b) The merry-go-round slows from 1.57 rad/s to 0 in 20 s. Thus
ω1 = 0 = ω 0 +
at
rω
(2.5 m)(1.57 rad/s)
t1 ⇒ at = − 0 = −
= –0.197 m/s 2
r
t1
20 s
During these 20 s, the wheel turns through angle
θ1 = θ 0 + ω 0t1 +
at 2
0.197 m/s 2
t1 = 0 + (1.57 rad/s) (20 s) −
(20 s) 2 = 15.6 rad
2r
2(2.5 m)
In terms of revolutions, θ1 = (15.6 rad)(1 rev/2π rad) = 2.49 rev.
4.36.
Model:
Visualize:
Model the particle on the crankshaft as being in nonuniform circular motion.
Solve: (a) The initial angular velocity is ω 0 = 2500 rpm × (1 min/60 s) × (2π rad/rev) = 261.8 rad/s. The
crankshaft slows from 261.8 rad/s to 0 in 1.5 s. Thus
ω1 = 0 = ω 0 +
at
rω
(0.015 m)(261.8 rad/s)
t1 ⇒ at = − 0 = −
= –2.618 m/s 2 = −2.6 m/s 2
r
t1
1.5 s
(b) During these 1.5 s, the crankshaft turns through angle
θ1 = θ 0 + ω 0t1 +
at 2
2.618 m/s 2
t1 = 0 + (261.8 rad/s) (1.5 s) −
(1.5 s) 2 = 196 rad
2r
2(0.015 m)
In terms of revolutions, θ1 = (196 rad)(1 rev/π rad) = 31.2 rev.
4.37. Model: The fan is in nonuniform circular motion.
Visualize:
Solve:
⎛ min ⎞
Note 1800 rev/min ⎜
⎟ = 30 rev/s. Thus
⎝ 60 s ⎠
ω f = ωi + αΔt ⇒ 30 rev/s = 0 rev/s + α ( 4.0 s ) ⇒ α =7.5 rev/s 2 . This can be expressed as
2π rad ⎞
2
⎟ = 47 rad/s .
⎝ rev ⎠
Assess: An increase in the angular velocity of a fan blade by 7.5 rev/s each second seems reasonable.
( 7.5 rev/s ) ⎛⎜
4.38. Model: The wheel is in nonuniform circular motion.
Visualize:
Solve:
(a) Express ωi in rad/s:
⎛ min ⎞⎛ 2π rad ⎞
⎟⎜
⎟ ⇒ 5.2 rad/s
⎝ 60 s ⎠⎝ rev ⎠
ωi = ( 50 rev/min ) ⎜
After 10 s, ω f = ωi + αΔt ⇒ ω f = 5.2 rad/s + ( 0.50 rad/s 2 ) (10 s ) = 55 rad/s. Converting to rpm,
2
( 55 rad/s ) ⎛⎜
60 s ⎞⎛ rev ⎞
⎟⎜
⎟ = 53 rpm
⎝ min ⎠⎝ 2π rad ⎠
(b) In 10 s, the wheel has turned a number of radians
θ f = θi + ω0 Δt + 12 αΔt 2 ⇒ θ f = 0 rad/sec + (5.2 rad/s)(10 s) + 12 (0.50 rad/s 2 )(10 s) 2 = 77 radians.
Converting, 77 rad = 12.3 revolutions.
Assess: Making a bicycle wheel turn just over 12 revolutions in 10 s when it is initially turning almost one
revolution per second to begin with seems attainable by a cyclist.
4.39.
Model:
Visualize:
We will assume that constant-acceleration kinematic equations in a plane apply.
G
(a) The particle’s position r0 = (9.0 ˆj ) m implies that at t0 the particle’s coordinates are x0 = 0 m and
G
y0 = 9.0 m. The particle’s position r1 = (20iˆ) m at time t1 implies that x1 = 20 m and y1 = 0 m. This is the
Solve:
position where the wire hoop is located. Let us find the time t1 when the particle crosses the hoop at x1 = 20 m.
From the vx -versus-t curve and using the relation x1 = x0 + area of the vx -t graph, we get
20 m = 0 m + area of the vx -t graph = area of the vx -t graph
From Figure P4.39 we see that the area of the vx -t graph equals 20 m when t = t1 = 3 s.
(b) We can now look at the y-motion to find a y . Note that the slope of the vx -t graph (that is, a y ) is negative
and constant, and we can determine a y by substituting into y3 = y0 + v0 y (t3 − t0 ) + 12 a y (t3 − t0 ) 2 :
0 m = 9 m + 0 m + 12 a y (3 s − 0 s) 2 ⇒ a y = −2 m/s 2
Therefore,
v4 y = v0 y + a y (t4 − t0 ) = 0 m/s + (−2 m/s 2 )(4 s − 0 s) = −8 m/s
(c)
4.40.
Solve: From the expression for R, Rmax = v02 /g . Therefore,
R=
Assess:
Rmax v02 sin 2θ
1
=
⇒ sin 2θ = ⇒ θ = 15° and 75°
2
g
2
The discussion of Figure 4.22 explains why launch angles θ and (90° − θ ) give the same range.
4.41.
Model:
Visualize:
Solve:
Assume particle motion in a plane and constant-acceleration kinematics for the projectile.
(a) We know that v0 y = v0 sin θ , a y = − g , and v1 y = 0 m/s. Using v1y2 = v02y + 2a y ( y1 − y0 ) ,
0 m 2 /s 2 = v02 sin 2 θ + 2 ( − g ) h ⇒ h =
v02 sin 2 θ
2g
(b) Using Equation 4.19 and the above expression for θ = 30.0° :
( 33.6 m/s ) sin 2 30.0° = 14.4 m
2
h=
( x2 − x0 ) =
2 ( 9.8 m/s 2 )
( 33.6 m/s ) sin ( 2 × 30.0° ) = 99.8 m
v02 sin 2θ
=
g
( 9.8 m/s2 )
2
For θ = 45.0° :
( 33.6 m/s ) sin 2 45.0° = 28.8 m
2
h=
2 ( 9.8 m/s 2 )
( x2 − x0 ) =
( 33.6 m/s ) sin ( 2 × 45.0° ) = 115.2 m
2
( 9.8 m/s )
2
For θ = 60.0° :
( 33.6 m/s ) sin 2 60.0° = 43.2 m
2
h=
( x2 − x0 ) =
Assess:
2 ( 9.8 m/s 2 )
( 33.6 m/s ) sin ( 2 × 60.0° ) = 99.8 m
2
2 ( 9.8 m/s 2 )
The projectile’s range, being proportional to sin(2θ ), is maximum at a launch angle of 45°, but the
maximum height reached is proportional to sin 2 (θ ). These dependencies are seen in this problem.
4.42.
Model:
Visualize:
Solve:
Assume the particle model for the projectile and motion in a plane.
(a) Using y2 = y0 + v0 y ( t2 − t0 ) + 12 a y ( t2 − t0 ) ,
2
y2 = 0 m + ( 30 m/s ) sin 60° ( 7.5 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 7.5 s − 0 s ) = −80.8 m
2
Thus the launch point is 81 m higher than where the projectile hits the ground.
(b) Using v12y = v02y + 2a y ( y1 − y0 ) ,
0 m 2 /s 2 = ( 30sin 60° m/s ) + 2 ( −9.8 m/s 2 ) ( y1 − 0 m ) ⇒ y1 = 34.4 m , or y1 = 34 m
2
(c) The x-component is v2 x = v0 x = v0 cos60° = ( 30 m/s ) cos60° = 15 m/s. The y-component is
v2 y = v0 y + a y ( t2 − t0 ) = v0 sin 60° − g ( t2 − t1 ) = ( 30 m/s ) sin 60° − ( 9.8 m/s 2 ) ( 7.5 s − 0 s ) = −47.52 m/s
⇒v=
(15 m/s ) + ( −47.52 m/s ) = 50 m/s
θ = tan −1
2
v2 y
v2 x
= tan −1
2
47.52
= 73° below + x
15
Assess: An angle of 73° made with the ground, as the projectile hits the ground 81 m below its launch point, is
reasonable in view of the fact that the projectile was launched at an angle of 60°.
4.43. Model:
plane.
Visualize:
Solve:
Assume the particle model and motion under constant-acceleration kinematic equations in a
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ,
2
0 m = 1.80 m + v0 sin 40° ( t1 − 0 s ) + 12 ( −9.8 m/s 2 ) ( t1 − 0 s )
2
= 1.80 m + ( 7.713 m/s ) t1 − ( 4.9 m/s 2 ) t12 ⇒ t1 = − 0.206 s and 1.780 s
The negative value of t1 is unphysical for the current situation. Using t1 = 1.780 s and x1 = x0 + v0 x ( t1 − t0 ) , we
get
x1 = 0 + ( v0 cos 40° m/s )(1.780 s − 0 s ) = (12.0 m/s ) cos 40° (1.78 s ) = 16.36 m
(b) We can repeat the calculation for each angle. A general result for the flight time at angle θ is
)
(
t1 = 12sin θ + 144sin 2 θ + 35.28 / 9.8 s
and the distance traveled is x1 = (12.0)cosθ × t1. We can put the results in a table.
θ
t1
x1
40.0°
42.5°
45.0°
47.5°
1.780 s
1.853 s
1.923 s
1.990 s
16.36 m
16.39 m
16.31 m
16.13 m
Maximum distance is achieved at θ ≈ 42.5°.
Assess: The well-known “fact” that maximum distance is achieved at 45° is true only when the projectile is
launched and lands at the same height. That isn’t true here. The extra 0.03 m = 3 cm obtained by increasing the
angle from 40.0° to 42.5° could easily mean the difference between first and second place in a world-class meet.
4.44.
Model:
Visualize:
The golf ball is a particle following projectile motion.
(a) The distance traveled is x1 = v0 xt1 = v0 cosθ × t1. The flight time is found from the y-equation, using the fact
that the ball starts and ends at y = 0 :
y1 − y0 = 0 = v0 sin θ t1 − 12 gt12 = (v0 sin θ − 12 gt1 ) t1 ⇒ t1 =
2v0 sin θ
g
Thus the distance traveled is
x1 = v0 cosθ ×
2v0 sin θ 2v02sin θ cosθ
=
g
g
For θ = 30°, the distances are
( x1 )earth =
2v02sin θ cosθ 2(25 m/s) 2sin30°cos30°
=
= 55.2 m
g earth
9.80 m/s2
( x1 ) moon =
2v02sin θ cosθ 2v02sin θ cosθ
2v02sin θ cosθ
=
=
6
×
= 6( x1 )earth = 331.2 m
1
g moon
g earth
6 g earth
The golf ball travels 331.2 m − 55.2 m = 276 m farther on the moon than on earth.
(b) The flight times are
(t1 )earth =
2v0 sin θ
= 2.55 s
g earth
(t1 ) moon =
2v0 sin θ 2v0 sin θ
= 1
= 6(t1 )earth = 15.30 s
g moon
g
6 earth
The ball spends 15.30 s − 2.55 s = 12.75 s longer in flight on the moon.
4.45.
Model:
Visualize:
Solve:
The particle model for the ball and the constant-acceleration equations of motion are assumed.
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ,
2
h = 0 m + ( 30 m/s ) sin 60° ( 4 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 4 s − 0 s ) = 25.5 m
2
The height of the cliff is 26 m.
(b) Using ( v y2 ) = v y2 + 2a y ( ytop − y0 ) ,
top
0 m /s = ( v0 sin θ ) + 2 ( − g ) ( ytop ) ⇒ ytop
2
2
2
( v sinθ ) = ⎡⎣( 30 m/s ) sin 60°⎤⎦ = 34.4 m
= 0
2
2
2g
2 ( 9.8 m/s 2 )
The maximum height of the ball is 34 m.
(c) The x and y components are
v1 y = v0 y + a y ( t1 − t0 ) = v0 sin θ − gt1 = ( 30 m/s ) sin 60° − ( 9.8 m/s 2 ) × ( 4.0 s ) = −13.22 m/s
v1x = v0 y = v0 cos60° = ( 30 m/s ) cos60° = 15.0 m/s
⇒ v1 = v12x + v12y = 20.0 m/s
The impact speed is 20 m/s.
Assess: Compared to a maximum height of 34.4 m, a height of 25.5 for the cliff is reasonable.
4.46.
Model:
assumed.
Visualize:
Solve:
The particle model for the ball and the constant-acceleration equations of motion in a plane are
The initial velocity is
v0 x = v0 cos5.0° = ( 20 m/s ) cos5.0° = 19.92 m/s
v0 y = v0 sin 5.0° = ( 20 m/s ) sin 5.0° = 1.743 m/s
The time it takes for the ball to reach the net is
x1 = x0 + v0 x ( t1 − t0 ) ⇒ 7.0 m = 0 m + (19.92 m/s )( t1 − 0 s ) ⇒ t = 0.351 s
The vertical position at t1 = 0.351 s is
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )
2
= ( 2.0 m ) + (1.743 m/s )( 0.351 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 0.351 s − 0 s ) = 2.01 m
2
Thus the ball clears the net by 1.01 m.
Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by
approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial
vertical velocity and the ball clears by a larger distance. The above result is reasonable.
4.47. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are
assumed.
Visualize:
Solve:
(a) The time for the ball to fall is calculated as follows:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )
2
⇒ 0 m = 4 m + 0 m + 12 ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 0.9035 s
2
Using this result for the horizontal velocity:
x1 = x0 + v0 x ( t1 − t0 ) ⇒ 25 m = 0 m + v0 x ( 0.9035 s − 0 s ) ⇒ v0 x = 27.7 m/s
The friend’s pitching speed is 28 m/s.
(b) We have v0 y = ± v0 sin θ , where we will use the plus sign for up 5° and the minus sign for down 5°. We can
write
y1 = y0 ± v0 sin θ ( t1 − t0 ) −
g
g
2
( t1 − t0 ) ⇒ 0 m = 4 m ± v0 sinθ t1 − t12
2
2
Let us first find t1 from x1 = x0 + v0 x ( t1 − t0 ) :
25 m = 0 m + v0 cosθ t1 ⇒ t1 =
25 m
v0 cosθ
Now substituting t1 into the y-equation above yields
⎛ 25 m ⎞ g ⎛ 25 m ⎞
0 m = 4 m ± v0 sin θ ⎜
⎟− ⎜
⎟
⎝ v0 cosθ ⎠ 2 ⎝ v0 cosθ ⎠
2
g ( 25 m ) ⎪⎧
1
⎪⎫
⎨
⎬ = 22.3 m/s and 44.2 m/s
2
2cos θ ⎩⎪ 4 m ± ( 25 m ) tan θ ⎭⎪
2
⇒ v02 =
The range of speeds is 22 m/s to 44 m/s, which is the same as 50 mph to 92 mph.
Assess: These are reasonable speeds for baseball pitchers.
4.48.
Model:
Visualize:
Solve:
We will use the particle model and the constant-acceleration kinematic equations in a plane.
The x- and y-equations of the ball are
x1B = x0B + ( v0B ) x ( t1B − t0B ) + 12 ( aB ) x ( t1B − t0B ) ⇒ 65 m = 0 m + ( v0B cos30° ) t1B + 0 m
2
2
y1B = y0B + ( v0B ) y ( t1B − t0B ) + 12 ( aB ) y ( t1B − t0B ) ⇒ 0 m = 0 m + ( v0B sin 30° ) t1B + 12 ( − g ) t1B
2
From the y-equation,
v0B =
gt1B
( 2sin 30° )
Substituting this into the x-equation yields
2
g cos30° t1B
2sin 30°
⇒ t1B = 2.77 s
65 m =
For the runner:
t1R =
20 m
= 2.50 s
8.0 m/s
Thus, the throw is too late by 0.27 s.
Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable.
4.49.
Model:
Visualize:
Use the particle model for the ball and the constant-acceleration kinematic equations.
Solve: (a) The distance from the ground to the peak of the house is 6.0 m. From the throw position this
distance is 5.0 m. Using the kinematic equation v12y = v02y + 2a y ( y1 − y0 ) ,
0 m 2 /s 2 = v02y + 2 ( −9.8 m/s 2 ) ( 5.0 m − 0 m ) ⇒ v0 y = 9.899 m/s
The time for up and down motion is calculated as follows:
y2 = y0 + v0 y ( t2 − t0 ) + 12 a y ( t2 − t0 ) ⇒ 0 m = 0 m + ( 9.899 m/s ) t2 − 12 ( 9.8 m/s 2 ) t22 ⇒ t2 = 0 s and 2.02 s
2
The zero solution is not of interest. Having found the time t2 = 2.02 s, we can now find the horizontal velocity
needed to cover a displacement of 18.0 m:
x2 = x0 + v0 x ( t2 − t0 ) ⇒ 18.0 m = 0 m + v0 x ( 2.02 s − 0 s ) ⇒ v0 x = 8.911 m/s
⇒ v0 =
(8.911 m/s ) + ( 9.899 m/s ) = 13.3 m/s
2
2
G
(b) The direction of v0 is given by
v
θ = tan −1 0 y = tan −1
v0 x
9.899
= 48°
8.911
Assess: Since the maximum range corresponds to an angle of 45°, the value of 48° corresponding to a range of
18 m and at a modest speed of 13.3 m/s is reasonable.
4.50.
Model:
equations.
Visualize:
Solve:
We will assume a particle model for the sand, and use the constant-acceleration kinematic
Using the equation x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ,
2
x1 = 0 m + ( v0 cos15° )( t1 − 0 s ) + 0 m = ( 60 m/s ) (cos15°)t1
We can find t1 from the y-equation, but note that v0 y = −v0 sin15° because the sand is launched at an angle below
horizontal.
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 3.0 m − ( v0 sin15° ) t1 − 12 gt12
2
= 3.0 m − ( 6.0 m/s ) (sin15°)t1 − 12 ( 9.8 m/s 2 ) t12
⇒ 4.9t12 + 1.55t1 − 3.0 = 0 ⇒ t1 = 0.6399 s and − 0.956 s (unphysical)
Substituting this value of t1 in the x-equation gives the distance
d = x1 = ( 6.0 m/s ) cos15° ( 0.6399 s ) = 3.71 m
4.51.
Model: We will assume a particle model for the cannonball, and apply the constant-acceleration
kinematic equations.
Visualize:
Solve:
(a) The cannonball that was accidentally dropped can be used to find the height of the wall:
2
y1A = y0A + ( v0A ) y ( t1A − t0A ) + 12 ( aA ) y ( t1A − t0A ) ⇒ 0 m = y0A + 0 m − 12 gt1A
2
⇒ y0A = 12 ( 9.8 m/s 2 ) (1.5 s ) = 11.03 m
2
For the cannonball that was shot:
(v0S ) x = v0S cos30° = (50 m/s)cos30° = 43.30 m/s
(v0S ) y = v0s sin 30° = (50 m/s)sin 30° = 25.0 m/s
We can now find the time it takes the cannonball to hit the ground:
y2S = y0S + (v0S ) y (t2S − t0S ) + 12 (aS ) y (t2S − t0S ) 2
⇒ 0 m = (11.03 m) + (25.0 m/s)t2S −
(9.8 m/s 2 ) 2
t2S
2
2
⇒ (4.9 m/s 2 )t2S
− (25.0 m/s)t2S − (11.03 m) = 0 ⇒ t2S = 5.51 s
There is also an unphysical root t2S = −0.41 s. Using this time t2S , we can now find the horizontal distance from
the wall as follows:
x2s = x0s + (v0s ) x (t2s − t0s ) = 0 m + (43.30 m/s)(5.51 s) = 239 m
The cannonball hits the ground 2.4 × 10 2 m from the castle wall.
(b) At the top of the trajectory (v1S ) y = 0 m/s. Using (v1S ) 2y = (v0S ) 2y − 2 g ( y1S − y0S ),
0 m2 /s 2 = (25.0 m/s)2 − 2(9.8 m/s 2 )( y1s − 11.03 m) ⇒ y1s = 42.9 m
The maximum height above the ground is 43 m.
Assess: In view of the fact that the cannonball has a speed of approximately 110 mph, a distance of 239 m for
the cannonball to hit the ground is reasonable.
4.52.
Model:
Visualize:
Solve:
We will use the particle model and the constant-acceleration kinematic equations for the car.
(a) The initial velocity is
v0 x = v0 cosθ = (20 m/s)cos 20° = 18.79 m/s
v0 y = v0 sin θ = (20 m/s)sin 20° = 6.840 m/s
Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,
0 m = 30 m + (6.840 m/s)(t1 − 0 s) + 12 (−9.8 m/s 2 )(t1 − 0 s)2 ⇒ 4.9t12 − 6.840t1 − 30 = 0
The positive root to this equation is t1 = 3.269 s. The negative root is physically unreasonable in the present case.
Using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 )2 , we get
x1 = 0 m + (18.79 m/s)(3.269 s − 0 s) + 0 = 61.4 m
The car lands 61 m from the base of the cliff.
(b) The components of the final velocity are v1x = v0 x = 18.79 m/s and
v1 y = v0 y + a y (t1 − t0 ) = 6.840 m/s − (9.8 m/s 2 )(3.269 s − 0 s) = −25.2 m/s
⇒ v = (18.79 m/s) + (−25.2 m/s)2 = 31.4 m/s
The car’s impact speed is 31 m/s.
Assess: A car traveling at 45 mph and being driven off a 30-m high cliff will land at a distance of
approximately 200 feet (61.4 m). This distance is reasonable.
4.53.
Model:
Visualize:
Solve:
Use the particle model for the cat and apply the constant-acceleration kinematic equations.
The relative velocity of the cat from the mouse’s reference frame is
(4.0cos30° − 1.5) m/s = 1.964 m/s = v0
Thus, v0 x = v0 = 1.964 m/s and v0 y = v0 sin 30° = (4.0 m/s)sin 30° = 2.0 m/s . The time for the cat to land on the
floor is found as follows:
y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 0 m + (2.0 m/s)t1 + 12 (−9.8 m/s 2 )t12
⇒ t1 = 0 s (trivial solution) and 0.408 s
The horizontal distance covered in time t1 is
x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (1.964 m/s)(0.408 s) = 0.802 m
That is, the cat should leap when he is 80 cm behind the mouse.
4.54.
Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the
staple gun be S′. Frame S′ moves relative to S with velocity Vx .
Solve:
The velocity vx of the parts to be stapled in frame S is +3.0 m/s. Also Vx = −1.0 m/s. Using
vx = v′x + Vx , we get
v′x = vx − Vx = +3.0 m/s − (−1.0 m/s) = 4.0 m/s
That is, the velocity of the parts in the frame of the staple gun is 4.0 m/s. In this frame 10 staples are fired per
second. That is,
4.0 m/s =
distance between two staples
distance
=
⇒ Distance = 0.4 m
1
time between two staples
s
10
Assess: Note that the staple gun is in frame S′ and we had to find the velocity of the moving parts in this frame
to solve the problem.
G
Model: If a frame S′ is in motion with velocity V relative to another frame S and has a displacement
G
G
G
G
G
R relative to S, the positions and velocities (r and v ) in S are related to the positions and velocities (r ′ and v′)
G
G
G
G G
in S ′ as r = r ′ + R and v = v′ + V . In the present case, ship A is frame S and ship B is frame S′. Both ships have
a common origin at t = 0 s. The position and velocity measurements are made in S and S′ relative to their
origins.
4.55.
Solve: (a) The velocity vectors of the two ships are:
G
G
vA = (20 mph)[cos30°iˆ − sin 30° ˆj ] = (17.32 mph)iˆ − (10.0 mph) ˆj
G
vB = (25 mph)[cos 20°iˆ + sin 20° ˆj ] = (23.49 mph)iˆ + (8.55 mph) ˆj
G
Since r = v Δt ,
G
G
K
As rA = rB + R,
G G
rA = vA (2 h) = (34.64 miles)iˆ − (20.0 miles) ˆj
G G
rB = vB (2 h) = (46.98 miles)iˆ + (17.10 miles) ˆj
G G G
R = rA − rB = (−12.34 miles)iˆ − (37.10 miles) ˆj ⇒ R = 39.1 miles
The distance between the ships two hours after they depart is 39 miles.
G
G
G
(b) Because vA = vB + V ,
G G G
V = vA − vB = −(6.17 mph)iˆ − (18.55 mph) ˆj ⇒ V = 19.5 mph
The speed of ship A as seen by ship B is 19.5 mph.
Assess: The value of the speed is reasonable.
Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S
with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for
both frames.
Solve: (a) The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he
needs to point at angle θ west of north. In frame S′, the water frame, his velocity is
G
v′ = (3.0 m/s, θ west of north) = (−3.0sin θ m/s)iˆ + (3.0cosθ m/s)ˆj
G
G G G
We can find his velocity in earth frame S from the Galilean transformation v = v′ + V , with V = (2.0 m/s)iˆ.
Thus
G
v = ((−3.0sin θ + 2.0) m/s)iˆ + (3.0cosθ m/s)ˆj
4.56.
In order to go straight north in the earth frame, the kayaker needs vx = 0. This will be true if
sin θ =
2.0
3.0
⎛ 2.0 ⎞
⇒ θ = sin −1 ⎜
⎟ = 41.8°
⎝ 3.0 ⎠
Thus he must paddle in a direction 42° west of north.
(b) His northward speed is v y = 3.0 cos(41.8°) m/s = 2.236 m/s. The time to cross is
t=
The kayaker takes 45 s to cross.
100 m
= 44.7 s
2.236 m/s
4.57. Model: Let Mike’s frame be S and Nancy’s frame be S′. Frame S′ moves relative to S with velocity
V. x is the horizontal direction and y is the vertical direction for motion. The frames S and S′ coincide at t = 0 s.
G G G
Solve: (a) According to the Galilean transformation of velocity v = v′ + V . Mike throws the ball with velocity
G
G
v = (22 m/s)cos63°iˆ + (22 m/s)sin63° ˆj, and V = (30 m/s)iˆ. Thus in Nancy’s frame
G G G
v′ = v − V = (22cos63° − 30)iˆ m/s + (22sin 63°) ˆj m/s = (−20.0iˆ + 19.6ˆj ) m/s
θ = tan −1
v′y
19.6 m/s
= tan −1
= 44.4°
20.0 m/s
v′x
The direction of the angle is 44.4° above the − x′ axis (in the second quadrant).
(b) In Nancy’s frame S′ the equation x′ = x0′ + v′x (t ′ − t0′ ) + 12 a′x (t ′ − t0′ ) 2 becomes
x′ = −(20.0 m/s)t ′
and the equation y′ = y0′ + v′y (t ′ − t0′ ) + 12 a′y (t ′ − t0′ ) 2 becomes
y′ = 0 m + (19.6 m/s)t ′ − 12 (9.8 m/s 2 )t ′2 = (19.6 m/s)t ′ − (4.9 m/s 2 )t ′2
Model: Let the earth frame be S and a frame attached to the sailboat be S′. Frame S′ moves relative to
G
S with velocity V = 8.0iˆ mph.
G G G
G
Solve: (a) From Equation 6.24, v = v′ + V , where v′ = (12.0 ) cos 45°iˆ mph + (12.0 ) sin 45° ˆj mph is the velocity
4.58.
of the wind in S′, or the apparent wind velocity. Thus,
G
v = (12.0 ) cos 45°iˆ mph + (12.0 ) sin 45° ˆj mph + ( 8.0 ) iˆ mph
(
)
= ( 8.0 + 12.0cos45° ) iˆ mph + (12.0sin45° ) ˆj mph = 16.48iˆ + 8.48ˆj mph
θ = tan −1
8.48 mph
= 27.2°
16.48 mph
The wind speed is v = 18.5 mph and the direction is from 27.2° south of west.
G
G
(b) In this case, v′ = − (12.0 mph ) cos 45°iˆ − (12.0 mph ) sin 45° ˆj and V = ( 8.0 mph ) iˆ. So,
G
v = ⎡⎣ − (12.0 mph ) cos 45° + 8.0 mph ⎤⎦ iˆ − (12.0 mph ) sin 45° ˆj = −0.485iˆ − 8.485 ˆj
θ = tan −1
8.485 mph
= 86.7°
0.485 mph
The speed is 8.5 mph and the direction is from 86.7° north of east or from 3.3° east of north.
Model: Let the ground frame be S and the car frame be S′. S′ moves relative to S with a velocity V
along the
x-direction.
G G G
G
G
Solve: The Galilean transformation of velocity is v = v′ + V where v and v′ are the velocities of the raindrops
G
in frames S and S′. While driving north, V = ( 25 m/s ) iˆ and v = −vR cosθ ˆj − vR sin θ iˆ. Thus,
G G G
v′ = v − V = ( −vR sinθ − 25 m/s ) iˆ − vR cosθ ˆj
4.59.
Since the observer in the car finds the raindrops making an angle of 38° with the vertical, we have
vR sin θ + 25 m/s
= tan 38°
vR cosθ
G
G
While driving south, V = − ( 25 m/s ) iˆ, and v = −vR cosθ ˆj − vR sin θ iˆ. Thus,
G
v′ = ( −vR sin θ + 25 m/s ) iˆ − vR cosθ ˆj
Since the observer in the car finds the raindrops falling vertically straight, we have
−vR sinθ + 25 m/s
= tan 0° = 0 ⇒ vR sin θ = 25 m/s
vR cosθ
Substituting this value of vR sinθ into the expression obtained for driving north yields:
25 m/s + 25 m/s
50 m/s
= tan 38° ⇒ vR cosθ =
= 64.0 m/s
vR cosθ
tan 38°
Therefore, we have for the velocity of the raindrops:
( vR sinθ ) + ( vR cosθ ) = ( 25 m/s ) + ( 64.0 m/s ) ⇒ vR2 = 4721( m/s ) ⇒ vR = 68.7 m/s
2
2
2
tan θ =
2
vR sin θ 25 m/s
=
⇒ θ = 21.3°
vR cosθ 64 m/s
The raindrops fall at 69 m/s while making an angle of 21° with the vertical.
2
4.60. Model: Let the ground be frame S and the wind be frame S′. S′ moves relative to S. The ground
frame S has the x-axis along the direction of east and the y-axis along the direction of north.
Visualize:
Solve:
G G G
(a) The Galilean velocity transformation is v = v′ + V , where
G
V = ( 50 mph ) cos30°iˆ + ( 50 mph ) sin 30° ˆj
G
v′ = ( 200 mph ) cosθ iˆ − ( 200 mph ) sin θ ˆj
G
G
Thus, v = ⎡⎣( 50cos30° + 200cosθ ) iˆ + ( 50sin 30° − 200sinθ ) ˆj ⎤⎦ mph. Because v should have no j-component,
50sin 30° − 200sin θ = 0 ⇒ θ = 7.18°
G
(b) The pilot must head 7.18° south of east. Substituting this value of θ in the v equation gives
600 mi
G
v = ( 242 ) iˆ mph, along the direction of east. At a speed of 242 mph, the trip takes t =
= 2.48 h
242 mph
4.61.
Solve:
Model: We will use the particle model for the test tube which is in nonuniform circular motion.
(a) The radial acceleration is
2
rev 1 min 2π rad ⎞
⎛
4
2
×
×
ar = rω 2 = ( 0.1 m ) ⎜ 4000
⎟ = 1.75 × 10 m/s
min 60 s
1 rev ⎠
⎝
(b) An object falling 1 meter has a speed calculated as follows:
v12 = v02 + 2a y ( y1 − y0 ) = 0 m + 2 ( −9.8 m/s 2 ) ( −1.0 m ) ⇒ v1 = 4.43 m/s
When this object is stopped in 1 × 10−3 s upon hitting the floor,
v2 = v1 + a y ( t2 − t1 ) ⇒ 0 m s = −4.43 m s + a y (1 × 10−3 s ) ⇒ a y = 4.4 × 103 m/s 2
This result is one-fourth of the above radial acceleration.
Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to
large accelerations when they are stopped by hard surfaces.
4.62.
Solve:
Model: We will use the particle model for the astronaut undergoing nonuniform circular motion.
(a) The initial conditions are ω 0 = 0 rad/s, θ 0 = 0 rad, t0 = 0 s, and r = 6.0 m. After 30 s,
ω1 =
1 rev 1 rev 2π rad
=
×
×
= 4.83 rad/s
1.3 s 1.3 s
rev
Using these values at t1 = 30 s,
ω1 = ω 0 + ( at r )( t1 − t0 ) = 0 + ( at r ) t1
⎛ 1 ⎞
2
⇒ at = ( 6.0 m )( 4.83 rad/s ) ⎜
⎟ = 0.97 m/s
30
s
⎝
⎠
(b) The radial acceleration is
ar = rω12 = ( 6.0 m )( 4.83 rad/s )
Assess:
2
g
= 14.3g
9.8
( m/s2 )
The above acceleration is typical of what astronauts experience during liftoff.
4.63.
Model:
Visualize:
Model the car as a particle in nonuniform circular motion.
Note that the tangential acceleration stays the same at 1.0 m/s2. As the tangential velocity increases, the radial
acceleration increases as well. After a time t1 , as the car goes through an angle θ1 − θ 0 , the total acceleration will
increase to 2.0 m/s2. Our objective is to find this angle.
Solve: Using v1 = v0 + at ( t1 − t0 ) , we get
v1 = 0 m/s + (1.0 m/s 2 ) ( t1 − 0 s ) = (1.0 m/s 2 ) t1
rv 2 (1.0 m/s ) t1
t2
⇒ ar = 21 =
= 1 ( m/s 4 )
r
120 m
120
2 2 2
2
⎡ t2
⎤
⇒ atotal = 2.0 m/s = a + a = (1.0 m/s ) + ⎢ 1 ( m/s 4 ) ⎥ ⇒ t1 = 14.4 s
120
⎣
⎦
2
2
t
2 2
2
r
We can now determine the angle θ1 using
⎛a ⎞
⎝r⎠
θ1 = θ 0 + ω 0 ( t1 − t0 ) + 12 ⎜ t ⎟ ( t1 − t0 )
1 (1.0 m/s )
2
(14.4 s ) = 0.864 rad = 49.5°
2 (120 m )
2
= 0 rad + 0 rad +
2
The car will have traveled through an angle of 50°.
4.64. Model: The earth is a rigid, rotating, and spherical body.
Visualize:
Solve:
At a latitude of θ degrees, the radius is r = Re cos θ with Re = 6400 km = 6.400 × 106 m.
(a) In Miami θ = 26°, and we have r = (6.400 × 106 m)(cos 26°) = 5.752 × 106 m . The angular velocity of the
earth is
ω=
2π
2π
=
= 7.272 × 10−5 rad/s
T
24 × 3600 s
Thus, vstudent = rω = (5.752 × 106 m)(7.272 × 10−5 rad/s) = 418 m/s.
(b) In Fairbanks θ = 65°, so r = (6.400 × 106 m)cos 65° = 2.705 × 106 m and
vstudent = rω = (2.705 × 106 m)(7.272 × 10−5 rad/s) = 197 m/s.
4.65. Model:
The satellite is a particle in uniform circular motion.
Visualize:
Solve: (a) The satellite makes one complete revolution in 24 h about the center of the earth. The radius of the
motion of the satellite is
r = 6.37 × 106 m + 3.58 × 107 m = 4.22 × 107 m
The speed of the satellite is v =
( distance traveled ) = 2π r = 3.07 × 103 m/s.
24 h
( time taken )
(b) The acceleration of the satellite is centripetal, with magnitude
v 2 ( 3.07 × 10 m/s )
= 0.223 m/s 2
ar = =
4.22 × 107 m
r
Assess: The small centripetal acceleration makes sense when realized it is for an object traveling in a circle with
radius ≈ 26,400 miles.
3
2
4.66. Model: The magnetic computer disk is a rigid rotating body.
Visualize:
Solve:
Using the rotational kinematic equation ω f = ωi + αΔt , we get
ω1 = 0 rad + (600 rad/s 2 )(0.5 s − 0 s) = 300 rad/s
ω 2 = (300 rad/s) + (0 rad/s 2 )(1.0 s − 0.5 s) = 300 rad/s
The speed of the painted dot v2 = rω 2 = (0.04 m)(300 rad/s) = 12 m/s. The number of revolutions during the time
interval t0 to t2 is
1
2
1
θ 2 = θ1 + ω1 (t2 − t1 ) + α1 (t2 − t1 ) 2
2
1
2
θ1 = θ 0 + ω0 (t1 − t0 ) + α 0 (t1 − t0 ) 2 = 0 rad + 0 rad + (600 rad/s 2 )(0.5 s − 0 s) 2 = 75 rad
⎛ 1 rev ⎞
= 75 rad + (300 rad/s)(1.0 s − 0.5 s) + 0 rad = 225 rad = (225 rad) ⎜
⎟ = 35.8 rev
⎝ 2π rad ⎠
4.67. Model: The drill is a rigid rotating body.
Visualize:
The figure shows the drill’s motion from the top.
Solve: (a)
The
kinematic
equation
ωf = ωi + α (tf − ti )
becomes,
ωi = 2400 rpm = (2400)(2π )/60 = 251.3 rad/s, tf − ti = 2.5 s − 0 s = 2.5 s, and wf = 0 rad/s,
0 rad = 251.3 rad/s + α (2.5 s) ⇒ α = −100 rad/s 2
(b) Applying the kinematic equation for angular position yields:
1
2
θ f = θ i + ωi (tf − ti ) + α (tf − ti ) 2
1
= 0 rad + (251.3 rad/s)(2.5 s − 0 s) + (−100 rad/s2 )(2.5 s − 0 s)2
2
= 3.2 × 102 rad = 50 rev
after
using
4.68. Model: The turbine is a rigid rotating body.
The known values are ωi = 3600 rpm = (3600)(2π )/60 = 120π rad/s, ti = 0 s, tf = 10 min = 600 s,
ω f = 0 rad/s, and θ i = 0 rad Using the rotational kinematic equation ω f = ωi + α (tf − ti ), we get
Solve:
0 rad = (120π rad/s) + α (600 s − 0 s). Thus, α = −0.628 rad/s 2 . Now,
1
2
θ f = θ i + ωi (tf − ti ) + α (tf − ti ) 2
1
= 0 rad + (120π rad/s)(600 s − 0 s) + (−0.628 rad/s 2 )(600 s − 0 s) 2
2
= 113,100 rad = 1.80 × 104 rev
Assess:
18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable.
4.69. Visualize: Please refer to Figure P4.69.
Solve: Since ω f = ωi + (area under α vs t curve), at t = 3 s, the angular velocity is
ω f = 60 rpm + ( 4.0 rad/s 2 ) ( 2 s − 1 s )
⎛ 20 rev 60 s ⎞
= 60 rpm + ( 4 rad/s ) ⎜
×
⎟
⎝ 2π rad 1 min ⎠
= 60 rpm + 38 rpm = 98 rpm
4.69. Visualize: Please refer to Figure P4.69.
Since ω f = ωi + (area under α vs t curve), at t = 3 s, the angular velocity is
Solve:
ω f = 60 rpm + ( 4.0 rad/s 2 ) ( 2 s − 1 s )
⎛ 20 rev 60 s ⎞
= 60 rpm + ( 4 rad/s ) ⎜
×
⎟
⎝ 2π rad 1 min ⎠
= 60 rpm + 38 rpm = 98 rpm
4.70. Model: The car’s wheels are in nonuniform circular motion with constant angular acceleration.
Solve: The problem must be solved in two steps: first, the angular acceleration α is found for the case of
stopping in one revolution, then α is applied to the case of stopping with twice the initial angular velocity ωi .
To find α for the case of stopping in 1.0 revolutions, set Δθ = 1.0 rev = 2π rad in ω f2 = ω i2 + 2αΔθ ⇒
⎛ ω i2 ⎞
0 m = ωi2 + 2α (2π rad) ⇒ α = − ⎜
⎟
⎝ 4π rad ⎠
Now use the same relationship, with that value of α , but this time Δθ is unknown and the initial angular
velocity has been doubled.
⎛ −ωi2 ⎞
ωi2
2
2
Δθ ⇒ Δθ = 8π rad = 4.0 rev
0 m = ( 2ωi ) + 2 ⎜
⎟ Δθ ⇒ 0 m = 4ω i −
2π rad
⎝ 4π rad ⎠
Assess: Doubling the initial velocity quadruples the number of turns the wheels make before locking. This is
within reason.
4.71. Model: The bicycle wheel undergoes nonuniform circular motion with constant angular acceleration.
Visualize:
Solve:
First find the angular acceleration α , then use it to find θ f . Using kinematics,
ω f = ωi + αΔt ⇒ 0 rpm = 100 rpm + α (1.0 min) ⇒ α = −100 (rpm)/min = −100 rev/min 2
The minus sign indicates the wheel is slowing down.
The total number of revolutions the wheel makes while stopping is
θ f = 0 rev + (100 rpm)(1.0 min) + 12 (−100 rev/min 2 )(1.0 min) 2 = 50 rev
Assess: A total of 50 revolutions in 60 s is on average less than one revolution per second, which is quite
reasonable.
4.72. Model: Treat the rock as a particle in nonuniform circular motion with constant angular acceleration.
Visualize:
Solve:
(a) The angular acceleration α is
α=
at −1.00 m/s 2
=
= −1.667 rad/s 2
r
0.600 m
The minus sign indicates the angular acceleration is clockwise, as shown in the figure above, opposite to the
3.00 m/s
angular velocity. The initial angular velocity is ωi =
= 5.00 rad/s. Using angular kinematics, the
0.600 m
angular velocity at some time t after braking starts is
ωf ( t ) = ωi + 0αΔt = 5.00 rad/s − 1.667 rad/s ( t − 0 s )
At t = 1.5 s, ω f = 2.50 rad/s while α = −1.667 rad/s.
(b) The total acceleration of the rock is a = at2 + ar2 . The value of at = 1.00 m/s 2 is given. The radial
acceleration at a certain time t is ar =
multiplying through by r = 0.600 m:
vt2
. Since vt = ω r , we can use the expression for ω f (t ) to find vt (t ) by
r
ω f ( t ) r = ( 5.00 rad/s ) r − (1.667 rad/s ) rt ⇒ vt ( t ) = 3.00 m/s − (1.00m/s 2 ) t ⇒ t =
The value of vt for a = g is needed. Setting a = g and substituting ar =
3.00 m/s − vt
1.00 m/s 2
vt2
,
r
2
⎛ v2 ⎞
v4
g 2 = at2 + ⎜ t ⎟ ⇒ g 2 − at2 = t2 = ( 9.80 m/s 2 ) − (1.00 m/s 2 ) = 95.0 m 2 /s 4
r
⎝ r ⎠
⇒ vt2 = ( 0.600 m ) 95.0 m 2 /s 4
⇒ vt = 2.42 m/s
3.00 m/s − ( 2.42 m/s )
= 0.58 s.
1.00 m/s 2
Assess: The time of 0.58 s occurs early during the braking. This is reasonable since an acceleration equal to
that supplied by gravity is fairly strong.
Finally, t =
4.73. Model: The string is wrapped around the spool in such a way that it does not pile up on itself, and
unwinds without slipping.
Visualize:
Solve:
Since the string unwinds without slipping, the angular distance the spool turns as the string is pulled 1.0
Δx
1.0 m
m is Δθ =
=
= 33 radians.
r 3.0 × 10−2 m
The angular acceleration of the spool due to the pull on the string is
α=
at
1.5 m/s 2
=
= 50 rad/s 2
r 3.0 × 10−2 m
The angular velocity of the spool after pulling the string is found with kinematics.
ω f2 = ωi2 + 2αΔθ ⇒ ω f2 = 0 rad 2 /s 2 + 2 ( 50 rad/s 2 ) ( 33 rad )
⇒ ω f2 = 57 rad/s
Converting to revolutions per minute,
( 57 rad/s ) ⎛⎜
rev ⎞⎛ 60 s ⎞
2
⎟⎜
⎟ = 5.5 × 10 rpm
π
2
rad
min
⎝
⎠⎝
⎠
Assess: The angular speed of 57 rad/s ≈ 9 rev/s is reasonable for a medium-sized spool.
4.74.
Solve: (a) A golfer hits an iron shot with a new club as she approaches the green. She is pretty sure,
based on past experience, that she hit the ball with a speed of 50 m/s, but she is not sure at what angle the golf
ball took flight. She observed that the ball traveled 100 m before hitting the ground. What angle did she hit the
ball?
(b) From the second equation,
(4.9 m/s 2 )t12 − (50sinθ m/s) t1 = 0 ⇒ t1 = 0 s and t1 =
(50 m/s)sinθ
4.9 m/s 2
Using the above value for t1 in the first equation yields:
100 m =
(50cosθ )(50sinθ ) m 2 /s 2
4.9 m/s 2
⇒ 2cosθ sinθ = sin 2θ =
Assess:
9.8
= 0.392 ⇒ 2θ = 23.1 ⇒ θ = 11.5°
25
Although the original speed is reasonably high (50 m/s = 112 mph), the ball travels a distance of only
100 m, implying either a small launch angle around 10° or an angle closer to 80°. The calculated angle of 11.5° is
thus pretty reasonable.
4.75.
Solve: (a) A submarine moving east at 3.0 m/s sees an enemy ship 100 m north of its path. The
submarine’s torpedo tube happens to be stuck in a position pointing 45° west of north. The tube fires a torpedo
with a speed of 6.0 m/s relative to the submarine. How far east or west of the ship should the sub be when it
fires?
(b) Relative to the water, the torpedo will have velocity components
vx = −6.0cos 45° m/s + 3.0 m/s = −4.24 m/s + 3 m/s = −1.24 m/s
v y = +6.0cos 45° m/s = +4.2 m/s
The time to travel north to the ship is
100 m = ( 4.2 m/s ) t1 ⇒ t1 = 24 s
Thus, x = (1.24 m/s )( 24 s ) = −30 m. That is, the ship should be 30 m west of the submarine.
4.76.
Solve: (a) A 1000 kg race car enters a 50 m radius curve and accelerates around the curve for 10.0 s.
The forward force provided by the car’s wheels is 1500 N. After 10.0 s the car has moved 125 m around the
track. Find the initial and final angular velocities.
(b) From Newton’s second law,
Ft = mat ⇒ 1500 N = (1000 kg ) at ⇒ at = 1.5 m/s 2
θ f = θ i + ω it +
Δθ =
Δs 125 m
=
= 2.5 rad
r
50 m
at 2
1.5 m/s 2
2
t ⇒ 2.5 rad = 0 rad + ωi (10 s ) +
(10 s ) ⇒ ωi = 0.10 rad/s
2r
2 ( 50 m )
ωf = ωi +
at
1.5 m/s 2
t = 0.1 rad/s +
(10 s ) = 0.40 rad/s
r
50 m
4.77.
Solve: You decide to test fly your model airplane off of a 125 m tall building. The model’s engine
starts fine and gets the airplane moving at 4.0 m/s but quits just as it gets to the edge of the building. The model
proceeds to fall “like a rock.” How far from the edge of the building will it crash into the ground? (Assume
g = 10 m/s 2 for easier calculation.)
Visualize:
Using the equation y1 = y0 + ν 0 yt + 12 a y ( t0 − t1 ) , we get
2
y1 = −(5 m/s 2 ) t12 = −125 m ⇒ t1 =
The distance x1 = ( 4 m/s )( 5 s ) = 20 m.
125 m
= 25 s 2 = 5 s
5 m/s 2
4.78.
Model: The ions are particles that move in a plane. They have vertical acceleration while between the
acceleration plates, and they move with constant velocity from the plates to the tumor. The flight time will be so
small, because of the large speeds, that we’ll ignore any deflection due to gravity.
Visualize:
Solve: There’s never a horizontal acceleration, so the horizontal motion is constant velocity motion at
vx = 5.0 × 106 m/s. The times to pass between the 5.0-cm-long acceleration plates and from the plates to the
tumor are
0.050 m
= 1.00 × 10−8 s
5.0 × 106 m/s
1.50 m
t2 − t1 =
= 3.00 × 10−7 s
5.0 × 106 m/s
t1 − t0 = t1 =
Upon leaving the acceleration plates, the ion has been deflected sideways to position y1 and has velocity v1 y .
These are
y1 = y0 + v0 y t1 + 12 a y t12 = 12 a y t12
v1 y = v0 y + a yt1 = a yt1
In traveling from the plates to the tumor, with no vertical acceleration, the ion reaches position
y2 = y1 + v1 y (t2 − t1 ) = 12 a yt12 + (a yt1 )(t2 − t1 ) = ( 12 t12 + t1 (t2 − t1 ) ) a y
We know y2 = 2.0 cm = 0.020 m, so we can solve for the acceleration a y that the ion had while between the
plates:
y2
0.020 m
=1
= 6.6 × 1012 m/s2
2
−8
s) + (1.00 × 10−8 s)(3.00 × 10−7 s)
2 t1 + t1 (t 2 − t1 )
2 (1.00 × 10
ay = 1
2
Assess: This acceleration is roughly 1012 times larger than the acceleration due to gravity. This justifies our
assumption that the acceleration due to gravity can be neglected.
4.79.
Model: We will use the particle model for the ball’s motion under constant-acceleration kinematic
equations. Note that the ball’s motion on the smooth, flat board is a y = − g sin 20° = −3.352 m/s 2 .
Visualize:
Solve:
The ball’s initial velocity is
v0 x = v0 cosθ = ( 3.0 m/s ) cosθ
v0 y = v0 sinθ = ( 3.0 m/s ) sinθ
Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ,
2
2.5 m = 0 m + ( 3.0 m/s ) cosθ ( t1 − 0 s ) + 0 m ⇒ t1 =
( 2.5 m ) = 0.833 s
( 3.0 m/s ) cosθ cosθ
Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) and the above equation for t1 ,
2
⎛ 0.833 s ⎞ 1
2 ( 0.833 s )
0 m = 0 m + ( 3.0 m/s ) sinθ ⎜
⎟ − 2 ( 3.352 m/s )
cos 2 θ
⎝ cosθ ⎠
⇒ ( 2.5 m )
2
sinθ 1.164
=
⇒ 2.5sin θ cosθ = 1.164 ⇒ 2θ = 68.6° ⇒ θ = 34.3°
cosθ cos 2 θ
4.80.
Model:
Visualize:
Solve:
Use the particle model for the arrow and the constant-acceleration kinematic equations.
Using v1 y = v0 y + a y ( t1 − t0 ) , we get
v1 y = 0 m/s − gt1 ⇒ v1 y = − gt1
Also using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ,
2
60 m = 0 m + v0 xt1 + 0 m ⇒ v0 x =
60 m
= v1x
t1
Since v1 y / v1x = − tan 3° = − 0.0524, using the components of v0 gives
− gt1
( 60 m/ t1 )
= − 0.0524 ⇒ t1 =
( 0.0524 )( 60 m ) = 0.566 s
( 9.8 m/s )
2
Having found t1 , we can go back to the x-equation to obtain v0 x = 60 m/0.566 s = 106 m/s.
Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of
106 m/s or 237 mph for the arrow is understandable.
4.81.
Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. We
will assume that the archer shoots from 1.75 m above the slope (about 5′ 9′′).
Visualize:
Solve:
For the y-motion:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ y1 = 1.75 m + ( v0 sin 20° ) t1 − 12 gt12
2
⇒ y1 = 1.75 m + ( 50 m/s ) sin 20°t1 − 12 gt12
For the x-motion:
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( v0 cos 20° ) t1 + 0 m = ( 50 m/s ) (cos 20°)t1
2
Because y1 x1 = − tan15° = − 0.268,
1.75 m + ( 50 m/s ) (sin 20°)t1 − 12 gt12
= − 0.268 ⇒ t1 = 6.12 s and −0.058 s (unphysical)
( 50 m/s ) (cos 20°)t1
Using t1 = 6.12 s in the x- and y-equations above, we get y1 = −77.0 m and x1 = 287 m. This means the distance
down the slope is
x12 + y12 =
( 287 m ) + ( −77.0 m ) = 297 m.
2
2
Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15° slope at an angle of
20° above the horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable
numbers.
4.82.
Model:
Visualize:
Treat the ball as a particle and apply the constant-acceleration equations of kinematics.
Solve: After the first bounce, the ball leaves the surface at 40° relative to the vertical or 50° relative to the
horizontal. We first calculate the time t1 between the second bounce and the first bounce as follows:
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ 3.0 m = 0 m + ( v0 cos50° ) t1 + 0 m ⇒ t1 =
2
3.0 m
v0 cos50°
In this time, the ball undergoes a vertical displacement of y1 − y0 = −(3.0 m) tan 20° = −1.092 m. Substituting
these values in the equation for the vertical displacement yields:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )
2
⎛ 3.0 m ⎞ 1
3.0 m ⎞
2 ⎛
−1.092 m = 0 m + ( v0 sin 50° ) t1 − 12 gt12 = ( v0 sin 50° ) ⎜
⎟ − 2 ( 9.8 m/s ) ⎜
⎟
⎝ v0 cos50° ⎠
⎝ v0 cos50° ⎠
⇒ −1.092 m − 3.575 m =
Assess:
−106.73 m3 /s 2
⇒ v0 = 4.78 m/s, or v0 = 4.8 m/s
v02
A speed of 4.8 m/s or 10.7 mph on the first bounce is reasonable.
2
4.83.
Model: Treat the skateboarder as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The skateboarder’s final velocity at the top of the ramp is her initial velocity as she
becomes airborne.
Solve: Without friction, the skateboarder’s acceleration on the ramp is a0 = − g sin 30° = −4.90 m/s 2 . The length
of the ramp is s1 = (1.0 m) / sin 30° = 2.0 m. We can use kinematics to find her speed at the top of the ramp:
v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1
⇒ v1 = (7.0 m/s)2 + 2(−4.90 m/s 2) (2.0 m) = 5.4 m/s
This is the skateboarder’s initial speed into the air, giving her velocity components v1x = v1 cos30° = 4.7 m/s and
v1 y = v1 cos30° = 2.7 m/s. We can use the y-equation of projectile motion to find her time in the air:
y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 1.0 m + (2.7 m/s) t2 − (4.90 m/s 2) t22
This quadratic equation has roots t2 = −0.253 s (unphysical) and t2 = 0.805 s. The x-equation of motion is thus
x2 = x1 + v1xt2 = 0 m + (4.7 m/s) t2 = 3.8 m
She touches down 3.8 m from the end of the ramp.
4.84.
Model:
motion.
Visualize:
Solve:
Use the particle model for the motorcycle daredevil and apply the kinematic equations of
We need to find the coordinates of the landing ramp ( x1 , y1 ). We have
x1 = x0 + v0 x ( t1 − t0 ) = 0 m + ( 40 m/s ) t1
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) = 0 m + 0 m + 12 ( −9.8 m/s 2 ) t12 = − ( 4.9 m/s 2 ) t12
2
G
This means we must find t1. Since v1 makes an angle of 20° below the horizontal we can find v1y as follows:
v1 y
v1x
= tan 20° ⇒ v1 y = −v1x tan 20° = − ( 40 m/s ) tan 20° = −14.56 m/s
We now use this value of v1y , a y = −9.8 m/s 2 , and v0y = 0 m/s in the following equation to obtain t1:
v1 y = v0 y + a y ( t1 − t0 ) ⇒ −14.56 m/s = 0 m/s − ( 9.8 m/s 2 ) t1 ⇒ t1 = 1.486 s
Now, we are able to obtain x1 and y1 using the above x- and y-equations:
x1 = ( 40 m/s )(1.486 s ) = 59.4 m
y1 = − ( 4.9 m/s 2 ) (1.486 s ) = −10.82 m
2
That is, the landing ramp should be placed 10.8 m lower and 59 m away from the edge of the horizontal
platform.
4.85. Model: The train and projectile are treated in the particle model. The height of the cannon above the
tracks is ignored.
Visualize:
Solve:
In the ground reference frame, the projectile is launched with velocity components
( v0 x )P = v0 cosθ + vtrain
( v ) = v sinθ
0y P
0
While the projectile is in free fall, vfy = viy − g Δt. The time for the projectile to rise to the highest point is
(with vfy = 0) Δt =
viy
g
. So the time to vertically rise and fall is
t1 = 2Δt = 2
( v ) = 2v sinθ
0y P
g
0
g
During this time the projectile travels a horizontal distance
( x1 )P = ( v0 x )P t1 =
2v0 2
2v v
sinθ cosθ + 0 train sinθ
g
g
During the same time, the train travels a horizontal distance
( x1 )T = ( v0 x )T t1 +
1 2 2v0vtrain
2v 2 a
at1 =
sin θ + 02 sin 2 θ
2
g
g
The range R is the difference between the two horizontal distances:
R = ( x1 )P − ( x1 )T =
2v0 2 ⎛
a 2 ⎞
⎜ sin θ cosθ − sin θ ⎟
g ⎝
g
⎠
Note that the range is independent of vtrain the train’s steady motion. This makes sense, since the train and
projectile share that motion when the projectile is launched.
2v 2
dR
= 0. Thus (ignoring the constant 0 )
Maximizing the range R requires
g
dθ
dR
a
a
= cos 2 θ − sin 2 θ − ( 2sinθ cosθ ) = cos 2θ − sin 2θ = 0
dθ
g
g
Solving for θ ,
sin 2θ
g
1
⎛g⎞
= tan 2θ = ⇒ θ = tan −1 ⎜ ⎟
a
cos 2θ
2
⎝a⎠
Note that
a > 0 (train speeding up) gives θ < 45D
and
a < 0 (train slowing down) gives θ > 45D
⎛g⎞
since tan −1 ⎜ ⎟ will be in the 2nd quadrant.
⎝a⎠
Assess: As a check, see what the angle θ is for the limiting case in which the train does not accelerate:
1
1
a = 0 ⇒ θ = tan −1 ( ∞ ) = × 90° = 45°
2
2
This is the expected answer.
4.86.
Model: Let the earth be frame S and the river be frame S′. Assume the river flows toward the east,
which is the x and x′-axis.
Visualize:
Solve:
In frame S′, the river’s frame, the child is at rest. The boat can go directly to the child at angle
θ = tan (200 /1500) = 7.595°. The boat’s speed is 8.0 m/s, so the components of the boat’s velocity in S′ are
−1
v′x = −(8.0 m/s)cos7.595° = −7.93 m/s
v′y = (8.0 m/s)sin 7.595° = 1.06 m/s
→
The river flows with velocity V = 2.0iˆ m/s relative to the earth. In the earth’s frame, which is also the frame of
the riverbank and the boat dock, the boat’s velocity is
vx = v′x = Vx = −5.93 m/s and v y = v′y + Vy = 1.06 m/s
Thus the boat’s angle with respect to the riverbank is θ = tan −1 (5.93/1.06) = 10.1°.
Assess: The boat, like the child, is being swept downstream. This moves the boat’s angle away from the shore.
4.87. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the
Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S.
G
Frame S′ moves relative to frame S with velocity V .
Visualize:
G G G
G
According to the Galilean transformation of velocity v = v′ + V , where v is the velocity of Val’s plane
G
G
relative to the earth, v′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane
relative to the earth. We have
G
v = − ( 500 mph ) cos30°iˆ + ( 500 mph ) sin 30° ˆj
Solve:
G
V = − ( 500 mph ) cos 20°iˆ − ( 500 mph ) sin 20° ˆj
G G G
v′ = v − V = ( 36.8 mph ) iˆ + ( 421 mph ) ˆj
⎛ 421 ⎞
⎟ = 85°
⎝ 36.8 ⎠
θ = tan −1 ⎜
The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east.
Thus the angle between the fuselage and the direction of motion is
φ = 180° − 30° − 85° = 65°
4.87.
Model: The airplanes are modeled as particles, and are undergoing relative motion according to the
Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S.
G
Frame S′ moves relative to frame S with velocity V .
Visualize:
G G G
G
According to the Galilean transformation of velocity v = v′ + V , where v is the velocity of Val’s plane
G
G
relative to the earth, v′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane
relative to the earth. We have
G
v = − ( 500 mph ) cos30°iˆ + ( 500 mph ) sin 30° ˆj
Solve:
G
V = − ( 500 mph ) cos 20°iˆ − ( 500 mph ) sin 20° ˆj
G G G
v′ = v − V = ( 36.8 mph ) iˆ + ( 421 mph ) ˆj
⎛ 421 ⎞
⎟ = 85°
⎝ 36.8 ⎠
θ = tan −1 ⎜
The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east.
Thus the angle between the fuselage and the direction of motion is
4.88. Model: The ball is a particle launched into projectile motion by the wheel.
Visualize:
Solve: The initial velocity of the projectile is the tangential velocity at the point of release, and the direction is tangential
to the wheel. The strategy is to find the initial velocity vi = ω f r , then use ω f to find the required angular acceleration α .
Since the release point is Δθ =
11
1
of a complete circle, ϕ = ( 2π rad ) = 30° remains. In the coordinate system of the
12
12
figure,
xi = −20sin30°cm = 10.0 cm
yi = 20cos30°cm = 17.3 cm
xf = 100 cm
yf = 0
ay = − g
vi x = vi cosθ
viy = vi sinθ
ax = 0
The launch angle ϕ is identified by calculating the angles in the right triangle highlighted in the figure and realizing the
initial velocity is tangential to the circle. Thus the launch angle is also ϕ .
We will use Equations 4.17 for the position of a projectile as a function of time, and use the fact that the time to travel the
horizontal distance to the cup is the same for the vertical motion. First, find the flight time required for the ball to hit the
cup from the horizontal motion:
100 cm = −10.0 cm + ( vi cos30° ) t ⇒ t =
110 cm 127 cm
=
vi cos30°
vi
Substitute this time in the equations for the vertical motion:
⎛ 127 cm ⎞ 1 ⎛ 127 cm ⎞
0 cm = 17.3 cm + vi sin 30° ⎜
⎟− g⎜
⎟
⎝ vi ⎠ 2 ⎝ vi ⎠
Solving for the required initial velocity, v =
2
(127 cm ) g . The centimeter units cancel, and v = 2.78 m/s.
i
2 ( 80.8 cm )
Now it is time to consider the angular motion. The required final angular velocity
v
r
ωf = i =
2.78 m/s
= 13.9 rad/s
0.20 m
Using ω f2 = ωi2 + 2αΔθ ,
ω f2 − ωi2
(13.9 rad/s ) = 16.8 rad/s 2
=
⎛ 11
⎞
2 Δθ
2 ⎜ ( 2π ) rad ⎟
2
α=
⎝ 12
Assess:
⎠
An angular acceleration of 16.8 rad/s 2 ≈ 2.7 rev/s 2 seems reasonable for a spring-loaded wheel.
4-1
Kinematics in Two Dimensions
4-2
5.1. Visualize:
Assess:
walls.
Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse
5.2. Visualize:
5.3. Visualize:
5.4. Model: Assume friction is negligible compared to other forces.
Visualize:
5.5. Visualize:
Assess: The bow and archer are no longer touching the arrow, so do not apply any forces after the arrow is
released.
5.6. Model: An object’s acceleration is linearly proportional to the net force.
Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F ∝ a
and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2, four rubber bands will produce an
acceleration of 2.4 m/s 2 .
(b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F = ma,
2 F = (2m) a ⇒ a = F/m = 0.6 m/s 2
5.7. Solve: Let the object have mass m and each rubber band exert a force F. For two rubber bands to
accelerate the object with acceleration a, we must have a =
acceleration 3a to a mass
2F
. We will need N rubber bands to give
m
1
m. Find N:
2
3a =
Three rubber bands are required.
NF
⎛ 2 F ⎞ 2 NF
⇒ 3⎜
⇒ N = 3.
⎟=
m
m
⎝ m ⎠
2
1
5.8. Visualize: Please refer to Figure EX5.8.
Solve:
Mass is defined to be
m=
1
slope of the acceleration-versus-force graph
A larger slope implies a smaller mass. We know m2 = 0.20 kg, and we can find the other masses relative to m2
by comparing their slopes. Thus
m1 1/slope 1 slope 2
1
2
=
=
=
= = 0.40
m2 1/slope 2 slope 1 5 2 5
⇒ m1 = 0.40m2 = 0.40 × 0.20 kg = 0.08 kg
Similarly,
m3 1/slope 3 slope 2
1
5
=
=
=
= = 2.50
m2 1/slope 2 slope 3 2 5 2
⇒ m3 = 2.50m2 = 2.50 × 0.20 kg = 0.50 kg
Assess: From the initial analysis of the slopes we had expected m3 > m2 and m1 < m2 . This is consistent with
our numerical answers.
5.9. Visualize:
Solve:
Please refer to Figure EX5.9.
Mass is defined to be
m=
1
slope of the acceleration-versus-force graph
Thus
−1
⎛ 5a1 ⎞
3
⎜
⎟
9
m1 ( slope of line 1)
3N ⎠
⎝
=
=
=5=
m2 ( slope of line 2 )−1 ⎛ 3a1 ⎞−1 5 25
⎜
⎟
3
⎝ 5N ⎠
−1
The ratio of masses is
Assess:
m1 9
=
m2 25
More rubber bands produce a smaller acceleration on object 2, so it should be more massive.
5.10
Solve:
Use proportional reasoning. Given that distance traveled is proportional to the square of the
d
time, d ∝ t , so 2 should be constant. We have
t
2.0 furlongs
x
=
2
2
( 2.0 s )
( 4.0 s )
Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs.
Assess: A longer time should result in a longer distance traveled.
2
5.11 Solve: Use proportional reasoning. Let T = period of the pendulum, L = length of pendulum. We are
T
should be constant. We have
L
3.0 s
x
=
2.0 m
3.0 m
Solving, the period of the 3.0 m long pendulum is x = 3.7 s.
Assess: Increasing the length increases the period, as expected.
told T ∝ L , so
5.12. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when
the net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes
acceleration. That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity
G
that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force Fnet on an
object of mass m causes the object to undergo an acceleration:
G
G F
a = net
m
The acceleration vector and the net force vector must point in the same direction.
5.13. Visualize:
Solve:
(a) Newton’s second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg) a, hence a = 4 m/s 2 .
(b) When F = 1 N, we have 1 N = (0.5 kg) a, hence a = 2 m/s 2 .
After repeating this procedure at various points, the above graph is obtained.
5.14 Solve: Newton’s second law tells us that F = ma. Compute F for each case:
(a) F = (0.200 kg)(5 m/s 2 ) = 1N .
(b) F = (0.200 kg)(10 m/s 2 ) = 2 N .
Assess: To double the acceleration we must double the force, as expected.
5.15. Visualize: Please refer to Figure EX5.15.
Solve: Newton’s second law is F = ma. We can read a force and an acceleration from the graph, and hence
find the mass. Choosing the force F = 1 N gives us a = 4 m/s 2 . Newton’s second law yields m = 0.25 kg.
5.16. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3
gives us no information on laptops, but does give the weight of a one-pound object. Place a pound weight in one
hand and the laptop on the other. The sensation on your hand is the weight of the object. The sensation from the
laptop is about five times the sensation from the pound weight. So we conclude the weight of the laptop is about
five times the weight of the one-pound object or about 25 N.
(b) According to Table 5.3, the propulsion force on a car is 5000 N. A bicycle (including the rider) is about 100
kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of
a bicycle is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as
follows:
F ( bicycle ) =
1
1
5000 N
= 100 N
( mass of car ) × ( acceleration of car ) =
10
5
50
So we would roughly estimate the propulsion force of a bicycle to be 100 N.
5.17. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3
gives us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand
and a pencil on the other hand. The sensation on your hand is the weight of the object. The sensation from the
quarter is about the same as the sensation from the pencil. So they both have about the same weight. We can
estimate the weight of the pencil to be 0.05 N.
(b) According to Table 5.3, the propulsion force on a car is 5000 N. The mass of a sprinter is about 100 kg. This
is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a
sprinter is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as
follows:
F ( sprinter ) =
1
1
5000 N
= 100 N
( mass of car ) × ( acceleration of car ) =
10
5
50
So, we would roughly estimate the propulsion force of a sprinter to be 100 N.
Assess: This is the same estimated number as we obtained in Exercise 5.16. This is reasonable since in both the
cases the propulsion force comes from a human and it probably does not matter how the human is providing that
force.
5.18. Visualize:
G
G G
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite
direction so that the sum of all the three forces is zero.
5.19. Visualize:
G
G G
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite
direction so that the sum of all three forces is zero.
5.20. Visualize:
G
G G
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite
direction so that the sum of all the three forces is zero.
5.21. Visualize:
Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force
directed down is labeled as a gravitational force, and the force directed up is labeled as a tension. With zero net
force the acceleration is zero. So, a possible description is: “An object hangs from a rope and is at rest.” Or, “An
object hanging from a rope is moving up or down with a constant speed.”
5.22. Visualize:
Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward.
G
G
G
There is a force labeled FG directed down, a force Fthrust directed up, and a force D directed down. So a possible
description is: “A rocket accelerates upward.”
5.23. Visualize:
G
The free-body diagram shows three forces. There is a gravitational force FG , which is down. There is a
G
G
G
normal force labeled n, which is up. The forces FG and n are shown with vectors of the same length so they
are equal in magnitude and the net
G vertical force is zero. So we have an object on the ground which is not moving
vertically. There is also a force f k to the left. This must be a frictional force and we need to decide whether it is
G
static or kinetic friction. The frictional force is the only horizontal force so the net horizontal force must be f k .
This means there is a net force to the left producing an acceleration to the left. This all implies motion and
therefore the frictional force is kinetic. A possible description is: “A baseball player is sliding into second base.”
Solve:
5.24. Visualize:
Assess:
you.
G
G
There is a gravitational force FG . You are touching the park bench, so it exerts a contact force n on
5.25. Visualize:
Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not include
either of these forces. The only remaining forces are the weight and the normal force.
5.26. Visualize:
Assess:
Since the velocity is constant, the acceleration is zero, and the net force is zero.
5.27. Visualize:
Assess: The problem uses the word “sliding.” Any real situation involves friction with the surface. Since we
are not told to neglect it, we show that force.
5.28. Visualize:
Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with
each step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down.
G
G
Since F = ma the force is in the same direction as the acceleration and must be directed down.
Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector
decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is
also directed down. As in part (a) the net force must be directed down.
5.29. Visualize:
The velocity vector in figure (a) is shown downward and to the left. So movement is downward and to the left.
The velocity vectors get successively longer, which means the speed is increasing. Therefore the acceleration is
G
G
downward and to the left. By Newton’s second law F = ma , the net force must be in the same direction as the
acceleration. Thus, the net force is downward and to the left.
The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. The
velocity vector gets successively shorter, which means the speed is decreasing. Therefore the acceleration is
downward and to the left. From Newton’s second law, the net force must be in the direction of the acceleration
and so it is directed downward and to the left.
5.30. Visualize:
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the
value of the acceleration by the mass of the object.
5.31. Visualize:
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the
value of the acceleration by the mass of the object.
5.32. Visualize:
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing
the value of the force by the mass of the object.
5.33. Visualize:
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing
the value of the force by the mass of the object.
5.34. Model: Use the particle model for the object.
Solve:
(a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the
acceleration of the mass is 10 m/s 2 . According to Newton’s second law, F0 = m0 (10 m/s 2 ). The force then
becomes 12 F0 . Newton’s second law gives
1
1
F0 = m0 a = ⎡⎣ m0 (10 m s 2 ) ⎤⎦
2
2
This means a is 5 m/s 2 .
(b) The force is F0 and the mass is now 12 m0 . Newton’s second law gives
1
F0 = m0 a = m0 (10 m s 2 )
2
This means a = 20 m/s 2 .
(c) A similar procedure gives a = 10 m/s 2 .
(d) A similar procedure gives a = 2.5 m/s 2 .
5.35. Model: Use the particle model for the object.
Solve:
(a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the
acceleration of the mass is 8 m/s 2 . According to Newton’s second law, F0 = m0 (8 m/s 2 ). The force then becomes
2 F0 . Newton’s second law gives
2 F0 = m0 a = 2 ⎡⎣ m0 ( 8 m s 2 ) ⎤⎦
This means a is 16 m/s 2 .
(b) The force is F0 and the mass is now 2m0 . Newton’s second law gives
F0 = 2m0 a = m0 ( 8 m s 2 )
This means a = 4 m/s 2 .
(c) A similar procedure gives a = 8 m/s 2 .
(d) A similar procedure gives a = 32 m/s 2 .
5.36. Visualize:
Solve: (d) There are a normal force and a gravitational force which are equal and opposite, so this is an object
on a horizontal surface. The description could be: “A tow truck pulls a stuck car out of the mud.”
5.37. Visualize:
Solve: (d) There is a normal force and a gravitational force which are equal and opposite, so this is an object on
a horizontal surface, or at least balanced in the vertical direction. The description of this free-body diagram could
be: “A jet plane is flying at constant speed.”
5.38. Visualize:
Solve: (d) This is an object on a surface because FG = n. It must be moving to the left because the kinetic
friction is to the right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic
block to the left.”
5.39. Visualize:
Solve: (d) There is only a single force of weight. We are unable to tell the direction of motion. The description
could be: “Galileo has dropped a ball from the Leaning Tower of Pisa.”
5.40. Visualize:
Solve: (d) There is an object on an inclined surface. The net force is down the plane so the acceleration is down
the plane. The net force includes both the frictional force and the component of the gravitational force. The
direction of the force of kinetic friction implies that the object is moving upward. The description could be: “A
car is skidding up an embankment.”
5.41. Visualize:
Solve: (d) There is an object on an inclined surface with a tension force down the surface. There is a small
frictional force up the surface implying that the object is sliding down the slope. A description could be: “A sled
is being pulled down a slope with a rope that is parallel to the slope.”
5.42. Visualize:
Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no normal force so
the object is not on a surface. The description could be: “A rocket is fired at an angle to the horizontal and there
is no drag force.”
5.43. Visualize:
Tension is the only contact force. The downward acceleration implies that FG > T .
5.44. Visualize:
5.45. Visualize:
The normal force is perpendicular to the ground. The thrust force is parallel to the ground and in the direction of
acceleration. The drag force is opposite to the direction of motion.
5.46. Visualize:
The normal force is perpendicular to the hill. The frictional force is parallel to the hill.
5.47. Visualize:
The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed
upward opposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the
slope (that is, there is no acceleration away from the slope) the net force must be parallel to the slope.
5.48. Visualize:
As the rock slides there is kinetic friction between it and the rough concrete sidewalk. Since the rock stays on the
level surface, the net force must be along that surface, and is equal to the kinetic friction.
5.49. Visualize:
The drag force due to air is opposite the motion.
5.50. Visualize:
The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. There is a
force of the spring to the right which causes an acceleration.
5.51. Visualize:
There are no contact forces on the rock. The gravitational force is the only force acting on the rock.
5.52. Visualize:
The gymnast experiences the long range force of gravity. There is also a contact force from the trampoline that is
the normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline is
decreasing her speed, so the acceleration is upward and there is a net force upward. Thus the normal force must
be larger than the gravitational force. The actual behavior of the normal force over time will be complicated as it
involves the stretching of the trampoline and therefore tensions.
5.53. Visualize:
You can see from the motion diagram that the box accelerates to the right along with the truck. According to
G
G
Newton’s second law, F = ma , there must be a force to the right acting on the box. This is friction, but not
kinetic friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents
slipping. Were it not for static friction, the box would slip off the back of the truck. Static friction acts in the
direction needed to prevent slipping. In this case, friction must act in the forward (toward the right) direction.
5.54. Visualize:
You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows
G
G
down. According to Newton’s second law, F = ma , there must be a force to the left acting on the bag. This is
friction, but not kinetic friction. The bag is not sliding across the seat. Instead, it is static friction, the force that
prevents slipping. Were it not for static friction, the bag would slide off the seat as the car stops. Static friction
acts in the direction needed to prevent slipping. In this case, friction must act in the backward (toward the left)
direction.
5.55. Visualize: (a)
(b)
(c)
(d) The ball accelerates downward until the instant when it makes contact with the ground. Once it makes
contact, it begins to compress and to slow down. The compression takes a short but nonzero distance, as shown
in the motion diagram. The point of maximum compression is the turning point, where the ball has an
instantaneous speed of v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses
G
contact with the ground. The motion diagram shows that the acceleration vector a points upward the entireG time
that the ball is in contact with the ground. An upward acceleration implies that there is a net upward force Fnet on
the ball. The only two forces on the ball are the gravitational force downward and the normal force of the ground
upward. To have a net force upward requires n > FG . So the ball bounces because the normal force of the ground
exceeds the gravitational force, causing a net upward force during the entire time that the ball is in contact with
the ground. This net upward force slows the ball, turns it, and accelerates it upward until it loses contact with the
ground. Once contact with the ground is lost, the normal force vanishes and the ball is simply in free fall.
5.56. Visualize: (a)
You are sitting on a bench driving along to the right. Both you and the bench are moving with a constant speed. There is a
force on you due to gravity, which is directed down. There is a contact force between you and the bench, which is directed
up. Since you are not accelerating up or down the net vertical force on you is zero, which means the two vertical forces are
equal in magnitude. The statement of the problem gives no indication of any other contact forces. Specifically, we are told
that the bench is very slippery. We can take this to mean there is no frictional force. So our force diagram includes only the
normal force up, the gravitational force down, and no horizontal force.
(b) The above considerations lead to the free-body diagram that is shown.
(c) The car (and therefore the bench) slows down. Does this create any new force on you? No. The forces remain the same.
This means the pictorial representation and the free-body diagram are unchanged.
(d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and the road
exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remains unchanged.
There are no horizontal forces on you. You do not slow down and you continue at an unchanged velocity until something
in the picture changes for you (for example, you fall off the bench or hit the windshield).
(e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forces at all.
According to Newton’s first law if the net force on you is zero, then you continue to move in a straight line with a constant
velocity. That is what happens to you when the car slows down. You continue to move forward with a constant velocity.
The statement that you are “thrown forward” is misleading and incorrect. To be “thrown” there would need to be a net
force on you and there is none. It might be correct to say that the car has been “thrown backward” leaving you to continue
onward (until you part company with the bench).
(f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional force
between the bench and you. This force is certainly horizontal (parallel to the surface of the bench). Is the frictional force
directed forward (in the direction of motion) or backward? The car is slowing down and you are staying on the bench. That
means you are slowing down with the bench. Your velocity to the right is decreasing (you are moving right and slowing
down) so you are accelerating to the left. By Newton’s second law that means the force producing the acceleration must
be to the left. That force is the force of static friction and it is shown on the free-body diagram below. Of course, when the
car accelerates (increases in speed to the right) and you accelerate with it, then your acceleration is to the right and the
frictional force must be to the right.
5-1
Force and Motion
5-2
6.1.
Model:
Visualize:
Solve:
We can assume that the ring is a single massless particle in static equilibrium.
Written in component form, Newton’s first law is
( Fnet ) x = ΣFx = T1x + T2 x + T3 x = 0 N ( Fnet ) y = ΣFy = T1 y + T2 y + T3 y = 0 N
Evaluating the components of the force vectors from the free-body diagram:
T1x = −T1 T2 x = 0 N T3 x = T3 cos30°
T1 y = 0 N T2 y = T2 T3 y = −T3 sin 30°
Using Newton’s first law:
−T1 + T3 cos30° = 0 N T2 − T3 sin 30° = 0 N
Rearranging:
T1 = T3 cos30° = (100 N )( 0.8666 ) = 86.7 N T2 = T3 sin 30° = (100 N )( 0.5 ) = 50.0 N
G
Assess: Since T3 acts closer to the x-axis than to the y-axis, it makes sense that T1 > T2 .
6.2.
Model:
Visualize:
We can assume that the ring is a particle.
This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only
three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle θ .
G
Solve: Because the ring is in equilibrium it must obey Fnet = 0 N. This is a vector equation, so it has both xand y-components:
( Fnet ) x = T3 cosθ − T2 = 0 N ⇒ T3 cosθ = T2
( Fnet ) y = T1 − T3 sinθ = 0 N ⇒ T3 sinθ = T1
We have two equations in the two unknowns T3 and θ . Divide the y-equation by the x-equation:
T3 sin θ
T 80 N
= tanθ = 1 =
= 1.6 ⇒ θ = tan −1 (1.6 ) = 58°
T3 cosθ
T2 50 N
Now we can use the x-equation to find
T3 =
T2
50 N
=
= 94 N
cosθ cos58°
The tension in the third rope is 94 N directed 58° below the horizontal.
6.3.
Model: We assume the speaker is a particle in static equilibrium under the influence of three forces:
gravity and the tensions in the two cables.
Visualize:
Solve:
From the lengths of the cables and the distance below the ceiling we can calculate θ as follows:
sinθ =
2m
= 0.677 ⇒ θ = sin −1 0.667 = 41.8°
3m
Newton’s first law for this situation is
( Fnet ) x = ΣFx = T1x + T2 x = 0 N ⇒ −T1 cosθ + T2 cosθ = 0 N
( Fnet ) y = ΣFy = T1 y + T2 y + wy = 0 N ⇒ T1 sinθ + T2 sinθ − w = 0 N
The x-component equation means T1 = T2 . From the y-component equation:
( 20 kg ) ( 9.8 m/s ) 196 N
w
mg
2T1 sin θ = w ⇒ T1 =
=
=
=
= 147 N
2sin θ 2sin θ
2sin 41.8°
1.333
2
Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from
the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the
relatively large angle of suspension.
6.4.
Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by
the two ropes, with friction providing the force that resists the pullers.
Visualize:
Solve:
Since the sled is not accelerating, it is in dynamic equilibrium and Newton’s first law applies:
( Fnet ) x = ΣFx = T1x + T2 x + f kx = 0 N
( Fnet ) y = ΣFy = T1 y + T2 y + f ky = 0 N
From the free-body diagram:
⎛1 ⎞
⎛1 ⎞
T1 cos ⎜ θ ⎟ + T2 cos ⎜ θ ⎟ − f k = 0 N
⎝2 ⎠
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
T1 sin ⎜ θ ⎟ − T2 sin ⎜ θ ⎟ + 0 N = 0 N
⎝2 ⎠
⎝2 ⎠
From the second of these equations T1 = T2 . Then from the first:
2T1 cos10° = 1000 N ⇒ T1 =
1000 N 1000 N
=
= 508 N
2cos10° 1.970
Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two
add up to only slightly more than 1000 N, which makes sense because the angle at which the two players are
pulling is small.
6.5.
Solve:
Visualize: Please refer to the Figure EX6.5.
Applying Newton’s second law to the diagram on the left,
ax =
( Fnet ) x
m
=
4 N−2N
= 1.0 m/s 2
2 kg
ay =
( Fnet ) y
m
=
3 N−3 N
= 0 m/s 2
2 kg
For the diagram on the right:
ax =
( Fnet ) x
m
=
4 N−2 N
= 1.0 m/s 2
2 kg
ay =
( Fnet ) y
m
=
3 N −1 N − 2 N
= 0 m/s 2
2 kg
6.6.
Visualize: Please refer to Figure EX6.6.
Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice
that the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted. Applying
Newton’s second law,
ax =
( Fnet ) x
ay =
m
=
( Fnet ) y
m
5 N − 1 N − ( 3sin 20° ) N
= 1.49 m/s 2
2 kg
=
2.82 N − ( 3cos 20° ) N
= 0 m/s 2
2 kg
For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15° with the positive xaxis. The other 2-newton force makes an angle of 15° with the negative y-axis. The accelerations are
ax =
ay =
( Fnet ) x ( 2cos15° ) N + ( 2sin15° ) N − 3 N
=
= −0.28 m/s 2
m
( Fnet ) y
m
2 kg
=
1.414 N + ( 2sin15° ) N − ( 2cos15° ) N
= 0 m/s 2
2 kg
6.7.
Solve:
Visualize: Please refer to Figure EX6.7.
(a) Apply Newton’s second law in both the x and y directions.
( Fnet ) x = ( 5.0 N ) cos37° − 2.0 N = ( 5.0 kg ) ax
⇒ ax = 0.40 m/s 2
( Fnet ) y = 2.0 N + ( 5.0 N ) sin 37° − 5.0 N = ( 5.0 kg ) a y
⇒ a y = 0.0 m/s 2
(b) The angle that the 5.0 N force makes with the –y-axis is 37°. Apply Newton’s second law for both the x and y
direction.
( Fnet ) x = 3.0 N + ( 5.0 N ) sin 37° − 2.0 N = ( 5.0 kg ) ax
⇒ ax = 0.80 m/s 2
( Fnet ) y = 4.0 N − ( 5.0 N ) cos37° = ( 5.0 kg ) a y
⇒ a y = 0.0 m/s 2
Assess: The orientation of the coordinate axes is chosen for convenience, and does not always need to conform
to the horizontal and vertical.
6.8.
Visualize: Please refer to Figure EX6.8.
Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations.
For t between 0 s and 3 s,
ax =
Δvx 12 m/s − 0 s
=
= 4 m/s 2
3s
Δt
For t between 3 s and 6 s, Δvx = 0 m/s, so ax = 0 m/s 2 . For t between 6 s and 8 s,
Δvx 0 m/s − 12 m/s
=
= −6 m/s 2
Δt
2s
From Newton’s second law, at t = 1 s we have
ax =
Fnet = max = (2.0 kg)(4 m/s 2 ) = 8 N
At t = 4 s, ax = 0 m/s 2 , so Fnet = 0 N. At t = 7 s,
Fnet = max = (2.0 kg)(−6.0 m/s 2 ) = −12 N
Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and
negative signs are appropriate for an object first speeding up, then slowing down.
6.9.
Visualize: Please refer to Figure EX6.9. Positive forces result in the object gaining speed and negative
forces result in the object slowing down. The final segment of zero force is a period of constant speed.
Solve: We have the mass and net force for all the three segments. This means we can use Newton’s second law
to calculate the accelerations. The acceleration from t = 0 s to t = 3 s is
ax =
Fx
4N
=
= 2 m/s 2
m 2.0 kg
ax =
Fx
−2 N
=
= −1 m/s 2
m 2.0 kg
The acceleration from t = 3 s to t = 5 s is
The acceleration from t = 5 s to 8 s is ax = 0 m/s 2 . In particular, ax ( at t = 6 s ) = 0 m/s 2 .
We can now use one-dimensional kinematics to calculate v at t = 6 s as follows:
v = v0 + a1 ( t1 − t0 ) + a2 ( t2 − t0 )
= 0 + ( 2 m/s 2 ) ( 3 s ) + ( −1 m/s 2 ) ( 2 s ) = 6 m/s − 2 m/s = 4 m/s
Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the
G
positive F1. A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the
mass involved.
6.10.
Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless,
the tension in the rope is the only horizontal force.
Visualize:
Solve: (a) Since the box is at rest, ax = 0 m/s 2 , and the net force on the box must be zero. Therefore, according
to Newton’s first law, the tension in the rope must be zero.
(b) For this situation again, ax = 0 m/s 2 , so Fnet = T = 0 N.
(c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since
ax = 5.0 m/s 2 ,
Fnet = T = max = ( 50 kg ) ( 5.0 m/s 2 ) = 250 N
Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal
force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems
reasonable to accelerate a box of this mass at 5.0 m/s 2 .
6.11.
Model: We assume that the box is a point particle that is acted on only by the tension in the rope and
the pull of gravity. Both the forces act along the same vertical line.
Visualize:
Solve:
(a) Since the box is at rest, a y = 0 m/s 2 and the net force on it must be zero:
Fnet = T − FG = 0 N ⇒ T = FG = mg = ( 50 kg ) ( 9.8 m/s 2 ) = 490 N
(b) Since the box is rising at a constant speed, again a y = 0 m/s 2 , Fnet = 0 N, and T = FG = 490 N.
(c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since
a y = 5.0 m/s 2 ,
Fnet = T − FG = ma y = ( 50 kg ) ( 5.0 m/s 2 ) = 250 N
⇒ T = 250 N + w = 250 N + 490 N = 740 N
(d) The situation is the same as in part (c), except that the rising box is slowing down. Thus a y = −5.0 m/s 2 and
we have instead
Fnet = T − FG = ma y = ( 50 kg ) ( −5.0 m/s 2 ) = −250 N
⇒ T = −250 N + FG = −250 N + 490 N = 240 N
Assess: For parts (a) and (b) the zero accelerations immediately imply that the gravitational force on the box
must be exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the
gravitational force on the box but must also accelerate it upward, hence, T must be greater than FG . When the
box accelerates downward, the rope need not support the entire gravitational force, hence, T is less than FG .
6.12.
Model: We assume the rocket is a particle moving in a vertical straight line under the influence of only
two forces: gravity and its own thrust.
Visualize:
Solve:
(a) Using Newton’s second law and reading the forces from the free-body diagram,
Fthrust − FG = ma ⇒ Fthrust = ma + mg Earth = ( 0.200 kg ) (10 m/s 2 + 9.80 m/s 2 ) = 3.96 N
(b) Likewise, the thrust on the moon is (0.200 kg)(10 m/s 2 + 1.62 m/s 2 ) = 2.32 N.
Assess: The thrust required is smaller on the moon, as it should be, given the moon’s weaker gravitational pull.
The magnitude of a few newtons seems reasonable for a small model rocket.
6.13.
Solve:
Model: The astronaut is treated as a particle.
The mass of the astronaut is
m=
wearth
800 N
=
= 81.6 kg
g earth 9.80 m/s 2
Therefore, the weight of the astronaut on Mars is
wMars = mg Mars = ( 81.6 kg ) ( 3.76 m/s 2 ) = 307 N
Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than
to the earth, so the smaller weight on Mars makes sense.
6.14.
Solve:
Model: Use the particle model for the woman.
(a) The woman’s weight on the earth is
wearth = mg earth = ( 55 kg ) ( 9.80 m/s 2 ) = 540 N
(b) Since mass is a measure of the amount of matter, the woman’s mass is the same on the moon as on the earth.
Her weight on the moon is
wmoon = mg moon = ( 55 kg ) (1.62 m/s 2 ) = 89 N
Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to
the moon than to the earth. Thus the woman’s smaller weight on the moon makes sense.
6.15.
Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of
gravity and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 6.10.
Visualize:
Solve:
(a) The weight is
⎛ a ⎞
⎛
0⎞
w = mg ⎜1 + y ⎟ = mg ⎜1 + ⎟ = mg = ( 60 kg ) ( 9.80 m/s 2 ) = 590 N
g⎠
⎝ g⎠
⎝
(b) The elevator speeds up from v0 y = 0 m/s to its cruising speed at v y = 10 m/s. We need its acceleration before
we can find the apparent weight:
ay =
Δv 10 m/s − 0 m/s
=
= 2.5 m/s 2
Δt
4.0 s
The passenger’s weight is
⎛
⎛ a ⎞
2.5 m/s 2 ⎞
w = mg ⎜ 1 + y ⎟ = ( 590 N ) ⎜1 +
= (590 N)(1.26) = 740 N
2 ⎟
g ⎠
⎝
⎝ 9.80 m/s ⎠
(c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus,
w = mg = 590 N as in part (a).
Assess: The passenger’s weight is the gravitational force on the passenger in parts (a) and (c), since there is no
acceleration. In part (b), the elevator must not only support the gravitational force but must also accelerate him
upward, so it’s reasonable that the floor will have to push up harder on him, increasing his weight.
6.16.
Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward
pull of gravity and the upward force of the elevator floor.
Visualize: Please refer to Figure EX6.16. The graph has three segments corresponding to different conditions:
(1) increasing velocity, meaning an upward acceleration; (2) a period of constant upward velocity; and (3)
decreasing velocity, indicating a period of deceleration (negative acceleration).
Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph
and then apply Equation 6.10. The acceleration for the first segment is
v − v 8 m/s − 0 m/s
= 4 m/s 2
ay = 1 0 =
t1 − t0
2 s−0 s
⎛
⎛ a ⎞
4 m/s 2 ⎞
4 ⎞
⎛
⇒ w = mg ⎜1 + y ⎟ = mg ⎜1 +
= ( 75 kg ) ( 9.80 m/s 2 ) ⎜1 +
⎟ = 1035 N
2 ⎟
9.80
m/s
9.80
g
⎝
⎠
⎝
⎠
⎝
⎠
For the second segment, a y = 0 m/s 2 and the weight is
⎛ 0 m/s 2 ⎞
2
w = mg ⎜1 +
⎟ = mg = ( 75 kg ) ( 9.80 m/s ) = 740 N
g ⎠
⎝
For the third segment,
ay =
v3 − v2 0 m/s − 8 m/s
=
= −2 m/s 2
t3 − t 2
10 s − 6 s
⎛
−2 m/s 2 ⎞
⇒ w = mg ⎜1 +
= ( 75 kg ) ( 9.80 m/s 2 ) (1 − 0.2 ) = 590 N
2 ⎟
9.80
m/s
⎝
⎠
Assess: As expected, the weight is greater than the gravitational force on the passenger when the elevator is
accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration
the weight is the gravitational force. In all three cases the magnitudes are reasonable, given the mass of the
passenger and the accelerations of the elevator.
6.17.
Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during
the entire problem, we can use the model of kinetic friction.
Visualize:
Solve: The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the
vertical and horizontal directions:
( Fnet ) x = ΣFx = FB + FC − f k = 0 N ⇒ f k = FB + FC = 350 N + 385 N = 735 N
( Fnet ) y = ΣFy = n − FG = 0 N ⇒ n = FG = mg = ( 300 kg ) ( 9.80 m/s2 ) = 2.94 × 103 N
Then, for kinetic friction:
fk = μk n ⇒ μk =
fk
735 N
=
= 0.250
n 2.94 × 103 N
Assess: The value of μ k = 0.250 is hard to evaluate without knowing the material the floor is made of, but it
seems reasonable.
6.18.
Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the
farmer’s pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it
will be subject to kinetic friction.
Visualize:
Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that
n = FG = mg . The maximum friction force is
f smax = μ s mg = ( 0.8 )(120 kg ) ( 9.80 m/s 2 ) = 940 N
The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not
be able to budge the mule.
Assess: The farmer should have known better.
6.19.
Model:
Visualize:
We will represent the crate as a particle.
G G
(a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s 2 and is in dynamic
G
G
equilibrium. Thus Fnet = 0. It is tempting to think that the belt exerts a friction force on the crate. But if it did,
there would be a net force because there are no other possible horizontal forces to balance a friction force.
Because there is no net force, there cannot be a friction force. The only forces are the upward normal force and
the gravitational force on the crate. (A friction force would have been needed to get the crate moving initially,
but no horizontal force is needed to keep it moving once it is moving with the same constant speed as the belt.)
(b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the
crate must have a net horizontal force. So now there is a friction force, and the force points in the direction of the
crate’s motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate
relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on
the belt.
(c) The static friction force has a maximum possible value ( f s ) max = μ s n. The maximum possible acceleration of
the crate is
Solve:
amax =
( fs )max μs n
=
m
m
If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the
free-body diagram that n = FG = mg . Thus,
amax = μs g = (0.5)(9.80 m/s 2 ) = 4.9 m/s 2
6.20.
Model:
Visualize:
We assume that the truck is a particle in equilibrium, and use the model of static friction.
Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newton’s first law. The normal
force has no component in the x-direction, so we can ignore it here. For the other two forces:
( Fnet ) x = ΣFx = fs − ( FG ) x = 0 N ⇒ fs = ( FG ) x = mg sin θ = ( 4000 kg ) ( 9.80 m/s 2 ) ( sin15° ) = 10,145 N
Assess: The truck’s weight (mg) is roughly 40,000 N. A friction force that is ≈ 25% of the truck’s weight
seems reasonable.
6.21. Model: The car is a particle subject to Newton’s laws and kinematics.
Visualize:
Solve: Kinetic friction provides a horizontal acceleration which stops the car. From the figure, applying
Newton’s first and second laws gives
∑ F = − f = ma
x
k
x
∑ F = n − F = 0 ⇒ n = F = mg
y
G
G
Combining these two equations with f k = μ k n yields
ax = − μk g = − ( 0.50 ) ( 9.80 m/s 2 ) = −4.9 m/s
Kinematics can be used to determine the initial velocity.
vf2 = vi2 + 2aΔx ⇒ vi2 = −2ax Δx
Thus
vi = −2 ( −4.9 m/s 2 ) ( 65 m − 0 m ) = 25 m/s 2
Assess: The initial speed of 25 m/s 2 ≈ 56 mph is a reasonable speed to have initially for a vehicle to leave 65meter-long skid marks.
6.22.
Model: We assume that the plane is a particle accelerating in a straight line under the influence of two
forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional
kinematics.
Visualize:
Solve:
We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain:
a=
Δv 82 m/s − 0 m/s
=
= 2.34 m/s 2
35 s
Δt
( Fnet ) = ΣFx = Fthrust − f r = ma ⇒ Fthrust = f r + ma
For rubber rolling on concrete, μ r = 0.02 (Table 6.1), and since the runway is horizontal, n = FG = mg. Thus:
Fthrust = μ r FG + ma = μ r mg + ma = m ( μ r g + a )
= ( 75,000 kg ) ⎡⎣( 0.02 ) ( 9.8 m/s 2 ) + 2.34 m/s 2 ⎤⎦ = 190,000 N
Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that
190,000 N is enough to produce the required acceleration.
6.23.
Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow
speed and large mass). We can use the one-dimensional kinematic equations.
Visualize:
Solve: The locomotive is not accelerating in the vertical direction, so the free-body diagram shows us that
n = FG = mg . Thus,
f r = μ r mg = ( 0.002 )( 50,000 kg ) ( 9.80 m/s 2 ) = 980 N
From Newton’s second law for the decelerating locomotive,
−f
−980 N
ax = r =
= −0.01960 m/s 2
m 50,000 kg
Since we’re looking for the distance the train rolls, but we don’t have the time:
v 2 − v02 ( 0 m/s ) − (10 m/s )
=
= 2.55 × 103 m
v12 − v02 = 2ax ( Δx ) ⇒ Δx = 1
2a x
2 ( −0.01960 m/s 2 )
2
2
Assess: The locomotive’s enormous inertia (mass) and the small coefficient of rolling friction make this long
stopping distance seem reasonable.
6.24.
Model:
Visualize:
We can treat the sliding player as a particle experiencing kinetic friction.
Solve: We can assume a mass of 80 kg (which corresponds to a gravitational force of about 175 pounds). We
have no value for μk of cloth sliding on loose dirt. It is probably greater than μk for wood on wood, due to the
roughness of both surfaces. Let’s guess 0.40. From the free-body diagram, n = FG = mg , since there’s no vertical
acceleration. Thus,
f k = μ k mg = ( 0.40 )( 80 kg ) ( 9.80 m/s 2 ) = 314 N
Assess: A frictional force of 314 N would produce an acceleration of
a=
Fnet −314 N
=
= − 4.0 m/s 2
m
80 kg
A player running initially at 8 m/s would thus be brought to a stop in about 2 seconds, which seems somewhat
too long. Our estimate of μk is probably a bit low but 8 m/s is a large speed as well.
6.25.
Model:
Visualize:
We assume that the skydiver is shaped like a box and is a particle.
The skydiver falls straight down toward the earth’s surface, that is, the direction of fall is vertical. Since the
skydiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm × 40 cm. The
physical conditions needed to use Equation 6.16 for the drag force are satisfied. The terminal speed corresponds
to the situation when the net force acting on the skydiver becomes zero.
Solve: The expression for the magnitude of the drag with v in m/s is
D≈
1 2
Av = 0.25 ( 0.20 × 0.40 ) v 2 N = 0.020v 2 N
4
The gravitational force on the skydiver is FG = mg = ( 75 kg ) ( 9.8 m/s 2 ) = 735 N. The mathematical form of the
condition defining dynamical equilibrium for the skydiver and the terminal speed is
G
G
G
Fnet = FG + D = 0 N
2
⇒ 0.02vterm
N − 735 N = 0 N ⇒ vterm =
735
≈ 192 m/s
0.02
Assess: The result of the above simplified physical modeling approach and subsequent calculation, even if
approximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very
badly hurt at landing if the parachute does not open in time.
6.26.
Model:
Visualize:
We will represent the tennis ball as a particle.
The tennis ball falls straight down toward the earth’s surface. The ball is subject to a net force that is the resultant
of the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively.
Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the
rest of the motion.
Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is
G
G
G G
Fnet = FG + D = 0 N
Since only the vertical direction matters, one can write:
ΣFy = 0 N ⇒ Fnet = D − FG = 0 N
When this condition is satisfied, the speed of the ball becomes the constant terminal speed v = vterm . The
magnitudes of the gravitational and drag forces acting on the ball are:
FG = mg = m ( 9.80 m/s 2 )
1
2
2
2
2
= ( 0.25π )( 0.0325 m ) ( 26 m/s ) = 0.56 N
( Avterm
) = 0.25 (π R 2 ) vterm
4
The condition for dynamic equilibrium becomes:
0.56 N
= 57 g
( 9.80 m/s2 ) m − 0.56 N = 0 N ⇒ m = 9.80
m/s 2
D≈
Assess: The value of the mass of the tennis ball obtained above seems reasonable.
6.27. Visualize:
We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was
calculated as follows:
amax =
Fmax 10 N
=
= 2 m/s 2
m
5 kg
Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the
more general result that
v(t ) = v0 + area under the acceleration curve from 0 s to t
The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is 12 (4 s)
(2 m/s 2 ) = 4.0 m/s. We’ve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s, vx = 4.0 m/s.
6.28. Visualize:
The acceleration is ax = Fx m , so the acceleration-versus-time graph has exactly the same shape as the forceversus-time graph. The maximum acceleration is amax = Fmax m = ( 6 N ) ( 2 kg ) = 3 m/s 2 .
Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the
more general result that
v(t ) = v0 + area under the acceleration curve from 0 s to t
The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a
rectangle (3 m/s 2 × 2 s = 6 m/s) plus a triangle ( 12 × 3 m/s 2 × 2 s = 3 m/s ). Thus vx = 9 m/s at t = 4 s.
6.29.
Model:
Visualize:
Solve:
You can model the beam as a particle in static equilibrium.
Using Newton’s first law, the equilibrium equations in vector and component form are:
G
G G G G
Fnet = T1 + T2 + FG = 0 N
( Fnet ) x = T1x + T2 x + FGx = 0 N
( Fnet ) y = T1 y + T2 y + FGy = 0 N
Using the free-body diagram yields:
−T1 sinθ1 + T2 sin θ 2 = 0 N
T1 cosθ1 + T2 cosθ 2 − FG = 0 N
The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, T1 and
T2 . Substituting θ1 = 20°, θ 2 = 30°, and FG = mg = 9800 N, the simultaneous equations become
−T1 sin 20° + T2 sin 30° = 0 N
T1 cos 20° + T2 cos30° = 9800 N
You can solve this system of equations by simple substitution. The result is T1 = 6397 N and T2 = 4376 N.
Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left
rope than in the right rope.
6.30.
Model:
Visualize:
The plastic ball is represented as a particle in static equilibrium.
Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the
string’s tension plus two long-range forces. The equilibrium condition is
( Fnet ) x = Tx + ( Felec ) x = T sin θ − Felec = 0 N
( Fnet ) y = Ty + ( FG ) y = T cosθ − mg = 0 N
We can solve the y-equation to get
( 0.001 kg ) ( 9.8 m/s )
mg
=
= 0.0104 N
cosθ
cos 20°
2
T=
Substituting this value into the x-equation,
Felec = T sin θ = (1.04 × 10−2 N ) sin 20° = 0.0036 N
(b) The tension in the string is 0.0104 N.
6.31.
Model:
Visualize:
Solve:
The piano is in static equilibrium and is to be treated as a particle.
(a) Based on the free-body diagram, Newton’s second law is
( Fnet ) x = 0 N = T1x + T2 x = T2 cosθ 2 − T1 cosθ1
( Fnet ) y = 0 N = T1 y + T2 y + T3 y + FGy = T3 − T1 sinθ1 − T2 sinθ 2 − mg
Notice how the force components all appear in the second law with plus signs because we are adding forces. The
negative signs appear only when we evaluate the various components. These are two simultaneous equations in
the two unknowns T2 and T3 . From the x-equation we find
T cosθ1 ( 500 N ) cos15°
T2 = 1
=
= 533 N
cosθ 2
cos 25°
(b) Now we can use the y-equation to find
T3 = T1 sin θ1 + T2 sin θ 2 + mg = 5.25 × 103 N
6.32.
Model: We will represent Henry as a particle. His motion is governed by constant-acceleration
kinematic equations.
Visualize: Please refer to the Figure EX6.32.
Solve: (a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and
another acceleration as the elevator brakes from 10.0 s to 12.0 s. The weight is the same as the gravitational force
during constant velocity motion, so Henry’s weight w = FG = mg is 750 N. His weight is less than the gravitational
force on him during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the
elevator’s initial motion is down.
(b) Because the gravitational force on Henry is 750 N, his mass is m = FG /g = 76.5 kg.
(c) The apparent weight (see Equation 6.10) during vertical motion is given by
⎛ w
⎞
⎛
a⎞
w = mg ⎜1 + ⎟ ⇒ a = g ⎜
− 1⎟
g
F
⎝
⎠
⎝ G
⎠
During the interval 0 s ≤ t ≤ 2 s, the elevator’s acceleration is
⎛ 600 N ⎞
a = g⎜
− 1⎟ = −1.96 m/s 2
⎝ 750 N ⎠
At t = 2 s, Henry’s position is
1
1
2
2
y1 = y0 + v0 Δt0 + a ( Δt0 ) = a ( Δt0 ) = −3.92 m
2
2
and his velocity is
v1 = v0 + aΔt0 = aΔt0 = −3.92 m/s
During the interval 2 s ≤ t ≤ 10 s, a = 0 m/s 2 . This means Henry travels with a constant velocity v1 = −3.92 m/s.
At t = 10 s he is at position
y2 = y1 + v1Δt1 = −35.3 m
and he has a velocity v2 = v1 = −3.92 m/s. During the interval 10 s ≤ t ≤ 12.0 s, the elevator’s acceleration is
⎛ 900 N ⎞
a = g⎜
− 1⎟ = +1.96 m/s 2
⎝ 750 N ⎠
The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t = 12.0 s Henry is at
position
1
y3 = y2 + v2 (Δt2 ) + a (Δt2 ) 2 = −39.2 m
2
Thus Henry has traveled distance 39.2 m.
6.33.
Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single
vertical line: gravity and the supporting force of the elevator.
Visualize:
Solve:
(a) Before the elevator starts braking, Zach is not accelerating. His weight (see Equation 6.10) is
⎛ 0 m/s 2 ⎞
⎛
a⎞
2
w = mg ⎜1 + ⎟ = mg ⎜1 +
⎟ = mg = ( 80 kg ) ( 9.80 m/s ) = 784 N
g
g
⎝
⎠
⎝
⎠
Zach’s weight is 7.8 × 102 N.
(b) Using the definition of acceleration,
a=
Δv v1 − v0 0 − ( −10 ) m/s
=
=
= 3.33 m/s 2
Δt t1 − t0
3.0 s
⎛ 3.33 m/s 2 ⎞
⎛
a⎞
⇒ w = mg ⎜1 + ⎟ = ( 80 kg ) ( 9.80 m/s 2 ) ⎜1 +
= (784 N)(1 + 0.340) = 1050 N
2 ⎟
⎝ g⎠
⎝ 9.80 m/s ⎠
Now Zach’s weight is 1.05 × 103 N.
Assess: While the elevator is braking, it not only must support the gravitational force on Zach but must also
push upward on him to decelerate him, so his weight is greater than the gravitational force.
6.34.
Model: We can assume your body is a particle moving in a straight line under the influence of two
forces: gravity and the support force of the scale.
Visualize:
Solve:
The weight (see Equation 6.10) of an object moving in an elevator is
⎛
⎛ w
⎞
a⎞
w = mg ⎜1 + ⎟ ⇒ a = ⎜
− 1⎟ g
g
mg
⎝
⎠
⎝
⎠
When accelerating upward, the acceleration is
⎛ 170 lb ⎞
a =⎜
− 1⎟ ( 9.80 m/s 2 ) = 1.3 m/s 2
⎝ 150 lb ⎠
When braking, the acceleration is
⎛ 120 lb ⎞
a =⎜
− 1⎟ ( 9.80 m/s 2 ) = −2.0 m/s 2
⎝ 150 lb ⎠
Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire
State Building must be. Also note that we did not have to convert the units of the weights from pounds to
newtons because the weights appear as a ratio.
6.35.
Model: We can assume the foot is a single particle in equilibrium under the combined effects of
gravity, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the
leg itself. We can also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the
tension is the same everywhere in the rope. Because all pulleys are in equilibrium, their net force is zero. So they
do not contribute to T.
Visualize:
Solve:
(a) From the free-body diagram for the mass, the tension in the rope is
T = FG = mg = ( 6 kg ) ( 9.80 m/s 2 ) = 58.8 N
(b) Using Newton’s first law for the vertical direction on the pulley attached to the foot,
( Fnet ) y = ΣFy = T sin θ − T sin15° − ( FG )foot = 0 N
⇒ sin θ =
T sin15° + ( FG )foot
T
( 4 kg ) ( 9.80 m/s )
mfoot g
= 0.259 +
= 0.259 + 0.667 = 0.926
T
58.8 N
2
= sin15° +
⇒ θ = sin −1 0.926 = 67.8°
(c) Using Newton’s first law for the horizontal direction,
( Fnet ) x = ΣFx = T cosθ + T cos15° − Ftraction = 0 N
⇒ Ftraction = T cosθ + T cos15° = T ( cos67.8° + cos15° )
= (58.8 N)(0.3778 + 0.9659) = (58.8 N)(1.344) = 79.0 N
Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward
pull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The
magnitude of the traction force, roughly one-tenth of the gravitational force on a human body, seems reasonable.
6.36.
Model: We can assume the person is a particle moving in a straight line under the influence of the
combined decelerating forces of the air bag and seat belt or, in the absence of restraints, the dashboard or
windshield.
Visualize:
Solve: (a) In order to use Newton’s second law for the passenger, we’ll need the acceleration. Since we don’t
have the stopping time:
v12 = v02 + 2a ( x1 − x0 ) ⇒ a =
0 m /s − (15 m/s )
v12 − v02
=
= −112.5 m/s 2
2 ( x1 − x0 )
2 (1 m − 0 m )
2
2
2
⇒ Fnet = F = ma = ( 60 kg ) ( −112.5 m/s 2 ) = −6750 N
The net force is 6750 N to the left.
(b) Using the same approach as in part (a),
0 m /s − (15 m/s )
v12 − v02
= ( 60 kg )
= −1,350,000 N
2 ( x1 − x0 )
2 ( 0.005 m )
2
F = ma = m
2
2
The net force is 1,350,000 N to the left.
(c) The passenger’s weight is mg = (60 kg)(9.8 m/s 2 ) = 588 N. The force in part (a) is 11.5 times the passenger’s
weight. The force in part (b) is 2300 times the passenger’s weight.
Assess: An acceleration of 11.5g is well within the capability of the human body to withstand. A force of 2300
times the passenger’s weight, on the other hand, would surely be catastrophic.
6.37.
Model:
Visualize:
The ball is represented as a particle that obeys constant-acceleration kinematic equations.
Solve: This is a two-part problem. During part 1 the ball accelerates upward in the tube. During part 2 the ball
undergoes free fall (a = − g ). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from
the tube. The free-body diagram for part 1 shows two forces: the air pressure force and the gravitational force. We
need only the y-component of Newton’s second law:
ay = a =
( Fnet ) y
m
=
2N
Fair − FG Fair
=
−g=
− 9.80 m/s 2 = 30.2 m/s 2
0.05 kg
m
m
We can use kinematics to find the velocity v1 as the ball leaves the tube:
v12 = v02 + 2a ( y1 − y0 ) ⇒ v1 = 2ay1 = 2 ( 30.2 m/s 2 ) (1 m ) = 7.77 m/s
For part 2, free-fall kinematics v22 = v12 − 2 g ( y2 − y1 ) gives
y2 − y1 =
v12
= 3.1 m
2g
6.38.
Model:
Visualize:
We will represent the bullet as a particle.
Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then
we can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant
to the problem.) The kinematic equation is
( 400 m/s ) = −6.67 × 105 m/s 2
v02
=−
2Δx
2 ( 0.12 m )
G
Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood
exerts two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the
G
G
wood. Second, a retarding frictional force f k that stops the bullet. The only horizontal force is f k , which points
to the left and thus has a negative x-component. The x-component of Newton’s second law is
2
v12 = v02 + 2aΔx ⇒ a = −
( Fnet ) x = − f k = ma ⇒ f k = −ma = − ( 0.01 kg ) ( −6.67 × 105 m/s 2 ) = 6670 N
Notice how the signs worked together to give a positive value of the magnitude of the force.
(b) The time to stop is found from v1 = v0 + aΔt as follows:
v
Δt = − 0 = 6.00 × 10−4 s = 600 μs
a
(c)
Using the above kinematic equation, we can find the velocity as a function of t. For example at t = 60 μs,
vx = 400 m/s + (−6.667 × 105 m/s 2 )(60 × 10−6 s) = 360 m/s
6.39.
Model:
Visualize:
Solve:
Represent the rocket as a particle that follows Newton’s second law.
(a) The y-component of Newton’s second law is
ay = a =
( Fnet ) y
m
=
Fthrust − mg 3.0 × 105 N
=
− 9.80 m/s 2 = 5.2 m/s 2
20,000 kg
m
(b) At 5000 m the acceleration has increased because the rocket mass has decreased. Solving the equation of part
(a) for m gives
m5000 m =
Fthrust
3.0 × 105 N
=
= 1.9 × 104 kg
a5000 m + g 6.0 m/s 2 + 9.80 m/s 2
The mass of fuel burned is mfuel = minitial − m5000 m = 1.0 × 103 kg.
6.40.
Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of
friction μ s = 0.80 and a kinetic coefficient of friction μ k = 0.60.
Visualize:
Solve:
(a) While the block is at rest, Newton’s second law is
( Fnet ) x = T − fs = 0 N ⇒ T = fs
( Fnet ) y = n − FG ⇒ n = FG = mg
The static friction force has a maximum value ( fs )max = μs mg . The string tension that will cause the block to slip
and start to move is
T = μ s mg = ( 0.80 )( 2.0 kg ) ( 9.80 m/s 2 ) = 15.7 N
Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension for
motion.
(b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the xcomponent of Newton’s second law as follows:
( Fnet ) x = T − f k = max ⇒ ax =
T − fk
m
The kinetic friction force f k = μk mg . The acceleration of the block is
ax =
20 N − ( 0.60 )( 2 kg ) ( 9.8 m/s 2 )
2 kg
= 4.12 m/s 2
Using kinematics, the block’s speed after moving 1.0 m will be
v12 = 0 m 2 /s 2 + 2 ( 4.12 m/s 2 ) (1.0 m ) ⇒ v1 = 2.9 m/s
(c) The only difference in this case is the coefficient of kinetic friction whose value is 0.050 instead of 0.60. The
acceleration of the block is
ax =
20 N − ( 0.050 )( 2.0 kg ) ( 9.80 m/s 2 )
2.0 kg
= 9.51 m/s 2
The block’s speed after moving 1.0 m will be
v12 = 0 m 2 /s 2 + 2 ( 9.51 m/s 2 ) (1.0 m ) ⇒ v1 = 4.4 m/s
6.41.
Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of
two forces: the thrust of his jet skis and the resisting force of friction on the skis. We can use one-dimensional
kinematics.
Visualize:
Solve: (a) The friction force of the snow can be found from the free-body diagram and Newton’s first law,
since there’s no acceleration in the vertical direction:
n = FG = mg = ( 75 kg ) ( 9.80 m/s 2 ) = 735 N ⇒ f k = μ k n = ( 0.10 )( 735 N ) = 73.5 N
Then, from Newton’s second law:
F
− f k 200 N − 73.5 N
=
= 1.687 m/s 2
m
75 kg
( Fnet ) x = Fthrust − f k = ma0 ⇒ a0 = thrust
From kinematics:
v1 = v0 + a0t1 = 0 m/s + (1.687 m/s 2 ) (10 s ) = 16.9 m/s
(b) During the acceleration, Sam travels to
1
1
2
x1 = x0 + v0t1 + a0t12 = (1.687 m/s 2 ) (10 s ) = 84 m
2
2
After the skis run out of fuel, Sam’s acceleration can again be found from Newton’s second law:
−73.5 N
F
= −0.98 m/s 2
( Fnet ) x = − f k = −73.5 N ⇒ a1 = net =
m
75 kg
Since we don’t know how much time it takes Sam to stop:
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ x2 − x1 =
v22 − v12 0 m /s − (16.9 m/s )
=
= 145 m
2a1
2 ( −0.98 m/s 2 )
2
2
2
The total distance traveled is ( x2 − x1 ) + x1 = 145 m + 84 m = 229 m.
Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coasting
distance of nearly 150 m also seems possible, starting from a high speed, given that we’re neglecting air
resistance.
6.42.
Model: We assume Sam is a particle moving in a straight line down the slope under the influence of
gravity, the thrust of his jet skis, and the resisting force of friction on the snow.
Visualize:
Solve:
From the height of the slope and its angle, we can calculate its length:
h
h
50 m
= sin θ ⇒ x1 − x0 =
=
= 288 m
x1 − x0
sin θ sin10°
Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force:
( Fnet ) y = ΣFy = n − FG cosθ = 0 N ⇒ n = FG cosθ = mg cosθ = (75 kg)(9.80 m/s2 )(cos10°) = 724 N
One-dimensional kinematics gives us Sam’s acceleration:
v12 = v02 + 2ax ( x − x0 ) ⇒ ax =
v12 − v02
( 40 m/s ) − 0 m /s = 2.78 m/s 2
=
2 ( x1 − x2 )
2 ( 288 m )
2
2
2
Then, from Newton’s second law and the equation f k = μ k n :
( Fnet ) x = ΣFx = FG sinθ + Fthrust − f k = max
⇒ μk =
=
mg sin θ + Fthrust − ma
n
( 75 kg ) ( 9.80 m/s 2 ) ( sin10° ) + 200 N − ( 75 kg ) ( 2.78 m/s 2 )
724 N
Assess: This coefficient seems a bit high for skis on snow, but not impossible.
= 0.165
6.43.
Model:
only.
Visualize:
We assume the suitcase is a particle accelerating horizontally under the influence of friction
Solve: Because the conveyor belt is already moving, friction drags your suitcase to the right. It will accelerate
until it matches the speed of the belt. We need to know the horizontal acceleration. Since there’s no acceleration
in the vertical direction, we can apply Newton’s first law to find the normal force:
n = FG = mg = (10 kg ) ( 9.80 m/s 2 ) = 98.0 N
The suitcase is accelerating, so we use μ k to find the friction force
f k = μ k mg = ( 0.3)( 98.0 N ) = 29.4 N
We can find the horizontal acceleration from Newton’s second law:
f
m
( Fnet ) x = ΣFx = f k = ma ⇒ a = k =
29.4 N
= 2.94 m/s 2
10 kg
From one of the kinematic equations:
v 2 − v02 ( 2.0 m/s ) − ( 0 m/s )
=
= 0.68 m
v12 = v02 + 2a ( x1 − x0 ) ⇒ x1 − x0 = 1
2a
2 ( 2.94 m/s 2 )
2
2
The suitcase travels 0.68 m before catching up with the belt and riding smoothly.
Assess: If we imagine throwing a suitcase at a speed of 2.0 m/s onto a motionless surface, 0.68 m seems a
reasonable distance for it to slide before stopping.
6.44. Model: The box of shingles is a particle subject to Newton’s laws and kinematics.
Visualize:
Solve: Newton’s laws can be used in the coordinate system in which the direction of motion of the box of
JG
shingles defines the +x-axis. The angle that FG makes with the –y-axis is 25°.
( ∑ F ) = F sin25° − f = ma
G
x
k
( ∑ F ) = n − F cos25° = 0 ⇒ n = F cos25°
y
G
G
We have used the observation that the shingles do not leap off the roof, so the acceleration in the y-direction is
zero. Combining these equations with f k = μ k n and FG = mg yields
mg sin 25° − μ k mg cos 25° = ma
⇒ a = ( sin25° − μ k cos25° ) g = −0.743 m/s 2
where the minus sign indicates the acceleration is directed up the incline. The required initial speed to have the
box come to rest after 5.0 m is found from kinematics.
vf2 = vi2 + 2aΔx ⇒ vi2 = −2 ( −0.743 m/s 2 ) ( 5.0 m ) ⇒ vi = 2.7 m/s
Assess: To give the shingles an initial speed of 2.7 m/s requires a strong, determined push, but is not beyond
reasonable.
6.45.
Model:
Visualize:
We will model the box as a particle, and use the models of kinetic and static friction.
The pushing force is along the +x-axis, but the force of friction acts along the –x-axis. A component of the
gravitational force on the box acts along the –x-axis as well. The box will move up if the pushing force is at least
equal to the sum of the friction force and the component of the gravitational force in the x-direction.
Solve: Let’s determine how much pushing force you would need to keep the box moving up the ramp at steady
speed. Newton’s second law for the box in dynamic equilibrium is
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( Fpush ) x = 0 N − mg sinθ − f k + Fpush = 0 N
( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( Fpush ) y = n − mg cosθ + 0 N + 0 N = 0 N
The x-component equation and the model of kinetic friction yield:
Fpush = mg sinθ + f k = mg sinθ + μ k n
Let us obtain n from the y-component equation as n = mg cosθ , and substitute it in the above equation to get
Fpush = mg sinθ + μ k mg cosθ = mg (sinθ + μ k cosθ )
= (100 kg)(9.80 m/s 2 )(sin 20° + 0.60 cos 20°) = 888 N
The force is less than your maximum pushing force of 1000 N. That is, once in motion, the box could be kept
moving up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of static
friction applies. The analysis is the same except that the coefficient of static friction is used and we use the
maximum value of the force of static friction. Therefore, we have
Fpush = mg (sinθ + μ s cosθ ) = (100 kg)(9.80 m/s 2 )(sin 20° + 0.90 cos 20°) = 1160 N
Since you can push with a force of only 1000 N, you can’t get the box started. The big static friction force and
the weight are too much to overcome.
6.46.
Model: We will represent the wood block as a particle, and use the model of kinetic friction and
kinematics. Assume w sin θ > fs , so it does not hang up at the top.
Visualize:
The block ends where it starts, so x2 = x0 = 0 m. We expect v2 to be negative, because the block will be moving
in the
–x-direction, so we’ll want to take v2 as the final speed. Because of friction, we expect to find v2 < v0 .
G
→
Solve: (a) The friction force is opposite to v, so f k points down the slope during the first half of the motion
G
G
and up the slope during the second half. FG and n are the only other forces. Newton’s second law for the
upward motion is
ax = a0 =
( Fnet ) x
m
a y = 0 m/s 2 =
=
− FG sinθ − f k − mg sinθ − f k
=
m
m
( Fnet ) y
m
=
n − FG cosθ n − mg cosθ
=
m
m
The friction model is f k = μ k n. First solve the y-equation to give n = mg cosθ . Use this in the friction model to
get f k = μ k mg cosθ . Now substitute this result for f k into the x-equation:
a0 =
−mg sin θ − μ k mg cosθ
= − g ( sinθ + μ k cosθ ) = − ( 9.8 m/s 2 ) ( sin 30° + 0.20cos30° ) = −6.60 m/s 2
m
Kinematics now gives
v 2 − v02 0 m /s − (10 m/s )
=
= 7.6 m
v12 = v02 + 2a0 ( x1 − x0 ) ⇒ x1 = 1
2a0
2 ( −6.60 m/s 2 )
2
2
2
The block’s height is then h = x1 sinθ = (7.6 m)sin 30° = 3.8 m.
G
(b) For the return trip, f k points up the slope, so the x-component of the second law is
ax = a1 =
( Fnet ) x
m
=
− FG sin θ + f k −mg sin θ + f k
=
m
m
Note the sign change. The y-equation and the friction model are unchanged, so we have
a1 = − g ( sin θ − μ k cosθ ) = −3.20 m/s 2
The kinematics for the return trip are
v22 = v12 + 2a1 ( x2 − x1 ) ⇒ v2 = −2a1 x1 = 2 ( −3.20 m/s 2 ) ( −7.6 m ) = −7.0 m/s
Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction.
The final speed is v2 = 7.0 m/s.
6.47.
Model: We will model the sled and friend as a particle, and use the model of kinetic friction because
the sled is in motion.
Visualize:
The net force on the sled is zero (note the constant speed of the sled). That means the component of the pulling
force along the +x-direction is equal to the magnitude of the kinetic force of friction in the –x-direction. Also
note that ( Fnet ) y = 0 N, since the sled is not moving along the y-axis.
Solve:
Newton’s second law is
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( Fpull ) x = 0 N + 0 N − f k + Fpull cosθ = 0 N
( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( Fpull ) y = n − mg + 0 N + Fpull sin θ = 0 N
The x-component equation using the kinetic friction model f k = μ k n reduces to
μ k n = Fpull cosθ
The y-component equation gives
n = mg − Fpull sin θ
We see that the normal force is smaller than the gravitational force because Fpull has a component in a direction
opposite to the direction of the gravitational force. In other words, Fpull is partly lifting the sled. From the xcomponent equation, μ k can now be obtained as
μk =
Fpull cosθ
mg − Fpull sinθ
=
( 75 N )( cos30° )
= 0.12
( 60 kg ) ( 9.80 m/s 2 ) − ( 75 N )( sin 30° )
Assess: A quick glance at the various μ k values in Table 6.1 suggests that a value of 0.12 for μ k is
reasonable.
6.48.
Model: As long as the static force of friction between the box and the sled is sufficiently large for the
box not to slip, the acceleration a of the box is the same as the acceleration of the sled. We will therefore model
the box as a particle and use the model of static friction.
Visualize:
The force on the box that is responsible for its acceleration is the force of static friction provided by the sled
because the box would slide backward without this force.
Solve: Newton’s second law for the box is
( Fnet ) x = ΣFx = nx + ( ( FG )box ) + ( fs ) x = 0 N + 0 N + fs max = max = mamax
x
( Fnet ) y = ΣFy = n y + ( ( FG )box ) + ( fs ) y = n − mg + 0 N = ma y = 0 N
y
The above equations can be combined along with the friction model to get
amax =
fs max μs n μ s mg
=
=
= μs g
m
m
m
To find Tmax , we write Newton’s second law for the box-sled system, with total mass m + M , as
( Fnet ) x = ΣFx = N x + ( ( FG )sled+box ) + ( fs ) x + (Tmax ) x = 0 N + 0 N + 0 N + Tmax = (m + M )amax
x
( Fnet ) y = ΣFy = N y + ( ( FG )sled+box ) + ( fs ) y + (Tmax ) y = N − (m + M ) g + 0 N + 0 N = 0 N
y
Referring to Table 6.1 for the coefficient of friction, the x-component equation for the box-sled system yields
Tmax = (m + M )amax = ( m + M ) μs g = (0.50)(9.80 m/s 2 )(5.0 kg + 10 kg) = 74 N
That is, the largest tension force for which the box does not slip is 74 N.
Assess: If the box were glued to the sled, there would be no resistance to motion because the ice is frictionless,
and the tension force could be as large as one wants. On the other hand, if the friction between the box and the
sled were zero, one could not pull at all without causing the box to slide.
6.49.
Model: We will model the steel cabinet as a particle. It touches the truck’s steel bed, so only the steel
bed can exert contact forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet
is equal to the acceleration of the truck.
Visualize:
Solve: The shortest stopping distance is the distance for which the static friction force has its maximum value
( fs )max . Newton’s second law for the box and the model of static friction are
( Fnet ) x = ΣFx = nx + ( FG ) x + ( fs ) x = 0 N + 0 N − ( fs )max = max = ma ⇒ − ( f s )max = ma
( Fnet ) y = ΣFy = n y + ( FG ) y + ( f s ) y = n − mg + 0 N = 0 N ⇒ n = mg
− ( f s )max = − μs n = − μs mg = ma
These three equations can be combined together to get a = − μ s g . Because constant-acceleration kinematics
gives v12 = v02 + 2a ( x1 − x0 ) and μ s = 0.80 (Table 6.1), we find
v2 − v2
−v02
(15 m/s )
=
= 14.3 m
( x1 − x0 ) = 1 0 =
g
−
μ
2a
2
2
0.80
( s ) ( )( ) ( 9.80 m/s 2 )
2
Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance without
making the contents slide of about 14.3 m or approximately 47 feet looks reasonable.
6.50. Model: The antiques (mass = m) in the back of your pickup (mass = M ) will be treated as a particle.
The antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The
pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the
road.
Visualize:
Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare
that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M ,
Newton’s second law is
( Fnet ) x = ΣFx = N x + ( ( FG )PA ) + ( f ) x = 0 N + 0 N − f = ( m + M ) ax = (m + M )a
x
( Fnet ) y = ΣFy = N y + ( ( FG )PA ) + ( f ) y = N − (m + M ) g + 0 N = 0 N
y
The model of static friction is f = μ N , where μ is the coefficient of friction between the tires and the road.
These equations can be combined to yield a = − μ g . Since constant-acceleration kinematics gives
v12 = v02 + 2a( x1 + x0 ), we find
v12 − v02
v02
( 25 m/s )
⇒ μ min =
=
= 0.58
2 ( x1 − x0 )
2 g ( x1 − x0 ) ( 2 ) ( 9.8 m/s 2 ) ( 55 m )
2
a=
The truck cannot stop if μ is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and
0.80 respectively (see Table 6.1), are larger. So the truck can stop.
(b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for
μ min . This value is smaller than the given coefficient of static friction ( μs = 0.60) between the antiques and the
truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m.
Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According
to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle
traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to
stop the truck.
6.51.
Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we
will also use the model of kinetic friction.
Visualize:
Solve: The normal force due to the wall, Gwhich is perpendicular to the wall, is here to the right.
JG The
G box slides
G
down the wall at constant speed, so a = 0 and the box is in dynamic equilibrium. Thus, Fnet = 0. Newton’s
second law for this equilibrium situation is
( Fnet ) x = 0 N = n − Fpush cos 45°
( Fnet ) y = 0 N = f k + Fpush sin 45° − FG = f k + Fpush sin 45° − mg
The friction force is f k = μ k n. Using the x-equation to get an expression for n, we see that f k = μ k Fpush cos 45°.
Substituting this into the y-equation and using Table 6.1 to find μ k = 0.20 gives,
μ k Fpush cos 45° + Fpush sin 45° − mg = 0 N
⇒ Fpush =
( 2.0 kg ) ( 9.80 m/s 2 )
mg
=
= 23 N
μ k cos 45° + sin 45° 0.20cos 45° + sin 45°
6.52.
Model:
Visualize:
Use the particle model for the block and the model of static friction.
Solve: The block is initially at rest, so initially the friction force is static friction. If the 12 N push is too strong,
the box will begin to move up the wall. If it is too weak, the box will begin to slide down the wall. And if the
pushing force is within the proper range, the box will remain stuck in place. First, let’s evaluate the sum of all the
forces except friction:
ΣFx = n − Fpush cos30° = 0 N ⇒ n = Fpush cos30°
ΣFy = Fpush sin 30° − FG = Fpush sin 30° − mg = (12 N)sin 30° − (1 kg)(9.8 m/s 2 ) = −3.8 N
In the first equation we utilized the fact that any motion is parallel to the wall, so ax = 0 m/s 2 . These three forces
add up to −3.8 ˆj N. This means the static friction force will be able to prevent the box from moving if
fs = +3.8 ˆj N. Using the x-equation and the friction model we get
( fs )max = μs n = μs Fpush cos30° = 5.2 N
G
where we used μ s = 0.5 for wood on wood. The static friction force fs needed to keep the box from moving is
less than ( fs )max . Thus the box will stay at rest.
6.53.
Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a
contact force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction
and drag.
Visualize:
We choose a coordinate system such that the skier’s motion is along the +x-direction. While the forces of kinetic
G
G
friction f k and drag D act along the –x-direction opposing the motion of the skier, the gravitational force on the
skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces
along the +x-direction cancel out the forces along the –x-direction.
Solve: Newton’s second law and the models of kinetic friction and drag are
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( D) x = 0 N + mg sin θ − f k
1 2
Av = max = 0 N
4
( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( D) y = n − mg cosθ + 0 N + 0 N = 0 N
fk = μk n
These three equations can be combined together as follows:
(1/4) Av 2 = mg sin θ − f k = mg sin θ − μ k n = mg sinθ − μ k mg cosθ
12
⎛
sinθ − μ k cosθ ⎞
⇒ vterm = ⎜ mg
⎟
1
A
4
⎝
⎠
Using μ k = 0.06 and A = 1.8 m × 0.40 m = 0.72 m 2 , we find
⎡
⎛ sin 40° − 0.06cos 40° ⎞ ⎤
⎟⎥
vterm = ⎢( 80 kg ) ( 9.8 m/s 2 ) ⎜ 1
⎜ ( 4 kg m3 )( 0.72 m 2 ) ⎟ ⎥
⎢
⎝
⎠⎦
⎣
12
= 51 m/s
Assess: A terminal speed of 51 m/s corresponds to a speed of ≈100 mph. This speed is
reasonable but high due to the steep slope angle of 40° and a small coefficient of friction.
6.54. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity.
Visualize:
Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus
the drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot
straight up,
⎞
( ∑ F ) = ma = − ( F + D ) = − ⎛⎜ mg + 14 Av ⎞⎟ = −mg − 14 A ( 2v ) = −mg − A ⎛⎜ 4mg
⎟ = −5mg
A
2
2
y
G
⎝
term
⎠
⎝
⎠
Thus a = −5 g , where the minus sign indicates the downward direction. We have used Equations 6.16 for the
drag force and 6.19 for the terminal velocity.
(b) When the ball is shot straight down,
( ∑ F ) = ma = D − F = 14 A ( 2v ) − mg = 3mg
2
y
G
term
Thus a = 3 g , this time directed upward.
(c)
The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the
terminal velocity because the drag force decreases as the ball slows.
6.55. Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will
be assumed to have zero mass.
Visualize:
Solve:
Newton’s first law in component form is
( Fnet ) x = ΣFx = T1x + T2 x + FGx = −T1 sin 30° + T2 sin 60° + 0 N = 0 N
( Fnet ) y = ΣFy = T1 y + T2 y + FGy = −T1 cos30° + T2 cos60° − FG = 0 N
Using the x-component equation to obtain an expression for T1 and substituting into the y-component equation
yields:
T2 =
FG
500 lbs
=
= 250 lbs
( sin 60°)( cos30°) + cos60°
2
sin 30°
Substituting this value of T2 back into the x-component equation,
T1 = T2
sin 60°
sin 60°
= 250 lbs
= 433 lbs
sin 30°
sin 30°
We will now find a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of
433 lbs. Since the cross-sectional area of the rope is 14 π d 2 , we have
12
⎡
⎤
4 ( 433 lbs )
⎥
d =⎢
2
⎢⎣ π ( 4000 lbs/inch ) ⎥⎦
= 0.371 inch
Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to
a diameter of 3/8 of an inch will therefore be appropriate.
Assess: If only a single rope were used to hang the sculpture, the rope would have to support a gravitational
force of 500 lbs. The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps
from a diameter of 3/8 to 4/8 of an inch. Also note that the gravitational force on the sculpture is distributed in
the two ropes. It is the sum of the y-components of the tensions in the ropes that will equal the gravitational force
on the sculpture.
6.56. Model: We will model the container as a particle of mass m. The steel cable of the crane will be
assumed to have zero mass.
Visualize:
Solve: As long as the container is stationary or it is moving with a constant speed (zero acceleration), the net
force on the container is zero. In these cases, the tension in the cable is equal to the gravitational force on the
container:
T = mg = 44,000 N
The cable should safely lift the load. More tension is required to accelerate the load. Newton’s second law is
( Fnet ) y = ΣFy = ( FG ) y + (T ) y = − mg + T = ma y
The crane’s maximum acceleration is amax = 1.0 m/s 2 . So the maximum cable tension is
Tmax = mg + mamax = 48,600 N
This is less than the cable’s rating, so the cable must have been defective.
6.57.
Model:
Visualize:
Solve:
We will model the skier as a particle, and use the model of kinetic friction.
Your best strategy, if it’s possible, is to travel at a very slow constant speed
G
G
G
( aG = 0 so F = 0).
net
Alternatively, you want the smallest positive ax . A negative ax would cause you to slow and stop. Let’s find the
G
G
value of μ k that gives Fnet = 0.
Newton’s second law for the skier and the model of kinetic friction are
( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( D) x = 0 + mg sin θ − f k − D cosθ = 0 N
( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( D ) y = n − mg cosθ + 0 N − D sin θ = 0 N
fk = μk n
The x- and y-component equations are
f k = + mg sin θ − D cosθ n = mg cosθ + D sin θ
From the model of kinetic friction,
μk =
2
f k mg sin θ − D cosθ 75 kg ( 9.8 m/s ) sin15° − ( 50 N ) cos15°
=
=
= 0.196
n mg cosθ + D sin θ 75 kg ( 9.8 m/s 2 ) cos15° + ( 50 N ) sin15°
Yellow wax with μ k = 0.20 applied to skis will make the skis stick and hence cause the skier to stop. The skier’s
next choice is to use the green wax with μ k = 0.15.
6.58.
Model:
Visualize:
The ball hanging from the ceiling of the truck by a string is represented as a particle.
Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between
“at rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight
down.
(b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle.
(c) The ball moves with the truck, so its acceleration is 5 m/s 2 in the forward direction.
G
(d) The free-body diagram shows that the horizontal component of T provides a net force in the forward
direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck.
(e) Newton’s second law for the ball is
ax =
( Fnet ) x
m
=
Tx T sin10°
=
m
m
a y = 0 m 2 /s 2 =
( Fnet ) y
m
=
Ty + ( FG ) y
m
We can solve the second equation for the magnitude of the tension:
(1.0 kg ) ( 9.8 m/s )
mg
=
= 9.95 N
cos10°
cos10°
2
T=
Then the first equation gives the acceleration of the ball and truck:
ax =
T sin10° ( 9.95 N ) sin10°
=
=1.73 m/s 2
m
m
The truck’s velocity cannot be determined.
=
T cos10° − mg
m
6.59.
Model: You will be treated as a particle. You will experience your weight force and the force (P) of
the scale pulling up.
Visualize:
G
Solve: The weight of an object is the magnitude of the contact force supporting it. Here, the contact force is P,
so the weight is w = P. Newton’s second law is
⎛ a⎞
Fnet = ΣFy = Py + ( FG ) y = ma y ⇒ Fnet = P − FG = ma ⇒ w = P = mg + ma = mg ⎜1 + ⎟
⎝ g⎠
6.60. Solve: Using ax =
dvx
, we express Newton’s second law as a differential equation, which we then use
dt
to solve for vx .
Fx = m
dvx
F
ct
⇒ dvx = x dt = dt
dt
m
m
Integrating from the initial to final conditions for each variable of integration,
vx
∫ dvx =
v0 x
Thus
t
c
ct 2
t
dt
⇒
v
−
v
=
x
x
0
m ∫0
2m
vx = v0 x +
ct 2
2m
6.61. Model: The astronaut is a particle oscillating on a spring.
Solve:
(a) The position versus time function x(t) can be used to find the velocity versus time function v(t ) =
We have
v(t ) =
d
( 0.30 m ) sin ( (π rad/s ) t ) = ( 0.30π m/s ) cos ( (π rad/s ) t )
dt
{
}
This can then be used to find the acceleration a (t ) =
a (t ) =
dv
.
dt
dv
= − ( 0.30π 2 m/s 2 ) sin ( (π rad/s ) t )
dt
Newton’s second law yields a general expression for the force on the astronaut.
Fnet (t ) = ma (t ) = −(75 kg) ( 0.30π 2 m/s 2 ) sin ( (π rad/s ) t )
Evaluating this at t = 1.0 s gives Fnet (1.0 s ) = 0 N, since sin(π ) = 0.
(b) Evaluating at t = 1.5 s,
⎛ 3π ⎞
Fnet = −22.5π 2 N sin ⎜ ⎟ = 2.2 × 102 N
⎝ 2 ⎠
Assess:
The force of 220 N is only one third of the astronaut’s weight on earth, so is easy for her to withstand.
dx
.
dt
6.62. Solve: (a) The terminal velocity for a falling object is reached when the downward gravitational force
is balanced by the upward drag force.
FG = D
mg = bvterm = 6πη Rvterm
⇒ vterm =
mg
6πη R
⎛4
⎞
(b) The mass of the spherical sand grain of density p = 2400 kg/m3 is m = ρ ⎜ π R 3 ⎟ .
⎝3
⎠
2 ρ gR 2 2 ( 2400 kg/m )( 9.80 m/s )( 5.0 × 10
=
Thus
vterm =
⎛
9η
9
−3 Ns ⎞
⎜ 1.0 × 10
⎟
m2 ⎠
⎝
50 m
= 38 s.
The time required for the sand grain to fall 50 m at this speed is t =
1.3 m/s
Assess: The speed of 1.3 m/s for a sand grain falling through water seems about right.
3
2
−4
m)
2
= 1.3 m/s
6.63.
Solve: (a) A 1.0 kg block is pulled across a level surface by a string, starting from rest. The string has a
tension of 20 N, and the block’s coefficient of kinetic friction is 0.50. How long does it take the block to move
1.0 m?
(b) Newton’s second law for the block is
ax = a =
( Fnet ) x
m
=
T − fk T − μk n
=
m
m
a y = 0 m/s 2 =
( Fnet ) y
m
=
n − FG n − mg
=
m
m
where we have incorporated the friction model into the first equation. The second equation gives n = mg .
Substituting this into the first equation gives
a=
T − μ k mg 20 N − 4.9 N
=
= 15.1 m/s 2
m
1.0 kg
Constant acceleration kinematics gives
2 (1.0 m )
1
1
2 x1
2
2
x1 = x0 + v0 Δt + a ( Δt ) = a ( Δt ) ⇒ Δt =
=
= 0.36 s
a
2
2
15.1 m/s 2
6.64.
Solve: (a) A 15,000 N truck starts from rest and moves down a 15° hill with the engine providing a
12,000 N force in the direction of the motion. Assume the frictional force between the truck and the road is very
small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill?
(b) Newton’s second law is
ΣFy = ny + FGy + f y + E y = ma y = 0
ΣFx = nx + FGx + f x + Ex = max ⇒ 0 N + FG sin θ + 0 N + 12,000 N = ma
⇒a=
mg sin θ + 12,000 N (15,000 N ) sin15° + 12,000 N
=
= 10.4 m/s 2
m
(15,000 N/9.8 m/s2 )
where we have calculated the mass of the truck from the gravitational force on it. Using the constant-acceleration
kinematic equation vx2 − v02 = 2ax,
vx2 = 2ax x = 2 (10.4 m/s 2 ) ( 50 m ) ⇒ vx = 32 m/s
6.65. Solve: (a) A 1.0 kg box is pushed along an assembly line by a mechanical arm. The arm exerts a 20 N
force at a downward angle of 30°. The box has a coefficient of kinetic friction on the assembly line of 0.25. What
is the speed of the box after traveling 50 cm, starting from rest?
(b) Newton’s second law for the box is
ax = a =
( Fnet ) x
m
a y = 0 m/s 2 =
=
Fpush cos30° − f k
( Fnet ) y
m
=
Fpush cos30° − μ k n
m
m
n − FG − Fpush sin 30° n − mg − Fpush sin 30°
=
=
m
m
The second equation is solved to give an expression for n. Substituting into the first equation:
a=
Fpush cos30° − μ k ( Fpush sin 30° + mg )
m
=
( 20 N ) cos30° − 4.9 N = 12.4 m/s 2
1.0 kg
Using kinematics,
v12 = v02 + 2aΔx = 2aΔx ⇒ v1 = 2aΔx = 2 (12.4 m/s 2 ) ( 0.50 m ) = 3.5 m/s
6.66. Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How
far does the auto slide before coming to rest?
(b)
(c) Newton’s second law is
ΣFy = n y + ( FG ) y = n − mg = ma y = 0 N
ΣFx = −0.80n = max
The y-component equation gives n = mg = (1500 kg ) ( 9.8 m/s 2 ) . Substituting this into the x-component equation
yields
(1500 kg ) ax = −0.80 (1500 kg ) ( 9.8 m/s 2 ) ⇒ ax = ( −0.80 ) ( 9.8 m/s2 ) = −7.8 m/s 2
Using the constant-acceleration kinematic equation v12 = v02 + 2aΔx, we find
v2
( 40 m/s ) = 102 m
Δx = − 0 = −
2a
2 ( −7.8 m/s 2 )
2
6.67. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20° wooden incline by a rope that is connected
to an electric motor. The crate’s acceleration is measured to be 2.0 m/s 2 . The coefficient of kinetic friction
between the crate and the incline is 0.20. Find the tension T in the rope.
(b)
(c) Newton’s second law for this problem in the component form is
( Fnet ) x = ΣFx = T − 0.20n − (20 kg)(9.80 m/s 2 )sin 20° = (20 kg)(2.0 m/s 2 )
( Fnet ) y = ΣFy = n − (20 kg)(9.80 m/s 2 )cos 20° = 0 N
Solving the y-component equation, n = 184.18 N. Substituting this value for n in the x-component equation yields
T = 144 N.
6.68.
Solve: (a) You wish to pull a 20 kg wooden crate across a wood floor ( μ k = 0.20) by pulling on a rope
attached to the crate. Your pull is 100 N at an angle of 30° above the horizontal. What will be the acceleration of
the crate?
(b)
(c) Newton’s equations and the model of kinetic friction are
ΣFx = nx + Px + ( FG ) x + f x = 0 N + (100 N ) cos30° + 0 N − f k = (100 N ) cos30° − f k = max
ΣFy = n y + Py + ( FG ) y + f y = n + (100 N ) sin 30° − mg − 0 N = ma y = 0 N
fk = μk n
From the y-component equation, n = 150 N. From the x-component equation and using the model of kinetic
friction with μ k = 0.20,
(100 N)cos30° − (0.20)(150 N) = (20 kg) ax ⇒ ax = 2.8 m/s 2
6.69.
Model: Take the coin to be a particle held against the palm of your hand (which is like a vertical wall)
by friction. The friction needs to be vertical and equal to the weight of the coin to hold the coin in place.
Visualize:
The hand pushes on the coin giving a normal force to the coin and causing a friction force.
Solve: (a) We need a force (push) to create a normal force. Since the force accelerates the coin, a minimum
acceleration amin is needed.
(b) Newton’s second law and the model of friction are
ΣFy = ( f min ) y − ( FG ) y = f min − mg = 0 N
ΣFx = nmin = mamin f min = μ nmin
Since you do not want the coin to slip down the hand you need μs . Combining the above three equations yields
f min = mg ⇒ μ s nmin = mg ⇒ μ s mamin = mg ⇒ amin =
g
μs
=
9.8 m/s 2
= 12.3 m/s 2
0.80
6.70.
Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also
use the model of kinetic friction for the motion of the shuttle along the square steel rail.
Visualize:
Solve:
The upward tension component Ty = T sin 45° = 14.1 N is larger than the gravitational force on the
shuttle. Consequently, the elastic cord pulls the shuttle up against the rail and the rail’s normal force pushes
downward. Newton’s second law in component form is
( Fnet ) x = ΣFx = Tx + ( f k ) x + (n) x + ( FG ) x = T cos 45° − f k + 0 N + 0 N = max = max
( Fnet ) y = ΣFy = Ty + ( f k ) y + ( n) y + ( FG ) y = T sin 45° + 0 N − n − mg = ma y = 0 N
The model of kinetic friction is f k = μ k n. We use the y-component equation to get an expression for n and hence
f k . Substituting into the x-component equation and using the value of μ k in Table 6.1 gives us
T cos 45° − μ k (T sin 45° − mg )
m
( 20 N ) cos 45° − ( 0.60 ) ⎡⎣ + ( 20 N ) sin 45° − ( 0.800 kg ) ( 9.80 m/s 2 )⎤⎦
=
= 13.0 m/s 2
0.800 kg
ax =
Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in the xdirection is max = (0.800 kg)(13.0 m/s 2 ) = 10.4 N. This value for ma is reasonable since a part of the 14.1 N
tension force is used up to overcome the force of kinetic friction.
6.71.
Model: Assume the ball is a particle on a slope, and that the slope increases as the x-displacement
increases. Assume that there is no friction and that the ball is being accelerated to the right so that it remains at
rest on the slope.
Visualize: Although the ball is on a slope, it is accelerating to the right. Thus we’ll use a coordinate system
with horizontal and vertical axes.
Solve:
Newton’s second law is
ΣFx = n sin θ = max
ΣFy = n cosθ − FG = ma y = 0 N
Combining the two equations, we get
max =
FG
sin θ = mg tan θ ⇒ ax = g tan θ
cosθ
The curve is described by y = x 2 . Its slope a position x is tanθ, which is also the derivative of the curve. Hence,
dy
= tan θ = 2 x ⇒ ax = (2 x) g
dx
(b) The acceleration at x = 0.20 m is ax = (2)(0.20)(9.8 m/s 2 ) = 3.9 m/s 2 .
6.72.
Model:
Visualize:
We will represent the widget as a particle.
Please refer to Figure CP6.72.
Solve: (a) There are only two forces on the widget: the normal force of the table and the gravitational force.
(b) Newton’s second law along the y-axis is
( Fnet ) y = ny + ( FG ) y = ny − mg = ma y ⇒ ny = m ( a y + g )
We know what the a y -vs-t graph looks like. To get the ny -vs-t graph, we need to add g y to the graph, which
amounts to shifting the whole graph up by 9.8 m/s 2 , and multiplying by m = 5 kg.
(c) The normal force is negative for t > 0.75 s. Physically, this means that the normal force is pointed in the
downward direction. In other words, the table is pulling down on the widget rather than pushing up on the
widget. It can do this because the widget is glued to the table rather than simply sitting on the table.
(d) The weight is a maximum at t = 0 s, when the upward acceleration is maximum.
(e) The weight is zero at t = 0.75 s when a y = −9.8 m/s 2 = − g .
(f) If not glued down, the widget will fly off the table at t = 0.75 s, the instant at which its weight becomes zero.
The table is accelerating in the negative direction so quickly after t = 0.75 s that the widget can stay on only if
the table pulls downward on it. That is the significance of the negative value for ny . If the widget is not glued
down, the largest downward acceleration it can achieve is the free-fall acceleration.
6.73. Visualize:
Solve: (a) The horizontal velocity as a function of time is determined by the horizontal net force. Newton’s
second law as the x-direction gives
( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx
G
JJG
Note that D points opposite to v, so the angle θ with the x-axis is the same for both vectors, and the x
components of both vectors have the same cosθ term. As the particle changes direction as it falls, the evolution
of the horizontal motion depends only on the horizontal component of the velocity.
Thus
m
vx ( t )
Separating and integrating,
dvx
b
dvx
= −bvx
dt
t
∫ v = − m ∫ dt
v0
x
0
⎛ v (t ) ⎞
b
⇒ ln ⎜ x ⎟ = − t
v
m
⎝ 0 ⎠
Solving,
−
bt
vx ( t ) = v0e m = v0e
−
6πη Rt
m
1
(b) The time to reach v ( t ) = v0 is found by solving for the time when
2
6πη Rt
−
1
v0 = v0e m
2
Hence
t=
m ln ( 2 )
6πη R
With η = 1.0 × 10−3 Ns/m 2 , R = 2.0 × 10−2 m, and m = 0.033 kg, we get t = 61 s.
6πη R
v = 1.1× 10−2 s-1 vx . This is a small fraction of the
Assess: The magnitude of the acceleration is ax =
m
velocity, so a time of about one minute to slow to half the initial speed is reasonable.
(
)
6.74. Visualize:
dvx ⎛ dvx ⎞⎛ dx ⎞
dvx
.
=⎜
⎟⎜ ⎟ = vx
dt ⎝ dx ⎠⎝ dt ⎠
dx
(b) The horizontal motion is determined by using Newton’s second law in the horizontal direction. Using the
free-body diagram at a later time t,
( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx
G
JJG
Note that since D points opposite to v, the angle θ with the x-axis is the same for both vectors, and the xcomponents of both vectors have the same cosθ term. Thus
Solve:
(a) Using the chain rule, ax =
max = mvx
dvx
= −bvx
dx
vx ( x )
∫ dvx = −
Separating and integrating,
v0
x (t )
b
dx
m x∫0
⇒ vx ( x ) − v0 = −
b
( x ( t ) − x0 )
m
Solving with x0 = 0,
vx ( x ) = v0 −
b
6πη R
x = v0 −
x
m
m
(c) The marble stops after traveling a distance d when vx ( d ) = 0.
v0 =
6πη R
d
m
⇒d =
mv0
6πη R
Hence
Using v0 = 10 cm/s, R = 0.50 cm, m = 1.0 × 10−3 kg, and using η = 1.0 × 10−3 Ns/m,
(1.0 ×10 kg ) ( 0.10 m/s ) = 1.1 m
6π (1.0 × 10 Ns/m )( 5.0 × 10 m )
−3
d=
Assess:
−3
2
−3
The equation for d indicates that a marble with a faster initial velocity travels a further distance.
6.75.
Model:
Visualize:
We will model the object as a particle, and use the model of drag.
Solve: (a) We cannot use the constant-acceleration kinematic equations since the drag force causes the
acceleration to change with time. Instead, we must use ax = dvx /dt and integrate to find vx . Newton’s second
law for the object is
( Fnet ) x = ΣFx = nx + wx + ( f k ) x + ( D) x = 0 N + 0 N + 0 N −
1 2
dv
Avx = max = m x
4
dt
This can be written
dvx
A
=
dt
2
vx
4m
We can integrate this from the start (v0 x at t = 0) to the end (vx at t ):
∫
vx
v0 x
dvx
A t
1
1
A
=
dt ⇒ − +
=
t
2
∫
0
vx
4m
vx v0 x 4m
Solving for vx gives
vx =
v0 x
1 + Av0 xt 4m
(b) Using A = (1.6 m)(1.4 m) = 2.24 m 2 , v0 x = 20 m/s, and m = 1500 kg, we get
vx =
20 m/s
1
20
⎛
⎞ ⎛ 20 m/s ⎞
⇒t =⎜
− 1⎟
=
⎟⎜
2.24t ( 20 ) 1 + 0.007467t
0.007467
⎝
⎠ ⎝ vx
⎠
1+
4 × 1500
where t is in seconds. We can now obtain the time t for v = 10 m/s:
1
⎛
⎞ ⎛ 20 m/s ⎞
⎛ 20 ⎞
t =⎜
− 1⎟ = 133.93⎜ − 1⎟ = 134 s
⎟⎜
⎝ 0.007467 ⎠ ⎝ 10 m/s ⎠
⎝ 10 ⎠
When vx = 5 m/s, then t = 402 s.
(c) If the only force acting on the object was kinetic friction with, say, μ k = 0.05, that force
would be (0.05)(1500 kg) (9.8 m/s 2 ) = 735 N. The drag force at an average speed of 10 m/s
is D = 14 ( 2.24 )(10 )2 N = 56 N. We conclude that it is not reasonable to neglect the kinetic
friction force.
7.1.
Visualize:
Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on
the barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight
lifter and the gravitational force downward. The weight lifter has a downward contact force from the barbells and
an upward one from the surface. Gravity also acts on the weight lifter.
(b) The system is the weight lifter and barbell, as indicated in the figure.
(c)
7.2.
Visualize:
Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational
force on them. The interaction forces between the two are equal and opposite.
(b) The system consists of the soccer ball and bowling ball, as indicated in the figure.
(c)
Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on the
other are part of an action/reaction pair, and therefore have equal magnitudes. Each ball’s acceleration due to the
forces on it is determined by Newton’s second law, a = Fnet / m, which depends on the mass. Since the masses of
the balls are different, their accelerations are different.
7.3.
Visualize:
Solve: (a) Both the mountain climber and bag of supplies have a normal force from the surface on them, as well as a
gravitational force vertically downward. The rope has gravity acting on it, along with pulls on each end from the
mountain climber and supply bag. Both the mountain climber and supply bag also experience a frictional force with the
surface of the mountain. In the case of the motionless mountain climber it is static friction, but the sliding supply bag
experiences kinetic friction.
(b) The system consists of the mountain climber, rope, and bag of supplies, as indicated in the figure.
(c)
Assess: Since the motion is along the surface, it is convenient to choose the x-coordinate axis along the surface.
The free-body diagram of the rope shows pulls that are slightly off the x-axis since the rope is not massless.
7.4.
Visualize:
Solve: (a) The car and rabbit both experience a normal force and friction from the floor and a gravitational
force from the Earth. The push each exerts on the other is a Newton’s third law force pair.
(b) The system consists of the car and stuffed rabbit, as indicated in the figure.
(c)
7.5.
Visualize: Please refer to Figure EX7.5.
Solve: (a) Gravity acts on both blocks, and where Block A is in contact with the floor there is a normal force and
friction. The string tension is the same on both blocks since the rope and pulley are massless, and the pulley is
frictionless. There are two third law pairs of forces at the surface where the two blocks meet. Block B pushes
against Block A with a normal force, while Block A has a reaction force that pushes back against Block B. There is
also friction between the two blocks at the surface.
(b) A string that will not stretch constrains the two blocks to accelerate at the same rate but in opposite
directions. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The
system consists of the two blocks, as indicated in the figure.
(c)
Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is
convenient since the one component of a is zero, simplifying the mathematical expression of Newton’s second
law.
7.6.
Visualize: Please refer to Figure EX7.6.
Solve: (a) For each block, there is a gravitational force with the earth, a normal force and kinetic friction with
the surface, and a tension force due to the rope.
(b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates
down the incline with the same acceleration that Block B has up the incline. The system consists of the two
blocks, as indicated in the figure.
(c)
Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is
convenient since then one component of a is zero, simplifying the mathematical expression of Newton’s second
law.
7.7. Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be
denoted by A and C, respectively, and they are separate systems. The launch pad is a part of the environment.
Visualize:
Solve:
(a) Newton’s second law for the astronaut is
∑( F ) = n
on A y
C on A
− ( FG )A = mA aA = 0 N ⇒ nC on A = ( FG )A = mA g
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA g = ( 80 kg ) ( 9.8 m/s 2 ) = 7.8 × 102 N
(b) Newton’s second law for the astronaut is:
∑( F ) = n
on A y
C on A
− ( FG )A = mA aA ⇒ nC on A = ( FG )A + mA aA = mA ( g + aA )
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA ( g + aA ) = ( 80 kg ) ( 9.8 m/s 2 + 10 m/s 2 ) = 1.6 × 103 N
Assess:
This is a reasonable value because the astronaut’s acceleration is more than g.
7.8. (a) Visualize: The upper magnet is labeled U, the lower magnet L. Each magnet exerts a long-range
magnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In
addition, the table experiences a normal force due to the surface.
(b) Solve:
Each object is in static equilibrium with ( Fnet ) = 0. Start with the lower magnet. Because
FU on L = 3( FG ) L = 6.0 N, equilibrium requires nT on L = 4.0 N. For the upper magnet, FL on U = FU on L = 6.0 N
because these are an action/ reaction pair. Equilibrium for the upper magnet requires nT on U = 8.0 N. For the
table, action/reaction pairs are nL on T = nT on L = 4.0 N and nU on T = nT on U = 8.0. The table’s gravitational force is
( FG )T = 20 N, leaving nS on T = 24 N in order for the table to be in equilibrium. Summarizing,
Upper magnet
Table
Lower magnet
( FG ) U = 2.0 N
nT on U = 8.0 N
FL on U = 6.0 N
( FG )T = 20 N
nU on T = 8.0 N
nL on T = 4.0 N
nS on T = 24 N
( FG ) L = 2.0 N
nT on L = 4.0 N
FU on L = 6.0 N
Assess: The result nS on T = 24 N makes sense. The combined gravitational force on the table and two magnets
is 24 N. Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the
total gravitational force on the table and magnets.
7.9. Model: The car and the truck will be modeled as particles and denoted by the symbols C and T,
respectively. The surface of the ground will be denoted by the symbol S.
Visualize:
Solve:
(a) The x-component of Newton’s second law for the car is
∑( F ) = F
on C x
− FT on C = mC aC
S on C
The x-component of Newton’s second law for the truck is
∑( F ) = F
on T x
C on T
= mT aT
Using aC = aT = a and FT on C = FC on T , we get
(F
C on S
⎛ 1 ⎞
⎛ 1 ⎞
− FC on T ) ⎜
⎟ = a ( FC on T ) ⎜
⎟=a
⎝ mT ⎠
⎝ mC ⎠
Combining these two equations,
(F
C on S
⎛ 1 ⎞
⎛ 1
⎛ 1 ⎞
⎛ 1 ⎞
1 ⎞
− FC on T ) ⎜
+
⎟ = ( FC on T ) ⎜
⎟ = ( FC on S ) ⎜
⎟
⎟ ⇒ FC on T ⎜
m
m
m
m
⎝ T⎠
T ⎠
⎝ C⎠
⎝ C
⎝ mC ⎠
⎛ mT ⎞
⎛
⎞
2000 kg
⇒ FC on T = ( FC on S ) ⎜
⎟ = ( 4500 N ) ⎜
⎟ = 3000 N
⎝ 1000 kg + 2000 kg ⎠
⎝ mC + mT ⎠
(b) Due to Newton’s third law, FT on C = 3000 N.
7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless
and along with the earth it is a part of the environment. The three blocks are our three systems of interest.
Visualize:
The force applied on block 1 is FA on 1 = 12 N. The acceleration for all the blocks is the same and is denoted by a.
Solve: (a) Newton’s second law for the three blocks along the x-direction is
∑( F ) = F
on 1 x
A on 1
− F2 on 1 = m1a ∑ ( Fon 2 ) = F1 on 2 − F3 on 2 = m2 a ∑ ( Fon 3 ) = F2 on 3 = m3a
x
x
Adding these three equations and using Newton’s third law ( F2 on 1 = F1 on 2 and F3 on 2 = F2 on 3 ), we get
FA on 1 = ( m1 + m2 + m3 ) a ⇒ (12 N ) = (1 kg + 2 kg + 3 kg ) a ⇒ a = 2 m/s 2
Using this value of a, the force equation on block 3 gives
F2 on 3 = m3a = ( 3 kg ) ( 2 m/s 2 ) = 6 N
(b) Substituting into the force equation on block 1,
12 N − F2 on 1 = (1 kg ) ( 2 m/s 2 ) ⇒ F2 on 1 = 10 N
Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the
2 kg block exerts on the 1 kg block is reasonable.
7.11. Model: The block (B) and the steel cable (C), the two objects of interest to us, are treated like particles.
The motion of these objects is governed by the constant-acceleration kinematic equations.
Visualize:
Solve:
Using v12x = v02x + 2ax ( x1 − x0 ) ,
( 4.0 m/s ) = 0 m 2 /s2 + 2ax ( 2.0 m ) ⇒ ax = 4.0 m/s 2
2
From the free-body diagram on the block:
∑( F ) = F
on B x
C on B
= mBax ⇒ FC on B = ( 20 kg ) ( 4.0 m/s 2 ) = 80 N
Also, according to Newton’s third law FB on C = FC on B = 80 N. Newton’s second law on the cable is:
∑( F ) = F − F
on C x
ext
B on C
= mC ax ⇒ 100 N − 80 N = mC ( 4.0 m/s 2 ) ⇒ mC = 5 kg
7.12. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume
the pulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the
pulley. The system is the man and the block.
Visualize:
Solve: Clearly the entire system remains in equilibrium since mB > mM . The block would move downward but
it is already on the ground. From the free-body diagrams, we can write down Newton’s second law in the vertical
direction as
∑( F ) = T
on M y
R on M
− ( FG )M = 0 N ⇒ TR on M = ( FG )M = ( 60 kg ) ( 9.8 m/s 2 ) = 588 N
Since the tension is the same on both sides, TB on R = TM on R = T = 588 N.
7.13. Model: Together the carp (C) and the trout (T) make up the system that will be represented through the
particle model. The fishing rod line (R) is assumed to be massless.
Visualize:
Solve: Jimmy’s pull T2 is larger than the total weight of the fish, so they accelerate upward. They are tied together,
so each fish has the same acceleration a. Newton’s second law along the y-direction for the carp and the trout is
∑( F ) = T − T − ( F ) = m a ∑( F ) = T − ( F ) = m a
on C y
2
1
G C
C
on T y
1
G T
T
Adding these two equations gives
a=
T2 − ( FG )C − ( FG )T
( mC + mT )
=
60 N − (1.5 kg ) ( 9.8 m/s 2 ) − ( 3 kg ) ( 9.8 m/s 2 )
1.5 kg + 3.0 kg
= 3.533 m/s 2
Substituting this value of acceleration back into the force equation for the trout, we find that
T1 = mT ( a + g ) = ( 3 kg ) ( 3.533 m/s 2 + 9.8 m/s 2 ) = 40 N
( FG )T = mT g = ( 3 kg ) ( 9.8 m/s 2 ) = 29.4 N ( FG )C = mC g = (1.5 kg ) ( 9.8 m/s 2 ) = 14.7 N
Thus, T2 > T1 > ( FG )T > ( FG )C .
7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must
therefore consider both the block of ice and the rope as objects in the system.
Visualize:
Solve:
G
The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same
acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts
on the rear end of the rope.
Newton’s second law along the x-axis for the ice block and the rope is
∑( F ) = F
on I x
R on I
= mI a = (10 kg ) ( 2.0 m/s 2 ) = 20 N
∑( F ) = F − F
on R x
ext
I on R
= mR a ⇒ Fext − FR on I = mR a
⇒ Fext = FR on I + mR a = 20 N + ( 0.500 kg ) ( 2.0 m/s 2 ) = 21 N
7.15. Model: The hanging block and the rail car are objects in the systems.
Visualize:
Solve: The mass of the rope is very small in comparison to the 2000-kg block, so we will assume a massless
G
G
rope. In that case, the forces T1 and T1′ act as if they are an action/reaction pair. The hanging block is in static
G
equilibrium, with Fnet = 0 N, so T1′ = mblock g = 19,600 N. The rail car with the pulley is also in static equilibrium:
T2 + T3 − T1 = 0 N
Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right. Now,
T1 = T1′ = 19,600 N by Newton’s third law. Also, the cable tension is T2 = T3 = T . Thus, T = 12 T1′ = 9800 N.
7.16. Visualize:
Solve: The rope is treated as two 1.0-kg interacting objects. At the midpoint of the rope, the rope has a tension
TB on T = TT on B ≡ T . Apply Newton’s first law to the bottom half of the rope to find T.
( Fnet ) y = 0 = T − ( FG ) B
⇒ T = mB g = (1.0 kg ) ( 9.80 m/s 2 ) = 9.8 N
Assess: 9.8 N is half the gravitational force on the whole rope. This is reasonable since the top half is holding
up the bottom half of the rope against gravity.
7.17. Model: The two hanging blocks, which can be modeled as particles, together with the two knots where
rope 1 meets with rope 2 and rope 2 meets with rope 3 form a system. All the four objects in the system are in
static equilibrium. The ropes are assumed to be massless.
Visualize:
Solve: (a) We will
consider both the two hanging blocks and the two knots. The blocks are G in static
G
equilibrium with Fnet = 0 N. Note that there are three action/reaction pairs. For Block 1 and Block 2, Fnet = 0 N
and we have
T4′ = ( FG )1 = m1 g T5′ = ( FG )2 = m2 g
Then, by Newton’s third law:
T4 = T4′ = m1 g T5 = T5′ = m2 g
The knots are also in equilibrium. Newton’s law applied to the left knot is
( Fnet ) x = T2 − T1 cosθ1 = 0 N ( Fnet ) y = T1 sinθ1 − T4 = T1 sinθ1 − m1g = 0 N
The y-equation gives T1 = m1 g sin θ1 . Substitute this into the x-equation to find
T2 =
m1 g cosθ1 m1 g
=
sin θ1
tanθ1
Newton’s law applied to the right knot is
( Fnet ) x = T3 cosθ 3 − T2′ = 0 N ( Fnet ) y = T3 sinθ 3 − T5 = T3 sinθ 3 − m2 g = 0 N
These can be combined just like the equations for the left knot to give
T2′ =
m2 g cosθ 3
mg
= 2
sin θ 3
tan θ 3
G
G
But the forces T2 and T2′ are an action/reaction pair, so T2 = T2′. Therefore,
m1 g
mg
m
= 2 ⇒ tanθ 3 = 2 tanθ1 ⇒ θ 3 = tan −1 ( 2 tan 20° ) = 36°
tan θ1 tanθ 3
m1
We can now use the y-equation for the right knot to find T3 = m2 g sinθ 3 = 67 N.
7.18. Visualize: Please refer to Figure P7.18.
Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton’s third law force
pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these
massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs
when the static friction on B is at its maximum value. Applying Newton’s first law to the vertical forces on block
B gives nB = ( FG ) B = mB g . The static friction force on B is thus
( fs )B = μs nB = μs mB g.
Applying Newton’s first law to the horizontal forces on B gives ( fs )B = TA on B , and the same analysis of the
vertical forces on A gives TB on A = ( FG )A = mA g. Since TA on B = TB on A , we have ( f s )B = mA g , so
μs mB g = mA g
⇒ mA = μs mB = ( 0.60 )( 20 kg ) = 12.0 kg
7.19. Model:
Visualize:
The astronaut and the satellite, the two objects in our system, will be treated as particles.
Solve: The astronaut and the satellite accelerate in opposite directions for 0.50 s. The force on the satellite and the
force on the astronaut are an action/reaction pair, so both are 100 N. Newton’s second law for the satellite along the
x-direction is
∑( F ) = F
on S x
A on S
= mSaS ⇒ aS =
FA on S
mS
=
− (100 N )
= −0.156 m/s 2
640 kg
Newton’s second law for the astronaut along the x-direction is
∑( F ) = F
on A x
S on A
= mA aA ⇒ aA =
FS on A
mA
=
FA on S
mA
=
100 N
= 1.25 m/s 2
80 kg
Let us first calculate the positions and velocities of the astronaut and the satellite at t1 = 0.50 s under the
accelerations aA and aS :
x1A = x0A + v0A ( t1 − t0 ) + 12 aA ( t1 − t0 ) = 0 m + 0 m + 12 (1.25 m/s 2 ) ( 0.50 s − 0 s ) = 0.156 m
2
2
x1S = x0S + v0S ( t1 − t0 ) + 12 aS ( t1 − t0 ) = 0 m + 0 m + 12 ( −0.156 m/s 2 ) ( 0.50 s − 0 s ) = −0.020 m
2
2
v1A = v0A + aA ( t1 − t0 ) = 0 m/s + (1.25 m/s 2 ) ( 0.50 s − 0 s ) = 0.625 m/s
v1S = v0S + aS ( t1 − t0 ) = 0 m/s + ( −0.156 m/s 2 ) ( 0.5 s − 0 s ) = −0.078 m/s
With x1A and x1S as initial positions, v1A and v1S as initial velocities, and zero accelerations, we can now obtain
the new positions at ( t2 − t1 ) = 59.5 s :
x2A = x1A + v1A ( t2 − t1 ) = 0.156 m + ( 0.625 m/s )( 59.5 s ) = 37.34 m
x2S = x1S + v1S ( t2 − t1 ) = − 0.02 m + ( −0.078 m/s )( 59.5 s ) = −4.66 m
Thus the astronaut and the satellite are x2A − x2S = ( 37.34 m ) − ( −4.66 m ) = 42 m apart.
7.20. Model: The block (B) and the steel cable (C), the two objects in the system, are considered particles,
and their motion is determined by the constant-acceleration kinematic equations.
Visualize:
Solve:
Using v1x = v0 x + ax ( t1 − t0 ) ,
4.0 m/s = 0 m/s + ax ( 2.0 s − 0 s ) ⇒ ax = 2.0 m/s 2
Newton’s second law along the x-direction for the block is
∑( F ) = F
on B x
C on B
= mB ax = ( 20 kg ) ( 2.0 m/s 2 ) = 40 N
Fext acts on the right end of the cable and FB on C acts on the left end. According to Newton’s third law,
FB on C = FC on B = 40 N. The difference in tension between the two ends of the cable is thus
Fext − FB on C = 100 N − 40 N = 60 N
7.21. Model: The block (B) and the steel cable (C), the two objects in the system, are treated as particles,
and their motion is determined by constant-acceleration kinematic equations. We ignore the hanging shape of the
cable.
Visualize:
Solve:
Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ,
2
4.0 m = 0 m + 0 m + 12 ax ( 2.0 s − 0 s ) ⇒ ax = 2.0 m/s 2
2
Newton’s second law along the x-direction for the block is
∑( F ) = F
on B x
C on B
= mBax = ( 20 kg ) ( 2.0 m/s 2 ) = 40 N
Therefore, the change in tension (T) in the cable from one
Fext − FB on C = 100 N − 40 N = 60 N. The tension in the cable as a function of x is
T ( x ) = 40 N +
( 60 N ) x = 40 + 60 x N
(
)
end
to
the
other
1m
with x in meters. x = 0 m is where the cable attaches to the box and x = 1.0 m is at the right end of the cable.
is
7.22. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight
of this segment of rope is a downward force. It is balanced by the tension force at height y.
Solve: The mass m of this segment of rope is the same fraction of the total mass M = 2.2 kg as length y is a
fraction of the total length L = 3.0 m. That is, m/M = y/L, from which we can write the mass of the rope
segment
m=
M
y
L
This segment of rope is in static equilibrium, so the tension force pulling up on it is
T = FG = mg =
Mg
(2.2 kg)(9.8 m /s 2 )
y=
y = 7.19 y N
L
3.0 m
where y is in m. The tension increases linearly from 0 N at the bottom ( y = 0 m) to 21.6 N at the top ( y = 3 m).
This is shown in the graph.
7.23. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction.
Visualize:
Solve:
The acceleration constraint is ( aA ) x = ( aB ) x = ax . Newton’s second law on sled A is
G
G
∑ Fon A = nA − ( FG )A = 0 N ⇒ nA = ( FG )A = mA g ∑ Fon A = T1 on A − fA = mA ax
(
)
(
y
)
x
Using f A = μ k nA , the x-equation yields
T 1 on A − μ k nA = mA ax ⇒ 150 N − ( 0.1)(100 kg ) ( 9.8 m/s 2 ) = (100 kg ) ax ⇒ ax = 0.52 m/s 2
On sled B:
G
G
∑( F ) = n − ( F ) = 0 N ⇒ n = ( F ) = m g ∑( F ) = T − T
on B
y
B
G B
B
G B
B
on B
x
2
1 on B
− f B = mBax
T 1 on B and T 1 on A act as if they are an action/reaction pair, so T 1 on B = 150 N. Using f B = μ k nB = ( 0.10 )( 80 kg )
( 9.8 m/s ) = 78.4 N, we get
2
T2 − 150 N − 78.4 N = ( 80 kg ) ( 0.52 m/s 2 ) ⇒ T2 = 270 N
Thus the tension T2 = 2.7 × 102 N.
7.24. Model: The coffee mug (M) is the only object in the system, and it will be treated as a particle. The
model of friction and the constant-acceleration kinematic equations will also be used.
Visualize:
Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the
same. Using v12x = v02x + 2ax ( x1 − x0 ) , we get
0 m 2 /s 2 = ( 20 m/s ) + 2ax ( 50 m ) ⇒ ax = −4.0 m/s 2
2
The static force needed to stop the mug is
( Fnet ) x = − fs = max = ( 0.5 kg ) ( −4.0 m/s2 ) = −2.0 N ⇒ fs = 2.0 N
The maximum force of static friction is
( f s ) max = μ s n = μ s FG = μ s mg = ( 0.50 )( 0.50 kg ) ( 9.8 m/s 2 ) = 2.45 N
Since ( f s ) max < ( f s ) max , the mug does not slide.
7.25.
Visualize:
The car and the ground are denoted by C and S, respectively.
Solve:
(a) The car has an internal source of energy–fuel–that allows it to turn the wheels and exert the Gforce
G
FC on G . As the drive wheels turn they push backward against the ground. This is a static friction force FC on S
because
the wheels don’t slip against the ground. ByG Newton’s third law, the ground exerts a reaction force
G
FS on C . This reaction force is opposite in direction to FC on S , hence, is in the forward direction. This is the force
that accelerates the car. Houses do not have an internal source of energy that allows them to push sideways
against the ground. They also aren’t on wheels, which let the car slide across the ground with minimal friction.
(b) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although
that is not critical) and the nondrive wheels. For this car, two-thirds of the gravitational force rests on the front
G
wheels. Physically, force FS on C is a static friction force. The maximum acceleration of the car on the ground (or
concrete surface) occurs when the static friction reaches its maximum possible value.
F S on C = ( fs )max = μs nF = μs ( FG )F = (1.00 ) ( 23 ) (1500 kg ) ( 9.8 m/s 2 ) = 9800 N
⇒ amax =
FS on C
m
=
9800 N
= 6.533 m/s 2
1500 kg
7.26. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be
treated as particles. We will also use the constant-acceleration kinematic equations.
Visualize:
G
(a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by
G
Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the
starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in).
However, the very different masses of the two crafts means that the distances they each move will also be very
different. The pictorial representation shows that they meet at time t1 when xM1 = xm1. There’s only one force on
each craft, so Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair,
Solve:
FM on m = Fm on M = Ftractor beam = 4.0 × 104 N
The accelerations of the two craft are
aM =
Fm on M
M
=
G
FM on m −4.0 × 104 N
4.0 × 104 N
2
=
a
=
=
= −2.0 m/s 2
0.020
m/s
m
2.0 × 106 kg
m
2.0 × 104 kg
Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. The
accelerations are very different even though the forces are the same. Now we have a constant-acceleration
problem in kinematics. At a later time t1 the positions of the crafts are
xM1 = xM0 + vM0 ( t1 − t0 ) + 12 aM ( t1 − t0 ) = 12 aMt12
2
xm1 = xm0 + vm0 ( t1 − t0 ) + 12 am ( t1 − t0 ) = xm0 + 12 amt12
2
The craft meet when xM1 = xm1 , so
1
2
aMt12 = xm0 + 12 amt12 ⇒ t1 =
2 (10,000 m )
2 xm0
2 xm0
=
=
= 99.5 s
aM − am
aM + am
2.02 m/s 2
Knowing t1 , we can now find the starship’s position as it meets the shuttlecraft:
xM1 = 12 aMt12 = 99 m
The starship moves 99 m as it pulls in the shuttlecraft from 10 km away.
7.27. Model: The rock (R) and Bob (B) are the two objects in our system, and will be treated as particles.
We will also use the constant-acceleration kinematic equations.
Visualize:
G
Solve: (a) Bob exerts a forward force FB on R on the rock to accelerate it forward. The rock’s acceleration is
calculated as follows:
( 30 m/s ) = 450 m/s 2
v2
v = v + 2a0R Δx ⇒ aR = 1R =
2Δx 2 (1.0 m )
2
2
1R
2
0R
The force is calculated from Newton’s second law:
FB on R = mR aR = (.500 kg ) ( 450 m/s 2 ) =225 N
Bob exerts a force of 2.3 × 102 N on the rock.
G
G
G
(b) Because Bob pushes on the rock, the rock pushes back on Bob with a force FR on B . Forces FR on B and FB on R are
an action/reaction pair, so FR on B = FB on R = 225 N. The force causes Bob to accelerate backward with an acceleration
equal to
aB =
(F
) =−F
net on B x
R on B
mB
mB
=−
225 N
= −3.0 m/s 2
75 kg
This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time interval
by returning to the kinematics of the rock:
v
v1R = v0R + aR Δt = aR Δt ⇒ Δt = 1R = 0.0667 s
aR
At the end of this interval, Bob’s velocity is
v1B = v0B + aBΔt = aBΔt = −0.20 m/s
Thus his recoil speed is 0.20 m/s.
7.28. Model: The boy (B) and the crate (C) are the two objects in our system, and they will be treated in the
particle model. We will also use the static and kinetic friction models.
Visualize:
Solve: The fact that the boy’s feet occasionally slip means that the maximum force of static friction must exist
between the boy’s feet and the sidewalk. That is, f sB = μ sB nB . Also f kC = μ kC nC .
Newton’s second law for the crate is
∑( F ) = n − ( F ) = 0 N ⇒ n = m g ∑( F ) = F
on C y
C
G C
C
C
on C x
B on C
− f kC = 0 N ⇒ FB on C = f kC = μ kC nC = μ kC mC g
Newton’s second law for the boy is
∑( F ) = n − ( F ) = 0 N ⇒ n = m g ∑( F ) = f − F
on B y
B
G B
B
B
on B x
sB
C on B
= 0 N ⇒ FC on B = f sB = μsBnB = μ sBmB g
G
G
FC on B and FB on C are an action/reaction pair, so
FC on B = FB on C ⇒ μsBmB g = μ kC mC g ⇒ mC =
μsB mB ( 0.8 )( 50 kg )
=
= 2.0 × 102 kg
μ kC
( 0.2 )
7.29. Model: Assume package A and package B are particles. Use the model of kinetic friction and the
constant-acceleration kinematic equations.
Visualize:
Solve: Package B has a smaller coefficient of friction. It will try to overtake package A and push against it.
Package A will push back on B. The acceleration constraint is ( aA ) x = ( aB ) x = a.
Newton’s second law for each package is
∑( F ) = F
on A x
B on A
+ ( FG )A sin θ − f kA = mA a
⇒ FB on A + mA g sin θ − μ kA ( mA g cosθ ) = mA a
∑( F ) = −F
on B x
A on B
− f kB + ( FG )B sinθ = mBa
⇒ − FA on B − μ kB ( mB g cosθ ) + mB g sin θ = mB a
where we have used nA = mA cosθ g and nB = mB cosθ g . Adding the two force equations, and using
FA on B = FB on A because they are an action/reaction pair, we get
a = g sin θ −
( μ kA mA + μ kBmB )( g cosθ ) = 1.82 m/s2
mA + mB
Finally, using x1 = x0 + v0x ( t1 − t0 ) + 12 a ( t1 − t0 ) ,
2
2.0 m = 0 m + 0 m + 12 (1.82 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 1.48 s
2
7.30. Model: The two blocks form a system of interacting objects.
Visualize:
Please refer to Figure P7.30.
Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand
block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or
negative answer for the tension in the string.
Newton’s first law applied to the y-direction on Block L yields
( ∑ F ) = 0 = n − ( F ) cos 20° ⇒ n = m g cos 20°
L y
L
G L
L
L
Therefore
( f k )L = ( μ k )L mL g cos 20° = ( 0.20 )(1.0 kg ) ( 9.80 m/s 2 ) cos 20° = 1.84 N
A similar analysis of the vertical forces on Block R gives ( f k )R = 1.84 N as well. Using Newton’s second law in
the x-direction for Block L,
( ∑ FL ) x = mLa = TR on L − ( f k )L + ( FG )L sin 20° ⇒ mLa = TR on L − 1.84 N + mL g sin 20°.
For Block R,
( ∑ F ) = m a = ( F ) sin 20° − 1.84 N − T
R x
R
G R
L on R
⇒ mR a = mR g sin 20° − 1.84 N − TL on R
These are two equations in the two unknowns a and TL on R = TR on L ≡ T . Solving them, we obtain a = 2.12 m/s 2 and
T = 0.61 N.
Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks,
which is reasonable.
7.31. Model: The two ropes and the two blocks (A and B) will be treated as particles.
Visualize:
Solve:
(a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined
system is accelerating upward at a = 3.0 m/s 2 under the influence of a force F and the gravitational force
− Mg ˆj. Newton’s second law applied to the combined system is
( Fnet ) y = F − Mg = Ma ⇒ F = M ( a + g ) = 32 N
(b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only
on block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’s
second law to each system, starting at the top. Each has an acceleration a = 3.0 m/s 2 . For block A:
(F
) = F − m g −T
net on A y
A
1 on A
= mA a ⇒ T1 on A = F − mA ( a + g ) = 19.2 N
(c) Applying Newton’s second law to rope 1:
( Fnet on 1 ) y = TA on 1 − m1g − TB on 1 = m1a
G
G
G
TA on 1 and T1 on A are an action/reaction pair. But, because the rope has mass, the two tension forces TA on 1 and
G
TB on 1 are not the same. The tension at the lower end of rope 1, where it connects to B, is
TB on 1 = TA on 1 − m1 ( a + g ) = 16.0 N
(d) We can continue to repeat this procedure, noting from Newton’s third law that
T1 on B = TB on 1 and T2 on B = TB on 2
Newton’s second law applied to block B is
(F
) =T
net on B y
1 on B
− mB g − T2 on B = mB a ⇒ T2 on B = T1 on B − mB ( a + g ) = 3.2 N
7.32. Model: The two blocks (1 and 2) are the systems of interest and will be treated as particles. The ropes
are assumed to be massless, and the model of kinetic friction will be used.
Visualize:
Solve: (a) The separate free-body diagrams for the two blocks show that there are two action/reaction pairs.
G
G
Notice how block 1 both pushes down on block 2 (force n1′ ) and exerts a retarding friction force f 2 top on the top
surface of block 2. Block 1 is in static equilibrium (a1 = 0 m/s 2 ) but block 2 is accelerating. Newton’s second
law for block 1 is
(F
) = f −T
net on 1 x
1
rope
= 0 N ⇒ Trope = f1 ( Fnet on 1 ) = n1 − m1 g = 0 N ⇒ n1 = m1 g
y
Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block
2. The friction model means f1 = μ k n1 = μ k m1 g . Substitute this result into the x-equation to get the tension in the
rope:
Trope = f1 = μ k m1 g = 3.92 N
(b) Newton’s second law for block 2 is
ax = a =
(F
) =T
net on 2 x
m2
pull
− f 2 top − f 2 bot
m2
a y = 0 m/s 2 =
(F
)
net on 2 y
m2
=
n2 − n1′ − m2 g
m2
G
G
Forces n1 and n1′ are an action/reaction pair, so n1′ = n1 = m1 g . Substituting into the y-equation gives
n2 = ( m1 + m2 ) g . This is not surprising because the combined weight of both objects presses down on the
surface. The kinetic friction on the bottom surface of block 2 is then
f 2 bot = μ k n2 = μ k ( m1 + m2 ) g
G
G
The forces f1 and f 2 top are an action/reaction pair, so f 2 bot = f1 = μ k m1 g . Inserting these friction results into
the x-equation gives
a=
(F
) =T
net on 2 x
m2
pull
− μ k m1 g − μ k ( m1 + m2 ) g
m2
= 2.16 m/s 2
7.33. Model: The 3-kg and 4-kg blocks are to be treated as particles. The models of kinetic and static friction
and the constant-acceleration kinematic equations will be used.
Visualize:
Solve: Minimum time will be achieved when static friction is at its maximum possible value. Newton’s second
law for the 4-kg block is
∑( F ) = n
on 4 y
3 on 4
− ( FG )4 = 0 N ⇒ n3 on 4 = ( FG )4 = m4 g = ( 4.0 kg ) ( 9.8 m/s 2 ) = 39.2 N
⇒ f s4 = ( f s ) max = μs n3 on 4 = ( 0.60 )( 39.2 N ) = 23.52 N
Newton’s second law for the 3-kg block is
∑( F ) = n − n
on 3 y
3
4 on 3
− ( FG )3 = 0 N ⇒ n3 = n4 on 3 + ( FG )3 = 39.2 N + ( 3.0 kg ) ( 9.8 m/s 2 ) = 68.6 N
Friction forces f and fs4 are an action/reaction pair. Thus
∑ ( F ) = f − f = m a ⇒ f − μ n = m a ⇒ 23.52 N − ( 0.20 )( 68.6 N ) = ( 3.0 kg ) a ⇒ a = 3.267 m/s
on 3 x
s3
k3
3 3
s4
k 3
3
3 3
3
Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows:
x1 − x0 + v0 x ( t1 − t0 ) + 12 a ( t1 − t0 ) ⇒ 5.0 m = 0 m + 0 m + 12 ( 3.267 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 1.75 s
2
2
2
7.34. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a
massless rope and frictionless pulley.
Visualize:
Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration
constraint is a2 = a = −a1 , where a will have a positive value. There are two real action/reaction pairs. The two
tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a
frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the freeG
body diagrams, especially force f1′ exerted on block 2 by block 1. We have quite a few pieces of information to
include. First, Newton’s second law for blocks 1 and 2:
G
Fnet on 1 = f1 − T1 = μ k n1 − T1 = m1a1 = − m1a ( Fnet on 1 ) = n1 − m1 g = 0 N ⇒ n1 = m1 g
(
)
y
x
(F
) = T − f ′− f − T = T − f ′− μ n − T = m a = m a
( F ) = n − n′ − m g = 0 N ⇒ n = n′ + m g
net on 2 x
pull
net on 2 y
1
2
2
2
1
pull
2
1
k 2
2
2
1
2 2
2
2
We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law:
n1′ = n1 = m1 g f1′ = f1 = μ k n1 = μ k m1 g T1 = T2 = T
Knowing n1′, we can now use the y-equation of block 2 to find n2 . Substitute all these pieces into the two xequations, and we end up with two equations in two unknowns:
μ k m1 g − T = −m1a Tpull − T − μ k m1 g − μ k ( m1 + m2 ) g = m2 a
Subtract the first equation from the second to get
Tpull − μ k ( 3m1 + m2 ) g = ( m1 + m2 ) a ⇒ a =
Tpull − μ k ( 3m1 + m2 ) g
m1 + m2
= 1.77 m/s 2
7.35. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will
be used.
Visualize:
In the sled’s free-body diagram nS is the normal (contact) force on the sled due to the snow. Similarly f kS is the
force of kinetic friction on the sled due to snow.
Solve: Newton’s second law on the box in the y-direction is
nS on B − ( FG )B cos 20° = 0 N ⇒ nS on B = (10 kg ) ( 9.8 m/s 2 ) cos 20° = 92.09 N
G
The static friction force fS on B accelerates the box. The maximum acceleration occurs when static friction reaches
its maximum possible value.
( fs ) max = μSnS on B = ( 0.50 )( 92.09 N ) = 46.05 N
Newton’s second law along the x-direction thus gives the maximum acceleration
fS on B − ( FG )B sin 20° = mB a ⇒ 46.05 N − (10 kg ) ( 9.8 m/s 2 ) sin 20° = (10 kg ) a ⇒ a = 1.25 m/s 2
Newton’s second law for the sled along the y-direction is
nS − nB on S − ( FG )S cos 20° = 0 N
⇒ nS = nB on S + mS g cos 20° = ( 92.09 N ) + ( 20 kg ) ( 9.8 m/s 2 ) cos 20° = 276.27 N
Therefore, the force of friction on the sled by the snow is
f kS = ( μ k )nS = (0.06)(276.27 N) = 16.58 N
Newton’s second law along the x-direction is
Tpull − wS sin 20° − f kS − f B on S = mSa
The friction force f B on S = fS on B because these are an action/reaction pair. We’re using the maximum
acceleration, so the maximum tension is
Tmax − ( 20 kg ) ( 9.8 m/s 2 ) sin 20° − 16.58 N − 46.05 N = ( 20 kg ) (1.25 m/s 2 )
⇒ Tmax = 155 N
7.36. Model: The masses m and M are to be treated in the particle model. We will also assume a massless
rope and frictionless pulley, and use the constant-acceleration kinematic equations for m and M.
Visualize:
Solve:
Using y1 = y0 + v0 y ( t1 − t0 ) + 12 aM ( t1 − t0 ) ,
2
( −1 m ) = 0 m + 0 m + 12 aM ( 6.0 s − 0 s ) ⇒ aM = −0.0556 m/s 2
2
Newton’s second law for m and M is:
∑( F ) = T
on m y
R on m
− ( FG )m = mam
∑( F ) = T
on M y
R on M
− ( FG )M = MaM
The acceleration constraint is am = − aM . Also, the tensions are an action/reaction pair, thus TR on m = TR on M . With
these, the second law equations are
TR on M − Mg = MaM
TR on M − mg = −maM
Subtracting the second from the first gives
⎡ g + aM ⎤
− Mg + mg = MaM + maM ⇒ m = M ⎢
⎥
⎣ g − aM ⎦
⎡ 9.8 − 0.556 ⎤
= (100 kg ) ⎢
⎥ = 99 kg
⎣ 9.8 + 0.556 ⎦
Assess: Note that am = − aM = 0.0556 m/s 2 . For such a small acceleration, a mass of 99 kg for m compared to
M = 100 kg is understandable.
7.37. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the
G
rope is massless and the pulleys are frictionless. The block is kept in place by an applied force F .
Visualize:
Solve:
Since there is no friction on the pulleys, T2 = T3 and T2 = T5 . Newton’s second law for mass M is
T1 − FG = 0 N ⇒ T1 = Mg = (10.2 kg ) ( 9.8 m/s 2 ) = 100 N
Newton’s second law for the small pulley is
T
T2 + T3 − T1 = 0 N ⇒ T2 = T3 = 1 = 50 N = T5 = F
2
Newton’s second law for the large pulley is
T4 − T2 − T3 − T5 = 0 N ⇒ T4 = T2 + T3 + T5 = 150 N
7.38. Model: Assume the particle model for m1 , m2 , and m3 , and the model of kinetic friction. Assume the
ropes to be massless, and the pulleys to be frictionless and massless.
Visualize:
Solve:
Newton’s second law for m1 is T1 − ( FG )1 = m1a1. Newton’s second law for m2 is
∑ ( F ) = n − ( F ) = 0 N ⇒ n = ( 2.0 kg ) ( 9.8 m/s ) = 19.6 N
2
on m2
∑ (F
2
y
G 2
2
) = T2 − f k2 − T = m2 a2 ⇒ T2 − μ k n2 − T1 = (2.0 kg)a2
on m2 x
Newton’s second law for m3 is
∑( F ) = T − ( F ) = m a
on m3
y
2
G 3
3 3
Since m1 , m2 , and m3 move together, a1 = a2 = −a3 = a. The equations for the three masses thus become
T1 − ( FG )1 = m1a = (1.0 kg ) a T2 − μ k n2 − T1 = m2 a = ( 2.0 kg ) a T2 − ( FG )3 = −m3a = − ( 3.0 kg ) a
Subtracting the third equation from the sum of the first two equations yields:
− ( FG )1 − μ k n2 + ( FG )3 = ( 6.0 kg ) a
⇒ − (1.0 kg ) ( 9.8 m/s 2 ) − ( 0.30 )(19.6 N ) + ( 3.0 kg ) ( 9.8 m/s 2 ) = ( 6.0 kg ) a ⇒ a = 2.3 m/s 2
7.39. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction.
Visualize:
Solve: (a) If the mass m is too small, the hanging 2.0 kg mass will pull it up the slope. We want to find the
smallest mass that will stick because of friction. The smallest mass will be the one for which the force of static
friction is at its maximum possible value: f s = ( f s )max = μs n. As long as the mass m is stuck, both blocks are at
G
rest with Fnet = 0 N. Newton’s second law for the hanging mass M is
( Fnet ) y = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N
For the smaller mass m,
( Fnet ) x = Tm − fs − mg sinθ = 0 N ( Fnet ) y = n − mg cosθ ⇒ n = mg cosθ
G
G
For a massless string and frictionless pulley, forces Tm and TM act as if they are an action/reaction pair. Thus
Tm = TM . Mass m is a minimum when ( fs )max = μs n = μs mg cosθ . Substituting these expressions into the x-
equation,
TM − μ s mg cosθ − mg sin θ = 0 N ⇒ m =
TM
( μs cosθ + sinθ ) g
= 1.83 kg
(b) Because μ k < μS the 1.83 kg block will begin to slide up the ramp, and the 2.0 kg mass will begin to fall, if
the block is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the
pictorial representation, we chose different coordinate systems for the two masses. This gives block M an
acceleration with only a y-component and block m an acceleration with only an x-component. The magnitudes of
the accelerations are the same because the blocks are tied together. But block M has a negative acceleration
G
component a y (vector a points down) whereas block m has a positive ax . Thus the acceleration constraint is
(am ) x = −(aM ) y = a, where a will have a positive value. Newton’s second law for block M is
( Fnet ) y = T − Mg = M ( aM ) y = − Ma
For block m we have
( Fnet ) x = T − f k − mg sinθ = T − μ k mg cosθ − mg sinθ = m ( am ) x = ma
In writing these equations, we used Newton’s third law to write Tm = TM = T . Also, we noticed that the yequation and the friction model for block m don’t change, except for μ s becoming μ k , so we already know f k
from part (a). Notice that the tension in the string is not the gravitational force Mg. We have two equations in the
two unknowns T and a:
T − Mg = − Ma T − ( μ k cosθ + sin θ ) mg = ma
Subtracting the second equation from the first to eliminate T,
− Mg + ( μ k cosθ + sin θ ) mg = − Ma − ma = −( M + m)a
⇒a=
M − ( μ k cosθ + sin θ ) m
g = 1.32 m/s 2
M +m
7.40. Model: Assume the particle model for the two blocks.
Visualize:
Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released,
the blocks will
G
move one way or the other. As long as m is held, the blocks are in static equilibrium with Fnet = 0 N. Newton’s
second law for the hanging block M is
( Fnet on M ) = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N
y
By Newton’s third law, TM = Tm = T = 19.6 N is the tension in the string.
(b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends on
whether the acceleration a is positive or negative. The acceleration constraint is (am ) x = a = −(aM ) y . Newton’s
second law for each system gives
( Fnet on m ) = T − mg sin 35° = m ( am )x = ma ( Fnet on M ) = T − Mg = M ( aM ) y = −Ma
x
y
We have two equations in two unknowns. Subtract the second from the first to eliminate T:
M − m sin 35°
g = −0.481 m/s 2
− mg sin 35° + Mg = ( m + M ) a ⇒ a =
M +m
Since a < 0 m/s 2 , the box accelerates down the slope.
(c) It is now straightforward to compute T = Mg − Ma = 21 N. Notice how the tension is larger than when the
blocks were motionless.
7.41. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic and
static friction, and the constant-acceleration kinematic equations.
Visualize:
Solve:
(a) Using v12x = v02x + 2a ( x1 − x0 ) ,
0 m 2 /s 2 = ( 3.0 m/s ) + 2a ( x1 ) ⇒ ax1 = −4.5 m 2 /s 2
2
To find x1 , we must first find a. Newton’s second law for the book and the coffee cup is
∑ ( F ) = n − ( F ) cos 20° = 0 N ⇒ n = (1.0 kg ) ( 9.8 m/s ) cos 20° = 9.21 N
2
on B y
B
G B
B
∑ ( F ) = −T − f − ( F ) sin 20° = m a ∑ ( F ) = T − ( F ) = m a
on B x
k
G B
B B
on C y
G C
C C
The last two equations can be rewritten, using aC = aB = a, as
−T − μ k nB − mB g sin 20° = mB a T − mC g = mC a
Adding the two equations,
a ( mC + mB ) = − g ( mC + mB sin 20° ) − μ k ( 9.21 N )
⇒ (1.5 kg ) a = − ( 9.8 m/s 2 ) ⎣⎡0.500 kg + (1.0 kg ) sin 20° ⎦⎤ − ( 0.20 )( 9.21 N ) ⇒ a = −6.73 m/s 2
Using this value for a, we can now find x1 as follows:
x1 =
−4.5 m 2 /s 2 −4.5 m 2 /s 2
=
= 0.67 m
a
−6.73 m/s 2
(b) The maximum static friction force is ( f s ) max = μs nB = ( 0.50 )( 9.21 N ) = 4.60 N. We’ll see if the force f s
needed to keep the book in place is larger or smaller than ( f s ) max . When the cup is at rest, the string tension is
T = mC g . Newton’s first law for the book is
∑ ( F ) = f − T − w sin 20° = f − m g − m g sin 20° = 0
on B x
s
B
s
C
B
⇒ fs = ( M C + M B sin 20° ) g = 8.25 N
Because fs > ( f s ) max , the book slides back down.
7.42. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable.
Visualize:
Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2).
G
G
G
G
Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1
G
to allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet = 0 N
for both systems. Newton’s second law for the cable car is
) = T + F − m g sinθ = 0 N ( F
(F
net on 1 x
1
B
1
1
Newton’s second law for the counterweight is
(F
) = m g sinθ − T = 0 N ( F
net on 2 x
2
2
2
) = n − m g cosθ = 0 N
net on 1 y
1
1
1
) = n − m g cosθ = 0 N
net on 2 y
2
2
2
From the x-equation for the counterweight, T2 = m2 g sin θ 2 . By Newton’s third law, T1 = T2 . Thus the x-equation
for the cable car becomes
FB = m1 g sinθ1 − T1 = m1 g sin θ1 − m2 g sin θ 2 = 3770 N
(b) If the brakes fail, then FB = 0 N. The car will accelerate down the hill on one side while the counterweight
accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the
acceleration vectors. The constraint is a1x = a2 x = a, where a will have a negative value. Using T1 = T2 = T , the
two x-equations are
( Fnet on 1 ) x = T − m1g sinθ1 = m1a1x = m1a ( Fnet on 2 ) x = m2 g sinθ 2 − T = m2a2x = m2a
Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T:
m sin θ1 − m2 sinθ 2
g = −0.991 m/s 2
− m1 g sin θ1 + m2 g sin θ = ( m1 + m2 ) a ⇒ a = − 1
m1 + m2
Now we have a problem in kinematics. The speed at the bottom is calculated as follows:
v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2 ( −0.991 m/s 2 ) ( −400 m ) = 28.2 m/s
Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless
slope of 30° is reasonable.
7.43. Model: Assume the cable mass is negligible compared to the car mass and that the pulley is
frictionless. Use the particle model for the two cars.
Visualize: Please refer to Figure P7.43.
Solve: (a) The cars are moving at constant speed, so they are in dynamic equilibrium. Consider the descending
car D. We can find the rolling friction force on car D, and then find the cable tension by applying Newton’s first
law. In the y-direction for car D,
( Fnet ) y = 0 = nD − ( FG )D cos35°
⇒ nD = mD g cos35°
So the rolling friction force on car D is
( f R )D = μ R nD = μ R mD g cos35°
Applying Newton’s first law to car D in the x-direction,
( Fnet ) x = TA on D + ( f R )D − ( FG )D sin 35° = 0
Thus
TA on D = mD g sin 35° − μ R mD g cos35°
= (1500 kg ) ( 9.80 m/s 2 ) ( sin 35° − 0.020cos35° )
= 8.2 × 103 N
(b) Similarly, we find that for car A, ( f R )A = μ R mA g cos35°. In the x-direction for car A,
( Fnet ) x = Tmotor + TD on A − ( f R )A − ( FG )A sin 35° = 0
⇒ Tmotor = mA g sin 35° + μ R mA g cos35° − ( mD g sin 35° − μ R mD g cos35° )
Here, we have used TA on D = TD on A . If we also use m A = m D , then
Tmotor = 2μ R mA g cos35° = 4.8 × 102 N.
Assess: Careful examination of the free-body diagrams for cars D and A yields the observation that
Tmotor = 2( FR ) A in order for the cars to be in dynamic equilibrium. It is a tribute to the design that the motor must
only provide such a small force compared to the tension in the cable connecting the two cars.
7.44. Model: The painter and the chair are treated as a single object and represented as a particle. We assume
that the rope is massless and that the pulley is massless and frictionless.
Visualize:
Solve: If the painter pulls down on the rope with force F, Newton’s third law requires the rope to pull up on the
painter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tension
force F also pulls up on the painter’s chair. Newton’s second law for (painter + chair) is
2F − FG = ( mP + mC ) a
⇒ F = ( ) ⎡⎣( mP + mC ) a + ( mP + mC ) g ⎤⎦ = 12 ( mP + mC )( a + g )
1
2
= ( 12 ) ( 70 kg + 10 kg ) ( 0.20 m/s 2 + 9.8 m/s 2 ) = 4.0 × 102 N
Assess: A force of 400 N, which is approximately one-half the total gravitational force, is reasonable since the
upward acceleration is small.
7.45. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be
massless, so the tension in the rope is uniform.
Visualize:
Solve:
Newton’s second law for the tightrope walker is
FR on W − FG = ma ⇒ FR on W = m ( a + g ) = ( 70 kg ) ( 8.0 m/s 2 + 9.8 m/s 2 ) = 1.25 × 103 N
Now, Newton’s second law for the rope gives
∑ ( F ) = T sinθ + T sinθ − F
on R
y
W on R
= 0 N ⇒T =
FW on R
2sin10°
We used FW on R = FR on W because they are an action/reaction pair.
=
FR on W
2sin10°
=
1.25 × 103 N
= 3.6 × 103 N
2sin10°
7.46. Model: Use the particle model for the wedge and the block.
Visualize:
The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note:
n1 on 2 = n2 on 1.
Solve:
The y-component of Newton’s second law for block m2 is
∑( F ) = n
on 2 y
1 on 2
cosθ − ( FG )2 = 0 N ⇒ n1 on 2 =
m2 g
cosθ
Combining this equation with the x-component of Newton’s second law yields:
∑( F ) = n
on 2 x
1 on 2
sin θ = m2 a ⇒ a =
n1 on 2 sin θ
= g tan θ
m2
Now, Newton’s second law for the wedge is
∑( F ) = F − n
on 1 x
2 on 1
sinθ = m1a
⇒ F = m1a + n2 on 1 sin θ = m1a + m2 a = ( m1 + m2 )a = ( m1 + m2 ) g tan θ
7.47. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic
equations.
Visualize:
Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightens
G
his legs. He exerts a force FP on F on the floor. The player experiences a gravitational force FG as well as a
P
G
G
normal force by the floor nF on P . The only force that the floor experiences is the one exerted by the player FP on F .
G
G
G
(b) The player standing at rest exerts a force FP on F on the floor. The normal force nF on P is the reaction force to FP on F .
G
But nF on P = FP on F , so Fnet = 0 N. When the basketball player accelerates upward by straightening his legs, his
( )
speed has to increase from zero to v1y with which he leaves the floor. Thus, according to Newton’s second law,
there must be a net upward force on him during this time. This can be true only if nF on P > ( FG )P . In other words,
the player presses on the floor with a force FP on F larger than the gravitational force on him, which is equal to his
G
weight. The reaction force nF on P then exceeds his weight and accelerates him upward until his feet leave the
floor.
(c) The height of 80 cm = 0.80 m is sufficient to determine the speed v1y with which he leaves the floor. Once
his feet are off the floor, he is simply in free fall, with a1 = − g. From kinematics,
v22 y = v12y + 2a1 ( y2 − y1 ) ⇒ 0 m 2 /s 2 = v12y + 2 ( − g )( 0.80 m )
⇒ v1 y = 2g ( 0.80 m ) = 3.96 m/s
The basketball player reached v1 y = 4.0 m/s by accelerating from rest through a distance of 0.60 m.
(d) Assuming a0 to be constant during the jump, we find
v12y = v02y + 2a0 ( y1 − y0 ) = 0 m 2 /s 2 + 2a0 ( y1 − 0 m ) ⇒ a0 =
v12y
2 y1
( 3.96 m/s ) = 13.1 m/s 2
2 ( 0.60 m )
2
=
(e) The scale reads the value of nF on P , the force exerted by the scale on the player. Before jumping,
nF on P − ( FG )P = 0 N ⇒ nF on P = ( FG )P = mg = (100 kg ) ( 9.8 m/s 2 ) = 980 N
While accelerating up,
⎛ a ⎞
⎛ 13.1 ⎞
nF on P − mg = ma0 ⇒ nF on P = ma0 + mg = mg ⎜ 1 + 0 ⎟ = ( 980 N ) ⎜1 +
⎟ = 2290 N
9.8 ⎠
g⎠
⎝
⎝
After leaving the scale, nF on P = 0 N because there is no contact with the scale.
7.48.
A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg block
across a frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide?
Visualize:
Solve:
The 1.0 block is accelerated by static friction. It moves smoothly with the lower block if f s < ( fs ) max . It
slides if the force that would be needed to keep it in place exceeds ( f s ) max . Begin by assuming the blocks move
together with common acceleration a. Newton’s second is
∑ ( Fon 1 ) x = fs = m1a
Bottom block: ∑ ( Fon 2 ) x = Tpull − fs = m2 a
Top block:
Adding these two equations gives Tpull = (m1 + m2 )a, or a = (21.0 N)/(1.0 kg + 2.0 kg) = 7.0 m/s 2 . The static
friction force needed to accelerate the top block at 7.0 m/s2 is
fs m1a = (1.0 kg)(7.0 m/s 2 ) = 7.0 N
To find the maximum possible static friction force ( f s ) max = μ s n1 , the y-equation of Newton’s second law for the
top block shows that n1 = m1 g . Thus
( f s ) max = μs m1 g = (0.50)(1.0 kg)(9.80 m/s 2 ) = 4.9 N
Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block. It slides.
7.49.
A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients of
kinetic friction with the table are 0.3 for the 1.0 kg block and 0.5 for the 2.0 kg block. The 1.0 kg block is pushed
forward, against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before
stopping?
Visualize:
Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the
rear of the 2.0 kg block and, in reaction, the 2.0 kg block pushes backward against the 1.0 kg block. There’s no
vertical acceleration, so n1 = m1 g and n2 = m2 g , leading to f1 = μ1m1 g and f 2 = μ 2 m2 g . Newton’s second law
along the x-axis is
1 kg block:
2 kg block:
∑ (F ) = −F
on 1 x
2 on 1
− f1 = − F2 on 1 − μ1m1 g = m1a
∑ ( Fon 2 ) x = F1 on 2 − f 2 = F2 on 1 − μ 2m2 g = m2a
where we used a1 = a2 = a. Also, F1 on 2 = F2 on 1 because they are an action/reaction pair. Adding these two
equations gives
− ( μ1m1 + μ 2 m2 ) g = ( m1 + m2 )a
a=−
μ1m1 + μ 2 m2
m1 + m2
g =−
(0.3)(1.0 kg) + (0.5)(2.0 kg)
× 9.80 m/s2 = −4.25 m/s2
1.0 kg + 2.0 kg
We can now use constant-acceleration kinematics to find
v2
(2.0 m/s) 2
v12x = 0 = v02x + 2a ( x1 − x0 ) ⇒ x1 = − 0 x = −
= 0.47 m
2a
2( −4.25 m/s2)
7.50. Model:
Visualize:
Treat the ball of clay and the block as particles.
G
G
Solve: (a) Forces FC on B and FB on C are an action/reaction pair, so FB on C = FC on B . Note that aB ≠ aC because
the clay is decelerating while the block is accelerating. Newton’s second law in the x-direction is
∑ ( Fon C ) x = − FB on C = mC aC
Block: ∑ ( Fon B ) x = FC on B = FB on C = mB aB
Clay:
Equating the two expressions for FB on C gives
aC = −
mB
aB
mC
Turning to kinematics, the velocity of each after Δt is
(vC )1 = (vC )0 + aC Δt
(vB )1 = (vB )0 + aB Δt = aBΔt
But (vC )1 = (vB )1 because the clay and the block are moving together after Δt has elapsed. Equating these two
expressions gives (vC )0 + aC Δt = aBΔt , from which we find
aC = aB −
(vC )0
Δt
We can now equate the two expressions for aC :
−
mB
(v )
(vC )0 / Δt
(10 m/s)(0.01 s)
=
= 100 m/s2
aB = aB − C 0 ⇒ aB =
Δt
mC
1 + mB / mC 1 + (900 g)(100 g)
Then aC = −9aB = −900 m/s 2 . With the acceleration now known, we can use either kinematic equation to find
(vC )1 = (vB )1 = (100 m/s 2 )(0.010 s) = 1.0 m/s
(b) FC on B = mB aB = (0.90 kg)(100 m/s 2 ) = 90 N.
(c) FB on C = mC aC = (0.10 kg)(900 m/s 2 ) = 90 N.
Assess: The two forces are of equal magnitude, as expected from Newton’s third law.
7.51. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless
pulleys.
Visualize:
Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on m2 has to
be lengthened by one-half meter. Thus the acceleration constraint is a2 = − 12 a1.
Solve: Newton’s second law for block 1 is T = m1a1. Newton’s second law for block 2 is 2T − ( FG ) 2 = m2 a2 .
Combining these two equations gives
2 ( m1a1 ) − m2 g = m2 ( − 12 a1 ) ⇒ a1 [ 4m1 + m2 ] = 2m2 g ⇒ a1 =
2m2 g
4m1 + m2
where we have used a2 = − 12 a1.
Assess:
If m1 = 0 kg, then a2 = − g. This is what is expected for a freely falling object.
7.52. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless
pulleys.
Visualize:
For every one meter that the 1.0-kg block goes down, each rope on the 2.0-kg block will be shortened by onehalf meter. Thus the acceleration constraint is a1 = −2a2 .
Solve: Newton’s second law for the two blocks is
2T = m2 a2 T − ( FG )1 = m1a1
Since a1 = −2a2 , the above equations become
2T = m2 a2 T − m1 g = m1 ( −2a2 )
2 (1.0 kg ) ( 9.8 m/s )
2m1 g
a2
=
= 3.3 m/s 2
+ m1 ( 2a2 ) = m1 g ⇒ a2 =
2
m2 + 4m1
( 2.0 kg + 4.0 kg )
2
⇒ m2
Assess:
If m1 = 0 kg, then a2 = 0 m/s 2 , which is expected.
7.53. Model: The hamster of mass m and the wedge with mass M will be treated as objects 1 and 2,
respectively. They will be treated as particles.
Visualize:
The scale is denoted by the letter s.
G
Solve: (a) The reading of the scale is the magnitude of the force n2 that the scale exerts upward. There are two
G
action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with Fnet = 0 N.
G
Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force fs on 2
to prevent horizontal motion. We’ve included one just in case. Newton’s second law for the hamster is
(F
) = mg sinθ − f
net on 1 x
(F
2 on 1
= 0 N ⇒ f 2 on 1 = mg sinθ
) = n − mg cosθ = 0 N ⇒ n = mg cosθ
net on 1 y
1
1
For the wedge, we see from Newton’s third law that n1′ = n1 = mg cosθ and that f 2 on 1 = f1 on 2 = mg sin θ . Using
these equations, Newton’s second law for the wedge is
( Fnet on 2 ) x = f1 on 2 cosθ + fs on 2 − n1′ sinθ = mg sinθ cosθ + fs on 2 − mg cosθ sinθ = 0 N ⇒ fs on 2 = 0 N
(F
) = n − n′ cosθ − f
net on 2 y
2
1
1 on 2
sin θ − Mg = n2 − mg cos 2 θ − mg sin 2 θ − Mg = 0 N
⇒ n2 = mg ( cos 2 θ + sin 2 θ ) + Mg = ( M + m ) g = ( 0.800 kg + 0.200 kg ) ( 9.8 m/s 2 ) = 9.80 N
First we find that f s on 2 = 0 N, so no horizontal static friction is needed to prevent motion. More interesting, the
scale reading is ( M + m ) g which is the total gravitational force resting on the scale. This is the expected result.
(b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the
reading still ( M + m) g ? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now
has an acceleration. However, the acceleration is along the hamster’s x-axis, so a1 y = 0 m/s 2 . The hamster’s yequation is still
( Fnet on 1 ) y = n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ
We still have n1′ = n1 = mg cosθ , so the y-equation for block 2 (with a2 y = 0 m/s 2 ) is
(F
) = n − n′ cosθ − Mg = n − mg cos θ − Mg = 0 N
2
net on 2 y
2
1
2
⇒ n2 = mg cos 2 θ + Mg = ( M + m cos 2 θ ) g = 8.99 N
Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit
θ → 90°, in which case cosθ → 0. If the face of the wedge is vertical, then the hamster is simply in free fall and
can have no effect on the scale (at least until impact!). So for θ = 90° we expect the scale to record Mg only, and
that is indeed what the expression for n2 gives.
7.54. Model: The hanging masses m1 , m2 , and m3 are modeled as particles. Pulleys A and B are massless
and frictionless. The strings are massless.
Visualize:
Solve:
(a) The length of the string over pulley B is constant. Therefore,
( yB − y3 ) + ( yB − yA ) = LB ⇒ yA = 2 yB − y3 − LB
The length of the string over pulley A is constant. Thus,
( yA − y2 ) + ( yA − y1 ) = LA = 2 yA − y1 − y2
⇒ 2 ( 2 yB − y3 − LB ) − y1 − y2 = LA ⇒ 2 y3 + y2 + y1 = constant
This constraint implies that
2
dy3 dy2 dy1
+
+
= 0 m/s = 2v3 y + v2 y + v1 y
dt
dt
dt
Also by differentiation, 2a3 y + a2 y + a1 y = 0 m/s 2 .
(b) Newton’s second law for the masses m3 , m2 , m1 , and pulley A is
TB − m3 g = m3a3 y TA − m2 g = m2a2 y TA − m1g = m1a1 y
TB − 2TA = 0 N
The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint
are five equations for the five unknowns (two tensions and three accelerations). To solve for TA , multiply the m3
equation by 2, substitute 2TB = 4TA , then divide each of the mass equations by the mass. This gives the three
equations
4TA /m3 − 2 g = 2a3 y
TA /m 2 − g = a2 y
TA /m 1 − g = a1 y
If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus
( 4/m3 + 1/m2 + 1/m2 )TA − 4 g = 0 ⇒ TA =
4g
( 4/m3 + 1/m2 + 1/m2 )
(c) Using numerical values, we find TA = 18.97 N. Then
a1 y = TA /m1 − g = −2.2 m/s 2
a2 y = TA /m 2 − g = 2.9 m/s 2
a3 y = 2TA /m 3 − g = −0.32 m/s 2
(d) m3 = m1 + m2 , so it looks at first like m3 should hang in equilibrium. For it to do so, tension TB would need
to equal m3 g . However, TB is not (m1 + m2 ) g because masses m1 and m2 are accelerating rather than hanging
at rest. Consequently, tension TB is not able to balance the weight of m3 .
8.1.
Model: The model rocket and the target will be treated as particles. The kinematics equations in two
dimensions apply.
Visualize:
Solve:
For the rocket, Newton’s second law along the y-direction is
( Fnet ) y = FR − mg = maR
⇒ aR =
1
1
⎡15 N − ( 0.8 kg ) ( 9.8 m/s 2 ) ⎤ = 8.95 m/s 2
( FR − mg ) =
⎦
0.8 kg ⎣
m
Using the kinematic equation y1R = y0R + (v0R ) y ( t1R − t0R ) + 12 aR ( t1R − t0R ) ,
2
30 m = 0 m + 0 m + 12 ( 8.95 m/s 2 ) ( t1R − 0 s ) ⇒ t1R = 2.589 s
2
For the target (noting t1T = t1R ),
x1T = x0T + (v0T ) x ( t1T − t0T ) + 12 aT ( t1T − t0T ) = 0 m + (15 m/s )( 2.589 s − 0 s ) + 0 m = 39 m
2
You should launch when the target is 39 m away.
Assess: The rocket is to be fired when the target is at x0T . For a net acceleration of approximately 9 m/s 2 in
the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is
reasonable.
8.2.
Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply.
Air resistance is neglected.
Visualize:
The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s.
Solve: For the rocket, Newton’s second law along the y-direction is:
1
⎡( 8.0 N ) − ( 0.5 kg ) ( 9.8 m/s 2 ) ⎤ = 6.2 m/s 2
( Fnet ) y = FR − mg = maR ⇒ a y =
⎦
0.5 kg ⎣
Thus using y1 = y0 + (v0 ) y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ,
2
( 20 m ) = 0 m + 0 m + 12 ( 6.2 m/s2 ) ( t1R − 0 s ) ⇒ ( 20 m ) = ( 3.1 m/s2 ) t12 ⇒ t1 = 2.54 s
2
Since t1 is also the time for the rocket to move horizontally up to the hoop,
x1 = x0 + (v0 ) x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( 3.0 m/s )( 2.54 s − 0 s ) + 0 m = 7.6 m
2
Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained
value for the horizontal distance is reasonable.
8.3.
Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the
constant-acceleration equations of kinematics.
Visualize:
Solve:
(a) The time it will take the asteroid to reach the earth is
displacement 4.0 × 106 km
=
= 2.0 × 105 s = 56 h
velocity
20 km/s
(b) The angle of a line that just misses the earth is
tanθ =
⎛R⎞
R
⎛ 6400 km ⎞
⇒ θ = tan −1 ⎜ ⎟ = tan −1 ⎜
⎟ = 0.092°
6
y0
⎝ 4.0 × 10 km ⎠
⎝ y0 ⎠
(c) When the rocket is fired, the horizontal acceleration of the asteroid is
ax =
5.0 × 109 N
= 0.125 m/s 2
4.0 × 1010 kg
(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored
completely.) The velocity of the asteroid after the rocket has been fired for 300 s is
vx = v0 x + ax ( t − t0 ) = 0 m/s + ( 0.125 m/s 2 ) ( 300 s − 0 s ) = 37.5 m/s
After 300 s, the vertical velocity is v y = 2 × 104 m/s and the horizontal velocity is vx = 37.5 m/s. The deflection
due to this horizontal velocity is
tan θ =
That is, the earth is saved.
⎛ 37.5 m/s ⎞
vx
⇒ θ = tan −1 ⎜
⎟ = 0.107°
4
vy
⎝ 2 × 10 m/s ⎠
8.4.
Model: We are using the particle model for the car in uniform circular motion on a flat circular track.
There must be friction between the tires and the road for the car to move in a circle.
Visualize:
Solve:
The centripetal acceleration is
v 2 ( 25 m/s )
=
= 6.25 m/s 2
100 m
r
2
ar =
The acceleration points to the center of the circle, so the net force is
G
G
Fr = ma = (1500 kg ) ( 6.25 m/s 2 , toward center )
= ( 9380 N, toward center )
This force is provided by static friction
f s = Fr = 9.4 kN
8.5.
Model:
Visualize:
Solve:
We will use the particle model for the car which is in uniform circular motion.
The centripetal acceleration of the car is
v 2 (15 m/s )
=
= 4.5 m/s 2
50 m
r
The acceleration is due to the force of static friction.
fs = mar = (1500 kg ) ( 4.5m s 2 ) = 6750 N = 6.8 kN.
2
ar =
Assess:
The
force
of
friction
The model of static friction is ( fs )max = nμs = mg μs ≈ mg ≈ 15,000 N since μ s ≈ 1 for a dry road
surface. We see that f s < ( fs )max , which is reasonable.
is
8.6.
Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a
frictionless table.
Visualize:
Solve:
(a) The angular velocity and speed are
ω = 75
rev 2π rad
1min
×
= 471.2 rad/min vt = rω = ( 0.50 m )( 471.2 rad min ) ×
= 3.93 m/s
min 1 rev
60 s
The tangential velocity is 3.9 m/s.
(b) The radial component of Newton’s second law is
mv 2
∑F =T = r
r
Thus
T = ( 0.20 kg )
( 3.93 m/s ) = 6.2 N
2
0.50 m
8.7.
Solve:
Newton’s second law is Fr = mar = mrω 2 . Substituting into this equation yields:
ω=
Fr
=
mr
8.2 × 10−8 N
(9.1×10 kg )(5.3 × 10−11 m )
−31
= 4.37 × 1016 rad/s = 4.37 × 1016
Assess:
rad 1 rev
×
= 6.6 × 1015 rev/s
s 2π rad
This is a very high number of revolutions per second.
8.8.
Model:
Visualize:
The vehicle is to be treated as a particle in uniform circular motion.
On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary
centripetal acceleration. The vertical component of the normal force balances the gravitational force.
Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written
mv 2
∑ F = n sinθ = r ∑ F = n cosθ − mg = 0 ⇒ n cosθ = mg
r
z
Dividing the two equations and making the conversion 90 km h = 25 m/s yields:
( 25 m/s )
v2
=
= 0.128 ⇒ θ = 7.3°
rg ( 9.8 m/s 2 ) 500 m
2
tanθ =
Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our
experience as well.
8.9.
Model: The motion of the moon around the earth will be treated through the particle model. The circular
motion is uniform.
Visualize:
Solve:
The tension in the cable provides the centripetal acceleration. Newton’s second law is
⎛ 2π ⎞
∑ Fr = T = mrω 2 = mr ⎜ T ⎟
⎝ moon ⎠
2
2
⎡ 2π
1 day
1h ⎤
20
= ( 7.36 × 1022 kg )( 3.84 × 108 m ) ⎢
×
×
⎥ = 2.01 × 10 N
27.3
days
24
h
3600
s
⎣
⎦
Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the
large radius of circular motion around the earth.
8.10.
Model the ball as a particle in uniform circular motion. Rolling friction is ignored.
Solve:
The track exerts both an upward normal force and an inward normal force. From Newton’s second law,
Model:
Visualize:
2
⎡ 60 rev 2π rad 1min ⎤
Fnet = n2 = mrω 2 = ( 0.030 kg )( 0.20 m ) ⎢
×
×
= 0.24 N
1 rev
60 s ⎥⎦
⎣ min
8.11.
Model:
Visualize:
The satellite is considered to be a particle in uniform circular motion around the moon.
Solve: The radius of the moon is 1.738 × 106 m and the satellite’s distance from the center of the moon is the
same quantity. The angular velocity of the satellite is
ω=
2π
2π rad 1min
=
×
= 9.52 × 10−4 rad/s
T 110 min 60 s
and the centripetal acceleration is
ar = rω 2 = (1.738 × 106 m )( 9.52 × 10−4 rad/s ) = 1.58 m/s 2
2
The acceleration of a body in orbit is the local “g” experienced by that body.
8.12.
Model: The earth is considered to be a particle in uniform circular motion around the sun.
Solve: The earth orbits the sun in 365 days and is 1.5 × 1011 m from the sun. The angular velocity and
centripetal acceleration are
ω=
2π rad 1 day
1h
×
×
= 2.0 × 10−7 rad/s
365 days 24 h 3600 s
ar = g = rω 2 = (1.5 × 1011 m )( 2.0 × 10−7 rad/s ) = 6.0 × 10−3 m/s 2
2
Assess:
The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance.
8.13.
Model:
Visualize:
Use the particle model for the car which is undergoing circular motion.
Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of
the hill is
mv 2
∑ F = ( F ) − n = mg − n = ma = r
r
G r
r
r
n⎞
⎛
⇒ v2 = r ⎜ g − ⎟
m
⎝
⎠
Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road.
vmax = rg =
Assess:
( 50 m ) ( 9.8 m/s 2 ) = 22 m/s
A speed of 22 m/s is equivalent to 49 mph, which seems like a reasonable value.
8.14.
Model:
Visualize:
The passengers are particles in circular motion.
Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force
pointing up that provides the needed centripetal acceleration. The normal force on the passengers is their weight.
Ordinarily their weight is FG , so if their weight increases by 50%, n = 1.5 FG . Newton’s second law at the
bottom of the dip is
mv 2
∑ Fr = n − FG = (1.5 − 1) FG = 0.5mg = r
⇒ v = 0.5 gr = 0.5 ( 9.8 m/s 2 ) ( 30 m ) = 12.1 m/s
Assess:
A speed of 12.1 m/s is 27 mph, which seems very reasonable.
8.15.
Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing
uniform circular motion.
Visualize:
Notice that the r-axis points downward, toward the center of the circle.
G
K
Solve: The critical speed occurs when n goes to zero and FG provides all the centripetal force pulling the car
in the vertical circle. At the critical speed mg = mvc2 r , therefore vc = rg . Since the car’s speed is twice the
critical speed, vt = 2vc and the centripetal force is
∑ Fr = n + FG =
mv 2 m ( 4vc ) m ( 4rg )
=
=
= 4mg
r
r
r
Thus the normal force is n = 3 mg . Consequently, n FG = 3.
2
8.16.
Model:
Visualize:
Model the roller coaster car as a particle undergoing uniform circular motion along a loop.
Notice that the r-axis points downward, toward the center of the circle.
Solve: In this problem the normal force is equal to the gravitational force: n = FG = mg . We have
mv 2
∑ F = n + F = r = mg + mg ⇒ v = 2rg = 2 ( 20 m ) ( 9.8 m/s ) = 19.8 m/s
r
G
2
8.17.
Model:
Visualize:
Model the bucket of water as a particle in uniform circular motion.
Solve: Let us say the distance from the bucket handle to the top of the water in the bucket is 35 cm. This makes the
shoulder to water distance 65 cm + 35 cm = 1.00 m. The minimum angular velocity for swinging a bucket of water in
a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case,
n = 0 N when the bucket is in the top position of the circular motion. We get
mvc 2
∑ F = n + F = 0 N + mg = r = mrω
r
⇒ ωc = g / r =
G
2
c
9.8 m/s 2
1 rev
60 s
= 3.13 rad/s = 3.13 rad s ×
×
= 30 rpm
1.00 m
2π rad 1 min
8.18.
Model:
Visualize:
Solve:
Use the particle model for the car, which is undergoing nonuniform circular motion.
The car is in circular motion with radius r =
d
= 100 m. We require
2
1.5 m/s 2
1.5 m/s 2
=
= 0.122 s −1
r
100 m
The definition of the angular velocity can be used to determine the time Δt using the angular acceleration
a 1.5 m/s 2
α= t =
= 1.5 × 10−2 s −2 .
r
100 m
ω = ωi + αΔt
ar = ω 2 r = 1.5 m/s 2 ⇒ ω =
⇒ Δt =
ω − ωi 0.122 s −1 − 0 s −1
=
= 8.2 s
α
0.015 s −2
8.19.
Model:
Visualize:
Solve:
The train is a particle undergoing nonuniform circular motion.
(a) Newton’s second law in the vertical direction is
( Fnet ) y = n − FG = 0
from which n = mg . The rolling friction is f R = μ R n = μ R mg . This force provides the tangential acceleration
at = −
The angular acceleration is
fR
= −μR g
m
2
at − μ R g − ( 0.10 ) ( 9.8 m/s )
=
=
= −1.96 rad/s 2
r
r
0.50 m
⎛ rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞
(b) The initial angular velocity is 30 ⎜
⎟⎜
⎟⎜
⎟ = 3.14 rad/s. The time to come to a stop due to
⎝ min ⎠⎝ 60 sec ⎠⎝ rev ⎠
the rolling friction is
ω − ωi 0 − 3.14 rad/s
Δt = f
=
= 1.60 s
α
−1.96 rad/s 2
Assess: The original angular speed of π rad/s means the train goes around the track one time every 2 seconds,
so a stopping time of less than 2 s is reasonable.
α=
8.20.
Model: The object is treated as a particle in the model of kinetic friction with its motion governed by
constant-acceleration kinematics.
Visualize:
The velocity v1x as the object sails off the edge is related to the initial velocity v0 x by
Solve:
v = v + 2ax ( x1 − x0 ) . Using Newton’s second law to determine ax while sliding gives
2
1x
2
0x
∑ Fx = − f k = max ⇒ ∑ Fy = n − mg = 0 N ⇒ n = mg
Using this result and the model of kinetic friction ( f k = μ k n ) , the x-component equation can be written as
− μ k mg = max . This implies
ax = − μ k g = − ( 0.50 ) ( 9.8 m/s 2 ) = −4.9 m/s 2
Kinematic equations for the object’s free fall can be used to determine v1x :
y2 = y1 + v1 y ( t2 − t1 ) + 12 ( − g )( t2 − t1 ) ⇒ 0 m = 1.0 m + 0 m −
2
g
2
( t2 − t1 ) ⇒ ( t2 − t1 ) = 0.4518 s
2
x2 = x1 + v1x ( t2 − t1 ) = 2.30 m = 2.0 m + v1x ( 0.4518 s ) ⇒ v1x = 0.664 m/s
Having determined v1x and ax , we can go back to the velocity equation v12x = v02x + 2ax ( x1 − x0 ) :
( 0.664 m/s ) = v02x + 2 ( −4.9 m/s2 ) ( 2.0 m ) ⇒ v0 x = 4.5 m/s
2
Assess:
v0 x = 4.5 m/s is about 10 mph and is a reasonable speed.
8.21.
Model: The rocket and puck together make a particle moving on frictionless ice. The thrust of the
rocket motor is assumed to be constant.
Visualize:
Solve The kinematics equations can be used to examine the motion for each coordinate axis independently as a
function of time t, then combined to eliminate t.
In the x-direction,
x f = xi + v0 xt = 0 m + ( 2.0 m/s ) t
So t =
xf
2.0 m/s
. The acceleration of the rocket and puck in the y-direction is
ay =
( Fnet ) y
mrocket + mpuck
=
8.0 N
= 13.3 m/s 2
0.600 kg
In the y–direction,
1
1
yf = yi + v0 yt + a yt 2 = 0 m + ( 0 m/s ) t + (13.3 m/s 2 ) t 2
2
2
= ( 6.67 m/s 2 ) t 2
Substituting t from the equation for the x-direction,
2
⎛ x ⎞
2
yf = ( 6.67 m/s 2 ) ⎜
⎟ = 1.67 xf
2.0
m/s
⎝
⎠
A graph of this equation is shown in the figure.
Assess: When the puck has traveled 9 m in the x-direction it has traveled 15 m in the y-direction.
8.22.
Model: Treat Sam as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes
airborne.
Solve:
Sam’s acceleration up the slope is given by Newton’s second law:
( Fnet ) s = F − mg sin10° = ma0
a0 =
F
200 N
− g sin10° =
− (9.8 m/s 2 ) sin10° = 0.965 m/s 2
m
75 kg
The length of the slope is s1 = (50 m)/ sin10° = 288 m. His velocity at the top of the slope is
v12 = v02 + 2a0 ( s1 − s0 ) = 2a0 s1 ⇒ v1 = 2(0.965 m/s 2) (288 m) = 23.6 m/s
This is Sam’s initial speed into the air, giving him velocity components v1x = v1 cos10° = 23.2 m/s and
v1 y = v1 sin10° = 410 m/s. This is not projectile motion because Sam experiences both the force of gravity and the thrust
of his skis. Newton’s second law for Sam’s acceleration is
a1x =
a1 y =
( Fnet ) y
m
=
( Fnet ) x (200 N)cos10°
=
= 2.63 m/s 2
m
75 kg
(200 N)sin10° − (75 kg)(9.80 m/s 2 )
= −9.34 m/s 2
75 kg
The y-equation of motion allows us to find out how long it takes Sam to reach the ground:
y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 50 m + (4.10 m/s) t2 − (4.67 m/s 2) t22
This quadratic equation has roots t2 = −2.86 s (unphysical) and t2 = 3.74 s. The x-equation of motion—this time
with an acceleration—is
x2 = x1 + v1xt2 + 12 a1xt22 = 0 m + (23.2 m/s) t2 − 12 (2.63 m/s 2) t22 = 105 m
Sam lands 105 m from the base of the cliff.
8.23. Model: Treat the motorcycle and rider as a particle.
Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it
becomes airborne.
Solve:
The motorcycle’s acceleration on the ramp is given by Newton’s second law:
( Fnet ) s = − f r − mg sin 20° = − μ r n − mg sin 20° = − μ r mg cos 20° − mg sin 20° = ma0
a0 = − g ( μ r cos 20° + sin 20°) = −(9.8 m/s 2 )((0.02)cos 20° + sin 20°) = −3.536 m/s 2
The length of the ramp is s1 = (2.0 m)/ sin 20° = 5.85 m. We can use kinematics to find its speed at the top of the
ramp:
v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1
⇒ v1 = (11.0 m/s) 2 + 2(−3.536 m/s 2) (5.85 m) = 8.92 m/s
This is the motorcycle’s initial speed into the air, with velocity components v1x = v1 cos 20° = 8.38 m/s and
v1 y = v1 sin 20° = 3.05 m/s. We can use the y-equation of projectile motion to find the time in the air:
y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 2.0 m + (3.05 m/s) t2 − (4.90 m/s 2) t22
This quadratic equation has roots t2 = −0.399 s (unphysical) and t2 = 1.021 s. The x-equation of motion is thus
x2 = x1 + v1xt2 = 0 m + (8.38 m/s) t2 = 8.56 m
8.56 m < 10.0 m, so it looks like crocodile food.
8.24.
Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket.
Solve: (a) The acceleration of the rocket in the launch direction is obtained from Newton’s second law
G
G
F = ma :
140,700 N = ( 5000 kg ) a ⇒ a = 28.14 m/s 2
Therefore, ax = a cos 44.7° = 20.0 m/s 2 and a y = a sin 44.7° = 19.8 m/s 2 . The net acceleration in the y-direction
is thus
( anet ) y = a y − g = (19.8 − 9.8) m/s 2 = 10.0 m/s2
With this acceleration, we can write the equations for the x- and y-motions of the rocket.
y = y0 + v0 y ( t − t0 ) + 12 ( anet ) y ( t − t0 ) = 0 m + 0 m + 12 (10.0 m/s 2 ) t 2 = ( 5.00 m/s 2 ) t 2
2
x = x0 + v0 x ( t − t0 ) + 12 ( anet ) x ( t − t0 ) = 0 m + 0 m + 12 ( 20.0 m/s 2 ) t 2 = (10.0 m/s 2 ) t 2
2
From these two equations,
2
2
x (10.0 m/s ) t
=
=2
y ( 5.00 m/s 2 ) t 2
The equation that describes the rocket’s trajectory is y = 12 x.
(b) It is a straight line with a slope of 12 .
(c) In general,
v y = v0 y + ( anet ) y ( t1 − t0 ) = 0 + (10.0 m/s 2 ) t1
vx = v0 x + ( anet ) x ( t1 − t0 ) = 0 + ( 20.0 m/s 2 ) t1
v=
(10.0 m/s ) t + ( 20.0 m/s ) t = ( 22.36 m/s ) t
2 2 2
1
2 2 2
1
2
1
The time required to reach the speed of sound is calculated as follows:
330 m/s = ( 22.36 m/s 2 ) t1 ⇒ t1 = 14.76 s
We can now obtain the elevation of the rocket. From the y-equation,
y = ( 5.00 m/s 2 ) t12 = ( 5.00 m/s 2 ) (14.76 s ) = 1090 m
2
8.25.
Model: The hockey puck will be treated as a particle whose motion is determined by constantacceleration kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and
the second to free fall.
Visualize:
Solve:
Newton’s second law is:
Fx = max ⇒ ax =
Fx 2.0 N
=
= 2.0 m/s 2
m 1.0 kg
The kinematic equation v12x = v02x + 2ax ( x1 − x0 ) yields:
v12x = 0 m 2 /s 2 + 2 ( 2.0 m/s 2 ) ( 4.0 m ) ⇒ v1x = 4.0 m/s
Let us now find the time of free fall ( t2 − t1 ) :
y2 = y1 + v1 y ( t2 − t1 ) + 12 a1 y ( t2 − t1 )
2
⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s 2 ) ( t2 − t1 ) ⇒ ( t2 − t1 ) = 0.639 s
2
Having obtained v1x and ( t2 − t1 ) , we can now find ( x2 − x1 ) as follows:
x2 = x1 + v1x ( t2 − t1 ) + 12 ax ( t2 − t1 )
2
⇒ x2 − x1 = ( 4.0 m/s )( 0.639 s ) + 12 ( 2.0 m/s 2 ) ( 0.639 s ) = 3.0 m
2
Assess:
For a modest horizontal thrust of 2.0 N, a landing distance of 3.0 m is reasonable.
8.26.
Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration
kinematic equations.
Visualize:
Solve:
As the rocket is accidentally bumped v0 x = 0.5 m/s and v0 y = 0 m/s. On the other hand, when the
engine is fired
Fx = max ⇒ ax =
Fx
20 N
=
= 40 m/s 2
m 0.500 kg
(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ,
2
0 m = 40 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 2.857 s
The distance from the base of the wall is
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( 0.5 m/s )( 2.857 s ) + 12 ( 40 m/s 2 ) ( 2.857 s ) = 165 m
2
2
(b) The x- and y-equations are
y = y0 + v0 y ( t − t0 ) + 12 a y ( t − t0 ) = 40 − 4.9t 2
2
x = x0 + v0 x ( t − t0 ) + 12 ax ( t − t0 ) = 0.5t + 20t 2
2
Except for a brief interval near t = 0, 20t 2 0.5t. Thus x ≈ 20t 2 , or t 2 = x / 20. Substituting this into the yequation gives
y = 40 − 0.245 x
This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.
8.27.
Model:
Visualize:
Assume the particle model for the satellite in circular motion.
To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete
rotation.
Because
the
altitude
of
the
satellite
is
3.58 × 107 m,
r = 3.58 × 107 m,
re = 3.58 × 107 m + 6.37 × 106 m = 4.22 × 107 m.
Solve: (a) The period (T ) of the satellite is 24.0 hours.
(b) The acceleration due to gravity is
2
2
1 hr ⎞
⎛ 2π ⎞
⎛ 2π
7
2
×
g = ar = rω 2 = r ⎜
⎟ = ( 4.22 × 10 m ) ⎜
⎟ = 0.223 m/s
⎝ T ⎠
⎝ 24.0 hr 3600 s ⎠
(c) There is no normal force on a satellite, so the weight is zero. It is in free fall.
8.28.
Model:
earth rotates.
Visualize:
Treat the man as a particle. The man at the equator undergoes uniform circular motion as the
Solve: The scale reads the man’s weight FG = n, the force of the scale pushing up against his feet. At the north
pole, where the man is in static equilibrium,
nP = FG = mg = 735 N
At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The raxis points toward the center, so
∑ F = F − n = mω r ⇒ n = mg − mω r = n − mω r
2
r
G
E
2
E
2
P
The equator scale reads less than the north pole scale by the amount mω 2 r. The man’s angular velocity is that of
the equator, or
2π
2π rad
ω=
=
= 7.27 × 10−5 rad/s
T
24 hours × (3600 s/1 h)
Thus the north pole scale reads more than the equator scale by
Δw = (75 kg)(7.27 × 10−5 rad/s) 2 (6.37 × 106 m) = 2.5 N
Assess:
The man at the equator appears to have lost Δm = Δw/g ≈ 0.25 kg, or the equivalent of ≈ 12 lb.
8.29.
Model:
Visualize:
Solve:
Use the particle model for the (cart + child) system which is in uniform circular motion.
Newton’s second law along r and z directions can be written:
∑ F = T cos 20° − n sin 20° = ma ∑ F = T sin 20° − n cos 20° − mg = 0
r
r
z
The cart’s centripetal acceleration is
2
rev 1 min 2π rad ⎞
⎛
2
×
×
ar = rω 2 = ( 2.0cos20° m ) ⎜14
⎟ = 4.04 m/s
min 60 s
1 rev ⎠
⎝
The above force equations can be rewritten as
0.94T − 0.342n = ( 25 kg ) ( 4.04 m/s 2 ) = 101 N
0.342T + 0.94n = ( 25 kg ) ( 9.8 m/s 2 ) = 245 N
Solving these two equations yields T = 179 N for the tension in the rope.
Assess: In view of the child + cart weight of 245 N, a tension of 179 N is reasonable.
8.30.
Model:
Visualize:
Solve:
Model the ball as a particle which is in a vertical circular motion.
At the bottom of the circle,
∑ Fr = T − FG =
mv 2
( 0.500 kg ) v ⇒ v = 5.5 m/s
⇒ (15 N ) − ( 0.500 kg ) ( 9.8 m/s 2 ) =
r
(1.5 m )
2
8.31.
Model: We will use the particle model for the car, which is undergoing uniform circular motion on a
banked highway, and the model of static friction.
Visualize:
Note that we need to use the coefficient of static friction μ s , which is 1.0 for rubber on concrete.
Solve: Newton’s second law for the car is
mv 2
∑ F = f cosθ + n sinθ = r ∑ F = n cosθ − f sinθ − F = 0 N
r
s
z
s
G
Maximum speed is when the static friction force reaches its maximum value ( fs )max = μs n. Then
mv 2
n ( cos15° − μs sin15° ) = mg
r
Dividing these two equations and simplifying, we get
n ( μ s cos15° + sin15° ) =
μs + tan15° v 2
μ + tan15°
=
⇒ v = gr s
1 − μs tan15° gr
1 − μ s tan15°
=
Assess:
( 9.80 m/s ) ( 70 m ) ( (1 − 0.268) ) = 34 m/s
2
1.0 + 0.268
The above value of 34 m/s ≈ 70 mph is reasonable.
8.32.
Model:
Visualize:
Solve:
Use the particle model for the rock, which is undergoing uniform circular motion.
Newton’s second law is
mv 2
∑ F = T cos10° = r ∑ F = T sin10° − mg = 0 N
r
z
where the radius of the circular motion is r = (1.0 m ) cos10° = 0.985 m. Dividing these two equations, we get
tan10° =
⇒ω =
gr
⇒v=
v2
gr
=
tan10°
( 9.8 m/s ) ( 0.985 m ) = 7.40 m/s
2
tan10°
rad 1 rev
60 s
v 7.40 m/s
=
= 7.51 rad/s = 7.51
×
×
= 72 rpm
r 0.985 m
s 2π rad 1 min
8.33.
Model:
motion.
Visualize:
Use the particle model and static friction model for the coin, which is undergoing circular
Solve: The force of static friction is f s = μ s n = μs mg . This force is equivalent to the maximum centripetal force
that can be applied without sliding. That is,
v2
r
2
μs mg = m t = m ( rω max
) ⇒ ωmax =
= 7.23
μs g
r
=
( 0.80 ) ( 9.8 m/s 2 )
0.15 m
= 7.23 rad/s
rad 1 rev
60 s
×
×
= 69 rpm
s 2π rad 1 min
So, the coin will stay still on the turntable.
Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.
8.34.
Model:
Visualize:
Solve:
Use the particle model for the car, which is in uniform circular motion.
Newton’s second law is
∑ Fr = T sin 20° = mar =
mv 2
r
∑ F = T cos 20° − F = 0 N
z
G
These equations can be written as
T sin 20° =
mv 2
T cos 20° = mg
r
Dividing these two equations gives
tan 20° = v 2 rg ⇒ v = rg tan 20° =
( 4.55 m ) ( 9.8 m/s 2 ) tan 20° = 4.03 m/s
8.35.
Model:
Visualize:
Use the particle model for the ball in circular motion.
G
Solve: (a) The mass moves in a horizontal circle of radius r = 20 cm. The acceleration a and the net force
G
vector point to the center of the circle, not along the string. The only two forces are the string tension T , which
G
does point along the string, and the gravitational force FG . These are shown in the free-body diagram. Newton’s
second law for circular motion is
mv 2
∑ Fz = T cosθ − FG = T cosθ − mg = 0 N ∑ Fr = T sinθ = mar = r
From the z-equation,
( 0.500 kg ) ( 9.8 m/s )
mg
=
= 5.00 N
cosθ
cos11.54°
2
T=
(b) We can find the tangential speed from the r-equation:
v=
rT sin θ
= 0.63 m/s
m
The angular speed is
v
r
ω= =
0.63 ms
= 3.15 rad/s = 30 rpm
0.20 m
8.36.
Model:
Visualize:
Consider the passenger to be a particle and use the model of static friction.
Solve: The passengers stick to the wall if the static friction force is sufficient to support the gravitational force
on them: fs = FG The minimum angular velocity occurs when static friction reaches its maximum possible value
( fs )max = μs n. Although clothing has a range of coefficients of friction, it is the clothing with the smallest
coefficient ( μ s = 0.60) that will slip first, so this is the case we need to examine. Assuming that the person is
stuck to the wall, Newton’s second law is
∑ F = n = mω r
2
r
The minimum frequency occurs when
∑ Fz = fs − w = 0 ⇒ fs = mg
2
f s = ( f s )max = μ s n = μ s mrω min
Using this expression for fs in the z-equation gives
2
fs = μ s mrω min
= mg
⇒ ω min =
g
9.80 m/s 2
1 rev
60 s
=
= 2.56 rad/s = 2.56 rad/s ×
×
= 24 rpm
μs r
0.60(2.5 m)
2π rad 1 min
Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is
not necessary.
8.37.
Model:
Visualize:
Use the particle model for the marble in uniform circular motion.
Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support the
gravitational on it: f s = FG If mg > ( fs )max then static friction is not sufficient and the marble will slip down the
side as it rolls around the circumference. The r-equation of Newton’s second law is
∑
2
2π rad 1 min ⎞
⎛
×
Fr = n = mrω 2 = (0.010 kg)(0.060 m) ⎜150 rpm ×
⎟ = 0.148 N
1 rev
60 s ⎠
⎝
Thus the maximum possible static friction is ( fs )max = μs n = (0.80)(0.148 N) = 0.118 N. The friction force
needed to support a 10 g marble is f s = mg = 0.098 N. We see that f s < ( fs )max , therefore friction is sufficient
and the marble spins in a horizontal circle.
Assess: In reality, rolling friction will cause the marble to gradually slow down until ( fs )max < mg . At that
point, it will begin to slip down the inside wall.
8.38.
Model:
Visualize:
Use the particle model for the car and the model of kinetic friction.
Solve: We will apply Newton’s second law to all three cars.
Car A:
∑ F = n + ( f ) + ( F ) = 0 N − f + 0 N = ma
∑ F = n + ( f ) + y = n + 0 N − mg = 0 N
x
k x
x
y
G x
k y
y
k
x
y
The y-component equation means n = mg . Since f k = μ k n, we have f k = μ k mg . From the x-component
equation,
−f
− μ k mg
ax = k =
= − μ k g = −9.8 m/s 2
m
m
Car B: Car B is in circular motion with the center of the circle above the car.
mv 2
∑ Fr = nr + ( f k )r + ( FG )r = n + 0 N − mg = mar = r
∑ F = n + ( f ) + ( F ) = 0 N − f + 0 N = + ma
t
k t
t
G t
t
k
From the r-equation
n = mg +
⎛
mv 2
v2 ⎞
⇒ fk = μk n = μk m ⎜ g + ⎟
r
r ⎠
⎝
Substituting back into the t-equation,
2
⎛
f
μ m⎛
v2 ⎞
( 25 m/s ) ⎞⎟ = −12.9 m/s 2
at = − k = − k ⎜ g + ⎟ = − μ k ⎜ 9.8 m/s 2 +
⎜
m
m ⎝
r ⎠
200 m ⎟⎠
⎝
Car C: Car C is in circular motion with the center of the circle below the car.
mv 2
∑ F = n + ( f ) + ( F ) = −n + 0 N + mg = ma = r
r
k r
r
r
G r
∑ F = n + ( f ) + ( F ) = 0 N − f + 0 N = ma
t
t
k t
G t
k
t
From the r-equation n = m ( g − v 2 r ) . Substituting this into the t-equation yields
at =
− fk −μk n
=
= − μ k ( g − v 2 r ) = −6.7 m/s 2
m
m
8.39.
Model:
Visualize:
Solve:
Model the ball as a particle that is moving in a vertical circle.
(a) The ball’s gravitational force FG = mg = ( 0.500 kg ) ( 9.8 m/s 2 ) = 4.9 N.
(b) Newton’s second law at the top is
v2
∑ F = T + F = ma = m r
r
1
G
r
⎡ ( 4.0 m/s )2
⎤
⎛ v2
⎞
⇒ T1 = m ⎜ − g ⎟ = ( 0.500 kg ) ⎢
− 9.8 m/s 2 ⎥ = 2.9 N
⎢⎣ 1.02 m
⎥⎦
⎝ r
⎠
(c) Newton’s second law at the bottom is
∑ Fr = T2 − FG =
mv 2
r
2
⎡
⎛
( 7.5 m/s ) ⎤ = 32 N
v2 ⎞
⇒ T2 = m ⎜ g + ⎟ = ( 0.500 kg ) ⎢9.8 m/s 2 +
⎥
1.02 m ⎥⎦
r ⎠
⎢⎣
⎝
8.40.
Model:
Visualize:
Solve:
Use the particle model for yourself while in uniform circular motion.
(a) The speed and acceleration are
2π r 2π (15 m )
v 2 ( 3.77 m/s )
= 0.95 m/s 2
=
= 3.77 m/s ar = =
15 m
r
T
25 s
2
v=
So the speed is 3.8 m/s and the centripetal acceleration is 0.95 m/s 2 .
(b) The weight w = m, the normal force. On the ground, your weight is the same as the gravitational force FG .
Newton’s second law at the top is
mv 2
∑ Fr = FG − n = mar = r
2
⎛
⎛
( 3.77 m/s ) ⎞⎟ = m 8.85 m/s 2
v2 ⎞
⇒ n = w = m ⎜ g − ⎟ = m ⎜ 9.8 m/s 2 −
(
)
⎜
⎟
15 m
r ⎠
⎝
⎝
⎠
⇒
w 8.85 m s 2
=
= 0.90
FG 9.8 m s 2
(c) Newton’s second law at the bottom is
∑ Fr = n − FG = mar =
mv 2
r
⎛
(3.77 m/s2 ) ⎞⎟ = 10.75 m/s2 m
⎛
v2 ⎞
⇒ n = w = m ⎜ g + ⎟ = m ⎜ 9.8 m/s 2 +
(
)
⎜
⎟
r ⎠
15 m
⎝
⎝
⎠
⇒
w 10.75 m s 2
=
= 1.10
FG
9.8 m s 2
8.41.
Model:
Visualize:
Solve:
Model a passenger as a particle rotating in a vertical circle.
(a) Newton’s second law at the top is
mv 2
∑ F = n + F = ma = r
r
T
G
r
⇒ nT + mg =
mv 2
r
The speed is
v=
2π r 2π ( 8.0 m )
=
= 11.17 m/s
T
4.5 s
⎡ (11.17 m/s )2
⎤
⎛ v2
⎞
⇒ nT = m ⎜ − g ⎟ = ( 55 kg ) ⎢
− 9.8 m/s 2 ⎥ = 319 N
⎢⎣ 8.0 m
⎥⎦
⎝ r
⎠
That is, the ring pushes on the passenger with a force of 3.2 × 102 N at the top of the ride. Newton’s second law at the
bottom:
mv 2
mv 2
⎛ v2
⎞
∑ F = n − F = ma = r ⇒ n = r + mg = m ⎜ r + g ⎟
r
B
G
r
B
⎝
⎠
⎡ (11.17 m/s )2
⎤
= ( 55 kg ) ⎢
+ 9.8 m/s 2 ⎥ = 1397 N
⎢⎣ 8.0 m
⎥⎦
Thus the force with which the ring pushes on the rider when she is at the bottom of the ring is 1.4 kN.
(b) To just stay on at the top, nT = 0 N in the r-equation at the top in part (a). Thus,
2
⎛ 2π ⎞
r
mv 2
8.0 m
= mrω 2 = mr ⎜
mg =
= 2π
= 5.7 s
⎟ ⇒ Tmax = 2π
g
9.8 m/s 2
r
⎝ Tmax ⎠
8.42.
Model:
Visualize:
Solve:
Model the chair and the rider as a particle in uniform circular motion.
Newton’s second law along the r-axis is
∑ F = T + ( F ) = ma ⇒ T sinθ + 0 N = mrω
r
r
G r
2
r
Since r = L sin θ , this equation becomes
2
⎡ 2π rad ⎤
T = mLω 2 = (150 kg )( 9.0 m ) ⎢
⎥ = 3330 N
⎣ 4.0 s ⎦
Thus, the 3000 N chain is not strong enough for the ride.
8.43.
Model:
Visualize:
Model the ball as a particle in motion in a vertical circle.
G
Solve: If the ball moves in a complete circle, then there is a tension force T when the ball is at the top of the
circle. The tension force adds to the gravitational force to cause the centripetal acceleration. The forces are along
the r-axis, and the center of the circle is below the ball. Newton’s second law at the top is
( Fnet )r = T + FG = T + mg =
⇒ vtop = rg +
mv 2
r
rT
m
The tension T can’t become negative, so T = 0 N gives the minimum speed vmin at which the ball moves in a
circle. If the speed is less than vmin , then the string will go slack and the ball will fall out of the circle before it
reaches the top. Thus,
vmin = rg ⇒ ω min =
rg
vmin
=
=
r
r
g
=
r
( 9.8 m/s ) = 3.13 rad / s = 30 rpm
2
(1.0 m )
8.44.
Model: The ball is a particle on a massless rope in circular motion about the point where the rope is
attached to the ceiling.
Visualize:
Solve:
Newton’s second law in the radial direction is
( ∑ F ) = T − F = T − mg = mvr
r
2
G
Solving for the tension in the rope and evaluating,
2
⎛
⎛
( 5.5 m/s ) ⎞⎟ = 168 N
v2 ⎞
T = m ⎜ g + ⎟ = (10.2 kg ) ⎜ 9.8 m/s 2 +
⎜
4.5 m ⎟⎠
r ⎠
⎝
⎝
Assess: The tension in the rope is greater than the gravitational force on the ball in order to keep the ball
moving in a circle.
8.45.
Model:
Visualize:
Model the person as a particle in uniform circular motion.
Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the
r-axis is:
∑ F = n = mrω
2
r
To create “normal” gravity, the normal force by the inside surface of the space station equals mg. Therefore,
mg = mrω 2 ⇒ ω =
Assess:
2π
=
T
g
r
500 m
⇒ T = 2π
= 2π
= 45.0 s
r
g
9.8 m/s 2
This is a fast rotation. The tangential speed is
v=
2π r 2π ( 500 m )
=
= 70 m/s ≈ 140 mph
T
45 s
8.46. Model:
Visualize:
Masses m1 and m2 are considered particles. The string is assumed to be massless.
Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth,
G
G
it acts like a pulley. Thus tension forces T1 and T2 act as if they were an action/reaction pair. Mass m1 is in
circular motion of radius r, so Newton’s second law for m1 is
∑ Fr = T1 =
m1v 2
r
Mass m2 is at rest, so the y-equation of Newton’s second law is
∑F =T − m g = 0 N ⇒T = m g
y
2
2
2
2
Newton’s third law tells us that T1 = T2 . Equating the two expressions for these quantities:
m1v 2
m2 rg
= m2 g ⇒ v =
r
m1
8.47.
Model:
Visualize:
Model the ball as a particle swinging in a vertical circle, then as a projectile.
Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circularmotion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the
bottom of the circle. Since T > FG , there is a net force toward the center of the circle that causes the centripetal
acceleration. The r-equation of Newton’s second law is
( Fnet )r = T − FG = T − mg =
mv 2
r
r
0.60 m
⎡5.0 N − ( 0.10 kg ) ( 9.8 m/s 2 ) ⎤ = 4.91 m/s
(T − mg ) =
⎦
m
0.100 kg ⎣
G
As a projectile the ball starts at y0 = 1.4 m with v0 = 4.91iˆ m/s. The equation for the y-motion is
⇒ vbottom =
y1 = 0 m = y0 + v0 y Δt − 12 g ( Δt ) = y0 − 12 gt12
2
This is easily solved to find that the ball hits the ground at time
t1 =
2 y0
= 0.535 s
g
During this time interval it travels a horizontal distance
x1 = x0 + v0 xt1 = ( 4.91 m/s )( 0.535 s ) = 2.63 m
So the ball hits the floor 2.6 m to the right of the point where the string was cut.
8.48.
Model:
Visualize:
Solve:
Use the particle model for a ball in motion in a vertical circle and then as a projectile.
For the circular motion, Newton’s second law along the r-direction is
mvt2
∑F =T + F = r
r
G
Since the string goes slack as the particle makes it over the top, T = 0 N. That is,
FG = mg =
mvt2
⇒ vt = gr =
r
( 9.8 m/s ) ( 0.5 m ) = 2.21 m/s
2
The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be
found as follows:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 0.639 s
2
2
The place where the ball hits the ground is
x1 = x0 + v0 x ( t1 − t0 ) = 0 m + ( +2.21 m/s )( 0.639 s − 0 s ) = +1.41 m
The ball hits the ground 1.41 m to the right of the point beneath the center of the circle.
8.49.
Model:
Visualize:
Model the ball as a particle undergoing circular motion in a vertical circle.
Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final
circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic
equation
v12 = v02 + 2a y ( y1 − y0 ) ⇒ 0 m 2 s 2 = ( v0 ) + 2 ( −9.8 m/s 2 ) ( 4.0 m − 0 m ) ⇒ v0 = 8.85 m/s
2
This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using
the r-component of the net force on the ball:
⎛ v2 ⎞
( 8.85 m/s )
∑ Fr = T = m ⎜⎜ r0 y ⎟⎟ ⇒ T = ( 0.100 kg ) 0.60 m = 13.1 N
⎝
⎠
2
8.50.
Model:
Visualize:
Solve:
Model the car as a particle on a circular track.
(a) Newton’s second law along the t-axis is
∑ F = F = ma ⇒ 1000 N = (1500 kg ) a ⇒ a = 2 3 m/s
t
t
t
t
2
t
With this tangential acceleration, the car’s tangential velocity after 10 s will be
v1t = v0t + at ( t1 − t0 ) = 0 m s + ( 2 3 m/s 2 ) (10 s − 0 s ) = 20 3 m/s
The radial acceleration at this instant is
v 2 ( 20 3 m/s ) 16
ar = 1t =
=
m/s 2
r
25 m
9
The car’s acceleration at 10 s has magnitude
2
a1 = at2 + ar2 =
⎛
⎞
( 2 3 m/s ) + (16 9 m/s ) = 1.90 m/s θ = tan aa = tan ⎜ 162 39 ⎟ = 21°
2 2
2 2
2
−1
t
−1
⎝
⎠
r
where the angle is measured from the r-axis.
(b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value
( fs )max = μs n. That is,
mv22t
⇒ v2t = rg = ( 25 m ) ( 9.8 m/s 2 ) = 15.7 m/s
r
In the above equation, n = mg follows from Newton’s second law along the z-axis. The time when the car
begins to slide can now be obtained as follows:
∑F = ( f )
r
s max
= μ s n = μ s mg =
v2t = v0t + at ( t2 − t0 ) ⇒ 15.7 m/s = 0 m s + ( 2 3 m/s 2 ) ( t2 − 0 ) ⇒ t2 = 24 s
8.51.
Model:
Visualize:
Solve:
Model the steel block as a particle and use the model of kinetic friction.
G
(a) The components of thrust ( F ) along the r-, t-, and z-directions are
Fr = F sin 20° = ( 3.5 N ) sin 20° = 1.20 N Ft = F cos 20° = ( 3.5 N ) cos 20° = 3.29 N Fz = 0 N
Newton’s second law is
( Fnet )r = T + Fr = mrω 2 ( Fnet )t = Ft − f k = mat
( Fnet ) z = n − mg = 0 N
The z-component equation means n = mg . The force of friction is
f k = μ k n = μ k mg = ( 0.60 )( 0.500 kg ) ( 9.8 m/s 2 ) = 2.94 N
Substituting into the t-component of Newton’s second law
( 3.29 N ) − ( 2.94 N ) = ( 0.500 kg ) at ⇒ at = 0.70 m/s 2
Having found at , we can now find the tangential velocity after 10 revolutions = 20π rad as follows:
1⎛ a ⎞
θ1 = ⎜ t ⎟ t12 ⇒ t1 =
2⎝ r ⎠
2rθ1
= 18.95 s
at
⎛a ⎞
⎝r⎠
ω1 = ω 0 + ⎜ t ⎟ t1 = 6.63 rad/s
The block’s angular velocity after 10 s is 6.6 rad/s.
(b) Substituting ω1 into the r-component of Newton’s second law yields:
T1 + Fr = mrω12 ⇒ T1 + (1.20 N ) = ( 0.500 kg )( 2.0 m )( 6.63 rad/s ) ⇒ T1 = 44 N
2
8.52.
Model:
Visualize:
Solve:
Assume the particle model for a ball in vertical circular motion.
(a) Newton’s second law in the r- and t-directions is
( Fnet )r = T + mg cosθ = mar =
mvt2
( Fnet )t = −mg sinθ = mat
r
Substituting into the r-component,
( 20 N ) + ( 2.0 kg ) ( 9.8 m/s 2 ) cos30° = ( 2.0 kg )
vt2
⇒ vt = 3.85 m/s
( 0.80 m )
The tangential velocity is 3.8 m/s.
(b) Substituting into the t-component,
− ( 9.8 m/s 2 ) sin 30° = at ⇒ at = −4.9 m/s 2
The radial acceleration is
vt2 ( 3.85 m/s )
=
= 18.5 m/s 2
0.80 m
r
2
ar =
Thus, the magnitude of the acceleration is
a = ar2 + at2 =
(18.5 m/s ) + ( −4.9 m/s ) = 19.1 m/s
2 2
2 2
The angle of the acceleration vector from the r-axis is
φ = tan −1
The angle is below the r-axis.
at
4.9
= tan −1
= 14.8°
18.5
ar
2
8.53.
Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a
pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead
weight has a mass of 0.30 kg. What was the weight’s angular velocity in rad/s and in rpm?
60 N
rev
60 s
(b)
ω2 =
⇒ ω = 20 rad/s ×
×
= 191 rpm
2π rad min
( 0.3 kg )( 0.5 m )
8.54.
Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of
11,760 N?
(b) The weight is the normal force.
1500 kg v 2
2940 N =
⇒ v = 19.8 m/s
200 m
8.55.
Model:
Visualize:
Assume the particle model and apply the constant-acceleration kinematic equations.
Solve:
(a) Newton’s second law for the projectile is
G
−F
Fnet = − Fwind = max ⇒ ax =
x
m
where Fwind is shortened to F. For the y-motion:
( )
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 0 m + ( v0 sin θ ) t1 − 12 gt12 ⇒ t1 = 0 s and t1 =
2
2v0 sin θ
g
Using the above expression for t1 and defining the range as R we get from the x motion:
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 )
2
⎛ 2v sin θ ⎞ F ⎛ 2v0 sin θ ⎞
⎛ F⎞
⇒ x1 − x0 = R = v0 xt1 + 12 ⎜ − ⎟ t12 = ( v0 cosθ ) ⎜ 0
⎟−
⎜
⎟
g
g
⎝ m⎠
⎝
⎠ 2m ⎝
⎠
=
2
2v02
2v 2 F
cosθ sin θ − 0 2 sin 2 θ
g
mg
We will now maximize R as a function of θ by setting the derivative equal to 0:
dR 2v02
2 Fv02
cos 2 θ − sin 2 θ ) −
2sinθ cosθ = 0
=
(
dθ
g
mg 2
⎛ 2 Fv02 ⎞ ⎛ g ⎞
mg
sin 2θ ⇒ tan 2θ =
⇒ cos 2 θ − sin 2 θ = cos 2θ = ⎜
2 ⎟⎜
2 ⎟
F
⎝ mg ⎠⎝ 2v0 ⎠
Thus the angle for maximum range is θ = 12 tan −1 ( mg / F )
(b) We have
mg ( 0.50 kg ) ( 9.8 m/s )
=
= 8.167 ⇒ θ = 12 tan −1 ( 8.167 ) = 41.51°
F
0.60 N
The maximum range without air resistance is
2
R′ =
2v02 sin 45° cos 45° v02
=
g
g
Therefore, we can write the equation for the range R as
R = 2 R′ sin 41.51° cos 41.51° −
⇒
2F
R′ sin 2 41.51° = R′ ( 0.9926 − 0.1076 ) = 0.885 R′
mg
R
R′ − R
= 0.8850 ⇒
= 1 − 0.8850 = 0.115
R′
R′
Thus R is reduced from R′ by 11.5%.
Assess: The condition for maximum range ( tan 2θ = mg / F ) means 2θ → 90° as F → 0. That is, θ = 45°
when F = 0, as is to be expected.
8.56.
Visualize:
G ⎛1
⎞
From Chapter 6 the drag on a projectile is D = ⎜ Av 2 , direction opposite to motion ⎟ , where A is the
4
⎝
⎠
cross- sectional area. Using the free-body diagram above, apply Newton’s second law to each direction.
In the x-direction,
1
( Fnet ) x D cosθ
Av 2 cosθ
=−
=−4
ax =
m
n
m
1
( Fnet ) y
Av 2 sin θ
D sin θ − FG
=−
=−4
−g
ay =
m
m
m
Solve:
Since v = vx2 + v y2 , vx = v cosθ , and v y = v sin θ , we can rewrite these as
ax = −
Avx vx2 + v y2
A ( v cosθ ) v
=−
4m
4m
ay = −
Av y vx2 + v y2
A ( v sin θ ) v
− g = −g −
4m
4m
8.57.
Model:
Visualize:
Solve:
Use the particle model and apply the constant-acceleration kinematic equations of motion.
(a) The thrust of the rocket burn must be such that vx decreases from 2.0 km/s to 0 during the same
interval that v y increases from 0 to 1.0 km/s. The forward distance in which to accomplish this is
x1 = (500 km)cos30° = 433 km. We can find the burn time by combining two kinematic equations:
v1x = 0 = v0 x + axt1 ⇒ axt1 = −v0 x
x1 = 0 + v0 xt1 + 12 axt12 = v0 xt1 + 12 (−v0 x )t1 = 12 v0 xt1
⇒ t1 =
2 x1 2(433 km)
=
= 433 s
2.0 km/s
v0 x
The required acceleration is ax = −v0 y /t1 = −(2000 m/s)/(433 s) = −4.62 m/s. During this interval, v y increases
from 0 to 1.0 km/s, thus
v1 y = 0 + a yt1 ⇒ a y =
1000 m/s
= 2.31 m/s 2
433 s
The thrust force is given by Newton’s second law:
K
G
Fthrust = ma = (20,000 kg)(–4.62iˆ + 2.31 ˆj m/s 2 ) = (−92,400iˆ + 46,200 ˆj ) N
The magnitude of the thrust force is 103,300 N and it is at angle θ = tan −1 (46,200/92,400) = 26.6° below the +xaxis (i.e., to the right and down). In order to point the thrusters in this direction, you must rotate the rocket
counterclockwise 180° − θ = 153.4°. To make it home with one rocket burn, you need to rotate your rocket to
153.4° and fire with a thrust of 103,300 N for 433 s.
(b) The x- and y-positions during the burn are
x = (2000 m/s) t − 12 (4.62 m/s 2) t 2
y = 12 (2.31 m/s 2) t 2
The y-position at t1 = 433 s is 216,500 m = 216.5 km, 33.5 km away from the entrance. With the rocket off,
you’re now coasting straight toward the entrance at 1.0 km/s and will cover this distance in 33.5 s. Thus you pass
through the entrance at 433 s + 33.5 s = 466.5 s. The following table calculates your position at intervals of 50 s
and, for accuracy, at t1 = 433 s when you end the burn. The trajectory is a parabola that intersects the
x = 433 km line at y = 216.5 km, then a vertical line into the entrance.
8.58.
Model:
Visualize:
Solve:
Use the particle model for the ball, which is in uniform circular motion.
From Newton’s second law along r and z directions,
mv 2
∑ F = n cosθ = r ∑ F = n sinθ − mg = 0 ⇒ n sinθ = mg
r
z
Dividing the two force equations gives
tan θ =
gr
v2
From the geometry of the cone, tanθ = r y . Thus
r gr
=
⇒ v = gy
y v2
8.59.
Model:
Visualize:
Model the block as a particle and use the model of kinetic friction.
Solve: The only radial force is tension, so we can use Newton’s second law to find the angular velocity ω max at
which the tube breaks:
∑ F = T = mω r ⇒ ω
2
r
max
=
Tmax
50 N
=
= 9.12 rad/s
mr
(0.50 kg)(1.2 m)
The compressed air and friction exert tangential forces, and the second law along the tangential direction is
∑ F = F − f = F − μ n = F − μ mg = ma
t
t
k
t
k
t
k
t
4.0 N
F
− (0.60)(9.80 m/s 2 ) = 2.12 m/s 2
at = t − μ k g =
0.50 kg
m
The time needed to accelerate to 9.12 rad/s is given by
⎛a ⎞
⎝r⎠
ω1 = ω max = 0 + ⎜ t ⎟ t1 ⇒ t1 =
rω max (1.2)(9.12 rad/s)
=
= 5.16 s
at
2.12 m/s 2
During this interval, the block turns through angle
⎛ 2.12 m/s 2 ⎞
1 rev
⎛a ⎞
2
Δθ = θ1 − θ 0 = ω 0t1 + 12 ⎜ t ⎟ t12 = 0 + 12 ⎜
= 3.7 rev
⎟ (5.16 s) = 23.52 rad ×
1.2
m
2
π rad
⎝r⎠
⎝
⎠
8.60.
Model:
Visualize:
Solve:
Assume the particle model for a sphere in circular motion at constant speed.
(a) Newton’s second law along the r and z axes is:
mvt2
∑ Fz = T1 cos30° + T 2 cos60° − FG = 0 N
r
Since we want T1 = T2 = T , these two equations become
∑ Fr = T1 sin 30° + T2 sin 60° =
T ( sin 30° + sin 60° ) =
mvt2
T ( cos30° + cos60° ) = mg
r
Since sin 30° + sin 60° = cos30° + cos 60°,
mvt2
⇒ vt = rg
r
The triangle with sides L1 , L2 , and 1.0 m is isosceles, so L2 = 1.0 m and r = L2 cos30°. Thus
mg =
L2 cos 30° g =
(b) The tension is
(1.0 m ) cos 30° g = ( 0.866 m ) ( 9.8 m/s2 ) = 2.9 m/s
( 2.0 kg ) ( 9.8 m/s )
mg
=
= 14.3 N
cos30° + cos60°
0.866 + 0.5
2
T=
8.61.
Model:
Visualize:
Solve:
Use the particle model for a sphere revolving in a horizontal circle.
Newton’s second law in the r- and z-directions is
mvt2
∑ ( F ) = T cos30° + T cos30° = r ∑ ( F ) = T sin 30° − T sin 30° − F = 0 N
r
1
2
z
1
2
Using r = (1.0 m ) cos30° = 0.886 m, these equations become
mvt2
( 0.300 kg )( 7.5 m/s ) = 22.5 N
=
r cos30°
( 0.866 m )( 0.866 )
2
T1 + T2 =
T1 − T2 =
( 0.300 kg ) ( 9.8 m/s 2 )
mg
=
= 5.88 N
sin 30°
( 0.5)
Solving for T1 and T2 yields T1 = 14.2 N and T2 = 8.3 N.
G
8.62.
Use the particle model for the ball.
Solve:
(a) Newton’s second law along the r- and z-directions is
Model:
Visualize:
∑ F = n cosθ = mrω ∑ F = n sinθ − F = 0 N
2
r
z
G
Using FG = mg and dividing these equations yields:
tan θ =
g
R− y
=
rω 2
r
where you can see from the figure that tanθ = ( R − y ) r . Thus ω =
g
.
R− y
(b) ω will be minimum when ( R − y) is maximum or when y = 0 m. Then ω min = g / R .
(c) Substituting into the above expression,
ω=
g
9.8 m/s 2
rad 60 s
1 rev
=
= 9.9
×
×
= 95 rpm
R−y
0.20 m − 0.10 m
s 1 min 2π rad
8.63.
Model:
Visualize:
Use the particle model for the airplane.
G
Solve: In level flight, the lift force L balances the gravitational force. When turning, the plane banks so that
the radial component of the lift force can create a centripetal acceleration. Newton’s second law along the r- and
z-directions is
mvt2
∑ F = L sinθ = r ∑ F = L cosθ − mg = 0 N
r
z
These can be written:
sin θ =
mv 2
mg
cosθ =
L
rL
Dividing the two equations gives:
2
miles 1 hr 1610 m ⎤
⎡
⎢⎣ 400 hour × 3600 s × 1 mile ⎥⎦
v2
v2
tanθ =
⇒r=
=
= 18.5 km
gr
g tan θ
⎡⎣9.8 m s 2 ⎤⎦ tan10°
The diameter of the airplane’s path around the airport is 2 × 18.5 km = 37 km.
8.64.
Model:
Visualize:
Use the particle model for a small volume of water on the surface.
Solve: Consider a particle of water of mass m at point C on the surface. Newton’s second law along the r- and zdirections is
( Fnet )r = n cosθ = mrω 2 ⇒ cosθ =
mrω 2
n
( Fnet ) z = n sinθ − mg = 0 N ⇒ sinθ =
mg
n
Dividing both equations gives tan θ = g rω 2 . For a parabola z = ar 2 . This means
1
1
dz
= 2ar = slope of the curve at C = tan φ = tan ( 90° − θ ) =
⇒ tan θ =
tan θ
2ar
dr
Equating the two equations for tanθ , we get
1
g
ω2
= 2 ⇒a=
2ar rω
2g
Thus the surface is described by the equation
z=
ω2
2g
r2
which is the equation of a parabola.
8-1
9.1. Model: Model the car and the baseball as particles.
Solve:
(a) The momentum p = mv = (1500 kg )(10 m/s ) = 1.5 × 104 kg m/s.
(b) The momentum p = mv = ( 0.2 kg )( 40 m/s ) = 8.0 kg m/s.
9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle.
Solve:
From the definition of momentum,
pcar = pbicycle ⇒ mcar v car = mbicyclevbicycle ⇒ vbicycle =
⎛ 1500 kg ⎞
mcar
v car = ⎜
⎟ ( 5.0 m/s ) = 75 m/s
mbicycle
⎝ 100 kg ⎠
Assess: This is a very high speed (≈ 168 mph). This problem shows the importance of mass in comparing two
momenta.
9.3.
Solve:
Visualize: Please refer to Figure EX9.3.
The impulse J x is defined in Equation 9.6 as
tf
J x = ∫ Fx ( t ) dt = area under the Fx (t ) curve between ti and tf
ti
Jx =
1
( 4 ms )(1000 N ) + ( 6 − 4 ms )(1000 N ) = 4 Ns
2
9.4. Model: The particle is subjected to an impulsive force.
Visualize: Please refer to Figure EX9.4.
Solve: Using Equation 9.6, the impulse is the area under the curve. From 0 s to 2 ms the impulse is
∫ Fdt = ( −500 N ) ( 2 × 10 s ) = −0.5 N s
1
2
−3
From 2 ms to 8 ms the impulse is
∫ Fdt = ( +2000 N )(8 ms − 2 ms ) = +6.0 N s
1
2
From 8 ms to 10 ms the impulse is
∫ Fdt = ( −500 N )(10 ms − 8 ms ) = −0.5 N s
1
2
Thus, from 0 s to 10 ms the impulse is ( −0.5 + 6.0 − 0.5 ) N s = 5.0 N s.
9.5. Visualize: Please refer to Figure EX9.5.
Solve:
The impulse is defined in Equation 9.6 as
tf
J x = ∫ Fx ( t ) dt = area under the Fx (t ) curve between ti and tf
ti
⇒ 6.0 N s = 12 ( Fmax )( 8 ms ) ⇒ Fmax = 1.5 × 103 N
9.6. Model: Model the object as a particle and the interaction as a collision.
Visualize: Please refer to Figure EX9.6.
Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negative
momentum). Using the impulse-momentum theorem pfx = pix + J x ,
−2 kg m/s = +6 kg m/s + J x ⇒ J x = −8 kg m/s = −8 N s
Since J x = Favg Δt , we have
Favg Δt = −8 N s ⇒ Favg =
G
The force is F = ( 8 × 102 N, left ) .
−8 N s
= −8 × 102 N
10 ms
9.7. Model: Model the object as a particle and the interaction with the force as a collision.
Visualize: Please refer to Figure EX9.7.
Solve: Using the equations
tf
pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve
ti
( 2.0 kg ) vfx = ( 2.0 kg )(1.0 m/s ) + (area under the force curve)
⇒ vfx = (1.0 m/s ) +
1
(1.0 s )( 2.0 N ) = 2.0 m/s
2.0 kg
Becaue vfx is positive, the object moves to the right at 2.0 m/s.
Assess: For an object with positive velocity, a positive impulse increases the object’s speed. The opposite is
true for an object with negative velocity.
9.8. Model: Model the object as a particle and the interaction with the force as a collision.
Visualize: Please refer to Figure EX9.8.
Solve: Using the equations
tf
pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve
ti
( 2.0 kg ) vfx = ( 2.0 kg )(1.0 m/s ) + (area under the force curve)
⎛ 1 ⎞
⇒ vfx = (1.0 m/s ) + ⎜
⎟ ( −2.0 N )( 0.50 s ) = 0.50 m/s
⎝ 2.0 kg ⎠
Assess: For an object with positive velocity, a negative impulse slows the object. The opposite is true for an
object with negative velocity.
9.9. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum
theorem.
Visualize:
Note that the force of kinetic friction f k imparts a negative impulse to the sled.
Solve:
Using Δpx = J x , we have
tf
tf
ti
ti
pfx − pix = ∫ Fx (t ) dt = − f k ∫ dt = − f k Δt ⇒ mvfx − mvix = − μ k nΔt = − μ k mg Δt
We have used the model of kinetic friction f k = μ k n, where μ k is the coefficient of kinetic friction and n is the
normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken
out of the impulse integral. Thus,
Δt =
1
μk g
( vix − vfx ) =
1
( 0.25) ( 9.8 m/s 2 )
(8.0 m/s − 5.0 m/s ) = 1.22 s
9.10. Model: Use the particle model for the falling object and the impulse-momentum theorem.
Visualize:
Note that the object is acted on by the gravitational force, whose magnitude is mg.
Solve: Using the impulse-momentum theorem,
tf
pfy − piy = J y = ∫ Fy (t ) dt ⇒ mvfy − mviy = −mg Δt
ti
v −v
−5.5 m/s − ( −10.4 m/s )
⇒ Δt = iy fy =
= 0.50 s
9.8 m/s 2
g
Assess:
Since Fy = − mg is independent of time, we have taken it out of the impulse integral.
9.11. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision.
Visualize:
The force increases to Fmax during the first two ms, stays at Fmax for two ms, and then decreases to zero during
the last two ms. The graph shows that Fx is positive, so the force acts to the right.
Solve: Using the impulse-momentum theorem pfx = pix + J x ,
6 ms
( 0.06 kg )( 32 m/s ) = ( 0.06 kg )( −32 m/s ) + ∫ Fx ( t ) dt
0
The impulse is
6 ms
1
1
∫ F ( t ) dx = area under force curve = 2 F ( 0.002 s ) + F ( 0.002 s ) + 2 F ( 0.002 s ) = ( 0.004 s ) F
max
x
0
⇒ Fmax =
max
max
( 0.06 kg )( 32 m/s ) + ( 0.06 kg )( 32 m/s ) = 9.6 × 102 N
0.004 s
max
9.12.
Model: Model the ball as a particle, and its interaction with the wall as a collision in the impulse
approximation.
Visualize: Please refer to Figure EX9.12.
Solve: Using the equations
tf
pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve
ti
( 0.250 kg ) vfx = ( 0.250 kg )( −10 m/s ) + (500 N)(8.0 ms)
⎛ 4.0 N ⎞
⇒ vfx = ( −10 m/s ) + ⎜
⎟ = 6 m/s
⎝ 0.250 kg ⎠
Assess:
The ball’s final velocity is positive, indicating it has turned around.
9.13. Model: Model the glider cart as a particle, and its interaction with the spring as a collision.
Visualize:
Solve:
Using the impulse-momentum theorem pfx − pix = ∫ Fdt ,
( 0.6 kg )( 3 m/s ) − ( 0.6 kg )( −3 m/s ) = area under force curve = 12 ( 36 N )( Δt ) ⇒ Δt = 0.20 s
9.14. Model: Choose car + gravel to be the system. Ignore friction in the impulse approximation.
Visualize:
Solve:
There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This
means pfx = pix . Hence,
(10,000 kg + 4000 kg ) vfx = (10,000 kg )( 2.0 m/s ) + ( 4000 kg )( 0.0 m/s ) ⇒ vfx = 1.43 m/s
9.15. Model: Choose car + rainwater to be the system.
Visualize:
There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved.
Solve: Conservation of momentum is pfx = pix . Hence,
( mcar + mwater )( 20 m/s ) = ( mcar )( 22 m/s ) + ( mwater )( 0 m/s )
⇒ ( 5000 kg + mwater )( 20 m/s ) = ( 5000 kg )( 22 m/s ) ⇒ mwater = 5.0 × 102 kg
9.16. Model:
Visualize:
Choose skydiver + glider to be the system in the impulse approximation.
Note that there are no external forces along the x-direction (ignoring friction in the impulse approximation),
implying conservation of momentum along the x-direction.
Solve: The momentum conservation equation pfx = pix is
( 680 kg − 60kg )( vG ) x + ( 60 kg )( vD ) x = ( 680 kg )( 30 m/s )
Immediately after release, the skydiver’s horizontal velocity is still ( vD ) x = 30 m/s. Thus
( 620 kg )( vG ) x + ( 60 kg )( 30 m/s ) = ( 680 kg )( 30 m/s ) ⇒ ( vG ) x = 30 m/s
Assess:
The skydiver’s motion in the vertical direction has no influence on the glider’s horizontal motion.
9.17. Model: We will define our system to be bird + bug. This is the case of an inelastic collision because
the bird and bug move together after the collision. Horizontal momentum is conserved because there are no
external forces acting on the system during the collision in the impulse approximation.
Visualize:
Solve:
The conservation of momentum equation pfx = pix is
( m1 + m2 ) vfx = m1 ( vix )1 + m2 ( vix )2 ⇒ ( 300 g + 10 g ) vfx = ( 300 g )( 6.0 m/s ) + (10 g )( −30 m/s ) ⇒ vfx = 4.8 m/s
Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from both
sides of the equation. Note that (vix ) 2 is negative.
9.18. Model: The two cars are not an isolated system because of external frictional forces. But during the
collision friction is not going to be significant. Within the impulse approximation, the momentum of the
Cadillac + Volkswagen system will be conserved in the collision.
Visualize:
Solve:
The momentum conservation equation pfx = pix is
( mC + mVW ) vfx = mC ( vix )C + mVW ( vix )VW
⇒ 0 kg mph = ( 2000 kg )(1.0 mph ) + (1000 kg )( vix )VW ⇒ ( vix )VW = −2.0 mph
You need a speed of 2.0 mph.
9.19. Model: Because of external friction and drag forces, the car and the blob of sticky clay are not exactly
an isolated system. But during the collision, friction and drag are not going to be significant. The momentum of
the system will be conserved in the collision, within the impulse approximation.
Visualize:
Solve:
The conservation of momentum equation pfx = pix is
( mC + mB )( vf ) x = mB ( vix )B + mC ( vix )C
⇒ 0 kg m/s = (10 kg )( vix )B + (1500 kg )( −2.0 m/s ) ⇒ (vix ) B = 3.0 × 10 2 m/s
Assess: This speed of the blob is around 600 mph, which is very large. However, we must point out that a very
large speed is expected in order to stop a car with only 10 kg of clay.
9.20. Model: We will define our system to be archer + arrow. The force of the archer (A) on the arrow (a) is
equal to the force of the arrow on the archer. These are internal forces within the system. The archer
G is Gstanding
on frictionless ice, and the normal force by ice on the system balances the weight force. Thus Fext = 0 on the
system, and momentum is conserved.
Visualize:
The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment
pfx must also be zero.
Solve:
We have M A vA + ma va = 0 kg m/s. Therefore,
vA =
−ma va − ( 0.100 kg )(100 m/s )
=
= − 0.20 m/s
50 kg
mA
The archer’s recoil speed is 0.20 m/s.
Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the
arrow has forward momentum, the archer will have backward momentum.
9.21. Model: We will define our system to be Bob + rock. Bob’s (B) force on the rock (R) is equal to the
rock’s force on Bob. These are internal forces within the
Bob is standing on frictionless ice, and the
G
G system.
normal force by ice on the system balances the weight. Fext = 0 on the system, and thus momentum is conserved.
Visualize:
The initial momentum pix of the system is zero because Bob and the rock are at rest. Thus pfx = 0 kg m/s.
Solve: We have mBvB + mR vR = 0 kg m/s. Hence,
vB = −
⎛ 0.500 kg ⎞
mR
vR = − ⎜
⎟ ( 30 m/s ) = −0.20 m/s
mB
⎝ 75 kg ⎠
Bob’s recoil speed is 0.20 m/s.
Assess: Since the rock has forward momentum, Bob’s momentum is backward. This makes the total momentum
zero.
9.22. Model: We will define our system to be Dan + skateboard, and their interaction as an explosion.
While friction is present between the skateboard and the ground, it is negligible in the impulse approximation.
Visualize:
The system has nonzero initial momentum pix . As Dan (D) jumps backward off the gliding skateboard (S), the
skateboard will move forward in such a way that the final total momentum of the system pfx is equal to pix .
Solve: We have mS ( vfx )S + mD ( vfx )D = ( mS + mD ) vix . Hence,
( 5.0 kg )( 8.0 m/s ) + ( 50 kg )( vfx )D = ( 5.0 kg + 50 kg )( 4.0 m/s ) ⇒ ( vfx )D = 3.6 m/s
9.23. Model: We assume that the momentum is conserved in the collision.
Visualize: Please refer to Figure EX9.23.
Solve: The conservation of momentum equation yields
( pfx )1 + ( pfx )2 = ( pix )1 + ( pix )2 ⇒ ( pfx )1 + 0 kg m/s = 2 kg m/s − 4 kg m/s ⇒ ( pfx )1 = −2 kg m/s
( p ) + ( p ) = ( p ) + ( p ) ⇒ ( p ) − 1 kg m/s = 2 kg m/s + 1 kg m/s ⇒ ( p ) = 4 kg m/s
fy 1
fy 2
iy 1
iy 2
fy 1
Thus, the final momentum of particle 1 is ( −2iˆ + 4 ˆj ) kg m/s.
fy 1
9.24. Model: This problem deals with the conservation of momentum in two dimensions in an inelastic
collision.
Visualize:
Solve:
G
G
The conservation of momentum equation pbefore = pafter is
m1 ( vix )1 + m2 ( vix )2 = ( m1 + m2 ) vfx m1 ( viy ) + m2 ( viy ) = ( m1 + m2 ) vfy
1
2
Substituting in the given values,
(.02 kg )( 3.0 m/s ) + 0 kg m/s = (.02 kg + .03 kg ) vf cosθ
0 kg m/s + (.03 kg )( 2.0 m/s ) = (.02 kg + .03 kg ) vf sin θ
⇒ vf cosθ = 1.2 m/s vf sinθ = 1.2 m/s
⇒ vf =
v
(1.2 m/s ) + (1.2 m/s ) = 1.7 m/s θ = tan −1 y = tan −1 (1) = 45°
2
The ball of clay moves 45° north of east at 1.7 m/s.
2
vx
9.25. Model: Model the ball as a particle. We will also use constant-acceleration kinematic equations.
Solve:
(a) The momentum just after throwing is
p0 x = p0 cos30° = mv0 cos30° = ( 0.050 kg )( 25 m/s ) cos30° = 1.083 kg m/s
p0 y = p0 sin 30° = mv0 sin 30° = ( 0.050 kg )( 25 m/s ) sin 30° = 0.625 kg m/s
The momentum just after throwing is (1.08, 0.63) kg m/s.
The momentum at the top is
p1x = mv1x = mv0 x = 1.08 kg m/s p1 y = mv1 y = 0 kg m/s
Just before hitting the ground, the momentum is
p2 x = mv2 x = mv0 x = 1.08 kg m/s p2 y = mv2 y = − mv0 y = −0.63 kg m/s
(b) px is constant because no forces act on the ball in the x-direction. Mathematically,
Fx =
dpx
= 0 N ⇒ px = constant
dt
(c) The change in the y-component of the momentum during the ball’s flight is
Δp y = −0.625 kg m/s − 0.625 kg m/s = −1.250 kg m/s
From kinematics, the time to reach the top is obtained as follows:
v1 y = v0 y + a y ( t1 − t0 ) ⇒ t1 =
−v0 y
ay
=
− (12.5 m/s )
−9.8 m/s 2
= 1.276 s
The time of flight is thus Δt = 2t1 = 2 (1.276 s ) = 2.552 s. Multiplying this time by –mg, the y-component of the
weight, yields –1.25 kg m/s. This follows from the impulse-momentum theorem:
Δp y = − mg Δt
9.26. Model: Model the rocket as a particle, and use the impulse-momentum theorem. The only force acting
on the rocket is due to its own thrust.
Visualize: Please refer to Figure P9.26.
Solve: (a) The impulse is
J x = ∫ Fx dt = area of the Fx (t ) graph between t = 0 s and t = 30 s = 12 (1000 N )( 30 s ) = 1.5 × 104 N s
(b) From the impulse-momentum theorem, pfx = pix + 1.5 × 104 N s. That is, the momentum or velocity increases as
long as J x increases. When J x increases no more, the speed will be a maximum. This happens at t = 30 s. At this
time,
mvfx = mvix + 1.5 × 104 N s ⇒ ( 425 kg ) vfx = ( 425 kg )( 75 m/s ) + 1.5 × 10 4 N s ⇒ vfx = 110 m/s
9.27.
Solve: Using the equation
area under the force curve
m
2.0 s
1
2π t ⎞
= 0 m/s +
(10 N ) sin ⎛⎜
⎟ dt
0.250 kg ∫0
⎝ 4.0 s ⎠
v fx = vix +
2.0 s
⎞
⎛ 4.0 s ⎞ ⎛
⎛ 2π t ⎞
⎜
⎟
= ( 40 N/m ) ⎜
−
cos
⎟⎜
⎜
⎟
⎝ 2π ⎠ ⎝
⎝ 4.0 s ⎠ 0 ⎟⎠
⎛ 80
⎞
= −⎜
m/s ⎟ ( cos (π ) − cos ( 0 ) )
⎝π
⎠
= 25 m/s
⎛ 2π t ⎞
The force is applied for half the period of 4.0 s. During that time, sin ⎜
⎟ is positive, so an object
⎝ 4.0 s ⎠
initially at rest acquires a positive velocity.
Assess:
9.28. Model: Let the system be ball + racket. During the collision of the ball and racket, momentum is
conserved because all external interactions are insignificantly small.
Visualize:
Solve:
(a) The conservation of momentum equation pfx = pix is
mR ( vfx )R + mB ( vfx )B = mR ( vix )R + mB ( vix )B
(1.000 kg )( vfx )R + ( 0.060 kg )( 40 m/s ) = (1.000 kg )(10 m/s ) + ( 0.060 kg )( −20 m/s ) ⇒ ( vfx )R = 6.4 m/s
(b) The impulse on the ball is calculated from ( pfx ) B = ( pix ) B + J x as follows:
( 0.060 kg )( 40 m/s ) = ( 0.060 kg )( −20 m/s ) + J x ⇒ J x = 3.6 N s = ∫ Fdt = Favg Δt
⇒ Favg =
3.6 Ns
= 3.6 × 102 N
10 ms
Let
us
now
compare
this
force
with
the
gravitational
ball ( FG )B = mB g = ( 0.060 kg ) ( 9.8 m/s 2 ) = 0.588 N. Thus, Favg = 612( FG ) B .
force
on
the
Assess: This is a significant force and is reasonable because the impulse due to this force changes the direction
as well as the speed of the ball from approximately 45 mph to 90 mph.
9.29. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor.
We will also use constant-acceleration kinematic equations. Ignore any forces other than the interaction between
the floor and the ball during the collision in the impulse approximation.
Visualize:
Solve:
To find the ball’s velocity just before and after it hits the floor:
v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2 ( −9.8 m/s 2 ) ( 0 − 2.0 m ) ⇒ v1 y = −6.261 m/s
v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2 ( −9.8 m/s 2 ) (1.5 m − 0 m ) ⇒ v2 y = 5.422 m/s
The force exerted by the floor on the ball can be found from the impulse-momentum theorem:
mv2 y = mv1 y + ∫ Fdt = mv1 y + area under the force curve
⇒ ( 0.200 kg )( 5.422 m/s ) = − ( 0.200 kg )( 6.261 m/s ) + 12 Fmax ( 5.0 × 10−3 s )
⇒ Fmax = 9.3 × 102 N
Assess:
A maximum force of 9.3 × 102 N exerted by the floor is reasonable.
9.30. Model: Model the rubber ball as a particle that is subjected to an impulsive force when it comes in
contact with the floor. We will also use constant-acceleration kinematic equations and the impulse-momentum
theorem.
Visualize:
Solve: (a) To find the magnitude and direction of the impulse that the floor exerts on the ball, we use the
impulse-momentum theorem:
pfy = piy + J y ⇒ J y = p2 y − p1 y = m ( v2 y − v1 y )
Let us now find v1y and v2 y by using kinematics. For the falling and rebounding ball,
v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2 ( −9.8 m/s 2 ) ( 0 m − 1.8 m ) ⇒ v1 y = −5.940 m/s
v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2 ( −9.8 m/s 2 ) (1.2 m − 0 m ) ⇒ v2 y = 4.850 m/s
Going back to the impulse-momentum equation, we find
J y = ( 0.040 kg ) ⎡⎣ 4.850 m/s − ( −5.940 m/s ) ⎤⎦ = 0.432 N s
The impulse is 0.43 N s upward.
(b) As the ball compresses, the force of contact increases and the ball slows to v y = 0 m/s. Then in
decompression the ball is accelerated upward. To a good approximation, the force due to the floor as a function
of time is shown in the figure.
(c) For a rubber ball, Δt is likely in the range 5 to 10 ms. For 10 ms,
J y = Favg Δt = 0.432 N s ⇒ Favg ≈ 40 N
The force is ≈80 N if Δt = 5 ms. Altogether, 40 to 80 N is a reasonable estimate.
9.31. Model: Model the cart as a particle rolling down a frictionless ramp. The cart is subjected to an
impulsive force when it comes in contact with a rubber block at the bottom of the ramp. We will use the impulsemomentum theorem and the constant-acceleration kinematic equations.
Visualize:
Solve:
From the free-body diagram on the cart, Newton’s second law “before the collision” is
∑ ( F ) = F sinθ = ma ⇒ a =
x
G
x
x
mg sin θ
g
= g sin 30° =
m
2
Using this acceleration, we can find the cart’s speed just before its contact with the rubber block:
v12x = v02x + 2ax ( x1 − x0 ) = 0 m 2 /s 2 + 2 ( 12 g )(1.0 m − 0 m) ⇒ v1x = 3.13 m/s
Now we can use the impulse-momentum theorem to obtain the velocity just after the collision:
mv2 x = mv1x + ∫ Fx dt = mv1x + area under the force graph
⇒ ( 0.500 kg ) v2 x = ( 0.500 kg )( 3.13 m/s ) − 12 ( 200 N ) ( 26.7 × 10−3 s ) ⇒ v2 x = −2.21 m/s
Note that the given force graph is positive, but in this coordinate system the impulse of the force is to the left up
the slope. That is the reason to put a minus sign while evaluating the ∫ Fx dt integral.
We can once again use a kinematic equation to find how far the cart will roll back up the ramp:
v32x = v22x + 2ax ( x3 − x2 ) ⇒ ( 0 m/s ) = ( −2.21 m/s ) + 2 ( − 12 g ) ( x3 − x2 ) ⇒ ( x3 − x2 ) = 0.50 m
2
2
9.32.
Solve: Using Newton’s second law for the x–direction, Fx =
Fx =
dpx
. Therefore,
dt
d
( 6t 2 kg m/s ) = 12t N
dt
Assess: The x-component of the net force on an object is equal to the time rate of change of the x-component of
the object’s momentum.
9.33.
Visualize:
Solve:
Using Newton’s second law for the y-direction and the chain rule,
( Fnet ) y =
dp y
dt
=
⎛ dv ⎞
d
dm
mv y ) =
vy ) + m ⎜ y ⎟
(
(
dt
dt
⎝ dt ⎠
= ( −0.50 kg/s )(120 m/s ) + ( 48 kg ) (18 m/s 2 )
= 8.0 × 102 N
Assess:
Since the rocket is losing mass,
dm
< 0. The time derivative of the velocity is the acceleration.
dt
9.34. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with
perfectly inelastic collisions. Momentum is conserved in the collisions of this problem in the impulse
approximation, in which we ignore external forces during the time of the collision.
Visualize:
Solve:
In the collision between the three-car train and the single car:
mv1x + ( 3m ) v2 x = 4mv3 x ⇒ v1x + 3v2 x = 4v3 x ⇒ ( 4.0 m/s ) + 3( 2.0 m/s ) = 4v3 x ⇒ v3 x = 2.5 m/s
In the collision between the four-car train and the stationary car:
( 4m ) v3 x + mv4 x = ( 5m ) v5 x ⇒ 4v3 x + 0 m/s = 5v5 x ⇒ v5 x =
4v3 x
= ( 0.8 )( 2.5 m/s ) = 2.0 m/s
5
9.35.
Model: The dart and cork are particles in free fall until they have a perfectly inelastic head-on
collision. Ignore air friction.
Visualize:
Solve:
The positions of the dart and cork just before the collision are
1
( y1 )D = ( y0 )D + ( v0 )D t1 − gt12
2
1 2
( y1 )C = ( y0 )C − gt1
2
In the particle model, the dart and cork have no physical size, so ( y1 )D = ( y1 )C . Hence
1
1
1
9.8 m/s 2 ) t12 = 3 m − ( 9.8 m/s 2 ) t12 ⇒ t1 = s
(
2
2
3
At this time, the velocities of the dart and cork are
1
( v1 )D = ( v0 )D − gt1 = 9.0 m/s − ( 9.8 m/s 2 ) ⎛⎜ s ⎞⎟ = 5.73 m/s
⎝3 ⎠
1
( v1 )C = ( v0 )C − gt1 = − ( 9.8 m/s 2 ) ⎜⎛ s ⎟⎞ = −3.27 m/s
⎝3 ⎠
These are the initial velocities to use with momentum conservation in a perfectly inelastic collision.
pfy = piy
0 m + ( 9.0 m/s ) t1 −
( mD + mC ) v = mD ( v1 )D + mC ( v1 )C
Thus
v=
( 0.030 kg )( 5.73 m/s ) + ( 0.020 kg )( −3.27 m/s ) = 2.13 m/s, up
( 0.030 kg + 0.20 kg )
Assess: The heavier, faster upward-going dart has more momentum than the falling cork, so the total
momentum is upwards.
Model: Model the earth (E) and the asteroid (A) as particles. Earth + asteroid is our system. Since the
two stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in
the collision since no significant external force acts on the system.
Visualize:
9.36.
Solve:
(a) The conservation of momentum equation pfx = pix is
mA ( vix )A + mE ( vix )E = ( mA + mE ) vfx
⇒ (1.0 × 1013 kg )( 4.0 × 104 m s ) + 0 kg m/s = (1.0 × 1013 kg + 5.98 × 1024 kg ) vfx ⇒ vfx = 6.7 × 10−8 m/s
(b) The speed of the earth going around the sun is
2π r 2π (1.50 × 10 m )
=
= 3.0 × 104 m/s
T
3.15 × 107 s
11
vE =
Hence, vfx vE = 2 × 10−12 = 2 × 10−10%.
Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.
9.37. Model: Model the skaters as particles. The two skaters, one traveling north (N) and the other traveling
west (W), are a system. Since the two skaters hold together after the “collision,” this is a case of a perfectly
inelastic collision in two dimensions. Momentum is conserved since no significant external force in the x-y plane
acts on the system in the “collision.”
Visualize:
Solve:
(a) The x-component of the conservation of momentum is
( mN + mW ) vfx = mN ( vix ) N + mW ( vix )W ⇒ ( 75 kg + 60 kg ) vfx = 0 kg m/s + ( 60 kg )( −3.5 m/s ) ⇒ vfx = −1.556 m/s
The y-component of the conservation of momentum is
( mN + mW ) vfy = mN ( viy )N + mW ( viy )W ⇒ ( 75 kg + 60 kg ) vfy = ( 75 kg )( 2.5 m/s ) + 0 kg m/s
⇒ vfy = 1.389 m/s ⇒ vf =
( vfx ) + ( vfy ) = 2.085 m/s
2
2
The time to glide to the edge of the rink is
radius of the rink
25 m
=
= 12.0 s
vf
2.085 m/s
(b) The location is θ = tan −1 vfy vfx = 42° north of west.
Assess:
A time of 12.0 s in covering a distance of 25 m at a speed of ≈ 2 m/s is reasonable.
9.38. Model: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier
and lighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each
other. During the “explosion,” the total momentum of the system is conserved.
Visualize:
Solve:
The initial momentum is zero. Thus the conservation of momentum equation pfx = pix is
mH ( vfx )H + mL ( vfx )L = 0 kg m/s ⇒ ( 75 kg )( vfx )H + ( 50 kg )( vfx )L = 0 kg m/s
Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find
( vfx )H = 30 m 20 s = 1.5 m/s. Thus,
( 75 kg )(1.5 m/s ) + ( 50 kg )( vfx )L = 0 kg m/s ⇒ ( vfx )L = −2.25 m/s
Thus, the time for the lighter skater to reach the edge is
30 m
30 m
=
= 13.3 s
2.25
m/s
v
( fx )L
Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time to
reach the edge of the ice rink.
9.39. Model: This problem deals with a case that is the opposite of a collision. Our system is comprised of
three coconut pieces that are modeled as particles. During the blow up or “explosion,” the total momentum of the
system is conserved in the x-direction and the y-direction.
Visualize:
Solve:
The initial momentum is zero. From pfx = pix , we get
+ m1 ( vfx )1 + m3 ( vf )3 cosθ = 0 kg m/s ⇒ ( vf )3 cosθ =
−m1 ( vfx )1
m3
=
−m ( −20 m/s )
= 10 m/s
2m
=
− m ( −20 m/s )
= 10 m/s
2m
From pfx = pix , we get
+ m2 ( vfy ) + m3 ( vf )3 sin θ = 0 kg m/s ⇒ ( vf )3 sin θ =
2
⇒ ( vf )3 =
− m2 ( vfy )
m3
2
(10 m/s ) + (10 m/s ) = 14.1 m/s θ = tan −1 (1) = 45°
The velocity is 14.1 m/s at 45° east of north.
2
2
9.40. Model: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m2
(moving west), together are our system. This is an isolated system because any frictional force during the brief
collision period is going to be insignificant. Within the impulse approximation, the momentum of our system will
be conserved in the collision.
Visualize:
Note that m1 = m2 = m.
Solve:
The equation pfx = pix yields:
m1 ( vfx )1 + m2 ( vfx )2 = m1 ( vix )1 + m2 ( vix )2 ⇒ m1 ( vf )1 cosθ + 0 kg m/s = m1 ( vix )1 + m2 ( vix )2
⇒ ( vf )1 cosθ = ( vix )1 + ( vix )2 = 2.0 m/s − 1.0 m/s = 1.0 m/s
The equation pfy = piy yields:
+ m1 ( vfy ) sin θ + m2 ( vfy ) = 0 kg m/s ⇒ ( vf )1 sin θ = − ( vfy ) = −1.41 m/s
1
2
⇒ ( vf )1 =
2
(1.0 m/s ) + ( −1.41 m/s ) = 1.73 m/s
2
2
⎛ 1.41 m/s ⎞
⎟ = 55°
⎝ 1.0 m/s ⎠
θ = tan −1 ⎜
The angle is below +x axis, or south of east.
9.41. Model: This is a two-part problem. First, we have an inelastic collision between the wood block and
the bullet. The bullet and the wood block are an isolated system. Since any external force acting during the
collision is not going to be significant (the impulse approximation), the momentum of the system will be
conserved. The second part involves the dynamics of the block + bullet sliding on the wood table. We treat the
block and the bullet as particles.
Visualize:
Solve:
The equation pfx = pix gives
( mB + mW ) vfx = mB ( vix )B + mW ( vix )W
⇒ ( 0.010 kg + 10 kg ) vfx = ( 0.010 kg )( vix )B + (10 kg )( 0 m/s ) ⇒ vfx =
1
( vix )B
1001
From the model of kinetic friction,
f k = − μ k n = − μ k ( mB + mW ) g = ( mB + mW ) ax ⇒ ax = − μ k g
Using the kinematic equation v12x = v22x + 2ax ( x1 − x0 ) ,
2
2
⎛ 1 ⎞
v12x = v22x − 2 μ k g ( x1 − x0 ) ⇒ 0 m 2 s 2 = vf2x − 2μ k g x1 ⇒ ⎜
⎟ ( vix )B = 2 μ k g x1
⎝ 1001 ⎠
⇒ ( vix )B = 1001 2 μ k g x1 = 1001 2 ( 0.20 ) ( 9.8 m/s 2 ) ( 0.050 m ) = 443 m/s
The bullet’s speed is 4.4 × 102 m/s.
Assess: The bullet’s speed is reasonable (≈ 900 mph).
9.42. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus
(B). Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since no
significant external force acts on the system. The second part involves the dynamics of the Fred + Brutus system
sliding on the ground.
Visualize:
Note that the collision is head-on and therefore one-dimensional.
Solve: The equation pfx = pix is
( mF + mB ) vfx = mF ( vix )F + mB ( vix )B ⇒ ( 60 kg + 120 kg ) vfx = ( 60 kg )( −6.0 m/s ) + (120 kg )( 4.0 m/s )
⇒ vfx = 0.667 m/s
The positive value indicates that the motion is in the direction of Brutus.
The model of kinetic friction yields:
f k = − μ k n = − μ k ( mF + mB ) g = ( mF + mB ) ax ⇒ ax = − μ k g
Using the kinematic equation v12x = v02x + 2ax ( x1 − x0 ) , we get
v12x = v02x − 2 μ k g x1 ⇒ 0 m 2 s 2 = vf2x − 2 ( 0.30 ) ( 9.8 m/s 2 ) x1
⇒ 0 m 2 s 2 = ( 0.667 m/s ) − ( 5.9 m/s 2 ) x1 ⇒ x1 = 7.6 cm
2
They slide 7.6 cm in the direction Brutus was running.
Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction.
A stopping distance of 7.6 cm is reasonable.
9.43. Model: Model the package and the rocket as particles. This is a two-part problem. First we have an
inelastic collision between the rocket (R) and the package (P). During the collision, momentum is conserved
since no significant external force acts on the rocket and the package. However, as soon as the package + rocket
system leaves the cliff they become a projectile motion problem.
Visualize:
Solve: The minimum velocity after collision that the package + rocket must have to reach the explorer is v0 x ,
which can be found as follows:
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ −200 m = 0 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 6.389 s
2
With this time, we can now find v0 x using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) . We obtain
2
30 m = 0 m + v0 x ( 6.389 s ) + 0 m ⇒ v0 x = 4.696 m/s = vfx
We now use the momentum conservation equation pfx = pix which can be written
( mR + mP ) vfx = mR ( vix )R + mP ( vix )P
⇒ (1.0 kg + 5.0 kg )( 4.696 m/s ) = (1.0 kg )( vix )R + ( 5.0 kg )( 0 m/s ) ⇒ ( vix )R = 28 m/s
9.44.
Model:
Visualize:
Solve:
The clay balls undergo a perfectly inelastic collision in the impulse approximation.
(a) The total momentum in reference frame S is
P = ( px )1 + ( px )2 = m1 ( vx )1 + m2 ( vx )2
= ( 0.020 kg )(12 m/s ) + 0 kg m/s
= 0.24 kg m/s
(b) In the S′ frame, the total momentum is zero.
P′ = ( px )1′ + ( px )2′ = 0
⇒ m1 ( vx )1′ = − m2 ( vx )2′
Equation 4.24 relates velocities in the two reference frames.
(v ) ′ = (v ) −V
x 1
x 1
x
( vx )2′ = ( vx )2 − Vx
Thus
m1 ( ( vx )1 − Vx ) = − m2 ( ( vx )2 − Vx )
⇒ Vx =
m1 ( vx )1 + m2 ( vx )2
m1 + m2
= 4.0 m/s
(c) Since the total momentum is zero, the final velocity of the resulting 60 g clay ball must be ( vx )f ′ = 0 m/s.
(d) Using Equation 4.24 again,
( vx )f = ( vx )f ′ + Vx = 0 m/s + 4.0 m/s = 4.0 m/s
Assess: The final velocity of both balls in the S frame is less than the original velocity of the smaller ball, and
is in the same direction. The S′ frame is called the “center of mass frame.” Some collisions are more easily
studied in the center of mass frame, such as high energy particle collisions at particle accelerator facilities.
9.45.
Model: This is a two-part problem. First, we have an explosion that creates two particles. The
momentum of the system, comprised of two fragments, is conserved in the explosion. Second, we will use
kinematic equations and the model of kinetic friction to find the displacement of the lighter fragment.
Visualize:
The initial momentum is zero. Using momentum conservation pfx = pix during the explosion,
Solve:
mH ( v1x )H + mL ( v1x )L = mH ( v0 x )H + mL ( v0 x )L ⇒ 7m ( v1x )H + m ( v1x )L = 0 kg m/s ⇒ ( v1x )H = − ( 17 ) ( v1x )L
Because mH slides to x2H = −8.2 m before stopping, we have
f k = μ k nH = μ k wH = μ k mH g = mH aH ⇒ aH = μ k g
Using kinematics,
( v2 x )H = ( v1x )H + 2aH ( x2H − x1H ) ⇒ 0 m 2 s 2 = ( 17 ) ( v1x )L + 2μ k g ( −8.2 m − 0 m ) ⇒ ( v1x )L = −88.74 μ k m/s
2
2
2
2
How far does mL slide? Using the information obtained above in the following kinematic equation,
( v2 x )L = ( v1x )L + 2aL ( x2L − x1L ) ⇒ 0 m 2 s 2 = μ k ( 88.74 ) − 2μ k gx2L ⇒ x2L = 4.0 × 102 m
2
Assess:
2
2
Note that aH is positive, but aL is negative, and both are equal in magnitude to μ k g . Also, x2H is
negative but x2L is positive.
9.46. Model: We will model the two fragments of the rocket after the explosion as particles. We assume the
explosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use
kinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we
will apply conservation of momentum to the system (that is, the two fragments) in the explosion. In the third
part, we will again use kinematic equations to find the velocity of the heavier fragment just after the explosion.
Visualize:
Solve:
The rocket accelerates for 2.0 s from rest, so
v1 y = v0 y + a y ( t1 − t0 ) = 0 m/s + (10 m/s 2 ) ( 2 s − 0 s ) = 20 m/s
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) = 0 m + 0 m + 12 (10 m/s 2 ) ( 2 s ) = 20 m
2
2
At the explosion the equation pfy = piy is
mL ( v2 y ) + mH ( v2 y ) = ( mL + mH ) v1 y ⇒ ( 500 kg ) ( v2 y ) + (1000 kg ) ( v2 y ) = (1500 kg )( 20 m/s )
L
H
L
H
To find (v2 y ) H we must first find (v2 y ) L , the velocity after the explosion of the upper section. Using kinematics,
( v ) = ( v ) + 2 ( −9.8 m/s ) ( y − y ) ⇒ ( v ) = 2 ( 9.8 m/s ) ( 530 m − 20 m ) = 99.98 m/s
2
2
3y L
2y L
2
2
3L
2L
2y L
Now, going back to the momentum conservation equation we get
( 500 kg )( 99.98 m/s ) + (1000 kg ) ( v2 y )H = (1500 kg )( 20 m/s ) ⇒ ( v2 y )H = −20 m/s
The negative sign indicates downward motion.
9.47. Model: Let the system be bullet + target. No external horizontal forces act on this system, so the
horizontal momentum is conserved. Model the bullet and the target as particles. Since the target is much more
massive than the bullet, it is reasonable to assume that the target undergoes no significant motion during the brief
interval in which the bullet passes through.
Visualize:
Solve: (a) By assuming that the target has negligible motion during the interval in which the bullet passes
through, the time is that needed to slow from 1200 m/s to 900 m/s in a distance of 30 cm. We’ll use kinematics to
first find the acceleration, then the time.
(v1x ) B2 = (v0 x ) B2 + 2ax Δx
⇒ ax =
⇒ Δt =
(v1x ) B2 − (v0 x ) B2 (900 m/s)2 − (1200 m/s)2
=
= −1.05 × 106 m/s 2
2Δx
2(0.30 m)
(v1x ) B = (v0 x ) B + ax Δt
(v1x ) B − (v0 x ) B 900 m/s − 1200 m/s
=
= 2.86 × 10−4 s = 286 μ s
ax
−1.05 × 106 m/s 2
The average force on the bullet is Favg = m ax = 26 kN.
(b) Now we can use the conservation of momentum equation p1x = p0 x to find
mT (v1x )T + mB (v1x ) B = mT (v0 x )T + mB (v0 x ) B = 0 + mB (v0 x ) B
⇒ (v1x )T =
mB (v0 x ) B − mB (v1x ) B 0.025 kg
=
( (1200 m/s) − (900 m/s) ) = 0.021 m/s
mT
350 kg
9.48. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem.
First, we have a collision between the bullet and the first block (A). Momentum is conserved since no external
force acts on the system (bullet + block A). The second part of the problem involves a perfectly inelastic
collision between the bullet and block B. Momentum is again conserved for this system (bullet + block B).
Visualize:
Solve:
For the first collision the equation pfx = pix is
mL ( v1x )L + mA ( v1x )A = mL ( v0 x )L + mA ( v0 x )A
⇒ ( 0.010 kg )( v1x )L + ( 0.500 kg )( 6.0 m/s ) = ( 0.010 kg )( 400 m/s ) + 0 kg m/s ⇒ ( v1x )L = 100 m/s
The bullet emerges from the first block at 100 m/s. For the second collision the equation pfx = pix is
( mL + mB ) v2 x = mL ( v1x )L ⇒ ( 0.010 kg + 0.500 kg ) v2 x = ( 0.010 kg )(100 m/s ) ⇒ v2 x = 1.96 m/s
9.49. Model: Model Brian (B) along with his wooden skis as a particle. The “collision” between Brian and
Ashley lasts for a short time, and during this time no significant external forces act on the Brian + Ashley
system. Within the impulse approximation, we can then assume momentum conservation for our system. After
finding the velocity of the system immediately after the collision, we will apply constant-acceleration kinematic
equations and the model of kinetic friction to find the final speed at the bottom of the slope.
Visualize:
Solve:
Brian skiing down for 100 m:
( v1x )B = ( v0 x )B + 2ax ( x1B − x0B ) = 0 m 2 s 2 + 2ax (100 m − 0 m ) ⇒ ( v1x )B = ( 200 m ) ax
2
2
To obtain ax , we apply Newton’s second law to Brian in the x and y directions as follows:
∑ ( F ) = w sinθ − f = m a ∑ ( F ) = n − w cosθ = 0 N ⇒ n = w cosθ
on B x
B
k
B x
on B y
B
From the model of kinetic friction, f k = μ k n = μ k wB cosθ . The x-equation thus becomes
wB sin θ − μ k wB cosθ = mB ax
⇒ ax = g ( sin θ − μ k cosθ ) = ( 9.8 m/s 2 ) ⎡⎣sin 20° − ( 0.060 ) cos 20°⎤⎦ = 2.80 m/s 2
Using this value of ax , ( v1x )B =
( 200 m ) ( 2.80 m/s 2 ) = 23.7 m/s. In the collision with Ashley the conservation
of momentum equation pfx = pix is
( mB + mA ) v2 x = mB ( v1x )B ⇒ v2 x =
mB
80 kg
( v1x )B =
( 23.66 m/s ) = 14.56 m/s
mB + mA
80 kg + 50 kg
Brian + Ashley skiing down the slope:
v32x = v22x + 2ax ( x3 − x2 ) = (14.56 m/s ) + 2 ( 2.80 m/s 2 ) (100 m ) ⇒ v3 x = 27.8 m/s
2
That is, Brian + Ashley arrive at the bottom of the slope with a speed of 27.8 m/s. Note that we have used the
same value of ax in the first and the last parts of this problem. This is because ax is independent of mass.
Assess:
A speed of approximately 60 mph on a ski slope of 200 m length and 20° slope is reasonable.
9.50. Model: Model the spy plane (P) and the rocket (R) as particles. The plane and the rocket undergo a
perfectly inelastic collision. During the brief collision time, momentum for the plane + rocket system will be
conserved because no significant external forces act on this system during the collision. After finding the velocity
of the system immediately after the collision, we will apply projectile equations to find where the system will hit
the ground.
Visualize:
Solve:
For the collision we have pfx = pix and pfy = piy . This means the magnitude of the final momentum is
pf =
⇒ vf =
( pfx ) + ( pfy ) = ( pix ) + ( piy ) ⇒ ( mR + mP ) vf =
2
2
1
2
2
⎡⎣ mR ( vix )R ⎤⎦ + ⎡ mP ( viy ) ⎤
P⎦
⎣
2
2
⎡(1280 kg )( 725 m/s ) ⎤⎦ + ⎡⎣( 575 kg )( 450 m/s ) ⎤⎦ = 519.4 m/s
2
(1280 kg + 575 kg ) ⎣
2
⎛ ( 575 kg )( 450 m/s ) ⎞
⎛ pfy ⎞
−1
⎟⎟ = 15.58° north of east
⎟ = tan ⎜⎜
⎝ pfx ⎠
⎝ (1280 kg )( 725 m/s ) ⎠
θ = tan −1 ⎜
We call this direction the x′ axis in the x-y plane. Let us now look at the falling plane + rocket system in the
x′-z plane where z is the vertical axis perpendicular to the x-y plane. We have
z0 = 2700 m
z1 = 0 m
v0 z = 0 m/s
x0′ = 0 m
t0 = 0 s
v0 x′ = 519.4 m/s
Using the kinematic equation z1 = z0 + v0 z ( t1 − t0 ) + 12 az ( t1 − t0 ) , we can find the time to fall:
2
0 m = 2700 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 23.47 s
In this time t1 , the wreckage travels horizontally to
x1′ = x0′ + v0 x′ ( t1 − t0 ) = 0 m + ( 519.4 m/s )( 23.47 s ) = 12,192 m = 12.19 km
The enmeshed plane + rocket system lands 12.19 km from the collision point at an angle of 15.58° north of east.
9.51. Model: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector
G
quantity, so the direction of the initial velocity vector v1 establishes the direction of the momentum vector. The
final momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue
to move along this line and have no y-components of momentum. The missing third piece cannot have a ycomponent of momentum if momentum is to be conserved, so it must move along the x-axis—either straight
forward or straight backward. We can use conservation laws to find out.
Visualize:
Solve:
From the conservation of mass, the mass of piece 3 is
m3 = mtotal − m1 − m2 = 7.0 × 105 kg
To conserve momentum along the x-axis, we require
[ pi = mtotalvi ] = [ pf = p1f + p2f + p3f = m1v1f + m2v2f + p3f ]
⇒ p3f = mtotalvi − m1v1f − m2v2f = +1.02 × 1013 kg m/s
Because p3f > 0, the third piece moves in the +x-direction, that is, straight forward. Because we know the mass
m3 , we can find the velocity of the third piece as follows:
v3f =
p3f 1.02 × 1013 kg m/s
=
= 1.46 × 107 m/s
m3
7.0 × 105 kg
9.52.
Model: The two railcars make up a system. The impulse approximation is used while the spring is
expanding, so friction can be ignored. The spring is massless.
Visualize:
Solve:
holds.
Since the cars are at rest initially, the total momentum of the system is zero. Conservation of momentum
0 = m1 ( vfx )1 + m2 ( vfx )2
We are only told that the relative velocity of the two cars after the spring expands is 4.0 m/s, so
( vfx )2 − ( vfx )1 = 4.0 m/s
The positive-going car 2 velocity is written first to ensure that the relative velocity is positive. Substitute
( vfx )2 = ( vfx )1 + 4.0 m/s into the conservation of momentum equation, then solve for ( vfx )1 .
0 = m1 ( vfx )1 + m2 ( ( vfx )1 + 4.0 m/s )
⇒ ( vfx )1 = −
Assess:
m2 ( 4.0 m/s )
( 90 tons )( 4.0 m/s ) = −3.0 m/s
=−
( m1 + m2 )
( 30 tons + 90 tons )
The other more massive railcar has a velocity ( vfx )2 = ( vfx )1 + 4.0 m/s = 1.0 m/s. A slower speed for the
more massive car makes sense.
9.53. Model: This is a three-part problem. In the first part, the shell, treated as a particle, is launched as a
projectile and reaches its highest point. We will use constant-acceleration kinematic equations for this part. The
shell, which is our system, then explodes at the highest point. During this brief explosion time, momentum is
conserved. In the third part, we will again use the kinematic equations to find the horizontal distance between the
landing of the lighter fragment and the origin.
Visualize:
Solve:
The initial velocity is
v0 x = v cosθ = (125 m/s ) cos55° = 71.7 m/s
v0 y = v sinθ = (125 m/s ) sin 55° = 102.4 m/s
At the highest point, v1 y = 0 m/s and v1x = 71.7 m/s. The conservation of momentum equation pfx = pix is
mL ( v1x )L + mH ( v1x )H = ( mL + mH ) v1x
The heavier particle falls straight down, so (v1x ) H = 0 m/s. Thus,
(15 kg )( v1x )L + 0 kg m s = (15 kg + 60 kg )( 71.7 m/s ) ⇒ ( v1x )L = 358 m/s
That is, the velocity of the smaller fragment immediately after the explosion is 358 m/s and this velocity is in the
horizontal x-direction. Note that (v1 y ) L = 0 m/s. To find x2 , we will first find the displacement x1 − x0 and then
x2 − x1. For x1 − x0 ,
v1 y = v0 y + a y ( t1 − t0 ) ⇒ 0 m s = (102.4 m/s ) + ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 10.45 s
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ x1 − x0 = ( 71.7 m/s )(10.45 s ) + 0 m = 749 m
2
For x2 − x1 :
x2 = x1 + ( v1x ) L ( t2 − t1 ) + 12 ax ( t2 − t1 ) ⇒ x2 − x1 = ( 358 m/s )(10.45 s ) + 0 m=3741 m
2
⇒ x2 = ( x2 − x1 ) + ( x1 − x0 ) = 3741 m + 749 m = 4490 m = 4.5 km
Assess: Note that the time of ascent to the highest point is equal to the time of descent to the ground, that is,
t1 − t0 = t2 − t1.
9.54. Model: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, and
momentum is conserved in the collision between the proton and the gold atom.
Visualize:
Solve:
The conservation of momentum equation pfx = pix is
mG ( vfx )G + mP ( vfx )P = mP ( vix )P + mG ( vix )G
⇒ (197 u )( vfx )G + (1 u ) ( −0.90 × 5.0 × 107 m/s ) = (1 u ) ( 5.0 × 107 m/s ) + 0 u m/s
⇒ ( vfx )G = 4.8 × 105 m/s
9.55. Model: Model the proton (P) and the target nucleus (T) as particles. The proton and the target nucleus
make our system and in the collision between them momentum is conserved. This is due to the impulse
approximation because the collision lasts a very short time and the external forces acting on the system during
this time are not significant.
Visualize:
Solve:
The conservation of momentum equation pfx = pix is
mT ( vfx )T + mP ( v fx ) = mT ( vix )T + mP ( vix )P
P
⇒ mT ( 3.12 × 10 m/s ) + (1 u ) ( −0.75 × 2.5 × 106 m/s ) = 0 u m/s + (1 u ) ( 2.5 × 106 m/s )
5
⇒ mT = 14.0 u
Assess:
This is the mass of the nucleus of a nitrogen atom.
9.56. Model: This problem deals with a case that is the opposite of a collision. It is a case of an “explosion”
in which a 214 Po nucleus (P) decays into an alpha-particle (A) and a daughter nucleus (N). During the
“explosion” or decay, the total momentum of the system is conserved.
Visualize:
Solve: Conservation of mass requires the daughter nucleus to have mass mN = 214 u − 4 u = 210 u. The
conservation of momentum equation pfx = pix is
mN ( vfx ) N + mA ( vfx )A = ( mN + mA ) vP ⇒ ( 210 u )( vfx ) N + ( 4 u ) ( −1.92 × 107 m / s ) = 0 u m/s
⇒ ( vfx ) N = 3.66 × 105 m/s
9.57. Model: The neutron’s decay is an “explosion” of the neutron into several pieces. The neutron is an
isolated system, so its momentum should be conserved. The observed decay products, the electron and proton,
move in opposite directions.
Visualize:
Solve:
(a) The initial momentum is pix = 0 kg m/s. The final momentum pfx = meve + mpvp is
2.73 × 10−23 kg m/s − 1.67 × 10−22 kg m/s = −1.40 × 10−22 kg m/s
No, momentum does not seem to be conserved.
(b) and (c) If the neutrino is needed to conserve momentum, then pe + pP + pneutrino = 0 kg m/s. This requires
pneutrino = − ( pe + pP ) = +1.40 × 10−22 kg m/s
The neutrino must “carry away” 1.40 × 10−22 kg m/s of momentum in the same direction as the electron.
9.58. Model: Model the two balls of clay as particles. Our system comprises these two balls. Momentum is
conserved in the perfectly inelastic collision.
Visualize:
Solve:
The x-component of the final momentum is
pfx = pix = m1 ( vix )1 + m2 ( vix )2
= ( 0.020 kg )( 2.0 m/s ) − ( 0.030 kg )(1.0 m/s ) cos30° = 0.0140 kg m/s
The y-component of the final momentum is
pfy = piy = m1 ( viy ) + m2 ( viy )
1
2
= ( 0.02 kg )( 0 m/s ) − ( 0.03 kg )(1.0 m/s ) sin 30° = −0.0150 kg m/s
⇒ pf =
( 0.014 kg m/s ) + ( −0.015 kg m/s ) = 0.0205 kg m/s
2
2
Since pf = ( m1 + m2 ) vf = 0.0205 kg m/s, the final speed is
vf =
0.0205 kg m/s
= 0.41 m/s
( 0.02 + 0.03) kg
and the direction is
θ = tan −1
pfy
pfx
= tan −1
0.015
= 47° south of east
0.014
9.59. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and
particle 3 (moving southeast). The three stick together during their collision, which is perfectly inelastic. The
momentum of the system is conserved.
Visualize:
Solve:
The three initial momenta are
G
G
pi1 = m1vi1 = ( 0.020 kg )( 2.0 m/s ) ˆj = 0.040ˆj kg m/s
(
)
G
G
pi2 = m2vi2 = ( 0.030 kg ) −3.0 m/s iˆ = −0.090iˆ kg m/s
(
)
G
G
pi3 = m3vi3 = ( 0.040 kg ) ⎡⎣( 4.0 m/s ) cos 45°iˆ − ( 4.0 m/s ) sin 45° ˆj ⎤⎦ = 0.113iˆ − 0.113 ˆj kg m/s
G
G
G
G
Since pf = pi = pi1 + pi2 + pi3 , we have
( m1 + m2 + m3 ) vf = ( 0.023iˆ − 0.073 ˆj ) kg m/s ⇒ vf = ( 0.256iˆ − 0.811 ˆj ) m/s
G
G
⇒ vf =
( 0.256 m/s ) + ( −0.811 m/s ) = 0.85 m/s
θ = tan −1
2
vfy
vfx
= tan −1
2
0.811
= 72° below +x
0.256
9.60. Model: Model the truck (T) and the two cars (C and C′) as particles. The three forming our system
stick together during their collision, which is perfectly inelastic. Since no significant external forces act on the
system during the brief collision time, the momentum of the system is conserved.
Visualize:
Solve:
The three momenta are
G
G
piT = mT viT = ( 2100 kg )( 2.0 m/s ) iˆ = 4200iˆ kg m/s
G
G
piC = mCviC = (1200 kg )( 5.0 m/s ) ˆj = 6000ˆj kg m/s
G
G
piC′ = mC′viC′ = (1500 kg )(10 m/s ) iˆ = 15,000iˆ kg m/s
(
)
G
G G
G
G
pf = pi = piT + piC + piC′ = 19,200iˆ + 6000ˆj kg m/s
⇒ pf = ( mT + mC + mC′ ) vf =
⇒ vf = 4.2 m/s θ = tan −1
(19,200 kg m/s ) + ( 6000 kg m/s )
2
py
px
= tan −1
2
6000
= 17.4° above +x
19,200
Assess: A speed of 4.2 m/s for the entangled three vehicles is reasonable since the individual speeds of the cars
and the truck before entanglement were of the same order of magnitude.
9.61. Model: The 14 C atom undergoes an “explosion” and decays into a nucleus, an electron, and a neutrino.
Momentum is conserved in the process of “explosion” or decay.
Visualize:
Solve:
G
G
The conservation of momentum equation pf = pi = 0 kg m/s is
G
G G
G
G
G G
G
pe + pn + pN = 0 N ⇒ pN = − ( pe + pn ) = − meve − mn vn
(
)
= − ( 9.11 × 10−31 kg )( 5.0 × 107 m/s ) iˆ − ( 8.0 × 10−24 kg m/s ) ˆj = − 45.55 × 10−24iˆ + 8.0 × 10−24 ˆj kg m/s
⇒ pN = mN vN =
( 45.55 × 10 ) + (8.0 × 10 ) kg m/s
−24 2
−24
2
⇒ ( 2.34 × 10−26 kg ) vN = 4.62 × 10−23 kg m/s ⇒ vN = 1.97 × 103 m/s
9.62. (a) A 100 g ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for the
force of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball in
contact with the bat?
(b)
(c) The solution is Δt = 0.100 s = 10 ms.
9.63. (a) A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling to
the right at 3.0 m/s. The balls stick and move forward at 4.0 m/s. What was the speed of the 200 g ball of clay?
(b)
(c) The solution is ( vix )2 = 6.0 m/s.
9.64. (a) A 2000 kg auto traveling east at 5.0 m/s suffers a head-on collision with a small 1000 kg auto
traveling west at 4.0 m/s. They lock bumpers and stick together after the collision. What will be the speed and
direction of the combined wreckage after the collision?
(b)
(c) The solution is vfx = 2.0 m/s along +x direction.
9.65. (a) A 150 g spring-loaded toy is sliding across a frictionless floor at 1.0 m/s. It suddenly explodes into
two pieces. One piece, which has twice the mass of the second piece, continues to slide in the forward direction
at 7.5 m/s. What is the speed and direction of the second piece?
(b)
(c) The solution is (vfx )1 = −12 m/s. The minus sign tells us that the second piece moves backward at 12 m/s.
9.66. Model:
Visualize:
Solve:
The cart + man (C + M) is our system. It is an isolated system, and momentum is conserved.
The conservation of momentum equation pfx = pix is
mM ( vfx )M + mC ( vfx )C = mM ( vix )M + mC ( vix )C
Note that (vfx ) M and (vfx )C are the final velocities of the man and the cart relative to the ground. What is given in
this problem is the velocity of the man relative to the moving cart. The man’s velocity relative to the ground is
( vfx )M = ( vfx )C − 10 m/s
With this form for (vfx ) M, we rewrite the momentum conservation equation as
mM ⎡⎣( vfx )C − 10 m/s ⎤⎦ + mC ( vfx )C = mM ( 5.0 m/s ) + mC ( 5.0 m/s )
⇒ ( 70 kg ) ⎡⎣( vfx )C − 10 m/s ⎤⎦ + (1000 kg )( vfx )C = (1000 kg + 70 kg )( 5.0 m/s )
⇒ ( vfx )C [1000 kg + 70 kg ] = (1070 kg )( 5.0 m/s ) + ( 70 kg )(10 m/s ) ⇒ ( vfx )C = 5.7 m/s
9.67. Model: Model Ann and cart as particles. The initial momentum is pi = 0 kg m/s in a coordinate
system attached to the ground. As Ann begins running to the right, the cart will have to recoil to the left to conserve
momentum.
Visualize:
Solve: The difficulty with this problem is that we are given Ann’s velocity of 5.0 m/s relative to the cart. If the
cart is also moving with velocity vcart then Ann’s velocity relative to the ground is not 5.0 m/s. Using the
Galilean transformation equation for velocity, Ann’s velocity relative to the ground is
( vfx )Ann = ( vfx )cart + 5.0 m/s
Now, the momentum conservation equation pix = pfx is
0 kg m/s = mAnn ( vfx )Ann + mcart ( vfx )cart ⇒ 0 kg m/s = ( 50 kg ) ⎡⎣( vfx )cart + 5.0 m/s ⎤⎦ + ( 500 kg )( vfx )cart
⇒ ( vfx )cart = −0.45 m/s
Using the recoil velocity (vfx )cart relative to the ground, we find Ann’s velocity relative to the ground to be
(vfx ) Ann = 5.00 m/s − 0.45 m/s = 4.55 m/s
The distance Ann runs relative to the ground is Δx = (vfx ) Ann Δt , where Δt is the time it takes to reach the end of
the cart. Relative to the cart, which is 15 m long, Ann’s velocity is 5 m/s. Thus Δt = (15 m ) ( 5 m/s ) = 3.00 s. Her
distance over the ground during this interval is
Δx = ( vfx )Ann Δt = ( 4.55 m/s )( 3.00 s ) = 13.6 m
9.68. Model:
Visualize:
Solve:
The projectile + wood ball are our system. In the collision, momentum is conserved.
The momentum conservation equation pfx = pix is
( mP + mB ) vfx = mP ( vix )P + mB ( vix )B ⇒ (1.0 kg + 20 kg ) vfx = (1.0 kg )( vix )P + 0 kg m/s
⇒ ( vix )P = 21vfx
We therefore need to determine vfx . Newton’s second law for circular motion is
T − FG = T − ( mP + mB ) g =
( mP + mB ) vf2x
r
Using Tmax = 400 N, this equation gives
400 N − (1.0 kg + 20 kg )( 9.8 m/s ) =
(1.0 kg + 20 kg ) vf2x ⇒ v
( fx )max = 4.3 m/s
2.0 m
Going back to the momentum conservation equation,
( vix )P = 21vfx = ( 21)( 4.3 m/s ) = 90 m/s
That is, the largest speed this projectile can have without causing the cable to break is 90 m/s.
9.69. Model: This is an “explosion” problem and momentum is conserved. The two-stage rocket is our
system.
Visualize:
Solve:
Relative to the ground, the conservation of momentum equation pfx = pix is
m1 ( vfx )1 + m2 ( vfx )2 = ( m1 + m2 ) v1x
⇒ 3 m 2 ( vfx )1 + m2 ( vfx )2 = ( 4 m 2 )(1200 m/s ) ⇒ 3( vfx )1 + ( vfx )2 = 4800 kg m/s
The fact that the first stage is pushed backward at 35 m/s relative to the second can be written
( vfx )1 = −35 m/s + ( vfx )2
Substituting this form of (vfx )1 in the conservation of momentum equation,
3 ⎡⎣ −35 m/s + ( vfx )2 ⎤⎦ + ( vfx )2 = 4800 kg m/s ⇒ ( vfx )2 = 1226 m/s
9.70. Model: Model the bullet and the vehicle as particles, and use the impulse-momentum theorem to find
the impulse provided to the vehicle by bullet(s). Because the final speed of the vehicle is small compared to the
bullet speed, and the mass of the bullet is so much smaller than the mass of the vehicle, we will assume that each
bullet exerts the same impulse on the vehicle.
Visualize:
Solve: For one bullet, the impulse-momentum theorem Δpx = J x allows us to find that the impulse exerted on
the bullet by the vehicle’s sail is
( J x ) B = mB (Δv) B = (0.020 kg) ( ( −200 m/s) − (400 m/s) ) = −12.0 kg m/s
From Newton’s third law, the impulse that the bullet exerts on the vehicle is ( J x ) V = −( J x ) B = 12.0 kg m/s. A
second use of the impulse momentum theorem allows us to find the vehicle’s increase in velocity due to one
bullet:
(Δv) V =
( J x ) V 12.0 kg m/s
=
= 0.120 m/s
mV
100 kg
This increase in velocity is independent of the vehicle’s speed, so as long as the impulse per bullet stays
essentially constant (which it does because the bullet speeds are so much larger than the vehicle speed), the
number of impacts needed to increase the vehicle speed to 12 m/s is
N bullets =
12 m/s
= 100 bullets
0.12 m/s per bullet
To reach this speed in 20 s requires the bullets to be fired at a rate of 5 bullets per second.
9.71. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved.
Visualize:
The fact that the bullet’s velocity relative to the rocket is 139,000 can be written (vf ) B =
(vf ) R + 139,000 m/s.
Solve:
Consider the firing of one bullet when the rocket has mass M and velocity vi . The conservation of
momentum equation pf = pi is
( M − 5kg)vf + (5 kg)(vf + 139,000 m/s) = Mvi
⇒ Δv = vf − vi = −
5 kg
(139,000 m/s)
M
The rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M will
not change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed per
bullet is Δv = −347.5 m/s = −1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000
km/h to 15,000 km/h. If you’re not sure that treating M as a constant is valid, you can calculate Δv for each bullet
and reduce M by 5 kg after each shot. The loss of mass causes Δv to increase slightly for each bullet. An eight-step
calculation then finds that 8 bullets will slow the rocket to 14,900 km/h. Seven bullets wouldn’t be enough, and
9 would slow the rocket far too much.
9.72.
Visualize:
Solve: Ladies and gentlemen of the jury, how far would the chair slide if it was struck with a bullet from my client’s
gun? We know the bullet’s velocity as it leaves the gun is 450 m/s. The bullet travels only a small distance to the chair, so
we will neglect any speed loss due to air resistance. The bullet and chair can be considered an isolated system during the
brief interval of the collision. The bullet embedded itself in the chair, so this was a perfectly inelastic collision. Momentum
conservation allows us to calculate the velocity of the chair immediately after the collision as follows:
pix = pfx ⇒ mB ( vi )B = ( mB + mC ) vf ⇒ vf =
mB ( vi )B
mB + mC
=
( 450 m/s )( 0.010 kg ) = 0.225 m/s
20.01 kg
This is the velocity immediately after the collision when the chair starts to slide but before it covers any distance. For the
purpose of the problem in dynamics, call this the initial velocity v0 . The free-body diagram of the chair shows three
forces. Newton’s second law applied to the chair (with the embedded bullet) is
ax = a =
( Fnet ) x
mtot
=
( Fnet ) y n − mtot g
− fk
μn
=
= − k a y = 0 m/s 2 =
mtot
mtot
mtot
mtot
where we’ve used the friction model in the x-equation. The y-equation yields n = mtot g , and the x-equation yields
a = − μ k g = −1.96 m/s 2 . We know the coefficient of kinetic friction because it is a wood chair sliding on a wood floor.
Finally, we have to determine the stopping distance of the chair. The motion of the chair ends with v1 = 0 m/s after sliding
a distance Δx, so
v2
( 0.225 m/s ) = 0.013 m = 1.3 cm
v12 = 0 m 2 /s 2 = v02 + 2aΔx ⇒ Δx = − 0 = −
2a
2 ( −1.96 m/s 2 )
2
If the bullet lost any speed in the air before hitting the chair, the sliding distance would be even less. So you can see that
the most the chair could slide if it had been struck by a bullet from my client’s gun, would be 1.3 cm. But in actuality, the
chair slid 3 cm, more than twice as far. The murder weapon, ladies and gentlemen, was a much more powerful gun than
the one possessed by my client. I rest my case.
9-1
10.1. Model: We will use the particle model for the bullet (B) and the running student (S).
Visualize:
Solve:
For the bullet,
1
1
K B = mBvB2 = (0.010 kg)(500 m/s) 2 = 1250 J
2
2
For the running student,
1
1
KS = mSvS2 = (75 kg)(5.5 m/s) 2 = 206 J
2
2
Thus, the bullet has the larger kinetic energy.
Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation
shows this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its
speed is more than 90 times larger.
10.2. Model: Model the hiker as a particle.
Visualize:
The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death
Valley is y0 = −8.5 m.
Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is
ΔU = U gf − U gi = mgyf − mgyi = mg ( yf − yi )
= (65 kg)(9.8 m/s 2 )[4420 m − (−85 m)] = 2.9 × 106 J
Assess:
Note that ΔU is independent of the origin of the coordinate system.
10.3. Model: Model the compact car (C) and the truck (T) as particles.
Visualize:
Solve:
For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,
mT
1
1
20,000 kg
K C = K T ⇒ mCvC2 = mT vT2 ⇒ vC =
vT =
(25 km/hr) = 112 km/hr
mC
2
2
1000 kg
Assess:
A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.
10.4.
Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic
and potential energy does not change as the car falls.
Visualize:
Solve:
(a) The kinetic energy of the car is
KC =
1
1
mCvC2 = (1500 kg)(30 m/s) 2 = 6.75 × 105 J
2
2
The car’s kinetic energy is 6.8 × 105 J.
(b) Let us relabel K C as K f and place our coordinate system at yf = 0 m so that the car’s potential energy U gf
is zero, its velocity is vf , and its kinetic energy is K f . At position yi , vi = 0 m/s or K i = 0 J, and the only
energy the car has is U gi = mgyi . Since the sum K + U g is unchanged by motion, K f + U gf = K i + U gi . This
means
K f + mgyf = K i + mgyi ⇒ K f + 0 = K i + mgyi
⇒ yi =
( K f − Ki )
(6.75 × 105 J − 0 J)
=
= 46 m
mg
(1500 kg)(9.8 m / s 2 )
(c) From part (b),
1 2 1 2
2
2
( K f − K i ) 2 mvf − 2 mvi ( vf − vi )
yi =
=
=
2g
mg
mg
Free fall does not depend upon the mass.
10.5.
Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not
change as the ball rises and falls.
Visualize:
The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c).
Solve: The quantity K + U g is the same during free fall: K f + U gf = Ki + U gi . We have
(a)
1 2
1
mv1 + mgy1 = mv02 + mgy0
2
2
⇒ y1 = ( v02 − v12 ) 2 g = [(10 m/s) 2 − (0 m/s) 2 ]/(2 × 9.8 m / s 2 ) = 5.10 m
5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground.
1 2
1
mv2 + mgy2 = mv02 + mgy0
(b)
2
2
Since y2 = y0 = 0, we get for the magnitudes v2 = v0 = 10 m / s.
(c)
1 2
1
mv3 + mgy3 = mv02 + mgy0 ⇒ v32 + 2 gy3 = v02 + 2 gy0 ⇒ v32 = v02 + 2 g ( y0 − y3 )
2
2
⇒ v32 = (10 m / s) 2 + 2(9.8 m / s 2 )[0 m − (−20 m)] = 492 m 2 / s 2
This means the magnitude of v3 is equal to 22 m/s.
Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which
it was tossed up, but in the opposite direction.
10.6. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the
ball, considered as a particle, does not change during its motion.
Visualize:
The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and (b).
Solve: The quantity K + U g is the same during free fall. Thus, K f + U gf = Ki + U gi .
1 2
1
mv1 + mgy1 = mv02 + mgy0 ⇒ v02 = v12 + 2 g ( y1 − y0 )
(a)
2
2
⇒ v02 = (0 m/s) 2 + 2(9.8 m/s 2 )(10 m − 1.5 m) = 166.6 m 2 /s 2 ⇒ v0 = 12.9 m/s
(b)
1 2
1
mv2 + mgy2 = mv02 + mgy0 ⇒ v22 = v02 + 2 g ( y0 − y2 )
2
2
2
2
⇒ v2 = 166.6 m /s 2 + 2(9.8 m/s 2 )(1.5 m − 0 m) ⇒ v2 = 14.0 m/s
Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is
reasonable. Also, note that mass does not appear in the calculations that involve free fall.
10.7. Model: Model the oxygen and the helium atoms as particles.
Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is
four times heavier than the helium atom, so mO = 4 mHe .
Solve:
The energy conservation equation K O = K He is
v
1
1
2
2
mOvO2 = mHevHe
⇒ (4 mHe )vO2 = mHevHe
⇒ He = 2.0
vO
2
2
Assess: The result vHe = 2vO , combined with the fact that mHe = 14 mO , is a consequence of the way kinetic
energy is defined: It is directly proportional to the mass and to the square of the speed.
10.8. Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of
the ball’s kinetic and gravitational potential energy, therefore, does not change during the rolling motion.
Visualize:
Solve:
Since the quantity K + U g does not change during rolling motion, the energy conservation equations
apply. For the linear segment the energy conservation equation K 0 + U g0 = K1 + U g1 is
1 2
1
1
1
1
mv1 + mgy1 = mv02 + mgy0 ⇒ m v12 + mg (0 m) = m(0 m/s) 2 + mgy0 ⇒ mv12 = mgy0
2
2
2
2
2
For the parabolic part of the track, K1 + U g1 = K 2 + U g 2 is
1 2
1
1
1
1
mv1 + mgy1 = mv22 + mgy2 ⇒ mv12 + mg (0 m) = m(0 m / s) 2 + mgy2 ⇒ mv12 = mgy2
2
2
2
2
2
Since from the linear segment we have 12 m v12 = mgy0 , we get mgy0 = mgy2 or y2 = y0 = 1.0 m. Thus, the ball rolls
up to exactly the same height as it started from.
Assess: Note that this result is independent of the shape of the path followed by the ball, provided there is no
rolling friction. This result is an important consequence of energy conservation.
10.9. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum
of the skateboarder’s kinetic and gravitational potential energy does not change during his rolling motion.
Visualize:
The vertical displacement of the skateboarder is equal to the radius of the track.
Solve: The quantity K + U g is the same at the upper edge of the quarter-pipe track as it was at the bottom. The
energy conservation equation K f + U gf = K i + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(3.0 m − 0 m) = 58.8 m 2 /s 2 ⇒ vi = 7.7 m/s
Assess:
Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion.
10.10. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck’s kinetic and
gravitational potential energy does not change during its sliding motion.
Visualize:
Solve:
The quantity K + U g is the same at the top of the ramp as it was at the bottom. The energy conservation
equation K f + U gf = K i + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
⇒ vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(1.03 m − 0 m) = 20.2 m 2 /s 2 ⇒ vi = 4.5 m/s
Assess: An initial push with a speed of 4.5 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seems
reasonable.
10.11. Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational
potential energy does not change as the pendulum swings from one side to the other.
Visualize:
The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts (a)
and (b).
Solve: (a) The quantity K + U g is the same at the lowest point of the trajectory as it was at the highest point.
Thus, K1 + U g1 = K 0 + U g0 means
1 2
1
mv1 + mgy1 = mv02 + mgy0 ⇒ v12 + 2 gy1 = v02 + 2 gy0
2
2
⇒ v12 + 2 g (0 m) = (0 m/s) 2 + 2 gy0 ⇒ v1 = 2 gy0
From the pictorial representation, we find that y0 = L − L cos30°. Thus,
v1 = 2 gL(1 − cos30°) = 2(9.8 m/s 2 )(0.75 m)(1 − cos30°) = 1.403 m/s
The speed at the lowest point is 1.40 m/s.
(b) Since the quantity K + U g does not change, K 2 + U g 2 = K1 + U g1. We have
1 2
1
mv2 + mgy2 = mv12 + mgy1 ⇒ y2 = ( v12 − v22 ) 2 g
2
2
⇒ y2 = [(1.403 m/s) 2 − (0 m/s) 2 ]/(2 × 9.8 m/s 2 ) = 0.100 m
Since y2 = L − L cosθ , we obtain
cosθ =
L − y2 (0.75 m) − (0.10 m)
=
= 0.8667 ⇒ θ = cos −1 (0.8667) = 30°
L
(0.75 m)
That is, the pendulum swings to the other side by 30°.
Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact
that the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy
bar chart in the figure.
10.12. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of
frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the
swing’s motion.
Visualize:
Solve:
The quantity K + U g is the same at the highest point of the swing as it is at the lowest point. That is,
K 0 + U g0 = K1 + U g1. It is clear from this equation that maximum kinetic energy occurs where the gravitational
potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the
swing and child will also be maximum. The above equation is
1 2
1
mv0 + mgy0 = mv12 + mgy1 ⇒ v12 = v02 + 2 g ( y0 − y1 )
2
2
2
⇒ v1 = (0 m/s) 2 + 2 g ( y0 − 0 m) ⇒ v1 = 2 gy0
We see from the pictorial representation that
y0 = L − L cos 45° = (3.0 m) − (3.0 m)cos 45° = 0.879 m
⇒ v1 = 2 gy0 = 2(9.8 m/s 2 )(0.879 m) = 4.2 m/s
Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.2 m/s is
reasonable.
10.13. Model: Model the car as a particle with zero rolling friction. The sum of the kinetic and gravitational
potential energy, therefore, does not change during the car’s motion.
Visualize:
Solve:
The initial energy of the car is
1
1
K 0 + U g0 = mv02 + mgy0 = (1500 kg)(10.0 m/s) 2 + (1500 kg)(9.8 m/s 2 )(10 m) = 2.22 × 105 J
2
2
The car increases its height to 15 m at the gas station. The conservation of energy equation K 0 + U g0 = K1 + U g1
is
1
1
2.22 × 105 J = mv12 + mgy1 ⇒ 2.22 × 105 J = (1500 kg)v12 + (1500 kg)(9.8 m/s 2 )(15 m)
2
2
⇒ v1 = 1.41 m/s
Assess: A lower speed at the gas station is reasonable because the car has decreased its kinetic energy and
increased its potential energy compared to its starting values.
10.14.
Model: Assume that the spring is ideal and obeys Hooke’s law. Also model the sled as a particle.
G
Visualize: The only horizontal force acting on the sled is Fsp .
Solve:
Applying Newton’s second law to the sled gives
∑ (F
) = Fsp = max ⇒ k Δy = max
on sled x
⇒ ax = k Δ x / m = (150 N/m)(0.20 m)/20 kg = 1.50 m/s 2
10.15. Model: Assume that the spring is ideal and obeys Hooke’s law.
Visualize: According to Hooke’s law, the spring force acting on a mass (m) attached to the end of a spring is
given as Fsp = k Δ x, where Δx is the change in length of the spring. If the mass m is at rest, then Fsp is also equal
to the gravitational force FG = mg .
Solve:
We have Fsp = k Δ x = mg . We want a 0.100 kg mass to give Δ x = 0.010 m. This means
k = mg / Δx = (0.100 kg)(9.8 N/m)/(0.010 m) = 98 N/m
10.16. Model: Assume an ideal spring that obeys Hooke’s law.
Visualize:
Solve:
(a) The spring force on the 2.0 kg mass is Fsp = − k Δy. Notice that Δy is negative, so Fsp is positive.
This force is equal to mg, because the 2.0 kg mass is at rest. We have − k Δy = mg . Solving for k:
k = −( mg / Δy ) = −(2.0 kg)(9.8 m/s 2 )/(−0.15 m − (−0.10 m)) = 392 N/m
The spring constant is 3.9 × 102 N/m.
(b) Again using − k Δy = mg:
Δy = − mg / k = −(3.0 kg)(9.8 m/s 2 )/(392 N/m)
y′ − ye = −0.075 m ⇒ y′ = ye − 0.075 m = −0.10 m − 0.075 m = −0.175 m = −17.5 cm
The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the
spring is negative because it is below the origin, but length must be a positive number.
10.17. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model the 5.0 kg mass as a
particle.
Visualize:
Solve:
We will use the subscript s for the scale and sp for the spring.
(a) The scale reads the upward force Fs on m that it applies to the mass. Newton’s second law gives
∑ (F
) = Fs on m − FG = 0 ⇒ Fs on m = FG = mg = (5.0 kg)(9.8 m/s 2 ) = 49 N
on m y
(b) In this case, the force is
∑ (F
) = Fs on m + Fsp − FG = 0 ⇒ 20 N + k Δy − mg = 0
on m y
⇒ k = (mg − 20 N)/ Δy = (49 N − 20 N)/0.02 m = 1450 N/m
The spring constant for the lower spring is 1.45 × 103 N/m.
(c) In this case, the force is
∑ ( Fon m ) y = Fsp − FG = 0 ⇒ k Δy − mg = 0
⇒ Δy = mg / k = (49 N)/(1450 N/m) = 0.0338 m = 3.4 cm
10.18. Model: Model the student (S) as a particle and the spring as obeying Hooke’s law.
Visualize:
Solve:
According to Newton’s second law the force on the student is
∑ (F ) = F
on S y
spring on S
− FG = ma y
⇒ Fspring on S = FG + ma y = mg + ma y = (60 kg)(9.8 m/s 2 + 3.0 m/s 2 ) = 768 N
Since Fspring on S = FS on spring = k Δy, k Δy = 768 N. This means Δy = (768 N)/(2500 N/m) = 0.31 m.
10.19. Model: Assume an ideal spring that obeys Hooke’s law.
Solve: The elastic potential energy of a spring is defined as U s = 12 k (Δs ) 2 , where Δs is the magnitude of the
stretching or compression relative to the unstretched or uncompressed length. We have Δs = 20 cm = 0.20 m
and k = 500 N/m. This means
1
1
U s = k (Δs ) 2 = (500 N/m)(0.20 m) = 10 J
2
2
Assess: Since Δs is squared, U s is positive for a spring that is either compressed or stretched. U s is zero
when the spring is in its equilibrium position.
10.20. Model: Assume an ideal spring that obeys Hooke’s law.
Solve:
The elastic potential energy of a spring is defined as U s = 12 k (Δs ) 2 , where Δs is the magnitude of the
stretching or compression relative to the unstretched or uncompressed length. ΔU s = 0 when the spring is at its
equilibrium length and Δs = 0. We have U s = 200 J and k = 1000 N/m. Solving for Δs :
Δs =
2U s / k = 2(200 J) / 1000 N/m = 0.632 m
10.21. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical
energy K + U s is conserved. Also model the book as a particle.
Visualize:
The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until
the spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium
position of the free end of the spring. The energy bar chart shows that the potential energy of the compressed
spring is entirely transformed into the kinetic energy of the book.
Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2
2
2
2
2
Using x2 = xe = 0 m and v1 = 0 m/s, this simplifies to
kx12
1 2 1
(1250 N/m)(0.040 m) 2
mv2 = k ( x1 − 0 m) 2 ⇒ v2 =
=
= 2.0 m/s
m
2
2
(0.500 kg)
Assess: This problem cannot be solved using constant-acceleration kinematic equations. The acceleration is not
a constant in this problem, since the spring force, given as Fs = − k Δx, is directly proportional to Δx or | x − xe |.
10.22. Model: Assume an ideal spring that obeys Hooke’s law. Since there is no friction, the mechanical
energy K + U s is conserved. Also, model the block as a particle.
Visualize:
The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at
the equilibrium position of the free end of the spring. This gives us x1 = xe = 0 cm and x2 = 2.0 cm.
Solve:
The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2
2
2
2
2
Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 0.020 m, we get
1
1
m
k ( x2 − xe ) 2 = mv12 ⇒ Δx = ( x2 − xe ) =
v1
2
2
k
That is, the compression is directly proportional to the velocity v1. When the block collides with the spring with
twice the earlier velocity (2v1 ), the compression will also be doubled to 2( x2 − xe ) = 2(2.0 cm) = 4.0 cm.
Assess: This problem shows the power of using energy conservation over using Newton’s laws in solving
problems involving nonconstant acceleration.
10.23. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke’s law. We will
also assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved.
Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the
spring in its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the
free end of the spring. This give x1 = xe = 0 and ( x2 − xe ) = 60 cm.
Solve:
The conservation of energy equation K 2 + U s2 = K1 + U s1 is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2
2
2
2
2
Using v2 = 0 m/s,( x2 − xe ) = 0.60 m, and x1 = xe = 0 m gives:
k
1
1
250 N/m
k ( x2 − xe ) 2 = m v12 ⇒ v1 =
( x2 − xe ) =
(0.60 m) = 3.0 m/s
m
2
2
10 kg
10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We will
also assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved.
Visualize:
The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet plane
lands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate
system at the right free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the
plane, x2 − xe = 30 m.
Solve:
The conservation of energy equation K 2 + U s2 = K1 + U s1 for the spring-jet plane system is
1 2 1
1
1
mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2
2
2
2
2
Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 30 m yields
k
1
1
60,000 N/m
k ( x2 − xe ) 2 = mv12 ⇒ v1 =
( x2 − x1 ) =
(30 m) = 60 m/s
m
2
2
15,000 kg
Assess:
A landing speed of 60 m/s or ≈120 mph is reasonable.
10.25. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum
and mechanical energy.
Visualize:
Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities
(vfx )1 and (vfx ) 2 are unknowns and can not be determined from one equation.
Solve:
Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2
Energy conservation:
1
1
1
1
m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22
2
2
2
2
These two equations can be solved for (vfx )1 and (vfx ) 2 , as shown by Equations 10.39 through 10.43, to give
(vfx )1 =
m1 − m2
50 g − 20 g
(vix )1 =
(2.0 m/s) = 0.86 m/s
m1 + m2
50 g + 20 g
(vfx ) 2 =
2m1
2(50 g)
(vix )1 =
(2.0 m/s) = 2.9 m/s
m1 + m2
50 g + 20 g
Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls
move along the +x-direction.
10.26. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The
collision obeys the momentum as well as the energy conservation law.
Visualize:
Solve:
Momentum conservation: mP (vix ) P + mC (vix ) C = mP (vfx ) P + mC (vfx ) C
1
1
1
1
mP (vix ) 2P + mC (vix )C2 = mP (vfx ) P2 + mC (vfx ) C2
2
2
2
2
These two equations can be solved, as described in the text through Equations 10.39 to 10.43:
Energy conservation:
(vfx ) P =
⎛ m − 12mP ⎞
mP − mC
7
7
mP + mC (vix ) P = ⎜ P
⎟ (2.0 × 10 m/s) = −1.69 × 10 m/s
mP + mC
m
12
m
+
P ⎠
⎝ P
(vfx )C =
⎛ 2mP ⎞
2mP
7
6
(vix ) P = ⎜
⎟ (2.0 × 10 m/s) = 3.1 × 10 m/s
mP + mC
⎝ mP + 12mP ⎠
After the elastic collision the proton rebounds at 1.69 × 107 m/s and the carbon atom moves forward at
3.08 × 106 m/s.
10.27. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external
force acts on the system (clay + brick). We also represent our system as a particle.
Visualize:
Solve:
(a) The conservation of momentum equation pfx = pix is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
Using (vix )1 = v0 and (vix ) 2 = 0, we get
vfx =
m1
0.050 kg
(vix )1 =
(vix )1 = 0.0476(vix )1 = 0.0476 v0
m1 + m2
(1.0 kg + 0.050 kg)
The brick is moving with speed 0.048v0 .
(b) The initial and final kinetic energies are given by
1
1
1
1
2
K i = m1 (vix )12 + m2 (vix ) 22 = (0.050 kg)v02 + (1.0 kg) ( 0 m/s ) = (0.025 kg)v02
2
2
2
2
1
1
K f = (m1 + m2 )vf2x = (1.0 kg + 0.050 kg)(0.0476) 2 v02 = 0.00119 v02
2
2
⎛ K − Kf ⎞
⎛ 0.00119 ⎞
The percent of energy lost = ⎜ i
⎟ × 100% = ⎜1 −
⎟ × 100% = 95%
K
0.025 ⎠
⎝
i
⎝
⎠
10.28. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed
whether the collision is perfectly elastic or perfectly inelastic.
Visualize:
Solve: In the case of a perfectly elastic collision, the two velocities (vfx )1 and (vfx ) 2 can be determined by combining
the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision
involves only one final velocity vfx and can be determined from just the momentum conservation equation.
(a)
Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2
1
1
1
1
m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22
2
2
2
2
These two equations can be solved as shown in Equations 10.39 through 10.43:
m − m2
(100 g) − (300 g)
(vfx )1 = 1
(vix )1 =
(10 m/s) = −5.0 m/s
m1 + m2
(100 g) + (300 g)
Energy conservation:
(vfx ) 2 =
2m1
2(100 g)
(vix )1 =
(10 m/s) = +5.0 m/s
m1 + m2
(100 g) + (300 g)
(b) For the inelastic collision, both balls travel with the same final speed vfx . The momentum conservation equation
pfx = pix is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
⎛
⎞
100 g
⇒ vfx = ⎜
⎟ (10 m/s) + 0 m/s = 2.5 m/s
⎝ 100 g + 300 g ⎠
Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0
m/s. In the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s.
10.29. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
The particle is released from rest at x = 1.0 m. That is, K = 0 at x = 1.0 m. Since the total energy is given by
E = K + U , we can draw a horizontal total energy (TE) line through the point of intersection of the potential
energy curve (PE) and the x = 1.0 m line. The distance from the PE curve to the TE line is the particle’s kinetic
energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but
the sum K + U does not change.
Solve: (a) We have E = 4.0 J and this energy is a constant. For x < 1.0, U > 4.0 J and, therefore, K must be
negative to keep E the same (note that K = E − U or K = 4.0 J − U ). Since negative kinetic energy is unphysical,
the particle can not move to the left. That is, the particle will move to the right of x = 1.0 m.
(b) The expression for the kinetic energy is E − U . This means the particle has maximum speed or maximum
kinetic energy when U is minimum. This happens at x = 4.0 m. Thus,
K max = E − U min = (4.0 J) − (1.0 J) = 3.0 J
1 2
2(3.0 J)
8.0 J
mvmax = 3.0 J ⇒ vmax =
=
= 17.3 m/s
2
m
0.020 kg
The particle possesses this speed at x = 4.0 m.
(c) The total energy (TE) line intersects the potential energy (PE) curve at x = 1.0 m and x = 6.0 m. These are
the turning points of the motion.
10.30. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U , we
can draw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particle’s
kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow
down, but the sum K + U does not change.
Solve: The kinetic energy is given by E − U , so we have
1 2
mv = E − U ⇒ v = 2( E − U )/ m
2
Using U B = 2.0 J, U C = 3.0 J, and U D = 0 J, we get
vB = 2(5.0 J − 2.0 J)/0.500 kg = 3.5 m/s
vD = 2(5.0 J − 0 J)/0.500 kg = 4.5 m/s
vC = 2(5.0 J − 3.0 J)/0.500 kg = 2.8 m/s
10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these
points. That is, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The distance from
the potential energy (PE) curve to the TE line is the particle’s kinetic energy. These values are transformed as the
position changes, but the sum K + U does not change.
Solve: The equation for total energy E = U + K means K = E − U , so that K is maximum when U is minimum.
We have
1 2
K max = mvmax
= 5.0 J − U min
2
⇒ vmax = 2(5.0 J − U min )/ m = 2(5.0 J − 1.0 J)/0.0020 kg = 63 m/s
10.32. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.
Visualize:
For the speed of the particle at A that is needed to reach B to be a minimum, the particle’s kinetic energy as it
reaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the
particle just makes it to the top with zero kinetic energy.
Solve: (a) The energy equation K A + U A = K top + U top is
1 2
mvA + U A = 0 J + U top
2
⇒ vA = 2(U top − U A )/ m = 2(5.0 J − 2.0 J)/0.100 kg = 7.7 m/s
(b) To go from point B to point A, K B + U B = K top + U top is
1 2
mvB + U B = 0 J + U top
2
⇒ vB = 2(U top − U B )/ m = 2(5.0 J − 0 J)/0.100 kg = 10.0 m/s
Assess:
The particle requires a higher kinetic energy to reach A from B than to reach B from A.
10.33. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic
and gravitational potential energy does not change as the vehicle coasts down the hill.
Visualize:
The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle
of the downward slope is given, since these are not needed to solve the problem. All we need is the change in
potential energy as you and your vehicle descend to the bottom of the hill. Also note that
35 km / hr = (35,000 m / 3600 s) = 9.722 m / s
Solve:
Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2
1
mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2
2
2
⇒ vf = vi2 + 2 gyi = (9.722 m/s) 2 + 2(9.8 m/s 2 )(15 m) = 19.7 m/s = 71 km/h
You are driving over the speed limit. Yes, you will get a ticket.
Assess: A speed of 19.7 m/s or 71 km/h at the bottom of the hill, when your speed at the top of the hill was 35
km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed
into the final kinetic energy.
10.34. Model: This is case of free fall, so the sum of the kinetic and gravitational potential energy does not
change as the cannon ball falls.
Visualize:
The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put
the origin of our coordinate system on the ground below the fortress.
Solve: Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2
1
mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2
2
2
vf = vi2 + 2 gyi = (80 m/s) 2 + 2(9.8 m/s 2 )(10 m) = 81 m/s
Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also
note that using the energy conservation equation, we can find only the magnitude of the final velocity, not the
direction of the velocity vector.
10.35. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not
change as the rock is thrown. Model the Frisbee and the rock as particles.
Visualize:
The coordinate system is put on the ground for this system, so that yf = 16 m. The rock’s final velocity vf must
be at least 5.0 m/s to dislodge the Frisbee.
Solve: (a) The energy conservation equation for the rock K f + U gf = K i + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi
2
2
This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to
dislodge the Frisbee. This equation is true for all angles that will take the rock to the Frisbee.
(b) Using the above equation we get
vf2 + 2 gyf = vi2 + 2 gyi
vi = vf2 + 2 g ( yf − yi ) = (5.0 m/s)2 + 2(9.8 m/s 2 )(16 m − 2.0 m) = 17.3 m/s
Assess: Kinetic energy is defined as K = 12 mv 2 and is a scalar quantity. Scalar quantities do not have directional
properties.
10.36. Model: For the ice cube sliding around the inside of a smooth pipe, the sum of the kinetic and
gravitational potential energy does not change.
Visualize:
We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is
10 cm, and therefore ytop = y2 = 2 R = 0.20 m. At an arbitrary angle θ , measured counterclockwise from the
bottom of the circle, y = R − R cosθ .
Solve: (a) The energy conservation equation K 2 + U g2 = K1 + U g1 is
1
1
⇒ mv22 + mgy2 = mv12 + mgy1
2
2
⇒ v2 = v12 + 2 g ( y1 − y2 ) = (3.0 m/s) 2 + 2(9.8 m/s 2 )(0 m − 0.20 m) = 2.25 m/s
(b) Expressing the energy conservation equation as a function of θ :
1
1
K (θ ) + U g (θ ) = K1 + U g1 ⇒ mv 2 (θ ) + mgy (θ ) = mv12 + 0 J
2
2
⇒ v (θ ) = v12 − 2 gy (θ ) = v12 − 2 gR (1 − cosθ )
Using v1 = 3.0 m/s, g = 9.8 m/s 2 , and R = 0.10 m, we get v(θ ) = 9 − 1.96(1 − cosθ ) (m/s)
(c) The accompanying figure shows a graph of v for a complete revolution (0° ≤ θ ≤ 360°).
Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy increases and kinetic
energy decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable since
some of the kinetic energy has been transformed into the marble’s potential energy.
10.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle.
Visualize:
Please refer to Figure P10.37.
Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber
band results from a negative displacement of the rock. That is, ( Fsp ) x = − kx, where x is the rock’s displacement
from the equilibrium position due to the spring force Fsp .
(b) Since the Fsp versus x graph is linear with a negative slope and can be expressed as Fsp = − kx, the rubber
band obeys Hooke’s law.
(c) From the graph, |ΔFsp | = 20 N for |Δx| = 10 cm. Thus,
k=
|ΔFsp |
|Δx|
=
20 N
= 200 N/m = 2.0 × 102 N/m
0.10 m
(d) The conservation of energy equation K f + U sf = Ki + U si for the rock is
1 2 1 2 1 2 1 2
1
1
1
1
mvf + kxf = mvi + kxi ⇒ mvf2 + k (0 m) 2 = m(0 m/s) 2 + kxi2
2
2
2
2
2
2
2
2
k
200 N/m
vf =
xi =
(0.30 m) = 19.0 m/s
m
0.050 kg
Assess: Note that xi is Δx, which is the displacement relative to the equilibrium position, and xf is the
equilibrium position of the rubber band, which is equal to zero.
10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is
no friction, hence the mechanical energy K + U s is conserved.
Visualize:
Note that xf = xe and xi − xe = Δx. The before-and-after pictorial representations show that we put the origin of
the coordinate system at the equilibrium position of the free end of the springs.
Solve: The conservation of energy equation K f + U sf = Ki + U si for the single spring is
1 2 1
1
1
mvf + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2
2
2
2
2
Using the value for vf given in the problem, we get
1 2
1
1
1
mv0 + 0 J = 0 J + k (Δx) 2 ⇒ mv02 = k (Δx ) 2
2
2
2
2
Conservation of energy for the two-spring case:
1
1
1
mVf2 + 0 J = 0 J + k ( xi − xe ) 2 + k ( xi − xe ) 2
2
2
2
Using the result of the single-spring case,
1
mVf2 = k (Δx ) 2
2
1
mVf2 = mv02 ⇒ Vf = 2v 0
2
Assess:
The block separates from the spring at the equilibrium position of the spring.
10.39. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is
no friction, hence the mechanical energy K + U s is conserved.
Visualize:
The springs in both cases have the same compression Δx. We put the origin of the coordinate system at the
equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two
connected springs for the two-spring case (b).
Solve: The conservation of energy for the single-spring case:
1
1
1
1
K f + U sf = K i + U si ⇒ mvf2 + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2
2
2
2
2
Using xf = xe = 0 m, vi = 0 m/s, and vf = v0, this equation simplifies to
1 2 1
mv0 = k (Δx) 2
2
2
Conservation of energy in the case of the two springs in series, where each spring compresses by Δx /2, is
1
1
1
1
K f + U sf = K i + U si ⇒ mVf2 + 0 = mvi2 + k (Δx /2) 2 + k (Δx /2) 2
2
2
2
2
Using xf = xe′ = 0 m and vi = 0 m/s, we get
1
1 ⎡1
⎤
mVf2 = ⎢ k (Δx) 2 ⎥
2
2 ⎣2
⎦
Comparing the two results we see that Vf = v0 / 2.
Assess: The block pushes on the spring until the spring has returned to its equilibrium length.
10.40. Model: Model the ball and earth as particles. An elastic collision between the ball and earth conserves
both momentum and mechanical energy.
Visualize:
The before-and-after pictorial representation of the collision is shown in the figure. The ball as it is dropped from
a height of 10 m has zero velocity. Its speed just before it hits earth can be found using kinematics. We can then
apply the conservation laws to find earth’s recoil velocity.
Solve: (a) To get the speed of the ball at collision:
(v1 ) B2 = (v0 ) B2 − 2 g ( y1 − y0 ) B = (0 m/s) − 2(9.8 m/s 2 )(−10.0 m)
⇒ (v1 ) B = 2(9.8 m/s 2 )(10 m) = 14.0 m/s
In an elastic collision, the velocity of the object that is struck is
(v2 ) E =
2mB
2(0.500 kg)
(v1 ) B =
(14.0 m / s) = 2.3 × 10−24 m/s
mE + mB
5.98 × 1024 kg
(b) The time it would take earth to move 1.0 mm at this speed is given by
speed =
distance
1.0 × 10−3 m
=
= 4.3 × 10 20 s = 1.36 × 1013 years
velocity 2.3 × 10−24 m/s
10.41. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore the
mechanical energy K + U s + U g is conserved.
Visualize:
The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate
system at the position where the ice cube has compressed the spring 10 cm. That is, y0 = 0.
Solve:
The energy conservation equation K 2 + U s2 + U g2 = K 0 + U s0 + U g0 is
1 2 1
1
1
mv2 + k ( xe − xe ) 2 + mgy0 = mv02 + k ( x − xe ) 2 + mgy0
2
2
2
2
Using v2 = 0 m / s, y0 = 0 m, and v0 = 0 m / s,
1
k ( x − xe ) 2
(25 N/m)(0.10 m) 2
mgy2 = k ( x − xe ) 2 ⇒ y2 =
=
= 25.5 cm
2
2 mg
2(0.050 kg)(9.8 m/s 2 )
The distance traveled along the incline is y2 / sin 30° = 25.5 cm/ sin 30° = 51 cm.
Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational
potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to
solve the problem.
10.42. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic
collisions. Kinetic energy is conserved only in a perfectly elastic collision.
Visualize:
Solve:
For a package with mass m the conservation of energy equation is
1
1
K1 + U g1 = K 0 + U g0 ⇒ m(v1 ) 2m + mgy1 = m(v0 ) m2 + mgy0
2
2
Using (v0 ) m = 0 m / s and y1 = 0 m,
1
m(v1 ) 2m = mgy0 ⇒ (v1 ) m = 2 gy0 = 2(9.8 m/s 2 )(3.0 m) = 7.668 m/s
2
(a) For the perfectly inelastic collision the conservation of momentum equation is
pfx = pix ⇒ ( m + 2m)(v2 )3m = m(v1 ) m + (2m)(v1 ) 2m
Using (v1 ) 2m = 0 m / s, we get
(v2 )3m = (v1 ) m / 3 = 2.56 m/s
The packages move off together at a speed of 2.6 m/s.
(b) For the elastic collision, the mass m package rebounds with velocity
m − 2m
1
(v1 ) m = − ( 7.668 m/s ) = −2.56 m/s
m + 2m
3
The negative sign with (v3 ) m shows that the package with mass m rebounds and goes to the position y4 . We can
(v3 ) m =
determine y4 by applying the conservation of energy equation as follows. For a package of mass m:
1
1
K f + U gf = K i + U gi ⇒ m(v4 ) 2m + mgy4 = m(v3 ) m2 + mgy3
2
2
Using (v3 )m = −2.55 m/s, y3 = 0 m, and (v4 ) m = 0 m/s, we get
1
mgy4 = m(−2.56 m/s) 2 ⇒ y4 = 33 cm
2
10.43. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between
the marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We
will also assume zero rolling friction between the marble and the incline.
Visualize:
This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the
marble’s speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the
horizontal surface just where the marble exits the incline. In the second part, we will consider the elastic collision
between the marble and the steel ball.
Solve: The conservation of energy equation K1 + U g1 = K 0 + U g0 gives us:
1
1
2
2
mM (v1 ) M
+ mM gy1 = mM (v0 ) M
+ mM gy0
2
2
1
2
(v1 ) M
= gy0 ⇒ (v1 )M = 2 gy0 . When the marble collides with the
2
steel ball, the elastic collision gives the ball velocity
Using (v0 ) M = 0 m / s and y1 = 0 m, we get
(v2 )S =
2mM
2mM
(v1 ) M =
2 gy0
mM + mS
mM + mS
Solving for y0 gives
2
⎤
1 ⎡ mM + mS
y0 =
(v2 )S ⎥ = 0.258 m = 25.8 cm
⎢
2 g ⎣ 2mM
⎦
10.44. Model: Assume an ideal spring that obeys Hooke’s law. Since this is a free-fall problem, the
mechanical energy K + U g + U s is conserved. Also, model the safe as a particle.
Visualize:
We have chosen to place the origin of our coordinate system at the free end of the spring, which is neither
stretched nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic
potential energy as the safe compresses the spring. The only two forces are gravity and the spring force, which
are both conservative, so energy is conserved throughout the process. This means that the initial energy—as the
safe is released—equals the final energy—when the safe is at rest and the spring is fully compressed.
Solve: The conservation of energy equation K1 + U g1 + U s1 = K 0 + U g0 + U s0 is
1 2
1
1
1
mv1 + mg ( y1 − ye ) + k ( y1 − ye ) 2 = mv02 + mg ( y0 − ye ) + k ( ye − ye ) 2
2
2
2
2
Using v1 = v0 = 0 m/s and ye = 0 m, the above equation simplifies to
1
mgy1 + ky12 = mgy0
2
⇒k =
2mg ( y0 − y1 ) 2(1000 kg)(9.8 m/s 2 )(2.0 m − (−0.50 m))
=
= 1.96 × 105 N/m
y12
( −0.50 m) 2
Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit
the spring.
10.45. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction and hence the
mechanical energy K + U s + U g is conserved.
Visualize:
Solve:
(a) When releasing the block suddenly, K 2 + U s2 + U g 2 = K1 + U s1 + U g1
1 2 1
1
1
mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1
2
2
2
2
Using v2 = 0 m/s, v1 = 0 m/s, and y1 = ye , we get
1
0 J + (490 N/m)( y2 − y1 ) 2 + mgy2 = 0 J + 0 J + mgy1 ⇒ (245 N/m)( y2 − y1 ) 2 = mg ( y1 − y2 )
2
⇒ (245 N/m)( y1 − y2 ) 2 = (5.0 kg)(9.8 m/s 2 )( y1 − y2 ) ⇒ ( y1 − y2 ) = 0.20 m
(b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium.
Fnet = k Δy − mg = 0 ⇒ Δy =
mg (5.0 kg)(9.8 m/s 2 )
=
= 0.10 m
k
490 N/m
(c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it
reaches 0.10 m. So the block continues on past the equilibrium point until all the gravitational potential energy is
stored in the spring.
10.46. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and thus the mechanical
energy K + U s + U g is conserved.
Visualize:
We place the origin of our coordinate system at the spring’s compressed position y1 = 0. The rock leaves the
spring with velocity v2 as the spring reaches its equilibrium position.
Solve: (a) The conservation of mechanical energy equation is
1 2 1
1
1
K 2 + U s 2 + U g 2 = K1 + U s1 + U g1
mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1
2
2
2
2
Using y2 = ye , y1 = 0 m, and v1 = 0 m/s, this simplifies to
1 2
1
mv2 + 0 J + mgy2 = 0 J + k ( y1 − ye ) 2 + 0
2
2
1
1
(0.400 kg)v22 + (0.400 kg)(9.8 m/s 2 )(0.30 m) = (1000 N/m)(0.30 m) 2 ⇒ v2 = 14.8 m/s
2
2
(b) Let us use the conservation of mechanical energy equation once again to find the highest position ( y3 ) of the
rock where its speed (v3 ) is zero:
1
1
K 3 + U g3 = K 2 + U g 2 ⇒ mv32 + mgy3 = mv22 + mgy2
2
2
1
v 2 (14.8 m/s) 2
⇒ 0 + g ( y3 − y2 ) = v22 ⇒ ( y3 − y2 ) = 2 =
= 11.2 m
2
2 g 2(9.8 m/s 2 )
If we assume the spring’s length to be 0.5 m, then the distance between ground and fruit is 11.2 m + 0.5 m = 11.7 m.
This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit, and the
contestants go hungry.
10.47. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical
energy K + U g + U s is conserved.
Visualize:
Place the origin of the coordinate system at the end of the unstretched spring, making ye = 0 m.
Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by
using this to find the spring constant.
Fsp = FG ⇒ − k ( y1 − ye ) = − ky1 = mg ⇒ k = −
mg
(0.10 kg)(9.8 m/s2 )
=−
= 9.8 N/m
y1
−0.10 m
Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the
end, the spring has maximum stretch and the clay is instantaneously at rest. Thus
K 2 + (U g ) 2 + (U s ) 2 = K 0 + (U g )0 + (U s )0 ⇒ 12 mv22 + mgy2 + 12 ky22 = 12 mv02 + mgy0 + 0 J
Since v0 = 0 m/s and v2 = 0 m/s, this equation becomes
2mg
2mgy0
y2 −
=0
k
k
y22 + 0.20 y2 − 0.10 = 0
mgy2 + 12 ky22 = mgy0 ⇒ y22 +
The numerical values were found using known values of m, g, k, and y0 . The two solutions to this quadratic
equation are y2 = 0.231 m and y2 = −0.432 m. The point we’re looking for is below the origin, so we need the
negative root. The distance of the pan from the ceiling is
L = y2 + 50 cm = 93 cm
10.48. Model: Assume an ideal spring that obeys Hooke’s law. Also assume zero rolling friction between
the roller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the
mechanical energy K + U s + U g is conserved.
Visualize:
We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and
touches the roller coaster car.
Solve: (a) The energy conservation equation for the car going to the top of the hill is
K 2 + U g 2 + U s2 = K 0 + U g0 + U s0
1 2
1
1
1
mv2 + mgy2 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2
2
2
2
2
Noting that y0 = 0 m, v2 = 0 m/s, v1 = 0 m/s, and x0 − xe = 2.0 m, we obtain
1
0 J + mgy2 + 0 J = 0 J + 0 J + k (2.0 m) 2
2
2mgy2
2(400 kg)(9.8 m/s 2 )(10 m)
⇒k =
=
= 1.96 × 104 N/m
(2.0 m) 2
(2.0 m) 2
We now increase this value for k by 10% for safety, giving a value of 2.156 × 104 N/m ≈ 2.2 × 104 N/m.
(b) The energy conservation equation K 3 + U g3 + U s3 = K 0 + U g 0 + U s0 is
1 2
1
1
1
mv3 + mgy3 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2
2
2
2
2
Using y3 = −5.0 m, v0 = 0 m / s, y0 = 0 m, and | x0 − xe | = 2.0 m, we get
1 2
1
mv3 + mg (−5.0 m) + 0 J = 0 J + 0 J + k ( x0 − xe ) 2
2
2
1
1
(400 kg)v32 − (400 kg)(9.8 m/s 2 )(5.0 m) = (2.156 × 104 N/m)(2.0 m) 2
2
2
⇒ v3 = 17.7 m/s
10.49. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not
change. Model Julie as a particle.
Visualize:
We place the coordinate system at the bottom of the ramp directly below Julie’s starting position. From
geometry, Julie launches off the end of the ramp at a 30º angle.
1
1
Solve: Energy conservation: K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0
2
2
Using v0 = 0 m/s, y0 = 25 m, and y1 = 3 m, the above equation simplifies to
1 2
mv1 + mgy1 = mgy0 ⇒ v1 = 2 g ( y0 − y1 ) = 2(9.8 m/s 2 )(25 m − 3 m) = 20.77 m/s
2
We can now use kinematic equations to find the touchdown point from the base of the ramp. First we’ll consider
the vertical motion:
1
1
y2 = y1 + v1 y (t2 − t1 ) + a y (t2 − t1 ) 2 0 m = 3 m + (v1 sin 30°)(t2 − t1 ) + (−9.8 m / s 2 )(t2 − t1 ) 2
2
2
(20.77 m/s)sin 30°
(3 m)
2
(t2 − t1 ) −
⇒ (t2 − t1 ) −
=0
(4.9 m/s 2 )
(4.9 m/s 2 )
(t2 − t1 ) 2 − (2.119 s)(t2 − t1 ) − (0.6122 s 2 ) = 0 ⇒ (t2 − t1 ) = 2.377 s
For the horizontal motion:
1
x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1 ) 2
2
x2 − x1 = (v1 cos30°)(t2 − t1 ) + 0 m = (20.77 m/s)(cos30°)(2.377 s) = 43 m
Assess: Note that we did not have to make use of the information about the circular arc at the bottom that
carries Julie through a 90° turn.
10.50. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and
the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations
apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid.
Visualize:
Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-andafter pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b),
respectively.
Solve: (a) For an elastic collision, the ball’s rebound velocity is
(vf ) b =
mb − mB
−80 g
(vi ) b =
(5.0 m/s) = −3.33 m/s
mb + mB
120 g
The ball’s speed is 3.3 m/s.
(b) An elastic collision gives the block speed
(vf ) B =
2mB
40 g
(vi ) b =
(5.0 m/s) = 1.667 m/s
mb + mB
120 g
To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the
block + spring system. That is K1 + U s1 = K 0 + U s0 :
1
1
1
1
mB (vf′ ) 2B + k ( x1 − x0 ) 2 = mB (vf ) 2B + k ( x0 − x0 ) 2
2
2
2
2
0 + k ( x1 − x0 ) 2 = mB (vf )B2 + 0
(x1 − x0 ) = (0.100 kg )(1.667 m/s )2 /(20 N/m ) = 11.8 cm
(c) Momentum conservation pf = pi for the perfectly inelastic collision means
(mB + mB )vf = mb (vi ) b + mB (vi ) B
(0.100 kg + 0.020 kg)vf = (0.020 kg)(5.0 m/s) + 0 m/ v ⇒ vf = 0.833 m/s
The maximum compression in this case can now be obtained using the conservation of energy equation
K1 + U s1 = K 0 + U s0 :
0 J + (1/ 2)k (Δx) 2 = (1/ 2)(mB + mb )vf 2 + 0 J
⇒ Δx =
mB + mb
0.120 kg
vf =
(0.833m/s) = 0.0645 m = 6.5 cm
k
20 N/m
10.51. Model: Assume an ideal spring that obeys Hooke’s law. We treat the bullet and the block in the
particle model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no
friction, the mechanical energy of the system (bullet + block + spring) is conserved.
Visualize:
We place the origin of our coordinate system at the end of the spring that is not anchored to the wall.
Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi . This means
⎛ m ⎞
(m + M )vf = m(vi ) m + M (vi ) M ⇒ (m + M )vf = mvB + 0 kg m/s ⇒ vf = ⎜
⎟ vB
⎝m+M ⎠
where we have used vB for the initial speed of the bullet. The mechanical energy conservation equation
K1 + U s1 = K e + U se as the bullet embedded block compresses the spring is:
1
1
1
1
m(v′f ) 2 + k ( x1 − xe ) 2 = (m + M )(vf ) 2 + k ( xe − xe ) 2
2
2
2
2
2
1
1
(m + M )kd 2
⎛ m ⎞ 2
0 J + kd 2 = (m + M ) ⎜
0
J
v
+
⇒
v
=
B
⎟ B
2
2
m2
⎝m+M ⎠
(b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,
vB = (0.0050 kg + 2.0 kg)(50 N/m)(0.10 m) 2 /(0.0050)2 = 2.0 × 102 m/s
(c) The fraction of energy lost is
1 2 1
2
2
mvB − (m + M )vf2
m +M ⎛ vf ⎞
m+M ⎛ m ⎞
2
2
=1−
=
1
−
⎜ ⎟
⎜
⎟
1 2
m ⎝ vB ⎠
m ⎝m+M ⎠
mvB
2
m
0.0050 kg
=1−
=1−
= 0.9975
m+M
(0.0050 kg + 2.0 kg)
That is, during the perfectly inelastic collision 99.75% of the bullet’s energy is lost. The energy is dissipated
inside the block. Although it is common to say, “The energy is lost to heat,” in the next chapter we’ll see that it is
more accurate to say, “The energy is transformed to thermal energy.”
10.52. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical
energy K + U g + U s is conserved.
Visualize:
We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched
nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m
and the block’s mass is M.
Solve: (a) The energy conservation equation K 2 + U s2 + U g2 = K1 + U s1 + U g1 becomes
1
1
1
1
(m + M )v22 + k ( y2 − ye )2 + (m + M ) g ( y2 − ye ) = (m + M )v12 + k ( y1 − ye )2 − (m + M ) g ( y1 − ye )
2
2
2
2
Noting v2 = 0 m/s, we can rewrite the above equation as
k (Δy2 ) 2 + 2( m + M ) g ( Δy2 + Δy1 ) = (m + M )v12 + k (Δy1 ) 2
Let us express v1 in terms of the bullet’s initial speed vB by using the momentum conservation equation pf = pi
which is (m + M )v1 = mvB + Mvblock . Since vblock = 0 m/s, we have
⎛ m ⎞
v1 = ⎜
⎟ vB
⎝m+M ⎠
We can also find the magnitude of y1 from the equilibrium condition k ( y1 − ye ) = Mg .
Δy1 =
Mg
k
With these substitutions for v1 and Δy1 , the energy conservation equation simplifies to
k (Δy2 ) 2 + 2(m + M ) g (Δy1 + Δy2 ) =
m 2vB2
⎛ Mg ⎞
+ k⎜
⎟
(m + M )
⎝ k ⎠
2
2
(m + M ) M 2 g 2
⎛m+M ⎞
⎛m+M ⎞
2
g
(
y
y
)
⇒ vB2 = 2 ⎜
Δ
+
Δ
−
+ k⎜
1
2
⎟
⎟ ( Δy 2 )
2
2
m
k
⎝ m ⎠
⎝ m ⎠
We still need to include the spring’s maximum compression (d) into this equation. We assume that
d = Δy1 + Δy2 , that is, maximum compression is measured from the initial position ( y1 ) of the block. Thus,
using Δy2 = d − Δy1 = (d − Mg / k ), we have
12
2 2
⎡ ⎛ m + M ⎞2
⎤
⎛m+M ⎞M g
⎛m+M ⎞
2
vB = ⎢ 2 ⎜
gd
−
+ k⎜
⎟
⎜
⎟
⎟ ( d − Mg / k ) ⎥
2
2
⎝ m ⎠ k
⎝ m ⎠
⎢⎣ ⎝ m ⎠
⎥⎦
(b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m,
2
2
⎛ 2.0 kg ⎞
⎛ 2.010 kg ⎞
2
2 2
vB2 = 2 ⎜
⎟ (9.8 m/s ) /(50 N/m)
⎟ (9.8 m/s )(0.45 m) − (2.010 kg) ⎜
0.010kg
0.010
kg
⎝
⎠
⎝
⎠
1
+ (50 N/m)(2.010 kg)
[0.45 m − (2.0 kg) × (9.8 m/s 2 )/50 N/m]2
(0.010 kg) 2
⇒ vB = 453 m/s
The bullet has a speed of 4.5 × 10 2 m/s.
Assess: This is a reasonable speed for the bullet.
10.53. Model: Because the track is frictionless, the sum of the kinetic and gravitational potential energy does
not change during the car’s motion.
Visualize:
We place the origin of the coordinate system at the ground level directly below the car’s starting position. This is
a two-part problem. If we first find the maximum speed at the top of the hill, we can use energy conservation to
find the maximum initial height.
Solve: Because its motion is circular, at the top of the hill the car has a downward-pointing centripetal accelG
eration ac = −(mv 2 / r ) ˆj. Newton’s second law at the top of the hill is
( Fnet ) y = n y + ( FG ) y = n − mg = m( ac ) y = −
⎛
mv 2
mv 2
v2 ⎞
⇒ n = mg −
= m⎜ g − ⎟
R
R
R⎠
⎝
If v = 0 m / s, n = mg as expected in static equilibrium. As v increases, n gets smaller—the weight of the car and
passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, so the
maximum speed vmax occurs when n → 0. We see that vmax = gR . Now we can use energy conservation to
relate the top of the hill to the starting height:
1
1
1
K f + U f = Ki + U i ⇒ mvf2 + mgyf = mv12 + mgy1 ⇒ mgR + mgR = 0 J + mghmax
2
2
2
where we used vf = vmax and yf = R. Solving for hmax gives hmax = 32 R.
(b) If R = 10 m, then hmax = 15 m.
10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to
go over the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitational
potential energy is conserved during the block’s motion. We will use this conservation equation in the second
part to find the minimum height the block must start from to make it around the loop.
Visualize:
We place the origin of our coordinate system directly below the block’s starting position on the frictionless track.
Solve: The free-body diagram on the block implies
mv 2
FG + n = c
R
For the block to just stay on track, n = 0. Thus the critical velocity vc is
FG = mg =
mvc2
⇒ vc2 = gR
R
The block needs kinetic energy 12 mvc2 = 12 mgR to go over the top of the loop. We can now use the conservation
of mechanical energy equation to find the minimum height h.
1
1
K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi
2
2
Using vf = vc = gR , yf = 2 R, vi = 0 m/s, and yi = h, we obtain
1
gR + g (2 R ) = 0 + gh ⇒ h = 2.5R
2
10.55. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless,
so that the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the
collision, as Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also
assume the spring to be an ideal one that obeys Hooke’s law.
Visualize:
We place the origin of our coordinate system directly below the bobsled’s initial position.
Solve: (a) Momentum conservation in Lisa’s collision with bobsled states p1 = p0 , or
(mL + mB )v1 = mL (v0 ) L + mB (v0 ) B ⇒ (mL + mB )v1 = mL (v0 ) L + 0
⎛ mL ⎞
⎛
⎞
40 kg
⇒ v1 = ⎜
⎟ (v0 ) L = ⎜
⎟ (12 m/s) = 8.0 m/s
m
m
40
kg
20
kg
+
+
⎝
⎠
B ⎠
⎝ L
The energy conservation equation: K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is
1
1
1
1
(mL + mB )v22 + k ( x2 − xe ) 2 + ( mL + mB ) gy2 = (mL + mB )v12 + k ( xe − xe ) 2 + (mL + mB ) gy1
2
2
2
2
Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m)sin 20° = 17.1 m, v1 = 8.0 m/s, and (mL mB ) = 60 kg, we get
1
1
0 J + (2000 N / m)( x2 − xe ) 2 + 0 J = (60 kg)(8.0 m / s) 2 + 0 J + (60 kg)(9.8 m / s 2 )(17.1 m)
2
2
Solving this equation yields ( x2 − xe ) = 3.5 m.
(b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical
energy that the speed at the bottom depends only on the vertical descent Δy. Only the ramp’s height h is important,
not its shape or angle.
10.56. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20
g ball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum.
Visualize:
The figure shows three distinct moments of time: the time before the collision, the time after the collision but
before the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our
coordinate system on the 100 g ball when it is hanging motionless.
Solve: For a perfectly elastic collision, the ball moves forward with speed
(v1 ) M =
2mm
1
(v0 ) m = (v0 ) m
mm + mM
3
In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings
up after the collision. That is, K 2 + U g 2 = K1 + U g1. We have
1
1
M (v2 ) 2M + Mgy2 = M (v1 ) 2M + Mgy1
2
2
Using (v2 ) M = 0 J, (v1 ) M =
(v0 ) m
, y1 = 0 m, and y2 = L − L cosθ , the energy equation simplifies to
3
g ( L − L cosθ ) =
1 (v0 ) 2m
2 9
⇒ (v0 ) m = 18 g L(1 − cosθ ) = 18(9.8 m/s 2 )(1.0 m)(1 − cos50°) = 7.9 m/s
10.57. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation
10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′
in which one ball is at rest. This allows us to apply Equations 10.43 to the case of a perfectly elastic collision in S′,
find the final velocities of the balls in S′, and then transform these velocities back to the lab frame S.
Visualize: Let S′ be the frame of the 200 g ball. Denoting masses as m1 = 100 g and m2 = 200 g, the initial
velocities in the S frame are (vix )1 = 4 m/s and (vix ) 2 = −3 m/s.
Figure (a) shows the before-collision situation as seen in frame S, and figure (b) shows the before-collision
situation as seen in frame S′. The after-collision velocities in S′ are shown in figure (c), and figure (d) indicates
velocities in S after they have been transformed to frame S from S′.
Solve: (a) In the S frame, (vix )1 = 4 m/s and (vix ) 2 = −3 m/s. S′ is the reference frame of the 200 g ball, so
V = −3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of
velocities v′ = v − V . So,
(vix )1′ = (vix )1 − V = 4 m/s − ( − 3 m/s) = 7 m/s
(vix )′2 = (vix ) 2 − V = −3 m/s − ( − 3 m/s) = 0 m/s
Figure (b) shows the “before” situation, where ball 2 is at rest.
Now we can use Equations 10.43 to find the after-collision velocities in frame S′:
(vfx )1′ =
m1 − m2
100 g − 200 g
7
(vix )1′ =
(7 m/s) = − m/s
m1 + m2
100 g + 200 g
3
(vfx )′2 =
2m1
2(100) g
14
(vix )1′ =
(7 m/s) = m/s
m1 + m2
100 g + 200 g
3
Finally, we need to apply the reverse Galilean transformation v = v′ + V , with the same V, to transform the aftercollision velocities back to the lab frame S:
7
(vfx )1 = (vfx )1′ + V = − m/s − 3 m/s = −5.33 m/s
3
14
(vfx ) 2 = (vfx )′2 + V = m/s − 3 m/s = 1.67 m/s
3
Figure (d) shows the “after” situation in the lab frame. The 100 g ball is moving left at 5.3 m/s; the 200 g ball is
moving right at 1.7 m/s.
(b) The momentum conservation equation pfx = pix for a perfectly inelastic collision is
(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
(0.100 kg + 0.200 kg)vfx = (0.100 kg)(4.0 m/s) + (0.200 kg)(−3.0 m/s) ⇒ vfx = −0.667 m/s
Both balls are moving left at 0.67 m/s.
10.58. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation
10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′
in which one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in S′. After
finding the final velocities of the balls in S′, we can then transform these velocities back to the lab frame S.
Visualize: Let S′ be the frame of the 400 g ball. Denoting masses as m1 = 100 g and m2 = 400 g, the initial
velocities in the S frame are (vix )1 = + 4.0 m/s and (vix ) 2 = +1.0 m/s.
Figures (a) and (b) show the before-collision situations in frames S and S′, respectively. The after-collision
velocities in S′ are shown in figure (c). Figure (d) indicates velocities in S after they have been transformed to S
from S′.
Solve: In frame S, (vix )1 = 4.0 m/s and (vix ) 2 = 1.0 m/s. Because S′ is the reference frame of the 400 g ball,
V = 1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of
velocities v′ = v − V . So,
(vix )1′ = (vix )1 − V = 4.0 m/s − 1.0 m/s = 3.0 m/s
(vix )′2 = (vix ) 2 − V = 1.0 m/s − 1.0 m/s = 0 m/s
Figure (b) shows the “before” situation in frame S′ where the ball 2 is at rest.
Now we can use Equations 10.43 to find the after-collision velocities in frame S′:
(vfx )1′ =
m1 − m2
100 g − 400 g
(vix )1′ =
(3.0 m/s) = −1.80 m/s
m1 + m2
100 g + 400 g
(vfx )′2 =
2m1
2(100 g)
(vix )1′ =
(3.0 m/s) = 1.20 m/s
m1 + m2
100 g + 400 g
Finally, we need to apply the reverse Galilean transformation v = v′ + V , with the same V, to transform the aftercollision velocities back to the lab frame S.
(vfx )1 = (vfx )1′ + V = −1.80 m/s + 1.0 m/s = −0.80 m/s
(vfx ) 2 = (vfx )′2 + V = 1.20 m/s + 1.0 m/s = 2.20 m/s
Figure (d) shows the “after” situation in frame S. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.2 m/s.
Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision
velocities.
10.59. Model: Use the model of the conservation of mechanical energy.
Visualize:
Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For
E = 12 J, this occurs at the points x = 1 m and x = 7 m.
(b) The equation for kinetic energy K = E − U gives the distance between the potential energy curve and total
energy line. U = 8 J at x = 2 m, so K = 12 J − 8 J = 4 J. The speed corresponding to this kinetic energy is
v=
2K
2(4 J)
=
= 4.0 m/s
m
0.5 kg
(c) Maximum speed occurs for minimum U. This occurs at x = 1 m and x = 4 m, where U = 0 J and K = 12 J.
The speed at these two points is
v=
2K
2(12 J)
=
= 6.9 m/s
m
0.500 kg
(d) The particle leaves x = 1 m with v = 6.9 m/s. It gradually slows down, reaching x = 2 m with a speed of
4.0 m/s. After x = 2 m, it speeds up again, returning to a speed of 6.9 m/s as it crosses x = 4 m. Then it slows
again, coming instantaneously to a halt (v = 0 m/s) at the x = 7 m turning point. Now it will reverse direction and
move back to the left.
(e) If the particle has E = 4 J it cannot cross the 8 J potential energy “mountain” in the center. It can either
oscillate back and forth over the range 1.0 m ≤ x ≤ 1.5 m or over the range 3 m ≤ x ≤ 5 m.
10.60. Model: We will use the conservation of mechanical energy.
Visualize:
The potential energy (U ) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the two
minima correspond to the nitrogen atom’s position on both sides of the plane containing the three hydrogens.
Solve: (a) At room temperature, the total energy line is below the “hill” in the center of the potential energy curve.
That is, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. There’s
a stable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, the
nitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-axis—that
is, toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B.
(b) The total energy line is now well above the “hill,” and the turning points of the nitrogen atom’s motion (where the
total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from one side
of the H3 plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it has
sufficient energy to get through.
dU
= 0.
dx
dU
1
= 0 = 1 + 2cos ( 2 x ) ⇒ cos ( 2 x ) = −
dx
2
1
⎛ 1⎞
⇒ x = cos −1 ⎜ − ⎟
2
⎝ 2⎠
10.61. Solve: (a) The equilibrium positions are located at points where
1
2π
4π
⎛ 1⎞
is in radians and x is in meters. The function cos −1 ⎜ − ⎟ may have values
and
. Thus
2
3
3
2
⎝
⎠
there are two values of x,
π
2π
x1 = and x2 =
3
3
within the interval 0 m ≤ x ≤ π m.
(b) A point of stable equilibrium corresponds to a local minimum, while a point of unstable equilibrium
corresponds to a local maximum. Compute the concavity of U(x) at the equilibrium positions to determine their
stability.
d 2U
= −4sin ( 2 x )
dx 2
⎛ 3⎞
d 2U
π
π d 2U
π
x = −4 ⎜⎜
x < 0, x1 =
At x1 = ,
is a local maximum, so x1 = is a point
⎟⎟ = −2 3. Since
2 ( 1)
2 ( 1)
dx
2
3
3
3 dx
⎝
⎠
of unstable equilibrium.
⎛
d 2U
2π d 2U
3⎞
2π
2π
− x2 ) = −4 ⎜⎜ −
,
At x2 =
> 0, x2 =
is a local minimum, so x2 =
is a
⎟⎟ = +2 3. Since
2 (
2
dx
dx
3
2
3
3
⎝
⎠
point of stable equilibrium.
Note that −
10.62. Model: The potential energy of two nucleons interacting via the strong force is
U = U 0 [1 − e − x / x0 ]
where x is the distance between the centers of the two nucleons, x0 = 2.0 × 10−15 m, and U 0 = 6.0 × 10−11 J.
Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called the
strong force. This force exists between nucleons at very small separations.
Solve: (a)
(b) For x = 5.0 × 10−15 m, U = 55.1 × 10−12 J. This energy is represented by a total energy line.
(c) Due to conservation of total energy, the potential energy when x = 5.0 × 10−15 J is transformed into kinetic
energy until x = twice the radius = 1.0 × 10 −15 m. At this separation, u = 15.7 × 10−12 J. Thus,
1 2 1 2
mv + mv + 15.7 × 10−12 J = 55.1 × 10−12 J ⇒ v = 1.53 × 108 m/s
2
2
Assess: A speed of 1.53 × 108 m/s is approximately 0.5 c where c is the speed of light. This speed is
understandable for the present model.
10.63. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring.
When the ball comes down it lands on and compresses the spring. If the spring has a spring constant of
k = 600 N/m, by how much is it compressed?
10.64. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initial
speed must the auto have had at the bottom of the hill?
(b)
We place the origin of our coordinate system at the bottom of the hill.
(c) The solution of the equation is vi = 14.9 m/s. This is approximately 30 mph and is a reasonable value for the
speed at the bottom of the hill.
10.65. (a) A spring gun is compressed 15 cm to launch a 200 g ball on a horizontal, frictionless surface. The ball has
a speed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun.
(b)
We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because
the surface is frictionless, the mechanical energy for the system (ball + spring) is conserved.
(c) The conservation of energy equation is
K f + U sf = K i + U si
1 2 1
1
1
mv1 + k (0 m) 2 = m(0 m/s) 2 + k (−0.15 m) 2
2
2
2
2
(0.200 kg)(2.0 m/s) 2 = k ( −0.15 m) 2
k = 36 N/m
10.66. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on a
frictionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of the
spring is firmly anchored to a fixed post on the surface. How far will the spring compress?
(b)
(c) Solving the conservation of momentum equation we get v1x = 1.0 m/s. Substituting this value into the
conservation of energy equation yields Δx2 = 32 cm.
10.67. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30° incline. A 500
g block is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with
which the block is launched up the incline?
(b) The origin is placed at the end of the uncompressed spring. This is the point from which the block is launched
as the spring expands.
(c) Solving the energy conservation equation, we get vf = 2.6 m/s.
10.68. Model: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + U s + U g is
conserved during the launch of the ball.
Visualize:
This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves
the spring. This will yield the ball’s kinetic energy as it leaves the spring.
Solve: Using the equations of kinematics,
1
x3 = x2 + v2 x (t3 − t2 ) + ax (t3 − t2 ) 2 ⇒ 5.0 m = 0 m + (v2 cos30°)(t3 − 0 s) + 0 m
2
(v2 cos30°)t3 = 5.0 m ⇒ t3 = (5.0 m/ v2 cos30°)
1
y3 = y2 + v2 y (t3 − t2 ) + a y (t3 − t2 ) 2
2
1
−1.5 m = 0 + (v2 sin 30°)(t3 − 0 s) + (9.8 m / s 2 )(t3 − 0 s) 2
2
⎛ 5.0 m ⎞
5.0 m ⎞
2 ⎛
Substituting the value for t3 , (−1.5 m) = (v2 sin 30°) ⎜
⎟ − ( 4.9 m/s ) ⎜
⎟
⎝ v2 cos30° ⎠
⎝ v2 cos30° ⎠
⇒ (−1.5 m) = + (2.887 m) −
2
163.33
⇒ v2 = 6.102 m/s
v22
The conservation of energy equation K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is
1 2 1
1
1
mv2 + k (0 m) 2 + mgy2 = mv12 + k ( Δs ) 2 + mgy1
2
2
2
2
Using y2 = 0 m, v1 = 0 m/s, Δs = 0.20 m, and y1 = −(Δs )sin g 30°, we get
1 2
1
mv2 + 0 J + 0 J = 0 J + k ( Δs ) 2 − mg (Δs )sin 30° (Δs ) 2 k = mv22 + 2mg (Δs )sin 30°
2
2
(0.20 m) 2 k = (0.020 kg)(6.102 m/s) 2 + 2(0.020 kg)(9.8 m/s 2 )(0.20)(0.5) ⇒ k = 19.6 N/m
Assess: Note that y1 = −( Δs )sin 30° is with a minus sign and hence the gravitational potential energy mgy1 is
− mg (Δs )sin 30°.
10.69. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go
over the top of the peg without the string going slack. Using the energy conservation equation, we will then
obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus
determining the angle θ .
Visualize:
We place the origin of our coordinate system on the peg. This choice will provide a reference to measure
gravitational potential energy. For θ to be minimum, the ball will just go over the top of the peg.
Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the
circle. Newton’s second law at this point is
mv 2
FG + T =
r
where T is the tension in the string. The critical speed vc at which the string goes slack is found when T → 0. In
this case,
mg =
mvC2
⇒ vC2 = gr = gL 3
r
The ball should have kinetic energy at least equal to
1 2 1
⎛ L⎞
mvC = mg ⎜ ⎟
2
2
⎝3⎠
for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get
the minimum angle θ . The equation for the conservation of energy is
1
1
K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi
2
2
Using vf = vc , yf = 13 L, vi = 0, and the above value for vC2 , we get
1
L
L
L
mg + mg = mgyi ⇒ yi =
2
3
3
2
That is, the ball is a vertical distance 12 L above the peg’s location or a distance of
⎛ 2L L ⎞ L
− ⎟=
⎜
⎝ 3 2⎠ 6
below the point of suspension of the pendulum, as shown in the figure on the right. Thus,
cosθ =
L /6 1
= ⇒ θ = 80.4°
L
6
10.70. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system,
so it is an isolated system. The interaction forces within the system are the spring force of the bungee cord and
the gravitational force. These are both conservative forces, so mechanical energy is conserved.
Visualize:
We can equate the system’s initial energy, as you step off the bridge, to its final energy when you reach the
lowest point. We do not need to compute your speed at the point where the cord starts to stretch. We do,
however, need to note that the end of the unstretched cord is at y0 = y1 − 30 m = 70 m, so U 2s = 12 k ( y2 − y0 ) 2 .
Also note that U1s = 0, since the cord is not stretched. The energy conservation equation is
1
K 2 + U 2g + U 2s = K1 + U1g + U1s ⇒ 0 J + mgy2 + k ( y2 − y0 ) 2 = 0 J + mgy1 + 0 J
2
Multiply out the square of the binomial and rearrange:
1
1
mgy2 + ky22 − ky0 y2 + ky02 = mgy1
2
2
⎛ 2mg
⎞
⎛ 2 2mgy1 ⎞
2
2
⇒ y2 + ⎜
− 2 y0 ⎟ y2 + ⎜ y0 −
⎟ = y2 − 100.8 y2 + 980 = 0
k
k
⎝
⎠
⎝
⎠
This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is a
height above the point where the cord started to stretch. So we find that your distance from the water when the
bungee cord stops stretching is 10.9 m.
10.71. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical
energy K + U g + U s is conserved.
Visualize:
We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use
lengths along the ramp with the variable s rather than x.
Solve: (a) The conservation of energy equation K 2 + U g2 + U s2 = K1 + U g1 + U s1 is
1 2
1
1
mv2 + mgy2 + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2
2
2
2
1
1
1
m(0 m/s) 2 + mg (0 m) + k ( Δs ) 2 = m(0 m/s) 2 + mg (4.0 m + Δs )sin 30° + 0 J
2
2
2
1
⎛1⎞
(250 N/m)( Δs ) 2 = (10 kg)(9.8 m/s 2 )(4.0 m + Δs ) ⎜ ⎟
2
⎝ 2⎠
This gives the quadratic equation:
(125 N/m)(Δs ) 2 − (49 kg ⋅ m/s 2 ) Δs − 196 kg ⋅ m 2 /s 2 = 0
⇒ Δs = 1.46 m and − 1.07 m (unphysical)
The maximum compression is 1.46 m.
(b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and the
block’s velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring
is neither compressed nor stretched.
1 2
1
1
1
mv + mgy + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2
2
2
2
2
1 2
1
mv + mg (−Δs sin 30°) + k ( Δs ) 2 = 0 J + mg (4.0 m sin 30°) + 0 J
2
2
1
1
2
k (Δs ) − ( mg sin 30°) Δs + mv 2 − mg sin 30°(4.0 m) = 0
2
2
To find the compression where v is maximum, take the derivative of this equation with respect to Δs :
1
1
dv
k 2(Δs ) − (mg sin 30°) + m 2v
−0=0
2
2
d Δs
Since
dv
= 0 at the maximum, we have
d Δs
Δs = (mg sin 30°)/ k = (10 kg)(9.8 m/s 2 )(0.5)/(250 N/m) = 19.6 cm
10.72. Model: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon
(C) fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion.
Visualize:
The before-and-after pictorial representation is shown, with the origin of the coordinate system located at the
spring’s free end when the spring is neither compressed nor stretched. This free end of the spring is just behind
the cannon.
Solve: The momentum conservation equation pfx = pix is
mB (vfx ) B + mC (vfx )C = mB (vix ) B + mC (vix )C
Since the initial momentum is zero,
(vfx ) B = −
⎛ 200 kg ⎞
mC
(vfx )C = − ⎜
⎟ (vfx )C = −20(vfx )C
mB
⎝ 10 kg ⎠
The mechanical energy conservation equation for the cannon + spring K f + U sf = Ki + U si is
1
1
1
1
1
m(vf )C2 + k (Δx) 2 = m(vi )C2 + 0 J ⇒ 0 J + k (Δx) 2 = m(vfx )C2
2
2
2
2
2
(20,000 N/m)
k
(0.50 m) = ±5.0 m/s
⇒ (vfx )C = ±
Δx = ±
200 kg
m
To make this velocity physically correct, we retain the minus sign with (vfx )C . Substituting into the momentum
conservation equation yields:
(vfx ) B = −20(−5.0 m/s) = 100 m/s
10.73. Model: This is a collision between two objects, and momentum is conserved in the collision. In
addition, because the interaction force is a spring force and the surface is frictionless, energy is also conserved.
Visualize:
Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximum
compression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is at
maximum compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2 . This is an
important observation.
Solve: First relate part 1 to part 2. Conservation of energy is
1
1
1
1
1
1
2
2
2
mA vA1
= mA vA2
+ mBvB2
+ k (Δxmax ) 2 = ( mA + mB )v22 + k (Δxmax ) 2
2
2
2
2
2
2
where Δxmax is the spring’s compression. U g is not in the equation because there are no elevation changes. Also note
that K 2 is the sum of the kinetic energies of all moving objects. Both v2 and Δxmax are unknowns. Now add the
conservation of momentum:
mA vA1
mA vA1 = mA vA 2 + mBvB2 = (mA + mB )v2 ⇒ v2 =
= 2.667 m/s
mA + mB
Substitute this result for v2 into the energy equation to find:
1
1
1
2
k (Δxmax ) 2 = mA vA1
− (mA + mB )v22
2
2
2
2
mAvA1
− (mA + mB )v22
= 0.046 m = 4.6 cm
k
Notice how both conservation laws were needed to solve this problem.
(b) Again, both energy and momentum are conserved between “before” and “after.” Energy is
1
1
1
1 2
2
2
2
2
mA vA1
= mA vA3
+ mBvB3
⇒ 16 = vA3
+ vB3
2
2
2
2
The spring is no longer compressed, so the energies are purely kinetic. Momentum is
mA vA1 = mA vA3 + mBvB3 ⇒ 8 = 2vA3 + vB3
⇒ Δxmax =
We have two equations in two unknowns. From the momentum equation, we can write vB3 = 2(4 − vA3 ) and use this
in the energy equation to obtain:
2
1
2
2
2
2
16 = vA3
+ ⋅ 4 ( 4 − vA3
= 3vA3
− 16vA3 + 32 ⇒ 3vA3
− 16vA3 + 16 = 0
)
2
This is a quadratic equation for vA3 with roots vA3 = (4 m/s, 1.33 m/s). Using vB3 = 2(4 − vA3 ), these two roots
give vB3 = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a “collision” in which A misses B, so each
keeps its initial speed. That’s not the situation here. We want the second pair of roots, from which we learn that
the blocks’ speeds after the collision are vA3 = 1.33 m/s and vB3 = 5.3 m/s.
10.74. Model: Mechanical energy and momentum are conserved during the expansion of the spring.
Visualize: Please refer to Figure CP10.74.
Solve: Example 10.8 is a very similar problem, except that the objects are initially at rest. We can use the
solution from Example 10.8 for this problem in a reference frame S′ in which the two carts are initially at rest,
then transform the answer to the frame S in which the carts are initially moving.
Thus in the S′ frame,
( vfx )′2 =
k ( Δxi )
m
m2 1 + 2
(
2
m1
)
m
m1
( vfx )1′ = − 2 ( vfx )2′
Let the 100 g cart be Cart 1 and the 300 g cart be Cart 2. With k = 120 N/m and Δxi = 4.0cm,
( vfx )2′ = 0.40 m/s, ( vfx )1′ = −1.2 m/s
An object at rest in the S′ frame is traveling to the right at 1.0 m/s in the S frame. The equation of transformation is
therefore
vx = vx′ + 1.0 m/s
In the S frame, the velocities of the carts are
( vfx )1 = −1.2 m/s + 1.0 m/s = −0.2 m/s
( vfx )2 = 0.40 m/s + 1.0 m/s = 1.4 m/s
Assess:
Cart 1 is moving slowly to the left while the heavier Cart 2 is moving quickly to the right.
10.75. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is
over, the balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not
change as the balls swing out.
Visualize:
In the pictorial representation we have identified before-and-after quantities for both the collision and the
pendulum swing. We have chosen to place the origin of the coordinate system at a point where the two balls at
rest barely touch each other.
Solve: As the ball with mass m1 , whose string makes an angle θ il with the vertical, swings through its
equilibrium position, it lowers its gravitational energy from m1 gy0 = m1 g ( L − L cosθ i1 ) to zero. This change in
potential energy transforms into a change in kinetic energy. That is,
1
m1 g ( L − L cosθ i1 ) = m1 (v1 )12 ⇒ (v1 )1 = 2 gL(1 − cosθ i1 )
2
Similarly, (v1 )2 = 2 gL(1 − cosθ i2 ). Using θ i1 = θ i2 = 45°, we get (v1 )1 = 2.396 m/s = (v1 ) 2 . Both balls are moving
at the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in
which ball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S′ in which ball 2
is at rest. Ball 2 is at rest in a frame that moves with ball 2, so choose S′ to have V = −2.396 m/s, with the minus
sign because this frame (like ball 2) is moving to the left. In this frame, ball 1 has velocity
(v1′ )1 = (v1 )1 − V = 2.396 m/s + 2.396 m/s = 4.792 m/s and ball 2 is at rest. The elastic collision causes the balls to
move with velocities
m − m2
1
(v2′ )1 = 1
(v1′ )1 = – (4.792 m/s)= − 1.597 m/s
m1 + m2
3
(v2′ ) 2 =
2m1
2
(v1′ )1 = (2.396 m/s)=3.194 m/s
m1 + m2
3
We can now use v = v′ + V to transform these back into the laboratory frame:
(v2 )1 = −1.597 m/s − 2.396 m/s = −3.99 m/s
(v2 ) 2 = 3.195 m/s − 2.396 m/s = 0.799 m/s
Having determined the velocities of the two balls after collision, we will once again use the conservation
equation Kf + U gf = Ki + U gi for each ball to solve for the θ f1 and θ f2 .
1
1
m1 (v3 )12 + m1 gL(1 − cosθ f 1 ) = m1 (v2 )12 + 0 J
2
2
Using (v3 )1 = 0, this equation simplifies to
1
1
gL(1 − cosθ f 1 ) = (−3.99 m/s) 2 ⇒ cosθ f 1 = 1 −
(−3.99 m/s) 2 ⇒ θ f 1 = 79.3°
2
2 gL
The 100 g ball rebounds to 79º. Similarly, for the other ball:
1
1
m2 (v3 ) 22 + m2 gL(1 − cosθ f 2 ) = m2 (v2 ) 22 + 0 J
2
2
Using (v3 ) 2 = 0, this equation becomes
⎛ 1 ⎞
2
cosθ f 2 = 1 − ⎜
⎟ (0.799) ⇒ θ f 2 = 14.7°
⎝ 2gL ⎠
The 200 g ball rebounds to 14.7°.
10.76. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic and
gravitational potential energy is conserved during motion.
Visualize:
Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry,
y1 = R cos φ .
Solve:
The energy conservation equation K1 + U1 = K 0 + U 0 is
1 2
1
1
mv1 + mgy1 = mv02 + mgy0 ⇒ mv12 + mgR cos φ = mgR ⇒ v1 = 2 gR (1 − cos φ )
2
2
2
G
(b) If the sled is on the hill, it is moving in a circle and the r-component of Fnet has to point to the center with
magnitude Fnet = mv 2 / R. Eventually the speed gets so large that there is not enough force to keep it in a circular
trajectory, and that is the point where it flies off the hill. Consider the sled at angle φ . Establish an r-axis
pointing toward the center of the circle, as we usually do for circular motion problems. Newton’s second law
along this axis requires:
mv 2
R
⎛
mv 2
v2 ⎞
⇒ n = mg cos φ −
= m ⎜ g cos φ − ⎟
R
R⎠
⎝
( Fnet ) r = FG cos φ − n = mg cos φ − n = mar =
The normal force decreases as v increases. But n can’t be negative, so the fastest speed at which the sled stays on
the hill is the speed vmax that makes n → 0. We can see that vmax = gR cos φ .
(c) We now know the sled’s speed at angle φ , and we know the maximum speed it can have while remaining on
the hill. The angle at which v reaches vmax is the angle φmax at which the sled will fly off the hill. Combining the
two expressions for v1 and vmax gives:
2 gR (1 − cos φ ) = gR cos φ ⇒ 2 R (1 − cos φmax ) = R cos φmax
⇒ cos φmax =
2
⎛ 2⎞
⇒ φmax = cos −1 ⎜ ⎟ = 48°
3
⎝3⎠
11.1. Visualize:
Please refer to Figure EX11.1.
G G
(a) A ⋅ B = AB cosα = (4)(5)cos 40° = 15.3.
Solve:
(b)
(c)
G G
C ⋅ D = CD cosα = (2)(4)cos120° = −4.0.
G G
E ⋅ F = EF cosα = (3)(4)cos90° = 0.
11.2. Visualize:
Please refer to Figure EX11.2.
G G
(a) A ⋅ B = AB cosα = (3)(4)cos110° = −4.1.
Solve:
(b)
(c)
G G
C ⋅ D = CD cosα = (4)(5)cos180° = −20.
G G
E ⋅ F = EF cosα = (4)(3)cos30° = 10.4.
11.3. Solve: (a)
(b)
G G
A ⋅ B = Ax Bx + Ay By = (3)( −2) + ( −4)(6) = −30.
G G
A ⋅ B = Ax Bx + Ay By = (2)(6) + (3)(−4) = 0.
11.4. Solve: (a)
(b)
G G
A ⋅ B = Ax Bx + Ay By = (4)(−3) + (−2)( −2) = −16.
G G
A ⋅ B = Ax Bx + Ay By = (−4)(−1) + (2)(−2) = 0.
G
G
G G
G
11.5. Solve: (a) The length of A is A = A = A ⋅ A = ( 3) + ( −4 ) = 25 = 5. The length of B is
B=
2
2
G G
( −2 ) + ( 6 ) = = 40 = 2 10. Using the answer A ⋅ B = −30 from Ex11.3(a),
2
2
G G
A ⋅ B = AB cosα
(
)
⇒ −30 = ( 5 ) 2 10 cosα
G G
(b) From EX11.3 (b), A ⋅ B = 0. Thus
⇒ α = cos
−1
( −0.9487 ) = 162°
cosα = 0
α = 90°
Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are
perpendicular.
G
G
G G
G
2
2
A = A = A ⋅ A = ( 4 ) + ( 2 ) = 20. The length of B is
G G
13. Using the answer A ⋅ B = −16 from EX11.4(a),
G G
A ⋅ B = AB cosα
11.6. Solve: (a) The length of A is
B=
( −3) + ( −2 ) =
2
2
⇒ −16 =
( 20 )( 13 ) cosα
⇒ α = cos −1 ( −0.9923) = 173°
G G
(b) From EX11.4(b), A ⋅ B = 0. Thus
cosα = 0
α = 90°
Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are
perpendicular.
G
G
G G
W = F ⋅ Δr = (6.0iˆ − 3.0 ˆj ) ⋅ (2.0 ˆj ) N ⋅ m = (12.0iˆ ⋅ ˆj − 6.0 ˆj ⋅ ˆj ) J = −6.0 J.
11.7. Solve: (a) W = F ⋅ Δr = (6.0iˆ − 3.0 ˆj ) ⋅ (2.0iˆ) N ⋅ m = (12.0iˆ ⋅ iˆ − 3.0 ˆj ⋅ iˆ) J = 12.0 J.
(b)
G
G
G
G G G
W = F ⋅ Δr = ( −4.0iˆ − 6.0 ˆj ) N ⋅ (−3.0iˆ + 2.0 ˆj ) m = (12.0 − 12.0) J=0 J.
11.8. Solve: (a) W = F ⋅ Δr = (−4.0iˆ − 6.0 ˆj ) N ⋅ (3.0iˆ) m = (−12.0iˆ ⋅ iˆ + 12.0 ˆj ⋅ iˆ) J= − 12.0 J.
(b)
11.9. Model: Use the work-kinetic energy theorem to find the net work done on the particle.
Visualize:
Solve:
From the work-kinetic energy theorem,
1
1
1
1
W = ΔK = mv12 − mv02 = m ( v12 − v02 ) = (0.020 kg)[(30 m/s) 2 − (−30 m/s) 2 ] = 0 J
2
2
2
2
Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is
done in bringing the particle to the original speed but in the opposite direction.
G
G
G
G
G
11.10. Model: Work done by a force F on a particle is defined as W = F ⋅ Δr , where Δr is the particle’s
displacement.
Visualize:
Solve:
(a) The work done by gravity is
G
G
Wg = FG ⋅ Δr = (− mgjˆ) N ⋅ (2.25 − 0.75) ˆj m = −(2.0 kg)(9.8 m/s 2 )(1.50 m) J = −29 J
G
G
(b) The work done by hand is WH = Fhand on book ⋅ Δr . As long as the book does not accelerate,
G
G
Fhand on book = − Fearth on book = −(− mgjˆ) = mgjˆ
⇒ WH = ( mgjˆ) ⋅ (2.25 − 0.75) ˆj m = (2.0 kg)(9.8 m/s 2 )(1.50 m)=29 J
G
G
G
11.11. Model: Model the piano as a particle and use W = F ⋅ Δr , where W is the work done by the force F
G
through the displacement Δr .
Visualize:
G
For the force FG :
G G G
G
W = F ⋅ Δr = FG ⋅ Δr = ( Fg )(Δr )cos0° = (2500 N)(5.00 m)(1) = 1.250 × 104 J
G
For the tension T1 :
G G
W = T1 ⋅ Δr = (T1 )(Δr )cos(150°) = (1830 N)(5.00 m)(−0.8660) = −7.92 × 103 J
G
For the tension T2 :
G G
W = T2 ⋅ Δr = (T2 )(Δr )cos(135°) = (1295 N)(5.00 m)(−0.7071) = −4.58 × 103 J
Solve:
Assess:
G
Note that the displacement Δr in all the above cases is directed downwards along − ˆj.
G
G
G
11.12. Model: Model the crate as a particle and use W = F ⋅ Δr , where W is the work done by a force F on
G
a particle and Δr is the particle’s displacement.
Visualize:
Solve:
G
For the force f k :
G
For the tension T1 :
G
For the tension T2 :
G G
W = f k ⋅ Δr = f k (Δr )cos(180°) = (650 N)(3.0 m)(−1) = −1.95 kJ
G G
W = T1 ⋅ Δr = (T1 )(Δr )cos 20° = (600 N)(3.0 m)(0.9397) =1.69 kJ
G G
W = T2 ⋅ Δr = (T2 )(Δr )cos30° = (410 N)(3.0 m)(0.866) = 1.07 kJ
G
Assess: Negative work done by the force of kinetic friction ( f k ) means that 1.95 kJ of energy has been
transferred out of the crate.
11.13. Model: Model the 2.0 kg object as a particle, and use the work-kinetic energy theorem.
Visualize: Please refer to Figure EX11.13. For each of the five intervals the velocity-versus-time graph gives
the initial and final velocities. The mass of the object is 2.0 kg.
Solve: According to the work-kinetic energy theorem:
1
1
1
W = ΔK = mvf2 = mvi2 = m ( vf2 − vi2 )
2
2
2
1
Interval AB: vi = 2 m/s, vf = −2 m/s ⇒ W = (2.0 kg)[(−2 m/s) 2 − (2 m/s) 2 ] = 0 J
2
1
Interval BC: vi = −2 m/s, vf = −2 m/s ⇒ W = (2.0 kg)[(−2 m/s) 2 − (−2 m/s) 2 ] = 0 J
2
1
Interval CD: vi = −2 m/s, vf = 0 m/s ⇒ W = (2.0 kg)[(0 m/s) 2 − (−2 m/s) 2 ] = −4.0 J
2
1
Interval DE: vi = 0 m/s, vf = 2 m/s ⇒ W = (2.0 kg)[(2 m/s) 2 − (0 m/s) 2 ] = +4.0 J
2
1
Interval EF: vi = 2 m/s, vf = 1 m/s ⇒ W = (2.0 kg)[(1 m/s)2 − (2 m/s) 2 ] = −3.0 J
2
Assess: The work done is zero in intervals AB and BC. In the interval CD + DE the total work done is zero. It
is not whether v is positive or negative that counts because K ∝ v 2 . What is important is the magnitude of v and
how v changes.
11.14. Model: Use the definition of work.
Visualize: Please refer to Figure EX11.14.
Solve: Work is defined as the area under the force-versus-position graph:
sf
W = ∫ Fs ds = area under the force curve
si
Interval 0–1 m: W = (4.0 N)(1.0 m − 0.0 m) = 4.0 J
Interval 1–2 m: W = (4.0 N)(0.5 m) + (− 4.0 N)(0.5 m) = 0 J
1
Interval 2–3 m: W = (−4.0 N)(1 m) = −2.0 J
2
11.15. Model: Use the work-kinetic energy theorem to find velocities.
Visualize: Please refer to Figure EX11.15.
Solve: The work-kinetic energy theorem is
x
f
1
1
ΔK = mvf2 − mvi2 = W = ∫ Fx dx = area under the force curve from xi to xf
2
2
xi
x
f
1
1
1
⇒ mvf2 − (0.500 kg)(2.0m/s) 2 = mvf2 − 1.0 J = ∫ Fx dx = area from 0 to x
2
2
2
0m
1
(0.500 kg)vf2 − 1.0 J = 12.5 J ⇒ vf = 7.35 m/s
2
1
At x = 2 m: (0.500 kg)vf2 − 1.0 J = 20 J ⇒ vf = 9.17 m/s
2
1
At x = 3 m: (0.500 kg)vf2 − 1.0 J = 22.5 J ⇒ vf = 9.70 m/s
2
At x =1 m:
11.16. Model: Use the work-kinetic energy theorem.
Visualize: Please refer to Figure EX11.16.
Solve: The work-kinetc energy theorem is
x
f
1
1
ΔK = mvf2 − mvi2 = W = ∫ Fx dx = area under the force curve from xi to xf
2
2
xi
x
f
1
1
1
⇒ mvf2 − (2.0 kg)(4.0 m/s) 2 = mvf2 − 16.0 J = ∫ Fx dx = area from 0 to x
2
2
2
0m
1
1
(2.0 kg)vf2 − 16.0 J = (10 N)(2 m) = 10 J ⇒ vf = 5.1 m/s
2
2
1
2
At x = 4 m : (2.0 kg)vf − 16.0 J = 0 J ⇒ vf = 4.0 m/s
2
At x = 2 m :
11.17. Model: Use the work-kinetic energy theorem.
Visualize: Please refer to Figure EX11.17.
Solve: The work-kinetic energy theorem is
xf
ΔK = W = ∫ Fx dx = area of the Fx -versus- x graph between xi and xf
xi
1 2 1 2 1
mvf − mvi = ( Fmax )(2 m)
2
2
2
Using m = 0.500 kg, vf = 6.0 m/s, and vi = 2.0 m/s, the above equation yields Fmax = 8.0 N.
Assess: Problems in which the force is not a constant can not be solved using constant-acceleration kinematic
equations.
11.18. Model: Use the definition Fs = − dU / ds.
Visualize: Please refer to Figure EX11.18.
Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and
x = 10 cm the slope is
slope = (U f − U i ) / ( xf − xi ) = (0 J − 10 J) / (0.10 m − 0.0 m) = −100 N
Thus, Fx = 100 N at x = 5 cm. The slope between x = 10 cm and x = 20 cm is zero, so Fx = 0 N at x = 15 cm.
Between 20 cm and 40 cm,
slope = (10 J − 0 J) / (0.40 m − 0.20 m) = 50 N
At x = 25 cm and x = 35 cm, therefore, Fx = −50 N.
11.19. Model: Use the definition Fs = − dU ds .
Visualize: Please refer to Figure EX11.19.
Solve: Fx is the negative of the slope of the potential energy graph at position x.
⎛ dU ⎞
Fx = − ⎜
⎟
⎝ dx ⎠
Between x = 0 m and x = 3 m, the slope is
slope = (U f − U i ) / ( xf − xi ) = (60 J − 0 J) / (3 m − 0 m) = 20 N
Thus, Fx = −20 N at x = 1 m. Between x = 3 m and x = 5 m, the slope is
slope = (U f − U i ) / ( xf − xi ) = ( 0 J − 60 J ) / ( 5 m − 3 m ) = −30 N
Thus, Fx = 30 N at x = 4 m.
11.20. Model: Use the negative derivative of the potential energy to determine the force acting on a particle.
Solve:
(a)
The graph of the potential energy U = 4 y 3 is shown.
(b) The y-component of the force is
Fy = −
dU
d
= − (4 y 3 ) = −12 y 2
dy
dy
At y = 0 m, Fy = 0 N; at y = 1 m, Fy = −12 N; and at y = 2 m, Fy = −48 N.
11.21. Model: Use the negative derivative of the potential energy to determine the force acting on a particle.
Solve:
(a)
The graph of the potential energy U = 10 / x is shown.
(b) The x-component of the force is
Fx = −
Fx = 2 m =
dU
d ⎛ 10 ⎞ 10
=− ⎜
⎟=
dx
dx ⎝ x ⎠ x 2
10
= 2.5 N
x2 x=2 m
Fx = 5 m =
10
= 0.40 N
x2 x =5 m
Fx =8 m =
10
= 0.156 N
x 2 x =8 m
11.22. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hooke’s law.
Visualize:
The quantity ( x − xe ) is the stretching relative to the spring’s equilibrium length. In the present case, bond
stretching is analogous to spring stretching.
Solve: (a) The kinetic energy of the carbon atom is
1
1
K = mv 2 = (2.0 × 10−26 kg)(500 m/s) 2 = 2.5 × 10−21 J
2
2
(b) The energy of the spring is given by
1
U s = k ( x − xe ) 2 = K
2
2K
2(2.5 × 10−21 J)
N
⇒k =
=
= 2.0
2
( x − xe )
(0.050 × 10−9 m) 2
m
11.23. Visualize: One mole of helium atoms in the gas phase contains N A = 6.02 × 1023 atoms.
Solve:
If each atom moves with the same speed v, the microscopic total kinetic energy will be
2 K micro
2(3700 J)
⎛1
⎞
K micro = N A ⎜ mv 2 ⎟ = 3700 J ⇒ v =
=
= 1360 m/s
−27
mN
2
(6.68
10
kg)(6.02 × 1023 )
×
⎝
⎠
A
11.24. Solve: (a) The car has an initial kinetic energy K i . That energy is transformed into thermal energy of
the car and the road surface. The gravitational potential energy does not change and no work is done by external
forces, so during the skid K i transforms entirely into thermal energy Eth . This energy transfer and
transformation is shown on the energy bar chart. Note that
1
1
K i = mv 2 = (1500 kg)(20 m/s) 2 = 3.0 × 105 J
2
2
(b) The change in the thermal energy of the car and the road surface is 3.0 × 105 J.
11.25. Visualize:
1
K i = K 0 = mv02 = 0 J U i = U g0 = mgy0 = (20 kg)(9.8 m/s 2 )(3.0 m) = 5.9 × 102 J
2
1
1
Wext = 0 J K f = K1 = mv12 = (20 kg)(2.0 m/s) 2 = 40 J U f = U g1 = mgy1 = 0 J
2
2
At the top of the slide, the child has gravitational potential energy of 5.9 × 102 J. This energy is transformed into
thermal energy of the child’s pants and the slide and the kinetic energy of the child. This energy transfer and
transformation is shown on the energy bar chart.
(b)
Solve:
(a)
The change in the thermal energy of the slide and of the child’s pants is 5.9 × 10 2 J − 40 J = 5.5 × 10 2 J.
11.26. Visualize: The system loses 400 J of potential energy. In the process of losing this energy, it does
400 J of work on the environment, which means Wext = −400 J. Since the thermal energy increases 100 J, we
have ΔEth = 100 J, which must have been 100 J of kinetic energy originally. This is shown in the energy bar
chart.
11.27. Visualize:
Note that the conservation of energy equation
K i + U i + Wext = K f + U f + ΔEth
requires that Wext be equal to +400 J.
11.28. Solve: Please refer to Figure EX11.28. The energy conservation equation yields
K i + U i + Wext = K f + U f + ΔEth ⇒ 4 J + 1 J + Wext = 1 J + 2 J + 1 J ⇒ Wext = −1 J
Thus, the work done to the environment is −1 J. In other words, 1 J of energy is transferred from the system into
the environment. This is shown in the energy bar chart.
11.29. Visualize: The tension of 20.0 N in the cable is an external force that does work on the block Wext =
(20.0 N)(2.00 m) = 40.0 J, increasing the gravitational potential energy of the block. We placed the origin of our
coordinate system on the initial resting position of the block, so we have U i = 0 J and U f = mgyf =
(1.02 kg)(9.8 m/s 2 )(2.00 m) = 20.0 J. Also, K i = 0 J, and ΔEth = 0 J. The energy bar chart shows the energy
transfers and transformations.
Solve:
The conservation of energy equation is
1
K i + U i + Wext = K f + U f + ΔEth ⇒ 0 J + 0 J + 40.0 J = mvf2 + 20.0 J + 0 J
2
⇒ vf = (20.0 J)(2) /1.02 kg = 6.26 m/s
11.30. Model: Model the elevator as a particle, and apply the conservation of energy equation.
Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we
say that the motor does the work of lifting the elevator.
(a) The energy conservation equation is K i + U i + Wext = K f + U f + ΔEth . Using K i = 0 J, K f = 0 J, and
ΔEth = 0 J gives
Wext = (U f − U i ) = mg ( yf − yi ) = (1000 kg)(9.8 m/s 2 )(100 m) = 9.80 × 105 J
(b) The power required to give the elevator this much energy in a time of 50 s is
P=
Wext 9.8 × 105 J
=
= 1.96 × 104 W
Δt
50 s
Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of
power to lift a mass of 1000 kg to a height of 100 m in 50 s.
11.31. Model: Model the steel block as a particle subject to the force of kinetic friction and use the energy
conservation equation.
Visualize:
G
G
G
Solve: (a) The work done on the block is Wnet = Fnet ⋅ Δr where Δr is the displacement. We will find the
displacement using kinematic equations and the force using Newton’s second law of motion. The displacement
in the x-direction is
1
Δx = x1 = x0 + v0 x (t1 − t0 ) + ax (t1 − t0 ) 2 = 0 m + (1.0 m/s)(3.0 s − 0 s) + 0 m = 3.0 m
2
G
Thus Δr = 3.0iˆ m.
The equations for Newton’s second law along the x and y components are
( Fnet ) y = n − FG = 0 N ⇒ n = FG = mg = (10 kg)(9.8 m/s2 ) = 98.0 N
G
G
( Fnet ) x = F − f k = 0 N ⇒ F = f k = μ k n = (0.6)(98.0 N) = 58.8 N
G
G
⇒ Wnet = Fnet ⋅ Δr = F Δ x cos0° = (58.8 N)(3.0 m)(1) = 176 J
(b) The power required to do this much work in 3.0 s is
P=
W 176 J
=
= 59 W
3.0 s
t
11.32. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of
the solar energy striking the earth is the power divided by area. We have
ΔE 150 × 106 J
W
=
= 41,667 W and intensity = 1000 2
Δt
3600 s
m
41,667 W
2
⇒ Area of solar collector =
= 41.7 m
1000 W/m 2
P=
11.33. Solve: The night light consumes more energy than the hair dryer. The calculations are
1.2 kW × 10 min = 1.2 × 103 × 10 × 60 J = 7.2 × 105 J
10 W × 24 hours = 10 × 24 × 60 × 60 J = 8.64 × 105 J
11.34. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy.
(b)
One kilowatt hour is energy
1 kwh = (1000 J/s)(3600 s) = 3.6 × 106 J
Thus
⎛ 3.6 × 106 J ⎞
9
500 kwh = (500 kwh) ⎜
⎟ = 1.8 × 10 J
⎝ 1 kwh ⎠
11.35. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and the
definition of power in terms of velocity.
Visualize:
Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’s
second law. We have
1
1
x = x0 + v0 x (t1 − t0 ) + ax (t1 − t0 ) 2 ⇒ 50 m = 0 m + 0 m + ax (7.0 s − 0 s) 2 ⇒ ax = 2.04 m/s 2
2
2
2
⇒ Fx = max = (50 kg)(2.04 m/s ) = 102 N
G G
G
(b) We obtain the sprinter’s power output by using P = F ⋅ v , where v is the sprinter’s velocity. At t = 2.0 s the
power is
P = ( Fx )[v0 x + ax (t − t0 )] = (102 N)[0 m/s + (2.04 m/s 2 )(2.0 s − 0 s)] = 0.42 kW
The power at t = 4.0 s is 0.83 kW, and at t = 6.0 s the power is 1.25 kW.
G
G
G
G
11.36. Model: Use the definition of work for a constant force F , W = F ⋅ Δs , where Δs is the
displacement.
G
Visualize: Please refer to Figure P11.36. The force F = (6iˆ + 8 ˆj ) N on the particle is a constant.
G
G
G
G
Solve: (a) WABD = WAB + WBD = F ⋅ (Δs ) AB + F ⋅ (Δs ) BD
= (6iˆ + 8ˆ j ) N ⋅ (3iˆ) m + (6iˆ + 8 ˆj ) N ⋅ (4 ˆj ) m = 18 J + 32 J = 50 J
G
G
G
G
(b) WACD = WAC + WCD = F ⋅ (Δs ) AC + F ⋅ (Δs )CD
= (6iˆ + 8ˆ j ) N ⋅ (4 ˆj ) m + (6iˆ + 8 ˆj ) N ⋅ (3 ˆj ) m = 32 J + 18 J = 50 J
G
G
(c) WAD = F ⋅ (Δs ) AD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ + 4 ˆj ) m = 18 J + 32 J = 50 J
The force is conservative because the work done is independent of the path.
11.37. Model:
The force is conservative, so it has a potential energy.
Visualize: Please refer to Figure P11.37 for the graph of the force.
Solve: The definition of potential energy is ΔU = −W (i → f ). In addition, work is the area under the force-versus-
displacement graph. Thus ΔU = U f − U i = − (area under the force curve). Since U i = 0 at x = 0 m, the potential
energy at position x is U ( x) = − (area under the force curve from 0 to x). From 0 m to 3 m, the area increases
linearly from 0 N m to −60 N m, so U increases from 0 J to 60 J. At x = 4 m, the area is −70 J. Thus U = 70 J at
x = 4 m, and U doesn’t change after that since the force is then zero. Between 3 m and 4 m, where F changes
linearly, U must have a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information
is shown on the potential energy graph below.
(b) Mechanical energy is E = K + U . From the graph, U = 20 J at x = 1.0 m.
The kinetic energy is K = 12 mv 2 = 12 (0.100 kg) (25 m/s) 2 = 31.25 J. Thus E = 51.25 J.
(c) The total energy line at 51.25 J is shown on the graph above.
(d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from the
graph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J.
The TE line crosses at the point where 20 x = 51.25, which is x = 2.56 m.
11.38. Model: Use the relationship between force and potential energy and the work-kinetic energy theorem.
Visualize:
Please refer to Figure P11.38. We will find the slope in the following x regions: 0 cm < x < 1 cm, 1 < x < 3 cm,
3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm.
Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m
dU
−4 J
=
= −400 N ⇒ Fx = +400 N
dx 0.01 m
Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown.
(b) Since W = ∫ xxif Fx dx = area of the Fx-versus-x graph between xi and xf, the work done by the force as the
particle moves from xi = 2 cm to xf = 6 cm is −2 J.
(c) The conservation of energy equation is K f + U f = Ki + U i . We can see from the graph that U i = 0 J and
U f = 2 J in moving from x = 2 cm to x = 6 cm. The final speed is vf = 10 m/s, so
2 J + 12 (0.010 kg)(10.0 m/s) 2 = 0 J + 12 (0.010 kg)vi 2 ⇒ vi = 22.4 m/s
11.39. Model: Use the relationship between a conservative force and potential energy.
Visualize: Please refer to Figure P11.39. We will obtain U as a function of x and Fx as a function of x by using
the calculus techniques of integration and differentiation.
Solve: (a) For the interval 0 m < x < 0.5 m, Fx = (4 x) N, where x is in meters. This means
dU
= − Fx = −4 x ⇒ U = −2 x 2 + C1 = −2 x 2
dx
where we have used U = 0 J at x = 0 m to obtain C1 = 0. For the interval 0.5 m < x < 1 m, Fx = ( −4 x + 4) N.
Likewise,
dU
= 4 x − 4 ⇒ U = 2 x 2 − 4 x + C2
dx
Since U should be continuous at the junction, we have the continuity condition
(−2 x 2 ) x = 0.5 m = (2 x 2 − 4 x + C2 ) x = 0.5 m ⇒ −0.5 = 0.5 − 2 + C2 ⇒ C2 = 1
U remains constant for x ≥ 1 m.
(b) For the interval 0 m < x < 0.5 m, U = +4 x, and for the interval 0.5 m < x < 1.0 m, U = −4 x + 4, where x is
in meters. The derivatives give Fx = −4 N and Fx = +4 N, respectively. The slope is zero for x ≥ 1 m.
dvx
, x = ∫ vx dt , K = 12 mvx2 , and F = max .
dt
Visualize: Please refer to Figure P11.40. We know ax = slope of the vx-versus-t graph and x = area under the
vx-versus-x graph between 0 and x.
Solve: Using the above definitions and methodology, we can generate the following table:
t(s)
ax (m/s2)
x(m)
K(J)
F(N)
11.40. Model: Use ax =
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
10
10
10
10
+10 or −10
−10
−10 or 0
0
0
0
1.25
5
11.25
20
28.75
0
6.25
25
56.25
100
56.25
35
40
45
25
25
25
5
5
5
5
−5 or +5
−5
−5 or 0
0
0
(e) Let J1 be the impulse from t = 0 s to t = 2 s and J 2 be the impulse from t = 2 s to t = 4 s. We have
2s
4s
0s
0s
J1 = ∫ Fx dt = (5 N)(2 s) = 10 N ⋅ s and J 2 = ∫ Fx dt = (−5 N)(1 s) = −5 N ⋅ s
(f) J = Δp = mvf − mvi ⇒ vf = vi + J / m
At t = 2 s, vx = 0 m/s + (10 N ⋅ s) / 0.5 kg = 0 m/s + 20 m/s = 20 m/s
At t = 4 s, vx = 20 m/s + ( −5 N ⋅ s) / 0.5 kg = 20 m/s − 10 m/s = 10 m/s
The vx-versus-t graph also gives vx = 20 m/s at t = 2 s and vx = 10 m/s at t = 4 s.
(g)
(h) From t = 0 s to t = 2 s, W = ∫ Fx dx = (5 N)(20 m) = 100 J
From t = 2 s to t = 4 s, W = ∫ Fx dx = (−5 N)(15 m) = −75 J
(i) At t = 0 s, vx = 0 m/s so the work-kinetic energy theorem for calculating vx at t = 2 s is
1
1
1
1
W = ΔK = mvf2 − mvi2 ⇒ 100 J = (0.5 kg)vx2 − (0.5 kg)(0 m/s) 2 ⇒ vx = 20 m/s
2
2
2
2
To calculate vx at t = 4 s, we use vx at t = 2 s as the initial velocity:
1
1
−75 J = (0.5 kg)vx2 − (0.5 kg)(20 m/s) 2 ⇒ vx = 10 m/s
2
2
Both of these values agree with the values on the velocity graph.
11.41. Model: Model the elevator as a particle.
Visualize:
Solve:
(a) The work done by gravity on the elevator is
Wg = −ΔU g = mgy0 − mgy1 = − mg ( y1 − y0 ) = −(1000 kg)(9.8 m/s 2 )(10 m) = −9.8 × 104 J
(b) The work done by the tension in the cable on the elevator is
WT = T (Δy )cos0° = T ( y1 − y0 ) = T (10 m)
To find T we write Newton’s second law for the elevator:
∑ F = T − F = ma ⇒ T = F + ma = m( g + a ) = (1000 kg)(9.8 m/s + 1.0 m/s )
2
y
G
y
G
y
y
= 1.08 × 104 N ⇒ WT = (1.08 × 10 4 N)(10 m) = 1.08 × 105 J
(c) The work-kinetic energy theorem is
1
1
Wnet = Wg + WT = ΔK = K f − K i = K f − mv02 ⇒ K f = Wg + WT + mv02
2
2
1
⇒ K f = ( −9.8 × 104 J) + (1.08 × 105 J) + (1000 kg)(0 m/s) 2 = 1.0 × 10 4 J
2
(d)
1
1
K f = mvf2 ⇒ 1.0 × 104 J = (1000 kg)v 2f ⇒ vf = 4.5 m/s
2
2
2
11.42. Model: Model the rock as a particle, and apply the work-kinetic energy theorem.
Visualize:
Solve:
(a) The work done by Bob on the rock is
1
1
1
1
WBob = ΔK = mv12 − mv02 = mv12 = (0.500 kg)(30 m / s) 2 = 225 J = 2.3 × 102 J
2
2
2
2
(b) For a constant force, WBob = FBob Δx ⇒ FBob = WBob / Δx = 2.3 × 102 N.
(c) Bob’s power output is PBob = FBob vrock and will be a maximum when the rock has maximum speed. This is just
as he releases the rock with vrock = v1 = 30 m/s. Thus, Pmax = FBob v1 = ( 225 J )( 30 m/s ) = 6750 W = 6.8 kW.
11.43. Model: Model the crate as a particle, and use the work-kinetic energy theorem.
Visualize:
Solve:
(a) The work-kinetic energy theorem is ΔK = 12 mv12 − 12 mv02 = 12 mv12 = Wtotal . Three forces act on the box,
so Wtotal = Wgrav + Wn + Wpush . The normal force is perpendicular to the motion, so Wn = 0 J. The other two forces
do the following amount of work:
G
G
Wpush = Fpush ⋅ Δr = Fpush Δx cos 20° = 137.4 J
G
G
Wgrav = FG ⋅ Δr = ( FG ) x Δ x = (− mg sin 20°)Δ x = −98.0 J
Thus, Wtotal = 39.4 J, leading to a speed at the top of the ramp equal to
v1 =
2Wtotal
2(39.4 J)
=
= 4.0 m/s
m
5.0 kg
(b) The x-component of Newton’s second law is
ax = a =
( Fnet ) x Fpush cos 20° − FG sin 20° Fpush cos 20° − mg sin 20°
=
=
= 1.35 m/s 2
m
m
m
Constant-acceleration kinematics with x1 = h / sin 20° = 5.85 m gives the final speed
v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(1.35 m / s 2 )(5.85 m) = 4.0 m/s
11.44. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy.
Visualize:
Solve:
(a) The conservation of energy equation is
K1 + U g1 + Δ Eth = K 0 + U g0 + Wext
The snow is frictionless, so ΔEth = 0 J. However, the wind is an external force doing work on Sam as he moves
down the hill. Thus,
Wext = Wwind = ( K1 + U g1 ) − ( K 0 + U g0 )
1
⎛1
⎞ ⎛1
⎞ ⎛1
⎞
= ⎜ mv12 + mgy1 ⎟ − ⎜ mv02 + mgy0 ⎟ = ⎜ mv12 + 0 J ⎟ − (0 J + mgy0 ) = mv12 − mgy0
2
⎝2
⎠ ⎝2
⎠ ⎝2
⎠
⇒ v1 = 2 gy0 +
2Wwind
m
We compute the work done by the wind as follows:
G
G
Wwind = Fwind ⋅ Δr = Fwind Δr cos160° = (200 N)(146 m)cos160° = −27,400 J
where we have used Δr = h / sin 20° = 146 m. Now we can compute
v1 = 2(9.8 m / s 2 )(50 m) +
2( −27,400 J)
= 15.7 m / s
75 kg
(b) We will use a tilted coordinate system, with the x-axis parallel to the slope. Newton’s second law for Sam is
( Fnet ) x FG sin 20° − Fwind cos 20° mg sin 20° − Fwind cos 20°
=
=
m
m
m
(75 kg)(9.8 m/s 2 )sin 20° − (200 N)cos 20°
=
= 0.846 m/s 2
75 kg
ax = a =
Now we can use constant-acceleration kinematics as follows:
v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(0.846 m/s 2 )(146 m) = 15.7 m/s
Assess:
We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because U g is
determined by its vertical position.
11.45. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kinetic
friction, and apply the work-kinetic energy theorem.
Visualize:
We define the x-axis along the floor and the y-axis perpendicular to the floor.
fk .
Newton’s
second
Solve: We
need
to
first
determine
law
in
the
y-direction
is
n + T sin 30° = FG = mg ⇒ n = mg − T sin 30° = (10 kg)(9.8 m/s ) − (30 N)(sin 30°) = 83.0 N. Using n and the model of
2
kinetic friction,
f k = μ k n = (0.2)(83.0 N) = 16.60 N. The net force on Paul and the mat is therefore
Fnet = T cos30° − f k = ( 30 N ) cos30° − 16.6 N = 9.4 N . Thus,
Wnet = Fnet Δr = (9.4 N)(3.0 m) = 28 J
G
G
G
The other forces n and FG make an angle of 90° with Δr and do zero work. We can now use the work-kinetic
energy theorem to find the final velocity as follows:
1
Wnet = K f − K i = K f − 0 J = mvf2 ⇒ vf = 2Wnet / m = 2(28 J)/10 kg = 2.4 m/s
2
Assess:
A speed of 2.4 m/s or 5.4 mph is reasonable for the present problem.
11.46. Model: Assume an ideal spring that obeys Hooke’s law. Model the box as a particle and use the
model of kinetic friction.
Visualize:
Solve:
When the horizontal surface is frictionless, conservation of energy means
1
1
1
k ( x0 − xe ) 2 = mv12x = K1 ⇒ K1 = (100 N/m)(0.20 m − 0 m) 2 = 2.0 J
2
2
2
That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface.
G
G
The net force on the box is Fnet = − f k = − μ k mgiˆ. The work-kinetic energy theorem is
G
G
Wnet = Fnet ⋅ Δr = K 2 − K1 = 0 J − 2.0 J = −2.0 J
⇒ (− μ k mg )( x2 − x1 ) = −2.0 J
⇒ ( x2 − x1 ) =
2.0 J
2.0 J
=
= 54 cm
( μ k )( mg ) (0.15)(2.5 kg)(9.8 m/s 2 )
Assess: Because the force of friction transforms kinetic energy into thermal energy, energy is transferred out of
the box into the environment. In response, the box slows down and comes to rest.
11.47. Model:
energy theorem.
Visualize:
Model the suitcase as a particle, use the model of kinetic friction, and use the work-kinetic
G
G
The net force on the suitcase is Fnet = f k .
Solve: The work-kinetic energy theorem is
G
1
1
1
1
G G G
Wnet = ΔK = mv12 − mv02 ⇒ Fnet ⋅ Δr = f k ⋅ Δr = 0 J − mv02 ⇒ ( f k )( x1 − x0 )cos180° = − mv02
2
2
2
2
1 2
v02
(1.2 m/s) 2
⇒ − μ k mg ( x1 − x0 ) = − mv0 ⇒ μ k =
=
= 0.037
2
2 g ( x1 − x0 ) 2(9.8 m/s 2 )(2.0 m − 0 m)
Assess: Friction transforms kinetic energy of the suitcase into thermal energy. In response, the suitcase slows
down and comes to rest.
11.48. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system
because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic
friction and the conservation of energy equation.
Visualize:
We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the
ramp and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the
truck is shown.
Solve: The conservation of energy equation is K1 + U g1 + Δ Eth = K 0 + U g0 + Wext . In the present case,
Wext = 0 J, v1x = 0 m/s, U g 0 = 0 J, v0 x = 35 m/s. The thermal energy created by friction is
ΔEth = ( f k )( x1 − x0 ) = ( μ k n)( x1 − x0 ) = μ k ( mg cos6.0°)( x1 − x0 )
= (0.40)(15,000 kg)(9.8 m/s 2 )(cos6.0°)( x1 − x0 ) = (58,478 J/m)( x1 − x0 )
Thus, the energy conservation equation simplifies to
1
0 J + mgy1 + (58,478 J/m)( x1 − x0 ) = mv02x + 0 J + 0 J
2
1
(15,000 kg)(9.8 m/s 2 )[( x1 − x0 )sin 6.0°] + (58,478 J/m)( x1 − x0 ) = (15,000 kg)(35 m/s) 2
2
⇒ ( x1 − x0 ) = 124 m
Assess:
A length of 124 m at a slope of 6º seems reasonable.
11.49. Model: We will use the spring, the package, and the ramp as the system. We will model the package
as a particle.
Visualize:
We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact
with the package to be shot.
Model: (a) The energy conservation equation is
K1 + U g1 + U s1 + Δ Eth = K 0 + U g 0 + U s0 + Wext
1 2
1
1
1
mv1 + mgy1 + k ( xe − xe ) 2 + Δ Eth = mv02 + mgy0 + k (Δx) 2 + Wext
2
2
2
2
Using y1 = 1 m, ΔEth = 0 J (note the frictionless ramp), v0 = 0 m/s, y0 = 0 m, Δx = 30 cm, and Wext = 0 J, we
get
1 2
1
mv1 + mg (1 m) + 0 J + 0 J = 0 J + 0 J + k (0.30 m)2 + 0 J
2
2
1
1
2
2
(2.0 kg)v1 + (2.0 kg)(9.8 m/s )(1 m) = (500 N/m)(0.30 m) 2
2
2
⇒ v1 = 1.70 m/s
(b) How high can the package go after crossing the sticky spot? If the package can reach y1 ≥ 1.0 m before
stopping (v1 = 0), then it makes it. But if y1 < 1.0 m when v1 = 0, it does not. The friction of the sticky spot
generates thermal energy
ΔEth = ( μ k mg )Δx = (0.30)(2.0 kg)(9.8 m/s 2 )(0.50 m) = 2.94 J
The energy conservation equation is now
1
2
mv12 + mgy1 + ΔEth = 12 k ( Δx) 2
If we set v1 = 0 m/s to find the highest point the package can reach, we get
y1 = ( 12 k (Δx) 2 − ΔEth ) / mg = ( 12 (500 N/m)(0.30 m) 2 − 2.94 J ) /(2.0 kg)(9.8 m/s 2 ) = 0.998 m
The package does not make it. It just barely misses.
11.50. Model: Model the two blocks as particles. The two blocks make our system.
Visualize:
We place the origin of our coordinate system at the location of the 3.0 kg block.
Solve: (a) The conservation of energy equation is K f + U gf + ΔEth = K i + U gi + Wext . Using ΔEth = 0 J and
Wext = 0 J we get
1
1
1
1
m2 (vf ) 22 + m3 (vf )32 + m2 g ( yf ) = m2 (vi ) 22 + m3 (vi )32 + m2 g ( yi )
2
2
2
2
Noting that (vf ) 2 = (vf )3 = vf and (vi ) 2 = (vi )3 = 0 m / s, this becomes
1
( m2 + m3 )vf2 = −m2 g ( yf − yi )
2
vf =
2m2 g ( yi − yf )
2(2.0 kg)(9.8 m/s 2 )(1.50 m)
=
= 3.4 m/s
m2 + m3
(2.0 kg + 3.0 kg)
(b) We will use the same energy conservation equation. However, this time
ΔEth = μ k (m3 g )(Δ x) = (0.15)(3.0 kg)(9.8 m/s 2 )(1.50 m) = 6.615 J
The energy conservation equation is now
1
1
1
1
m2vf2 + m3vf2 + m2 gyf + 6.615 J = m2 (vi ) 22 + m3 (v1 )32 + m2 gyi + 0 J
2
2
2
2
⎛
1
2 ⎞
(m2 + m3 )vf2 + 6.615 J = m2 g ( yi − yf ) ⇒ vf = ⎜
⎟[ m2 g ( yi − yf ) − 6.615 J]
+
2
m
⎝ 2 m3 ⎠
⎛ 2 ⎞
2
= ⎜
⎟[(2.0 kg)(9.8 m/s )(1.50 m) − 6.615 J] = 3.0 m/s
⎝ 5.0 kg ⎠
Assess:
A reduced speed when friction is present compared to when there is no friction is reasonable.
G
G
11.51. Model: Use the particle model, the definition of work W = F ⋅ Δs , and the model of kinetic friction.
Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline.
Solve:
(a)
G G
WT = T ⋅ Δr = (T )(Δx)cos18° = (120 N)(5.0 m)cos18° = 0.57 kJ
G
G
Wg = ( FG ⋅ Δr ) = mg (Δx)cos(120°) = (8 kg)(9.8 m/s 2 )(5.0 m)cos120° = −196 J
G G
Wn = (n ⋅ Δr ) = (n)(Δ x)cos90° = 0 J
(b) The amount of energy transformed into thermal energy is ΔEth = ( f k )(Δx) = ( μ k n)(Δx).
To find n, we write Newton’s second law as follows:
∑ F = n − F cos30° + T sin18° = 0 N ⇒ n = F cos30° − T sin18°
y
G
G
⇒ n = mg cos30° − T sin18° = (8.0 kg)(9.8 m/s )cos30° − (120 N)sin18° = 30.814 N
2
Thus, ΔEth = (0.25)(30.814 N)(5.0 m) = 38.5 J.
Assess: Any force that acts perpendicular to the displacement does no work.
11.52. Model: Assume the spring is ideal so that Hooke’s law is obeyed, and model the weather rocket as a
particle.
Visualize:
The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring is
anchored to the ground.
Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newton’s second
law as
( ∑ F ) = 0 N ⇒ F = F = mg ⇒ k Δy = mg
y
sp
G
mg (10.2 kg)(9.8 m/s 2 )
⇒ Δy =
=
= 20.0 cm
k
(500 N/m)
(b) The thrust does work. Using the energy conservation equation when y2 − y0 = 40 cm = 0.40 m:
K 2 + U g 2 + U sp 2 = K1 + U g1 + U sp1 + Wext
1
1
1
1
Wext = F ( y2 − y1 ) ⋅ mv22 + mgy2 + k ( y2 − ye ) 2 = mv12 + mgy1 + k ( y1 − ye ) 2 + (200 N)(0.60 m)
2
2
2
2
2
(5.10 kg)v2 + 40.0 J + 40.0 J = 0 − 20.0 J + 10.0 J + 120 J ⇒ v2 = 2.43 m/s
If the rocket were not attached to the spring, the energy conservation equation would not involve the spring
energy term U sp2 . That is,
K 2 + U g 2 = K1 + U g1 + U sp1 + Wext
1
(10.2 kg)v22 + (10.2 kg)(9.8 m/s 2 )(0.40 m) = 0 J − (10.2 kg)(9.8 m/s 2 )(0.20 m)
2
1
+ (500 N/m)(0.20 m) 2 + (200 N)(0.60 m)
2
⇒ (5.10 kg)v22 = 70.0 J ⇒ v2 = 3.70 m/s
11.53. Model: Use the particle model for the ice skater, the model of kinetic/static friction, and the workkinetic energy theorem.
Visualize:
Solve:
(a) The work-kinetic energy theorem is
1
1
ΔK = mv12 − mv02 = Wnet = Wwind
2
2
There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slipping
sideways on the ice, but this force is perpendicular to the motion and does not contribute to a change in thermal
G
G
energy. The angle between Fwind and Δr is θ = 135°, so
G
G
Wwind = Fwind ⋅ Δr = Fwind Δy cos135° = (4 N)(100 m)cos135° = −282.8 J
Thus, her final speed is
v1 = v02 +
2Wwind
= 2.16 m/s
m
(b) If the skates don’t slip, she has no acceleration in the x-direction and so ( Fnet ) x = 0 N. That is:
f s − Fwind cos 45° = 0 N ⇒ fs = Fwind cos 45° = 2.83 N
Now there is an upper limit to the static friction: f s ≤ ( fs ) max = μ s mg . To not slip requires
μs ≥
fs
2.83 N
=
= 0.0058
mg (50 kg)(9.8 m/s 2 )
Thus, the minimum value of μs is 0.0058.
Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down.
11.54. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hooke’s law, and the law of
conservation of energy.
Visualize:
Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved.
We’ve drawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure
distance along the slope. Both have the same origin which is at the point where the spring is not compressed.
Thus, the two axes are related by y = s sin θ . Also, this choice of origin makes the elastic potential energy simply
U s = 12 k ( s − s0 ) 2 = 12 ks 2 . Because energy is conserved, we can relate the initial point—with the spring
compressed—to the final point where the ice cube is at maximum height. We do not need to find the speed with
which it leaves the spring. We have
K 2 + U g 2 + U s2 = K1 + U g1 + U s1
1 2
1
1
1
mv2 + mgy2 + ks02 = mv12 + mgy1 + ks12
2
2
2
2
It is important to note that at the final point, when the ice cube is at y2, the end of the spring is only at s0. The
spring does not stretch to s2 , so U s2 is not 12 ks22 . Three of the terms are zero, leaving
1
ks 2
mgy2 = + mgy1 + ks12 ⇒ y2 − y1 = Δy = height gained = 1 = 0.255 m = 25.5 cm
2
2mg
The distance traveled is Δs = Δy / sin 30° = 51.0 cm.
(b) Using the energy equation and the expression for thermal energy:
K 2 + U g 2 + U s2 + ΔEth = K1 + U g1 + U s1 + Wext
ΔEth = f k Δs = μ k nΔs
From the free-body diagram,
G
( Fnet ) y = 0 N = n − mg cos30° ⇒ n = mg cos30°
Now, having found ΔEth = μ k (mg cos30°) Δs, the energy equation can be written
1
0 J + mgy2 + 0 J + μ k (mg cos30°)Δs = 0 J + mgy1 + ks12 + 0 J
2
1 2
⇒ mg ( y2 − y1 ) − ks1 + μ k mg cos30°Δs = 0
2
Using Δy = ( Δs )sin 30°, the above equation simplifies to
1
ks12
mg Δs sin 30° + μ k mg cos30°Δs = ks12 ⇒ Δs =
= 0.379 m = 37.9 cm
2
2mg (sin 30° + μ k cos30°)
11.55. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the box as a particle and apply the
energy conservation law. Box, spring, and the ground make our system, and we also use the model of kinetic
friction.
Visualize: We place the origin of the coordinate system on the ground directly below the box’s starting
position.
Solve:
(a) The energy conservation equation is
K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext
1 2
1
1
mv1 + mgy1 + 0 J + 0 J = mv02 + mgy0 + 0 J + 0 J ⇒ mv12 + 0 J = 0 J + mgy0
2
2
2
⇒ v1 = 2 gy0 = 2(9.8m/s 2 )(5.0 m) = 9.9 m/s
(b) The friction creates thermal energy. The energy conservation equation for this part of the problem is
1 2
1
mv2 + 0 J + 0 J + μ k mg ( x2 − x1 ) = mv12 + 0 J + 0 J + 0 J
2
2
1 2
1 2
1 2
1 2
mv2 + μ k n( x2 − x1 ) = mv1 ⇒ mv2 + μ k mg ( x2 − x1 ) = mv1
2
2
2
2
K 2 + U g 2 + U s 2 + ΔEth = K1 + U g1 + U s1 + Wext
⇒ v2 = v12 − 2μ k g ( x2 − x1 ) = (9.9 m/s) 2 − 2(0.25)(9.8 m/s 2 )(2.0 m) = 9.4 m/s
(c) To find how much the spring is compressed, we apply the energy conservation once again:
K 3 + U g3 + U s3 + ΔEth = K 2 + U g 2 + U s2 + Wext
1
1
0 J + 0 J + k ( x3 − x2 ) 2 + 0 J = mv22 + 0 J + 0 J + 0 J
2
2
Using v2 = 9.4 m/s, k = 500 N/m and m = 5.0 kg, the above equation yields ( x3 − x2 ) = Δx = 94 cm.
(d) The initial energy = mgy0 = (5.0 kg)(9.8 m / s 2 )(5.0 m) = 254 J. The energy transformed to thermal energy
during each passage is
μ k mg ( x2 − x1 ) = (0.25)(5.0 kg)(9.8 m / s 2 )(2.0 m) = 24.5 J
The number of passages is equal to 245 J / 24.5 J or 10.
11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and
apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We
also use the model of kinetic friction.
Visualize: We place the origin of the coordinate system on the ground directly below the end of the
compressed spring that is in contact with the student.
Solve:
(a) The energy conservation equation is
K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext
1 2
1
1
1
mv1 + mgy1 + k ( x1 − xe ) 2 + 0 J = mv02 + mgy0 + k ( x1 − x0 ) 2 + 0 J
2
2
2
2
Since y1 = y0 = 10 m, x1 = xe , v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and ( x1 − x0 ) = 0.5 m,
1 2 1
k
mv1 = k ( x1 − x0 ) 2 ⇒ v1 =
( x1 − x0 ) = 14.14 m/s
2
2
m
(b) Friction creates thermal energy. Applying the conservation of energy equation once again:
K 2 + U g 2 + U s2 + ΔEth = K 0 + U g 0 + U s0 + Wext
1 2
1
mv2 + mgy2 + 0 J + f k Δs = 0 J + mgy0 + k ( x1 − x0 ) 2 + 0 J
2
2
With v2 = 0 m/s and y2 = (Δs )sin 30°, the above equation is simplified to
1
mg (Δs )sin 30° + μ k nΔs = mgy0 + k ( x1 − x0 ) 2
2
From the free-body diagram for the physics student, we see that n = FG cos30° = mg cos30°. Thus, the
conservation of energy equation gives
1
Δs (mg sin 30° + μ k mg cos30°) = mgy0 + k ( x1 − x0 ) 2
2
Using m = 100 kg, k = 80,000 N/m, ( x1 − x0 ) = 0.50 m, y0 = 10 m, and μ k = 0.15, we get
1
mgy0 + k ( x1 − x0 ) 2
2
Δs =
= 32.1 m
mg (sin 30° + μ k cos30°)
Assess:
y2 = (Δs )sin 30° = 16.05 m, which is greater than y0 = 10 m. The higher value is due to the
transformation of the spring energy into gravitational potential energy.
11.57. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy
conservation law. The block and the incline comprise our system.
Visualize: We place the origin of the coordinate system directly below the block’s starting position at the same
level as the horizontal surface. On the horizontal surface the model of kinetic friction applies.
Solve:
(a) For the first incline, the conservation of energy equation gives
1 2
K1 + U g1 + ΔEth = K 0 + U g0 + Wext
mv1 + 0 J + 0 J = 0 J + mgy0 + 0 J ⇒ v1 = 2 gy0 = 2 gh
2
(b) The friction creates thermal energy. Applying once again the conservation of energy equation, we have
K 3 + U g3 + ΔEth = K1 + U g1 + Wext
1 2
1
mv3 + mgy3 + μ k mg ( x2 − x1 ) = mv12 + mgy1 + Wext
2
2
Using v3 = 0 m/s, y1 = 0 m, Wext = 0 J, v1 = 2 gh , and ( x2 − x1 ) = L, we get
1
mgy3 + μ k mgL = m(2 gh) ⇒ y3 = h − μ k L
2
Assess:
For μ k = 0, y3 = h which is predicted by the law of the conservation of energy.
11.58. Model: Assume an ideal spring, so Hooke’s law is obeyed.
Visualize:
Solve: For a conservative force the work done on a particle as it moves from an initial to a final position is
independent of the path. We will show that WA →C → B = WA → B for the spring force. Work done by a spring force
F = − kx is given by
xf
W = ∫ Fdx = − ∫ kx dx
xi
This means
xB
WA → B = − ∫ kx dx = −
xA
x
x
C
B
k 2
k
k
xB − xA2 ) , WA → C = − ∫ kx dx = − ( xC2 − xA2 ) , and WC → B = − ∫ kx dx = − ( xB2 − xC2 )
(
2
2
2
xA
xC
Adding the last two:
WA → C → B = WA → C + WC → B = −
k 2
( xC − xA2 + xB2 − xC2 ) = WA→ B
2
11.59. Model: A “sprong” that obeys the force law Fx = −q( x − xe )3 , where q is the sprong constant and xe
is the equilibrium position.
Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, xe = xf = 0 m.
Solve: (a) The units of q are N/m3 .
(b) A cubic curve rises more steeply than a parabola. The force increases by a factor of 8 every time x increases
by a factor of 2.
x
dU
qx 4
, we have U ( x) = − ∫ Fx dx = − ∫ (− qx 3 )dx =
.
0
4
dx
(d) Applying the energy conservation equation to the ball and sprong system:
K f + U f = Ki + U i
(c) Since Fx = −
1 2
qx 4
mvf + 0 J = 0 J + i
2
4
⇒ vf =
q x4
(40,000 N/m3 ) ( −0.10 m) 4
=
⋅
= 10 m/s
m 2
(0.020 kg)
2
11.60. Solve: (a) Because sin (cx) is dimensionless, F0 must have units of force in newtons.
(b) The product cx is an angle because we are taking the sine of it. An angle has no real physical units. If x has
units of m and the product cx is unitless, then c has to have units of m -1.
(c) The force is a maximum when sin(cx) = 1. This occurs when cx = π /2, or for xmax = π /2c.
(d) The graph is the first quarter of a sine curve.
(e) We can find the velocity vf at xf = xmax from the work-kinetic energy theorem:
1
1
1
1
2W
ΔK = mvf2 − mvi2 = mvf2 − mv02 = W ⇒ vf = v02 +
2
2
2
2
m
This is a variable force. As the particle moves from xi = 0 m to xf = xmax = π /2c, the work done on it is
xf
π / 2c
xi
0
W = ∫ F ( x) dx = F0 ∫
sin(cx) dx = −
F0
F ⎛
π
⎞ F
cos(cx)|π0 / 2 c = − 0 ⎜ cos − cos0 ⎟ = 0
2
c
c ⎝
⎠ c
Thus, the particle’s speed at xf = xmax = π /2c is vf = v02 + 2 F0 / mc .
11.61. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor.
Solve: (a) We might estimate y2 − y1 ≈ 4.0 m ≈ 12 ft ≈ y3 − y2 , thus, y3 − y1 ≈ 8.0 m.
(b) We might estimate the time to run up these two flights of stairs to be 20 s.
(c) Estimate your mass as m ≈ 70 kg ≈ 150 lb. Your power output while running up the stairs is
work done by you change in potential energy mg ( y3 − y1 )
=
=
time
time
time
(70 kg)(9.8 m/s 2 )(8.0 m)
⎛ 1 hp ⎞
=
≈ 270 W = (270 W) ⎜
⎟ ≈ 0.35 hp
20 s
⎝ 746 W ⎠
Assess:
Your estimate may vary, depending on your mass and how fast you run.
11.62. Solve: Power output during the push-off period is equal to the work done by the cat divided by the
time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat
by the floor, work done by the cat can be found using the work-kinetic energy theorem during the push-off
period, Wnet = Wfloor = ΔK . We do not need to explicitly calculate Wcat , since we know that the cat’s kinetic energy
is transformed into its potential energy during the leap. That is,
ΔU g = mg ( y2 − y1 ) = (5.0 kg)(9.8 m / s 2 )(0.95 m) = 46.55 J
Thus, the average power output during the push-off period is
P=
Wnet 46.55 J
=
= 0.23 kW
0.20 s
t
11.63. Solve: Using the conversion 746 W = 1 hp, we have a power of 1492 J/s. This means W = Pt =
(1492 J/s)(1 h) = 5.3712 × 106 J is the total work done by the electric motor in one hour. Furthermore,
Wmotor = −Wg = U gf − U gi = mg ( yf − yi ) = mg (10 m)
m=
Wmotor
5.3712 × 106 J
1 liter
=
= 5.481 × 104 kg = 5.481 × 104 kg ×
= 5.5 × 104 liters
g (10 m) (9.8 m/s 2 )(10 m)
1 kg
11.64. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is ΔU g = − mgh =
−(1.0 kg)(9.8 m/s 2 )(25 m) = −245 J ≈ −0.25 kJ
(b) The power required of the dam is
P=
W W
=
= 50 × 106 Watts ⇒ W = 50 × 106 J
t 1s
That is, 50 × 106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy,
(245 J)(0.80) = 196 J is converted to electrical energy. Thus, the amount of water needed per second is
(50 × 106 J)(1 kg /196 J) = 255,000 kg ≈ 2.6 × 105 kg.
11.65. Solve: The force required to tow a water skier at a speed v is Ftow = Av. Since power P = Fv, the
power required to tow the water skier is Ptow = Ftow v = Av 2 . We can find the constant A by noting that a speed of
v = 2.5 mph requires a power of 2 hp. Thus,
(2 hp) = A(2.5 mph) 2 ⇒ A = 0.32
hp
(mph) 2
Now, the power required to tow a water skier at 7.5 mph is
Ptow = Av 2 = 0.32
Assess:
hp
⋅ (7.5 mph) 2 = 18 hp
(mph) 2
Since P ∝ v 2 , a three-fold increase in velocity leads to a nine-fold increase in power.
11.66. Solve: By definition, the maximum power output of a horse is P ≈ 1 hp = 746 W. At maximum
G
G
speed, when the horse is running at constant speed, Fnet = 0. The propulsion force Fhorse , provided by the horse
pushing against the ground, is balanced by the drag force of air resistance: Fhorse = D. We learned in Chapter 6
that a reasonable model for drag is D ≈ 14 Av 2 , where A is the cross section area. Since the power needed for
force F to push an object at velocity v is P = Fv, we have
P ≈ 1 hp = 746 W = Fhorsev = ( 14 Av 2 ) v = 14 Av3
1/ 3
⎛ 4P ⎞
v≈⎜
⎟
⎝ A⎠
Assess:
1/ 3
⎛ 4(746 W) ⎞
=⎜
⎟
⎝ (0.5 m)(1.8 m) ⎠
= 15 m/s
15 m/s ≈ 30 mph is a reasonable top speed for a well-trained horse.
11.67. Solve: The net force on a car moving at a steady speed is zero. The motion is opposed both by rolling
friction and by air resistance. Thus the propulsion force provided by the drive wheels must be
Fcar = μ r mg + 14 Av 2 , where μ r is the rolling friction, m is the mass, A is the cross-section area, and v is the car’s
velocity. The power required to move the car at speed v is
1 3
Av
4
Since the maximum power output is 200 hp and 75% of the power reaches the drive wheels,
P = (200 hp)(0.75) = 150 hp. Thus,
P = Fcar v = μ r mgv +
⎛ 746 W ⎞
1
2
3
(150 hp) ⎜
⎟ = (0.02)(1500 kg)(9.8 m/s )v + (1.6 m)(1.4 m)v
1
hp
4
⎝
⎠
⇒ 0.56 v3 + 294 v − 111,900 = 0 ⇒ v = 55.5 m/s
The easiest way to solve this equation is through iterations by trial and error.
Assess: A speed of 55.5 m/s ≈ 110 mph is very reasonable.
11.68. Model: Use the model of static friction, kinematic equations, and the definition of power.
Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the gravitational force on the car is
(1480 kg)(9.8 m/s 2 ) = 14,504 N. The amount of that force on the drive wheels is (14,504)(2/3) = 9670 N. Because
the static friction of the tires on road pushes the car forward,
Fmax = fs max = μs n = μ s mg = (1.00)(9670 N) = mamax
⇒ amax =
9670 N
= 6.53 m/s 2
1480 kg
(b) Only 70% of the power generated by the motor is applied at the wheels.
P = Fvmax ⇒ vmax =
P (0.70)(161,882 W)
=
= 11.7 m/s
9670 N
F
(c) Using the kinematic equation, vmax = v0 + amax (tmin − t0 ) with v0 = 0 m/s and t0 = 0 s, we obtain
tmin =
Assess:
vmax 11.7 m/s
=
= 1.79 s
amax 6.53 m/s 2
An acceleration time of 1.79 s for the Porsche to reach a speed of ≈ 26 mph from rest is reasonable.
11.69. (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long lab
bench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the book’s final speed is
4.0 m/s, what is the tension in the string?
(b)
(c) The tension does external work Wext . This work increases the book’s kinetic energy and also causes an
increase ΔEth in the thermal energy of the book and the lab bench. Solving the equation gives T = 10.9 N.
11.70. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40° ramp from the back of a truck to the ground.
The coefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going
at the bottom of the ramp?
(b)
(c) v1 = 6.34 m/s.
11.71. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic
friction between the sled and the surface is μ k = 0.20, what speed will you be able to maintain?
(b)
(c) Fpush = (0.20)(30 kg)(9.8 m/s 2 ) = 58.8 N ⇒ 75 W = (58.8 N)v ⇒ v =
75 W
= 1.28 m/s
58.8 N
11.72. (a) A 1500 kg object is being accelerated upward at 1.0 m/s2 by a rope. How much power must the
motor supply at the instant when the velocity is 2.0 m/s?
(b)
(c) T = (1500 kg)(9.8 m/s 2 ) + 1500 kg(1.0 m/s 2 ) = 16,200 N=16.2 kN
P = T (2m/s) = (16,200 N)(2.0 m/s) = 32,400 W = 32 kW
11.73. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and
use the constant-acceleration kinematic equations.
Visualize:
We placed the origin of the coordinate system at the base of the frictionless ramp.
Solve: We’ll start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank.
From the vertical motion for jumping the shark tank,
1
y2 = y1 + v1 y (t2 − t1 ) + a y (t2 − t1 ) 2
2
1
⇒ 0 m = (2.0 m) + 0 m + (−9.8 m/s 2 )(t2 − t1 ) 2 ⇒ (t2 − t1 ) = 0.639 s
2
From the horizontal motion,
1
x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1 ) 2
2
5.0 m
⇒ ( x1 + 5.0 m) = x1 + v1 (0.639 s) + 0 m ⇒ v1 =
= 7.825 m/s
0.639 s
Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find
the minimum speed v0. We have
1
1
K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0
2
2
v0 = v12 + 2 g ( y1 − y0 ) = (7.825 m/s) 2 + 2(9.8 m/s 2 )(2.0 m) = 10.0 m/s
11.74. Model: Assume the spring to be ideal that obeys Hooke’s law, and model the block as a particle.
Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is in
contact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy
appears as kinetic energy of the block until the block begins to climb up the incline.
Solve: Although we could find the speed v1 of the block as it leaves the spring, we don’t need to. We can use
energy conservation to relate the initial potential energy of the spring to the energy of the block as it begins
projectile motion at point 2. However, friction requires us to calculate the increase in thermal energy. The energy
equation is
K 2 + U g2 + ΔEth = K 0 + U g0 + Wext ⇒ 12 mv22 + mgy2 + f k Δs = 12 k ( x0 − xe ) 2
The distance along the slope is Δs = y2 / sin 45°. The friction force is f k = μ k n, and we can see from the freebody diagram that n = mg cos 45°. Thus
1/ 2
⎡k
⎤
v2 = ⎢ ( x0 − xe ) 2 − 2 gy2 − 2 μ k gy2 cos 45°/ sin45° ⎥
⎣m
⎦
1/ 2
⎡1000 N/m
⎤
=⎢
(0.15 m) 2 − 2(9.8 m/s 2 )(2.0 m) − 2(0.20)(9.8 m/s 2 )(2.0 m) cos 45° / sin45° ⎥ = 8.091 m/s
0.20
kg
⎣
⎦
Having found the velocity v2 , we can now find ( x3 − x2 ) = d using the kinematic equations of projectile motion:
1
y3 = y2 + v2 y (t3 − t2 ) + a2 y (t3 − t2 ) 2
2
1
2.0 m = 2.0 m + (v2 sin 45°)(t3 − t2 ) + (−9.8 m / s 2 )(t3 − t2 ) 2
2
⇒ (t3 − t2 ) = 0 s and 1.168 s
Finally,
1
x3 = x2 + v2 x (t3 − t2 ) + a2 x (t3 − t2 ) 2
2
d = ( x3 − x2 ) = (v2 cos 45°)(1.168 s) + 0 m = 6.68 m
11.75. Model: Assume an ideal spring that obeys Hooke’s law, the particle model for the ball, and the model
of kinetic friction. The ball, the spring, and the barrel comprise our system.
Visualize:
We placed the origin
of the coordinate system on the free end of the spring which is in contact with the ball.
G
Solve: Force F does external work Wext . This work compresses the spring, gives the ball kinetic energy, and
is partially dissipated to thermal energy by friction.
(a) The energy equation for our system is
K1 + U s1 + ΔEth = K 0 + U s0 + Wext
1 2 1
1
1
mv1 + k ( x1 − x0 ) 2 + μ k mg ( x1 − x0 ) = mv02 + k ( xe − xe ) 2 + F ( x1 − x0 )
2
2
2
2
1
1
(1.0 kg)(2.0 m/s) 2 + (3000 N/m)(0.3 m) 2 + μ k mg (0.30 m) = 0 J + 0 J + F (0.30 m)
2
2
With μ k = 0.30 and m = 1.0 kg, we can solve this equation to obtain F = 460 N.
(b) Using the energy equation for our system once again, we have
1 2 1
1
1
mv2 + k ( xe − xe ) 2 + μ k mg ( x2 − x1 ) = mv12 + k ( x1 − x0 ) 2 + 0 J
2
2
2
2
1 2
1
mv2 + μ k mg ( x2 − x1 ) = 0 J+ k ( x1 − x0 ) 2
2
2
1
1
2
2
(1.0 kg)v2 + (0.30)(1.0 kg)(9.8 m/s )(1.5 m) = (3000 N/m)(0.30 m) 2
2
2
(0.5 kg)v22 + 4.41 J=135 J ⇒ v2 = 16.2 m/s
K 2 + U s2 + ΔEth = K1 + U s1 + Wext
11.76. Solve: (a)
The graph is a hyperbola.
(b) The separation for zero potential energy is r = ∞, since
U =−
Gm1m2
→ 0 J as r → ∞
r
This makes sense because two masses don’t interact at all if they are infinitely far apart.
(c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is
conserved.
The equations of energy and momentum conservation are
K f + U gf = K i + U gi
⎛ Gm1m2 ⎞ 1
⎛ Gm1m2 ⎞
1
1
1
2
2
m1v1f2 + m2v2f2 + ⎜ −
⎟ = m1v1i + m2v2i + ⎜ −
⎟
2
2
2
rf ⎠ 2
ri ⎠
⎝
⎝
⎛ 1 1⎞
1
1
m1v1f2 + m2v2f2 = Gm1m2 ⎜ − ⎟
2
2
⎝ rf ri ⎠
pf = pi ⇒ m1v1f + m2v2f = 0 kg m / s ⇒ v2f = −
m1
v1f
m2
Substituting this expression for v2f into the energy equation, we get
2
⎞
⎛ 1 1⎞
1
1 ⎛m
2Gm2 ⎛ 1 1 ⎞
m1v1f2 + m2 ⎜ 1 v1f ⎟ = Gm1m2 ⎜ − ⎟ ⇒ v1f2 =
⎜ − ⎟
2
2 ⎝ m2 ⎠
r
r
(1
m1 / m2 ) ⎝ rf ri ⎠
+
i ⎠
⎝ f
With G = 6.67 × 10−11 N ⋅ m(kg) −2 , rf = R1 + R2 = 18 × 108 m, ri = 1.0 × 1014 m, m1 = 8.0 × 1030 kg, and m2 = 2.0 × 1030 kg,
the above equation can be simplified to yield
v1f = 1.72 × 105 m/s, and v2f = −
⎛ 8.0 × 1030 kg ⎞
m1
5
5
v1f = ⎜
⎟ × (1.72 × 10 m/s ) = 6.89 × 10 m/s
30
m2
⎝ 2.0 × 10 kg ⎠
The speed of the heavier star is 1.7 × 105 m/s. That of the lighter star is 6.9 × 105 m/s.
11.77. Model: Model the lawnmower as a particle and use the model of kinetic friction.
Visualize:
We placed the origin of our coordinate system on the lawnmower and drew the free-body diagram of forces.
G
G
G
Solve: The normal force n, which is related to the frictional force, is not equal to FG . This is due to the presence of F .
G
G
The rolling friction is f r = μ r n, or n = f r μ r . The lawnmower moves at constant velocity, so Fnet = 0. The two
components of Newton’s second law are
( ∑ F ) = n − F − F sin 37° = ma = 0 N ⇒ f / μ − mg − F sin 37° = 0 N ⇒ f = μ mg + μ F sin 37°
( ∑ F ) = F cos37° − f = 0 N ⇒ F cos37° − μ mg − μ F sin 37° = 0 N
G
y
y
r
x
⇒F=
r
r
r
r
r
r
r
(0.15)(12 kg)(9.8 m/s 2 )
μ r mg
=
= 29.4 N
cos37° − μ r sin 37° (0.7986) − (0.15)(0.6018)
G G
Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P = F ⋅ v =
Fv cosθ = (24.9 N)(1.2 m/s)cos37° = 24 W.
11-1
12.1.
Model:
A spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
Solve: The speed v = rω , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2π /60 rad/s = 6π rad/s. Thus,
v = (0.70 m)(6π rad/s) = 13.2 m/s.
Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.
12.2.
Model:
Assume constant angular acceleration.
⎛ 2π rad ⎞⎛ min ⎞
(a) The final angular velocity is ω f = ( 2000 rpm ) ⎜
⎟⎜
⎟ = 209.4 rad/s. The definition of
⎝ rev ⎠⎝ 60 s ⎠
angular acceleration gives us
Solve:
α=
Δω ω f − ωi 209.4 rad/s − 0 rad/s
=
=
= 419 rad/s
Δt
Δt
0.50 s
The angular acceleration of the drill is 4.2 × 102 rad/s.
(b)
1
2
θ f = θ i + ωi Δt + α ( Δt ) = 0 rad + 0 rad +
2
⎛ rev ⎞
The drill makes (52.4 rad) ⎜
⎟ = 8.3 revolutions.
⎝ 2π rad ⎠
1
2
( 419 rad/s )( 0.50 s ) = 52.4 rad
2
12.3.
Model:
Visualize:
Solve:
Assume constant angular acceleration.
(a) Since at = rα , find α first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s,
α=
Δω 9.43 rad − 6.28 rad/s
=
= 0.314 rad/s 2
Δt
10 s
The angular acceleration of the sprocket and pedal are the same. So
at = rα = ( 0.18 m ) ( 0.314 rad/s 2 ) = 0.057 m/s2
(b) The length of chain that passes over the sprocket during this time is L = r Δθ . Find Δθ :
1
2
θ f = θ i + ωi Δt + α ( Δt )
2
θ f − θ i = Δθ = ( 6.28 rad/s )(10 s ) +
1
( 0.314 rad/s 2 ) (10 s )2 = 78.5 rad
2
The length of chain which has passed over the top of the sprocket is
L = (0.10 m)(78.5 rad) = 7.9 m
12.4.
Model:
Visualize:
Assume constant angular acceleration.
⎛ 2π rad ⎞⎛ min ⎞
The initial angular velocity is ωi = ( 60 rpm ) ⎜
⎟⎜
⎟ = 2π rad/s.
⎝ rev ⎠⎝ 60 s ⎠
The angular acceleration is
Solve:
α=
ωf − ωi
Δt
=
0 rad/s − 2π rad/s
= −0.251 rad/s 2
25 s
The angular velocity of the fan blade after 10 s is
ω f = ωi + α ( t − t0 ) = 2π rad/s+ ( −0.251 rad/s 2 ) (10 s − 0 s ) = 3.77 rad/s
The tangential speed of the tip of the fan blade is
vt = rω = ( 0.40 m )( 3.77 rad/s ) = 1.51 m/s
(b)
1
2
2
The fan turns 78.6 radians = 12.5 revolutions while coming to a stop.
θ f = θ i + ω i Δt + α ( Δt ) = 0 rad + ( 2π rad/s )( 25 s ) +
1
( −0.251 rad/s 2 ) ( 25 s )2 = 78.6 rad
2
12.5.
Model:
Visualize:
The earth and moon are particles.
Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the
distance from the center of the earth.
Solve:
24
22
8
m x + mM xM ( 5.98 × 10 kg ) ( 0 m ) + ( 7.36 × 10 kg )( 3.84 × 10 m )
xcm = E E
=
mE + mM
5.98 × 1024 kg + 7.36 × 10 22 kg
= 4.67 × 106 m
Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the
surface of the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the
denominator of the expression for xcm .
12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses mA , mB , and mC are
(0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm), respectively.
Solve: The coordinates of the center of mass are
xcm =
mA xA + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm)
=
= 5.0 cm
mA + mB + mC
(100 g + 200 g + 300 g)
ycm =
mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm)
=
= 3.3 cm
mA + mB + mC
(100 g + 200 g + 300 g)
12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses mA , mB , and mC are
(0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively.
Solve: The coordinates of the center of mass are
xcm =
mA xA + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm)
=
= 6.7 cm
mA + mB + mC
(200 g + 300 g + 100 g)
ycm =
mA yA + mB yB + mC yC (200 g)(10 cm) + (300 g)(10 cm) + (100 g)(0 cm)
=
= 8.3 cm
mA + mB + mC
(200 g + 300 g + 100 g)
12.8.
The balls are particles located at the ball’s respective centers.
Solve:
The center of mass of the two balls measured from the left hand ball is
Model:
Visualize:
xcm =
(100 g )( 0 cm ) + ( 200 g )( 30 cm ) = 20 cm
100 g + 200 g
The linear speed of the 100 g ball is
⎛ 2π rad ⎞⎛ min ⎞
v1 = rω = xcmω = ( 0.20 m )(120 rev/min ) ⎜
⎟⎜
⎟ = 2.5 m/s
⎝ rev ⎠⎝ 60 s ⎠
12.9.
Solve:
Model:
The earth is a rigid, spherical rotating body.
The rotational kinetic energy of the earth is K rot = 12 Iω 2 . The moment of inertia of a sphere about its
diameter (see Table 12.2) is I = 52 M earth R 2 and the angular velocity of the earth is
ω=
2π rad
= 7.27 × 10 −5 rad/s
24 × 3600 s
Thus, the rotational kinetic energy is
1⎛ 2
⎞
K rot = ⎜ M earth R 2 ⎟ω 2
2⎝ 5
⎠
1
= (5.98 × 1024 kg)(6.37 × 106 m) 2 (7.27 × 10−5 rad/s)2 = 2.57 × 1029 J
5
12.10.
Model: The triangle is a rigid body rotating about an axis through the center.
Visualize: Please refer to Figure EX12.10. Each 200 g mass is a distance r away from the axis of rotation,
where r is given by
0.20 m
0.20 m
= cos30° ⇒ r =
= 0.2309 m
r
cos30°
Solve: The moment of inertia of the triangle is I = 3 × mr 2 = 3(0.200 kg)(0.2309 m) 2 = 0.0320 kg m 2 . The
frequency of rotation is given as 5.0 revolutions per s or 10π rad/s. The rotational kinetic energy is
K rot. =
1 2 1
Iω = (0.0320 kg m 2 )(10.0π rad/s) 2 = 15.8 J
2
2
12.11.
Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve:
The speed of the point on the rim is given by vrim = Rω . The angular velocity ω of the disk can be
determined from its rotational kinetic energy which is K = 12 Iω 2 = 0.15 J. The moment of inertia I of the disk
about its center and perpendicular to the plane of the disk is given by
1
1
MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 × 10−5 kg m 2
2
2
2(0.15 J)
0.30 J
2
⇒ω =
=
⇒ ω = 61.237 rad/s
I
8.0 × 10−5 kg m 2
I=
Now, we can go back to the first equation to find vrim . We get vrim = Rω = (0.040 m)(61.237 rad/s) = 2.4 m/s.
12.12.
Model:
The baton is a thin rod rotating about a perpendicular axis through its center of mass.
1
Solve: The moment of inertia of a thin rod rotating about its center is I = ML2 . For the baton,
12
1
2
I = ( 0.400 kg )( 0.96 m ) = 0.031 kg m 2
12
The rotational kinetic energy of the baton is
2
K rot =
1 2 1
⎛
⎛ 2π rad ⎞⎛ min ⎞ ⎞
Iω = ( 0.031 kg m 2 ) ⎜ (100 rev/min ) ⎜
⎟⎜
⎟ ⎟ = 1.68 J
2
2
⎝ rev ⎠⎝ 60 s ⎠ ⎠
⎝
12.13.
Model: The structure is a rigid body rotating about its center of mass.
Visualize:
We placed the origin of the coordinate system on the 300 g ball.
Solve: First, we calculate the center of mass:
xcm =
(300 g)(0 cm) + (600 g)(40 cm)
= 26.67 cm
300 g + 600 g
Next, we will calculate the moment of inertia about the structure’s center of mass:
I = (300 g)( xcm ) 2 + (600 g)(40 cm − xcm )2
= (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2
Finally, we calculate the rotational kinetic energy:
2
1
1
⎛ 100 × 2π
⎞
K rot = Iω 2 = (0.032 kg m 2 ) ⎜
rad/s ⎟ = 1.75 J
2
2
⎝ 60
⎠
12.14.
Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the
rotation axis passes through mass A and is perpendicular to the page.
Visualize: Please refer to Figure EX12.14.
∑ mi xi = mA xA + mB xB + mC xC + mD xD
Solve: (a) xcm =
mA + mB + mC + mD
∑ mi
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
= 0.057 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
ycm = A A
mA + mB + mC + mD
=
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m)
= 0.057 m
700 g
(b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page
is
=
I A = ∑ mi ri 2 = mA rA2 + mB rB2 + mC rC2 + mD rD2
i
= (0.100 kg)(0 m) 2 + (0.200 kg)(0.10 m) 2 + (0.200 kg)(0.1414 m) 2 + (0.200 kg)(0.10 m) 2 = 0.0080 kg m 2
12.15.
Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
Solve:
(a) xcm = ∑
mi xi
∑m
i
=
mA xA + mB xB + mC xC + mD xD
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)
=
= 0.057 m
100 g + 200 g + 200 g + 200 g
m y + mB yB + mC yC + mD yD
ycm = A A
mA + mB + mC + mD
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m)
= 0.057 m
700 g
(b) The moment of inertia about a diagonal that passes through B and D is
=
I BD = mA rA2 + mC rC2
where rA = rC = (0.10 m)cos 45° = 7.07 cm and are the distances from the diagonal. Thus,
I BD = (0.100 kg) rA2 + (0.200 kg)rC2 = 0.0015 kg m 2
Assess:
Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
12.16.
Model: The three masses connected by massless rigid rods is a rigid body.
Visualize:
Solve:
(a)
Please refer to Figure EX12.16.
xcm =
∑ m x = (0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m) = 0.060 m
0.100 kg + 0.200 kg + 0.100 kg
∑m
i i
i
∑ m y = (0.100 kg)(0 m) + (0.200 kg) ( (0.10 m) − (0.06 m) ) + (0.100 kg)(0 m) = 0.040 m
0.100 kg + 0.200 kg + 0.100 kg
∑m
2
ycm =
2
i i
i
(b) The moment of inertia about an axis through A and perpendicular to the page is
I A = ∑ mi ri 2 = mB (0.10 m) + mC (0.10 m) 2 = (0.100 kg)[(0.10 m) 2 + (0.10 m) 2 ] = 0.0020 kg m 2
2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA
Assess:
( (0.10 m) − (0.06 m) ) = 0.00128 kg m
2
2
2
2
Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
12.17.
Model: The door is a slab of uniform density.
(a) The hinges are at the edge of the door, so from Table 12.2,
1
2
I = ( 25 kg )( 0.91 m ) = 6.9 kg m 2
3
(b) The distance from the axis through the center of mass along the height of the door is
⎛ 0.91 m
⎞
d =⎜
− 0.15 m ⎟ = 0.305 m. Using the parallel–axis theorem,
2
⎝
⎠
1
2
2
I = I cm + Md 2 = ( 25 kg )( 0.91 m ) + ( 25 kg )( 0.305 cm ) = 4.1 kg m 2
12
Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass.
Solve:
12.18.
Model: The CD is a disk of uniform density.
(a) The center of the CD is its center of mass. Using Table 12.2,
1
1
2
I cm = MR 2 = ( 0.021 kg )( 0.060 m ) = 3.8 × 10−5 kg m 2
2
2
(b) Using the parallel–axis theorem with d = 0.060 m,
Solve:
I = I cm + Md 2 = 3.8 × 10−5 kg m 2 + ( 0.021 kg )( 0.060 m ) = 1.14 × 10−4 kg m 2
2
12.19. Visualize:
G
Solve: Torque by a force is defined as τ = Fr sin φ where φ is measured counterclockwise from the r vector to
G
the F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to
the 20 N force:
(30 N)r1 sinφ1 + (20 N)r2 sinφ2 = (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°)
= ( − 0.60 N m) + (0.40 N m) = −0.20 N m
Assess:
A negative torque causes a clockwise acceleration of the pulley.
12.20. Visualize:
The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes
a net torque.
Solve: The distance between the lines of action is l = d cos30°. The net torque is given by
τ = lF = (d cos30°)F = (0.10 m)(0.866)(50 N) = 4.3 N m
12.21. Visualize:
Solve:
The net torque on the spark plug is
τ = Fr sin φ = −38 N m = F (0.25 m)sin(−120°) ⇒ F = 176 N
That is, you must pull with a force of 176 N to tighten the spark plug.
Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative
sign for the net torque.
12.22.
Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center.
Solve: The net torque on the axle is
τ = FA rA sin φA + FBrB sin φB + FC rC sin φC + FD rD sin φD
= (30 N)(0.10 m)sin(−90°) + (20 N)(0.050 m)sin 90° + (30 N)(0.050 m)sin135° + (20 N)(0.10 m)sin 0°
= −3 N m + 1 N m + 1.0607 N m = −0.94 N m
Assess: A negative torque means a clockwise rotation of the disk.
12.23.
Model: The beam is a solid rigid body.
Visualize:
G
The steel beam experiences a torque due to the gravitational force on the construction worker FG and the
C
G
gravitational force on the beam FG . The normal force exerts no torque since the net torque is calculated about the
( )
( )
B
point where the beam is bolted into place.
G
Solve: The net torque on the steel beam about point O is the sum of the torque due to FG and the torque due to
C
G
FG . The gravitational force on the beam acts at the center of mass.
( )
( )
B
τ = (( FG )C )(4.0 m)sin( −90°) + (( FG ) B )(2.0 m)sin(−90°)
= −(70 kg)(9.80 m/s 2 )(4.0 m) − (500 kg)(9.80 m/s 2 )(2.0 m) = −12.5 kN m
The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12.5
kN m.
12.24.
Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.
Visualize:
Solve:
(a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:
τ = τ ball + τ arm = ( mb g )rb sin 90° + (ma g )ra sin 90°
= (3.0 kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) = 34 N m
(b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line,
so
τ = τ ball + τ arm = (mb g )rb sin 45° + ( ma g ) ra sin 45°
= (3.0 kg)(9.8 m/s 2)(0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2)(0.35 m)(0.707) = 24 N m
12.25. Solve:
τ = Iα is the rotational analog
2
τ = (2.0 kg m )(4.0 rad/s 2 ) = 8.0 kg m 2 /s 2 = 8.0 N m.
of
Newton’s
second
law
F = ma.
We
have
12.26.
Visualize:
Since
α = τ /I , a graph of the angular acceleration looks just like the torque graph with
the numerical values divided by I = 4.0 kg m 2 .
Solve: From the discussion about Figure 4.47
ω f = ωi + area under the angular acceleration α curve between ti and tf
The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With ω1 = 0 rad/s, we have
ω f = 0 rad/s + 0.75 rad/s = 0.75 rad/s
12.27.
Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through
the center of mass. Assume that the size of the balls is small compared to 1 m.
Visualize:
We placed the origin of the coordinate system on the 1.0 kg ball.
Solve: The center of mass and the moment of inertia are
xcm =
(1.0 kg)(0 m) + (2.0 kg)(1.0 m)
= 0.667 m and ycm = 0 m
(1.0 kg + 2.0 kg)
I about cm = ∑ mi ri 2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg m 2
We have ω f = 0 rad/s, tf − ti = 5.0 s, and ωi = −20 rpm = −20(2π rad/60 s) = − 32 π rad/s, so ω f = ωi + α (tf − ti )
becomes
2π
⎛ 2π
⎞
0 rad/s = ⎜ −
rad/s ⎟ + α (5.0 s) ⇒ α =
rad/s 2
15
⎝ 3
⎠
Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s:
⎛2
⎝3
⎞⎛ 2π
⎞ 4π
rad/s 2 ⎟ =
N m = 0.28 N m
15
⎠⎝
⎠ 45
τ = I about cmα = ⎜ kg m 2 ⎟⎜
The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.
12.28.
Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is
perpendicular to the disk.
Solve: To determine the torque (τ) needed to take the plastic disk from ωi = 0 rad/s
to
ω f = 1800 rpm = (1800)(2π ) / 60 rad/s = 60π rad/s in tf − ti = 4.0 s, we need to determine the angular acceleration (α )
and the disk’s moment of inertia (I ) about the axle in its center. The radius of the disk is R = 10.0 cm. We have
1
1
I = MR 2 = (0.200 kg)(0.10 m) 2 = 1.0 × 10−3 kg m 2
2
2
ω − ωi 60π rad/s − 0 rad/s
ω f = ωi + α (tf − ti ) ⇒ α = f
=
= 15π rad/s 2
4.0 s
t f − ti
Thus, τ = Iα = (1.0 × 10−3 kg m 2 )(15π rad/s 2 ) = 0.047 N m.
12.29.
Model: The compact disk is a rigid body rotating about its center.
Visualize:
Solve:
(a) The rotational kinematic equation ω1 = ω 0 + α (t1 − t0 ) gives
200π
⎛ 2π ⎞
(2000 rpm) ⎜
rad/s 2
⎟ rad/s = 0 rad + α (3.0 s − 0 s) ⇒ α =
9
⎝ 60 ⎠
The torque needed to obtain this operating angular velocity is
⎛ 200π
⎞
rad/s 2 ⎟ = 1.75 × 10−3 N m
⎝ 9
⎠
τ = Iα = (2.5 × 10−5 kg m 2 ) ⎜
(b) From the rotational kinematic equation,
1
1 ⎛ 200π
θ1 = θ 0 + ω 0 (t1 − t0 ) + α (t1 − t0 ) 2 = 0 rad + 0 rad + ⎜
2
2⎝ 9
100π
= 100π rad =
revolutions = 50 rev
2π
Assess:
Fifty revolutions in 3 seconds is a reasonable value.
2
⎞
rad/s 2 ⎟ ( 3.0 s − 0 s )
⎠
12.30.
Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is
small compared to 60 cm.
Visualize: Please refer to Figure EX12.30.
Solve: We can determine the rocket’s angular acceleration from the relationship τ = Iα . The torque τ can be
found from the thrust (F) using τ = Fr sin φ . The moment of inertia (I) can be calculated from equations given in
Table 12.2. Specifically, I = I rod about one end + I rocket becomes
1
1
M rod L2 + ML2 = (0.100 kg)(0.60 m) 2 + (0.200 kg)(0.60 m) 2
3
3
= 0.012 kg m 2 + 0.072 kg m 2 = 0.0840 kg m 2
⇒α =
Assess:
τ
I
=
Fr sin φ (4.0 N)(0.60 m)sin(45°)
=
= 20 rads/s 2
I
0.0840 kg m 2
The rocket will accelerate counterclockwise since α is positive.
12.31.
Model: The rod is in rotational equilibrium, which means that τ net = 0.
Visualize:
As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the
figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the
right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and
forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up).
Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this
G
point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about
the left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational
force on the rod acts at the center of mass, so
τ net = 0 N m = τ pin − (0.40 m)(2.0 kg)(9.8 m/s 2) − (0.80 m)(0.50 kg)(9.8 m/s 2)
⇒ τ pin = 11.8 N m
12.32.
Model: The massless rod is a rigid body.
Visualize:
Solve:
G
To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τ net = 0 Nm). We have ( Fnet ) y = (40 N) − (100 N) + (60 N) = 0 N, so the object is in translational
equilibrium. Measuring τ net about the left end,
τ net = (60 N)(3.0 m)sin( +90°) + (100 N)(2.0 m)sin(−90°) = −20 N m
The object is not in equilibrium.
12.33.
Model: The object balanced on the pivot is a rigid body.
Visualize:
Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium.
G
Solve: There are three forces acting on the object: the gravitational force FG acting through the center of
1
G
mass of the long rod, the gravitational force FG acting through the center of mass of the short rod, and the
2
G
normal force P on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is
( )
( )
− ( FG )1 − ( FG )2 + P = 0 N ⇒ P = ( FG )1 + ( FG )2 = (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 ) = 49 N
Measuring torques about the left end, the equation for rotational equilibrium τ net = 0 Nm is
Pd − w1 (1.0 m) − w2 (1.5 m) = 0 Nm
⇒ (49 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s 2 )(1.5 m) = 0 N ⇒ d = 1.40 m
Thus, the pivot is 1.40 m from the left end.
12.34. Model:
Visualize:
Solve:
The see-saw is a rigid body. The cats and bowl are particles.
The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.
τ net = 0 = ( FG )1 ( 2.0 m ) − ( FG )2 ( d ) − ( FG ) B ( 2.0 m )
m2 gd = m1 g ( 2.0 m ) − mB g ( 2.0 m )
d=
( m1 − mB )( 2.0 m ) = ( 5.0 kg − 2.0 kg )( 2.0 m ) = 1.5 m
m2
4.0 kg
Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the
combined mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.
12.35.
Solve:
(a) According to Equation 12.35, the speed of the center of mass of the tire is
vcm = Rω = 20 m/s ⇒ ω =
vcm 20 m/s
⎛ 60 ⎞
2
=
= 66.67 rad/s = ( 66.7 ) ⎜
⎟ rpm = 6.4 × 10 rpm
R 0.30 m
2
π
⎝
⎠
(b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s.
(c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.
12.36.
Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass
distribution.
Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of
mass. The moment of inertia of the can about the center of mass is 12 MR 2, where R is the radius of the can. Also
vcm = Rω , where ω is the angular velocity of the can. The total kinetic energy of the can is
K = K cm + K rot =
=
1
1
1
1⎛1
⎞⎛ v ⎞
2
2
+ I cmω 2 = Mvcm
+ ⎜ MR 2 ⎟⎜ cm ⎟
Mvcm
2
2
2
2⎝ 2
⎠⎝ R ⎠
3
3
2
Mvcm
= (0.50 kg)(1.0 m/s) 2 = 0.38 J
4
4
2
12.37.
Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down.
Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation
equation will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means
Kf =
1
1
1⎛ 2
1
⎞
2
I cmω 2 + Mvcm
= Mgh ⇒ ⎜ MR 2 ⎟ω 2 + M ( Rω ) 2 = Mgh
2
2
2⎝ 5
2
⎠
7
⇒ MR 2ω 2 = Mg (2.1 m)sin 25°
10
⇒ω =
10
7
g (2.1 m)(sin 25°)
=
R2
10
7
g (2.1 m)(sin 25°)
= 88 rad/s
(0.04 m) 2
(b) From part (a)
K total =
1
1
7
1
1⎛ 2
1
⎞
2
I cmω 2 + Mvcm
= MR 2ω 2 and K rot = I cmω 2 = ⎜ MR 2 ⎟ω 2 = MR 2ω 2
2
2
10
2
2⎝ 5
5
⎠
⇒
1
MR 2ω 2 1 10 2
K rot
= 75
= × =
K total 10 MR 2ω 2 5 7 7
12.38. Visualize: Please refer to Figure EX12.38. To determine angle α , put the tails of the vectors
together.
G G
G G
Solve: (a) The magnitude of A × B is AB sin α = (6)(4)sin 45° = 17. The direction of A × B, using the right
G G
hand rule, is out of the page. Thus, A × B = (17, out of the page).
G G
G G G
(b) The magnitude of C × D is CD sin α = (6)(4)sin180° = 0. Thus C × D = 0.
12.39.
Visualize: Please refer to Figure EX12.39.
G G
G G
(a) The magnitude of A × B is AB sin α = (6)(4)sin 60° = 20.78. The direction of A × B is given by the
G
G G
G
right hand rule. To curl our fingers from A to B, we have to point our thumb out of the page. Thus, A × B =
(21, out of the page).
G G
(b) C × D = ((6)(4)sin 90°, into the page) = (24, into the page).
Solve:
12.40.
Solve:
(a) (iˆ × ˆj ) × iˆ = kˆ × iˆ = ˆj
(b) iˆ × ( ˆj × iˆ) = iˆ × (− kˆ) = −iˆ × kˆ = −(− ˆj ) = ˆj
(a) iˆ × (iˆ × ˆj ) = iˆ × kˆ = − ˆj
G
(b) (iˆ × ˆj ) × kˆ = kˆ × kˆ = 0
12.41.
Solve:
12.42.
Solve:
G G
(a) A × B = (3iˆ + ˆj ) × (3iˆ − 2 ˆj + 2kˆ)
= 9iˆ × iˆ − 6iˆ × ˆj + 6iˆ × kˆ + 3 ˆj × iˆ − 2 ˆj × ˆj + 2 ˆj × kˆ
= 0 − 6kˆ + 6(− ˆj ) + 3(−kˆ) − 0 + 2iˆ = 2iˆ − 6 ˆj − 9kˆ
(b)
G
G
G
G
12.43. Solve: (a) C × D = G0 implies that D must also be in the same or opposite direction as the C vector or
zero, because iˆ × iˆ = 0. Thus D = niˆ, where n could be any real number.
G G
G
G
(b) C × E = 6kˆ implies that E must be along the ĵ vector, because iˆ × ˆj = kˆ. Thus E = 2 ˆj.
G G
G
G
(c) C × F = −3 ˆj implies that F must be along the k̂ vector, because iˆ × kˆ = − ˆj. Thus F = 1kˆ.
12.44.
Solve:
G
G
G
τ = r × F = (5iˆ + 5 ˆj ) × (−10 ˆj ) N m
G
= [−50(iˆ × ˆj ) − 50( ˆj × ˆj )] N m = [−50(+ kˆ) − 0] N m = −50kˆ N m
12.45.
Solve:
G
G
G
τ = r × F = (5 ˆj ) × (−10iˆ + 10 ˆj ) N m = (−5 ˆj × iˆ + 50 ˆj × ˆj ) N m
G
= [−50( −kˆ) + 0] N m = 50kˆ N m
12.46.
Solve:
Visualize: Please refer to Figure EX12.46.
G G
G
L = r × mv = (1.0iˆ + 2.0 ˆj ) m × (0.200 kg)(3.0 m/s) cos45°iˆ − sin 45° ˆj
(
)
= (0.42iˆ × iˆ − 0.42iˆ × ˆj + 0.85 ˆj × iˆ − 0.85 ˆj × ˆj ) kg m 2 /s = −(1.27 kˆ) kg m 2 /s or (1.27 kg m 2 /s, into page)
12.47.
Solve:
Visualize: Please refer to Figure EX12.47.
G G
G
L = r × mv = (3.0iˆ + 2.0 ˆj ) m × (0.1 kg)(4.0ˆj ) m/s
= 1.20(iˆ × ˆj ) kg m 2 /s + 0.8( ˆj × ˆj ) kg m 2 /s = 1.20kˆ kg m 2 /s + 0 kg m 2 /s
= 1.20kˆ kg m 2 /s or (1.20 kg m 2 /s, out of page)
12.48.
Model: The bar is a rotating rigid body. Assume that the bar is thin.
Visualize: Please refer to Figure EX12.48.
Solve: The angular velocity ω = 120 rpm = (120)(2π ) / 60 rad/s = 4π rad/s. From Table 12.2, the moment of
inertial of a rod about its center is I = 121 ML2 . The angular momentum is
⎛1⎞
L = Iω = ⎜ ⎟ (0.50 kg)(2.0 m) 2 (4π rad/s) = 2.1 kg m 2 /s
⎝ 12 ⎠
If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the
page. Consequently,
G
L = (2.1 kg m 2 /s, out of the page)
12.49.
Model: The disk is a rotating rigid body.
Visualize: Please refer to Figure EX12.49.
Solve: From Table 12.2, the moment of inertial of the disk about its center is
The
angular
1
1
I = MR 2 = (2.0 kg)(0.020 m) 2 = 4.0 × 10−4 kg m 2
2
2
ω
is
600 rpm = 600 × 2π /60 rad/s = 20π rad/s.
velocity
Thus,
L = Iω = (4.0 × 10 kg m )(20π rad/s) = 0.025 kg m /s. If we wrap our right fingers in the direction of the
disk’s rotation, our thumb will point in the − x direction. Consequently,
G
L = −0.025 iˆ kg m 2 /s = (0.025 kg m 2 /s, into page)
−4
2
2
12.50. Model: The beach ball is a spherical shell.
Solve: From Table 12.2, the moment of inertia about a diameter of a spherical shell is
2
2
1
2
I = MR 2 = ( 0.100 kg )( 0.50 m ) =
kg m 2
3
3
60
Require
⎛ 1
⎞
L = 0.10 kg m 2 /s = Iω = ⎜
kg m 2 ⎟ω
60
⎝
⎠
⇒ ω = ( 60 rad/s )
⎛ rev ⎞ ⎛ 60 s ⎞
In rpm, this is ( 60 rad/s ) ⎜
⎟⎜
⎟ = 57 rpm.
⎝ 2π rad ⎠ ⎝ min ⎠
12.51.
Model: The wheel is a rigid rolling body.
Visualize:
Solve:
The front of the disk is moving forward at velocity vcm . Also, because of rotation the point is moving
downward at velocity vrel = Rω = vcm . So, this point has a speed
2
2
v = vcm
+ vcm
= 2vcm = 2(20 m/s) = 28 m/s
Assess:
The speed v is independent of the radius of the wheel.
12.52 Model: The triangle is a rigid body rotating about its center of mass perpendicular to the plane of the
triangle. The center of mass of any symmetrical object of uniform density is at the physical center of the object.
Visualize:
The distance to one tip of the triangle from the center of mass is given by r cos30° = (5.0/2) cm, which yields
r = (2.5 cm)/ cos30° = 2.9 cm.
Solve: The speed of the tip is
v = rω = (2.9 cm)(120 rpm) = (2.9 cm)(4π rad/s) = 36 cm/s
12.53. Visualize:
Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula
for the x component of the center of mass is
1
xcm =
x dm
M∫
The area of the steel plate is A = 12 (0.2 m)(0.3 m) = 0.030 m 2 . Mass dm in the strip is the same fraction of M as
dA is of A. Thus
dm dA
M
⎛ 0.800 kg ⎞
=
⇒ dm = dA = ⎜
dA = (26.67 kg/m 2 )l dx
2 ⎟
M
A
A
⎝ 0.030 m ⎠
The relationship between l and x is
l
x
2
=
⇒l = x
0.20 m 0.30 m
3
Therefore,
0.3 m
1
(17.78 kg/m 2 ) x3
(17.78 kg/m 2 ) (0.3 m) 2
2 ⎛ 2⎞ 2
xcm =
(26.67
kg/m
)
x
dx
=
=
= 20 cm
⎜
⎟
M∫
M
3 0m
0.8 kg
3
⎝ 3⎠
Due to symmetry ycm = 0 cm.
12.54.
Visualize:
Solve: Build the plate from the three shapes 1, 2, and 3. The center of mass of the complete shape is the center of
mass of the center of masses of the three smaller shapes.
m1 ( xcm )1 + m2 ( xcm )2 + m3 ( xcm )3
xcm =
m1 + m2 + m3
ycm =
m1 ( ycm )1 + m2 ( ycm )2 + m3 ( ycm )3
m1 + m2 + m3
The masses of each of the smaller shapes are a fraction of the larger shape’s mass by the ratio of areas. The
following table will be useful in the solution.
Shape
A
mass
xcm
ycm
2
All
M
xcm
ycm
32 cm
1
24 cm 2
24
32
M
−1.0 cm
0 cm
2
2.0 cm
2
2.0
32
M
2.0 cm
−2.5 cm
6.0 cm
2
6.0
32
M
2.0 cm
1.5 cm
3
Thus,
⎛ 24 ⎞
⎛ 2.0 ⎞
⎛ 6.0 ⎞
M ⎟ ( −1.0 cm ) + ⎜
M ⎟ ( 2.0 cm ) + ⎜
M ⎟ ( 2.0 cm )
⎜
32 ⎠
32
32
⎝
⎝
⎠
⎝
⎠
xcm =
= −0.25 cm
M
Similarly, ycm = 0.125 cm.
Assess: The plate’s center of mass of (−0.25 cm, 0.125 cm) is reasonable. The cutout means more of the mass
is left of center and above the center.
12.55.
Model:
The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.
Visualize:
Solve:
(a) From Table 12.2, the moment of inertia of a disk about its center is
1
1
I = MR 2 = (2.0 kg)(0.10 m) 2 = 0.010 kg m 2
2
2
(b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:
I = I center + Mh 2 = (0.010 kg m 2 ) + (2.0 kg)(0.10 m) 2 = 0.030 kg m 2
Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the
edge than about the center.
12.56. Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is
thin.
Visualize: Please refer to P12.56.
Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1 , and mass
m2 . Using Table 12.2 for the moment of inertia of the rod, we get
2
I rod = I rod about center + I m1 + I m2 =
=
Assess:
1
⎛ L⎞
⎛ L⎞
ML2 + m1 ⎜ ⎟ + m2 ⎜ ⎟
12
2
⎝ ⎠
⎝ 4⎠
1
1
1
L2 ⎛ M
m ⎞
ML2 + m1L2 + m2 L2 = ⎜ + m1 + 2 ⎟
12
4
16
4⎝ 3
4 ⎠
With m1 = m2 = 0 kg, I rod 121 ML2 , as expected.
2
12.57. Visualize:
We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of
the rod.
Solve: The moment of inertia can be calculated as follows:
x2
I = ∫ x 2 dm
x1
dm dx
M
=
⇒ dm = dx
M
L
L
L−d
L−d
⇒I =
and
3
M
M
1⎛ M ⎞
⎛M ⎞x
x 2 dx = ⎜ ⎟
= ⎜ ⎟[( L − d )3 − ( − d )3 ] = [( L − d )3 + d 3 ]
∫
L −d
3⎝ L ⎠
3L
⎝ L ⎠ 3 −d
For d = 0 m, I = 13 ML2 , and for d = 12 L,
I=
3
3
M ⎡⎛ L ⎞ ⎛ L ⎞ ⎤ 1
2
⎢⎜ ⎟ + ⎜ ⎟ ⎥ = ML
3L ⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎦⎥ 12
Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in
Table 12.2.
12.58. Visualize:
Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let
dA = 2π rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as
dA is of the total area A.
(a) The moment of inertia can be calculated as follows:
I disk = ∫ r 2 dm and dm =
M
M
dA =
( 2π r ) dr
A
π ( r22 − r12 )
r
⇒ I disk =
r
r2
2
M
2M 2
2M r 4
r 2 (2π r ) dr = 2 2 ∫ r 3dr = 2 2
2
2 ∫
π ( r2 − r1 ) r1
( r2 − r1 ) r1
( r2 − r1 ) 4 r1
=
M
2M
r 4 − r14 ) = ( r22 + r12 )
2
2 ( 2
2
4 ( r2 − r1 )
Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is I disk = 12 M ( R 2 + r 2 ).
(b) For r = 0 m, I disk = 12 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center.
Additionally, for r ≅ R, we have I = MR 2 . This is the expression for the moment of inertia of a cylindrical hoop
or ring about the center.
(c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we
choose the bottom of the incline as the zero of potential energy, and use vcm = ω R, the energy conservation
equation K f = U i is
v2
1 2 1
1⎛ M ⎞
1
2
2
Iω + Mvcm
= Mgh ⇒ ⎜ ⎟ ( R 2 + r 2 ) cm2 + Mvcm
= Mgyi = Mg (0.50 m)sin 20°
R
2
2
2⎝ 2 ⎠
2
2
2
1 2
1 r2 ⎞
2 ⎛ R +r ⎞
2 ⎛1
v
v
⇒ vcm
+
=
+
+ 2 ⎟ = 1.6759 m 2 /s 2
⎜
⎟
⎜
cm
cm
2
⎝ 4R ⎠ 2
⎝ 2 4 4R ⎠
(0.015 m) 2 ⎞
2 ⎛3
vcm
= 1.6759 m 2 /s 2 ⇒ vcm = 1.37 m/s
⎜ +
2 ⎟
4
4(0.020
m)
⎝
⎠
For a sliding particle on a frictionless surface K f = U i , so
v
1 2
mvf = mgyi ⇒ vf = 2 gyi = 2 g (0.50 m)sin 20° = 1.83 m/s ⇒ cm = 0.75
vf
2
That is, vcm is 75% of the speed of a particle sliding down a frictionless ramp.
12.59.
Model:
Visualize:
Solve:
The plate has uniform density.
The moment of inertia is
I = ∫ r 2 dm.
Let the mass of the plate be M. Its area is L2 . A region of area dA located at (x,y) has mass
M
M
dm = dA = 2 dx dy. The distance from the axis of rotation to the point (x, y) is r = x 2 + y 2 . With
A
L
L
L
L
L
− ≤ x ≤ and − ≤ y ≤ ,
2
2
2
2
L
2
I= ∫
L
2
⎛M ⎞
M
2
2
L
2
⎛ x3
⎞
L
2
∫ ( x + y ) ⎜⎝ L ⎟⎠ dx dy = L ∫ ⎜⎝ 3 + y x ⎟⎠
L
L
−
−
2
2
2
2
2
L
−
2
L
dy
−
L
L
2
L
⎛
⎞
M 2 ⎛ L3 y 2 L ⎛ − L3 y 2 L ⎞ ⎞
M 2 ⎛ L3
M ⎜ L4
y3 2 ⎟
2⎞
= 2 ∫ ⎜⎜ +
−⎜
−
⎟ ⎟⎟dy = 2 ∫ ⎜ + Ly ⎟dy = 2 ⎜ + L
2 ⎝ 24
2 ⎠⎠
3 − L ⎟⎟
L L ⎝ 24
L L ⎝ 12
L ⎜ 12
⎠
−
−
2 ⎠
⎝
2
2
=
⎛ L3 L3 ⎞ ⎞ 1 2
M ⎛ L4
⎜ + L ⎜ + ⎟ ⎟⎟ = ML
2 ⎜
L ⎝ 12
⎝ 24 24 ⎠ ⎠ 6
12.60.
Solve:
From Equation 12.16,
I = ∫ ( x 2 + y 2 )dm
A small region of area dA has mass dm, and
M
M
dA =
dx dy
A
A
The area of the plate is 12 (0.20 m)(0.30 m) = 0.030 m 2 . So
dm =
M ⎛ 0.800 kg ⎞
2
=⎜
⎟ = 26.67 kg m
A ⎝ 0.030 m 2 ⎠
1
1
1
The limits for x are 0 ≤ x ≤ 30 cm. For a particular value of x, − x ≤ y ≤ x. Note that ± is the slope of the
3
3
3
top and bottom edges of the triangle. Therefore,
30 cm
I=
1
x
3
∫ ∫ ( x + y )( 26.67 kg/m ) dx dy
0
2
2
2
1
− x
3
1
x
⎛
y3 ⎞ 3
= ( 26.67 kg/m ) ∫ ⎜ x 2 y + ⎟ dx
3 ⎠ −1 x
0 ⎝
30 cm
2
3
⎛2
2 x3 ⎞
= ( 26.67 kg/m 2 ) ∫ ⎜ x3 +
⎟dx
81 ⎠
0 ⎝3
30 cm
30 cm
⎛ 56 ⎞⎛ 1 ⎞
= ( 26.67 kg/m 2 ) ⎜ ⎟⎜ x 4 ⎟
⎝ 81 ⎠⎝ 4 ⎠ 0
= 0.037 kg m 2
12.61. G Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium
G
( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). We also apply the model of static friction.
Visualize:
G
Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other
G
G
G
hand, the floor exerts both the normal force n1 and the static frictional force f s . The gravitational force FG on the
ladder acts through the center of mass of the ladder.
G
G
Solve: The x- and y-components of Fnet = 0 N are
∑F = n − f = 0 N ⇒ f = n
x
2
s
s
2
∑F = n − F = 0 N ⇒ n = F
y
1
G
1
G
The minimum angle occurs when the static friction is at its maximum value f s max = μ s n1. Thus we have
n2 = f s = μs n1 = μ s mg . We choose the bottom corner of the ladder as a pivot point to obtain τ net , because two
forces pass through this point and have no torque about it. The net torque about the bottom corner is
τ net = d1mg − d 2 n2 = (0.5L cosθ min )mg − ( L sinθ min ) μ s mg = 0 N m
⇒ 0.5cosθ min = μs sin θ min ⇒ tanθ min =
0.5
μs
=
0.5
= 1.25 ⇒ θ min = 51°
0.4
12.62.
Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
Solve:
G
G
To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium (τ net = 0 Nm). The first condition is
∑ F = −( F )
y
G beam
− ( FG )student + F1 + F2 = 0 N
⇒ F1 + F2 = ( FG ) beam + ( FG )student = (100 kg + 80 kg)(9.80 m/s 2 ) = 1764 N
Taking the torques about the left end of the beam, the second condition is
−( FG ) beam (1.5 m) − ( FG )student (2.0 m) + F2 (3.0 m) = 0 N m
−(100 kg)(9.8 m/s 2 )(1.5 m) − (80 kg)(9.8 m/s 2 )(2.0 m) + F2 (3.0 m) = 0 N m
⇒ F2 = 1013 N
From F1 + F2 = 1764 N, we get F1 = 1764 N − 1013 N = 0.75 kN.
Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any
point on the body of interest.
12.63.
Model: The structure is a rigid body.
Visualize:
Solve:
We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv
because the pivot point passes through the line of application of Fh and Fv and therefore these forces do not
exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write
τ about left end = −( FG ) B (3.0 m) − ( FG ) W (4.0 m) + (T sin150°)(6.0 m) = 0 N m
Using ( FG ) B = (1450 kg)(9.8 m/s 2 ) and ( FG ) W = (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield
T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.
12.64. Model: G Model the beam as a rigid body. For the beam not to fall over, it must be both in translational
G
equilibrium ( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m).
Visualize:
The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting
G
G
on the beam. F1 and F2 are from the two supports, FG is the gravitational force on the beam, and FG is
( )
( )
b
B
the gravitational force on the boy.
Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium
is
−( FG ) b (2.5 m) + F2 (3.0 m) − ( FG ) B x = 0 N m
−(40 kg)(9.80 m/s 2 )(2.5 m) + F2 (3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m
The equation for translation equilibrium is
∑F = 0 N = F + F − (F ) − (F )
y
1
2
G b
G B
⇒ F1 + F2 = ( FG )b + ( FG )B = (40 kg + 20 kg)(9.8 m/s 2 ) = 588 N
Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2 , we can
simplify the torque equation to:
−(40 kg)(9.80 m/s 2 )(2.5 m) + (588 N)(3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m
⇒ x = 4.0 m
Thus, the distance from the right end is 5.0 m − 4.0 m = 1.0 m.
12.65.
Visualize:
Please refer to Figure P12.65.
Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of
bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a
base of support. To determine the relative positions of the bricks, work from the top down. The top brick can
extend past the second brick by L 2. For maximum extension, their combined center of gravity will be at the
edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the
fourth brick. The combined center of gravity of all four bricks will be over the edge of the table.
Measuring from the left edge of the brick 2, the center of gravity of the top two bricks is
⎛L⎞
m ⎜ ⎟ + mL
m1 x1 + m2 x2
3
2
( x12 )com =
= ⎝ ⎠
= L.
m1 + m2
2m
4
Thus the top two bricks can extend L 4 past the edge of the third brick. The top three bricks have a center of
mass
⎛L⎞
⎛ 3L ⎞
⎛ 5L ⎞
m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟
m1 x1 + m2 x2 + m3 x3
2⎠
4 ⎠
⎝
⎝
⎝ 4 ⎠ = 5 L.
=
( x123 )com =
m1 + m2 + m3
3m
6
Thus the top three bricks can extend past the edge of the fourth brick by L 6. Finally, the four bricks have a
combined center of mass at
⎛L⎞
⎛ 4L ⎞
⎛ 11L ⎞
⎛ 17 L ⎞
m⎜ ⎟ + m⎜
⎟ + m⎜
⎟ + m⎜
⎟
2⎠
6 ⎠
12 ⎠
⎝
⎝
⎝
⎝ 12 ⎠ = 7 L.
( x1234 )com =
4m
8
The center of gravity of all four bricks combined is 7 L 8 from the left edge of the bottom brick, so brick 4 can
extend L 8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge
is
L L L L 25
d max = + + + =
L.
8 6 4 2 24
Thus, yes, it is possible that no part of the top brick is directly over the table because d max > L.
Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net
gravitational torque is zero, and the bricks do not fall over.
12.66. Model:
Visualize:
The pole is a uniform rod. The sign is also uniform.
Solve:
The geometry of the rod and cable give the angle that the cable makes with the rod.
⎛ 250 ⎞
θ = tan −1 ⎜
⎟ = 51.3°
⎝ 200 ⎠
The rod is in rotational equilibrium about its left-hand end.
⎛1⎞
⎛1⎞
τ net = 0 = − (100 cm )( FG )P − ( 80 cm ) ⎜ ⎟ ( FG )S − ( 200 cm ) ⎜ ⎟ ( FG )S + ( 200 cm ) T sin 51.3°
⎝ 2⎠
⎝ 2⎠
= − (100 cm )( 5.0 kg ) ( 9.8 m/s 2 ) − mS ( 9.8 m/s 2 ) (140 cm ) + (156 cm ) T
With T = 300 N , mS = 30.6 kg.
Assess: A mass of 30.6 kg is reasonable for a sign.
12.67.
Model: The hollow cylinder is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the ground.
Solve: (a) Newton’s second law for the block is −( FG ) B + T = mBa y , where T is the tension in the string,
( FG ) B = mB g is the gravitational force on the block, and a y is the acceleration of block. The string tension
exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow
cylinder is
τ = −TR = Iα = I
ay
R
⇒T = −
Ia y
R2
where we used the acceleration constraint a y = α R. With this expression for T, Newton’s second law for the
block becomes
− mB g −
Ia
−mB g
= mB a y ⇒ a y =
2
R
(mB + I / R 2 )
The moment of inertia of a hollow cylinder is I = mC R 2 , so the equation for a y simplifies to
ay =
−mB g
−(3.0 kg)(9.8 m/s 2 )
=
= −5.88 m/s 2
mB + mC
3.0 kg + 2.0 kg
The speed of the block just before it hits the ground can now be found using kinematics:
v12 = v02 + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2(−5.88 m/s 2 )(0 m − 1.0 m) ⇒ v1 = 3.4 m/s
(b) The conservation of energy equation K1 + U g1 = K 0 + U g0 for the system (block + cylinder + earth) is
1
1
1
1
mBv12 + Iω12 + mB gy1 = mBv02 + Iω 02 + mB gy0
2
2
2
2
2
v
m ⎞
1
1
⎛m
mBv12 + ( mC R 2 ) 12 + 0 J = 0 J + 0 J + mB gy0 ⇒ v12 ⎜ B + C ⎟ = mB gy0
R
2
2
2 ⎠
⎝ 2
⇒ v12 =
2mB gy0 2(3.0 kg)(9.8 m/s 2 )(1.0 m)
=
= 11.76 m 2 /s 2 ⇒ v1 = 3.4 m/s
mB + mC
(3.0 kg) + (2.0 kg)
Assess: Newton’s second law and the conservation of energy method give the same result for the block’s final
velocity.
12.68.
Model: The bar is a solid body rotating through its center.
Visualize:
Solve:
(a) The two forces form a couple. The net torque on the bar about its center is
Iα
L
where F is the force produced by one of the air jets. We can find I and α as follows:
τ net = LF = Iα ⇒ F =
1
1
ML2 = (0.50 kg)(0.60 m) 2 = 0.015 kg m 2
12
12
ω1 = ω 0 + α (t1 − t0 ) ⇒ 150 rpm = 5.0π rad/s = 0 rad + α (10 s − 0 s) ⇒ α = 0.50π rad/s 2
I=
⇒F=
(0.015 kg m 2 )(0.5π rad/s 2 )
= 0.0393 N
(0.60 m)
The force F = 39 mN.
(b) The torque of a couple is the same about any point. It is still τ net = LF . However, the moment of inertia has
changed.
LF
1
1
where I = ML2 = (0.500 kg)(0.6 m) 2 = 0.060 kg m 2
3
3
I
(0.0393 N) × (0.60 m)
⇒α =
= 0.393 rad/s 2
0.060 kg m 2
τ net = LF = Iα ⇒ α =
Finally,
ω1 = ω 0 + α (t1 − t0 ) = 0 rad/s + (0.393 rad/s 2 )(10 s − 0 s)
= 3.93 rad/s =
(3.93)(60)
rpm = 37.5 rpm
2π
The angular speed is 38 rpm.
Assess: Note that ω ∝ α and α ∝ 1/ I . Thus, ω ∝ 1/ I . I about the center of the rod is 4 times smaller than I
about one end of the rod. Consequently, ω is 4 times larger.
12.69.
Model: The flywheel is a rigid body rotating about its central axis.
Visualize:
Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about
the axis of rotation is that of a disk:
1
1
I = MR 2 = (250 kg)(0.75 m) 2 = 70.31 kg m 2
2
2
The angular acceleration is calculated as follows:
τ net = Iα ⇒ α = τ net / I = (50 N m)/(70.31 kg m 2 ) = 0.711 rad/s 2
Using the kinematic equation for angular velocity gives
ω1 = ω 0 + α (t1 − t0 ) = 1200 rpm = 40 π rad/s = 0 rad/s + 0.711 rad/s 2 (t1 − 0 s)
⇒ t1 = 177 s
(b) The energy stored in the flywheel is rotational kinetic energy:
1
1
K rot = Iω12 = (70.31 kg m 2 )(40π rad/s) 2 = 5.55 × 105 J
2
2
5
The energy stored is 5.6 × 10 J.
energy delivered (5.55 × 105 J)/2
(c) Average power delivered =
=
= 1.39 × 105 W = 139 kW
time interval
2.0 s
(d) Because τ = Iα , ⇒ τ = I
− ω half energy ⎞
⎛ω
Δω
= I ⎜ full energy
⎟ . ω full energy =ω1 (from part (a)) = 40π rad/s. ω half energy
Δt
Δt
⎝
⎠
can be obtained as:
1 2
1
Iω half energy = K rot ⇒ ω half energy =
2
2
K rot
5.55 × 105 J
=
= 88.85 rad/s
I
70.31 kg m 2
Thus
⎛ 40 π rad/s − 88.85 rad/s ⎞
⎟ = 1.30 kN m
2.0 s
⎝
⎠
τ = (70.31 kg m 2 ) ⎜
12.70.
Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution
of a disk and that the string does not slip.
Visualize:
Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same.
Solve: Applying Newton’s second law to m1 , m2 , and the pulley yields the three equations:
T1 − ( FG )1 = m1a1
− ( FG )2 + T2 = m2 a2
T2 R − T1R − 0.50 N m = Iα
Noting that − a2 = a1 = a, I = 12 mp R 2 , and α = a/R, the above equations simplify to
T1 − m1 g = m1a
m2 g − T2 = m2 a
0.50 N m
⎛1
⎞⎛ a ⎞ 1 0.50 N m 1
T2 − T1 = ⎜ mp R 2 ⎟⎜ ⎟ +
= mp a +
R
2
0.060 m
⎝2
⎠⎝ R ⎠ R
Adding these three equations,
1 ⎞
⎛
(m2 − m1 ) g = a ⎜ m1 + m2 + mp ⎟ + 8.333 N
2 ⎠
⎝
(m − m1 ) g − 8.333 N (4.0 kg − 2.0 kg)(9.8 m/s 2 ) − 8.333 N
⇒a= 2
=
= 1.610 m/s 2
2.0 kg + 4.0 kg + (2.0 kg/2)
m1 + m2 + 12 mp
We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:
1
1
y1 = y0 + v0 (t1 − t0 ) + a2 (t1 − t0 ) 2 ⇒ 0 = 1.0 m + 0 + (−1.610 m/s 2 )(t1 − 0 s) 2
2
2
2(1.0 m)
⇒ t1 =
= 1.11 s
(1.610 m/s 2 )
12.71.
Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward
force on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force
on block m1 to the right, but a leftward force on the outer edge of the pulley.
Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T − m2 g = m2 a2 . Using the constraint
− a2 = + a1 = a, we have T = m1a and −T + m2 g = m2 a. Adding these equations, we get m2 g = (m1 + m2 )a, or
a=
m2 g
mm g
⇒ T = m1a = 1 2
m1 + m2
m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are
different. Newton’s second law for m1 and the pulley are:
T1 = m1a
and
T1R − T2 R = − Iα
We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα . Thus, T1 = m1a and
T2 − T1 =
I a aI
=
R R R2
Adding these two equations gives
I ⎞
⎛
T2 = a ⎜ m1 + 2 ⎟
R
⎝
⎠
Newton’s second law for m2 is T2 − m2 g = m2 a2 = − m2 a. Using the above expression for T2 ,
I ⎞
m2 g
⎛
a ⎜ m1 + 2 ⎟ + m2 a = m2 g ⇒ a =
R ⎠
m1 + m2 + I / R 2
⎝
Since I = 12 mp R 2 for a disk about its center,
a=
m2 g
m1 + m2 + 12 mp
With this value for a we can now find T1 and T2 :
T1 = m1a =
Assess:
m1m2 g
m1 + m2 + 12 mp
T2 = a (m1 + I / R 2 ) =
1
m2 g
1 ⎞ m2 ( m1 + 2 mp ) g
⎛
m
+
m
=
1
p
⎜
⎟
( m1 + m2 + 12 mp ) ⎝ 2 ⎠ m1 + m2 + 12 mp
For m = 0 kg, the equations for a, T1 , and T2 of part (b) simplify to
a=
m2 g
m1 + m2
These agree with the results of part (a).
and T1 =
m1m2 g
m1 + m2
and
T2 =
m1m2 g
m1 + m2
12.72.
Model: The disk is a rigid spinning body.
Visualize: Please refer to Figure P12.72. The initial angular velocity is 300 rpm or (300)(2π )/60 = 10π rad/s.
After 3.0 s the disk stops.
Solve: Using the kinematic equation for angular velocity,
ω1 = ω 0 + α (t1 − t0 ) ⇒ α =
ω1 − ω 0
t1 − t0
=
(0 rad/s − 10π rad/s) −10π
=
rad/s 2
(3.0 s − 0 s)
3
Thus, the torque due to the force of friction that brings the disk to rest is
τ = Iα = − fR ⇒ f = −
( 2 mR )α = − 1 (mR)α = − 1 (2.0 kg)(0.15 m) ⎛ −10 π rad/s2 ⎞ = 1.57 N
Iα
=−
⎜
⎟
R
R
2
2
3
⎝
⎠
1
2
The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.
12.73.
Model: The entire structure is a rigid rotating body. The two thrust forces are a couple that exerts a
torque on the structure about its center of mass. We will assume the thrust forces are perpendicular to the connecting
tunnel.
Visualize: Please refer to Figure 12.30. We chose a coordinate system in which m1 and m2 are on the x-axis
and m1 = 100,000 kg is at the origin. m3 is the mass of the tunnel, whose center is at x3 = 45 m.
Solve: (a) Assuming the center of mass of the tunnel is at the center of the tunnel, we get
xcm =
m1 x1 + m2 x2 + m3 x3
m1 + m2 + m3
xcm =
(1.0 × 105 kg)(0 m) + (2.0 × 105 kg)(90 m) + (5.0 × 104 kg)(45 m)
= 57.9 m
1.0 × 105 kg + 2.0 × 105 kg + 5.0 × 104 kg
The center of mass of the entire structure is 58 m from the 100,000 kg rocket.
(b) Initially, the angular velocity is zero. The structure’s angular velocity after 30 s is
ω1 = ω 0 + α (t1 − t0 ) = 0 rad/s + α (30 s − 0 s) = α (30 s)
The angular acceleration α can be found from τ = Iα , where τ is the net torque on the structure and I is its
moment of inertia. The two thrusts form a couple with torque
τ = lF = (90 m)(50,000 N) = 4.50 × 106 N m
2
2⎞
⎛1
I = I m1 + I m2 + I tunnel = m1 x12 + m2 x22 + ⎜ m3 ( 90 m ) + m3 ( 58 m − 45 m ) ⎟
12
⎝
⎠
1
2
5
2
5
2
4
= (1.0 × 10 kg)(57.9 m) + (2.0 × 10 kg)(32.1 m) + (5 × 10 kg)(90 m) 2 + ( 5 × 104 kg ) (13 m )
12
= 5.83 × 108 kg m 2
⇒α =
τ
I
=
4.5 × 106 N m
= 7.71 × 10−3 rad/s 2
5.83 × 108 kg m 2
⇒ ω1 = (30 s)(7.71 × 10−3 rad/s 2 ) = 0.23 rad/s
Assess: Note that the parallel axis theorem was used in finding the moment of inertia of
the tunnel.
12.74.
Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline
without slipping. We also assume that the coefficient of rolling friction is zero.
Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in
gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential
energy.
Solve: The conservation of energy equation K f + U gf = K i + U gi is
1
1
1
1
2
2
M (v1 )cm
+ I cm (ω1 ) 2 + Mgy1 = M (v0 )cm
+ I cm (ω 0 ) 2 + Mgy0
2
2
2
2
1
1
2
1
1
(v ) 2
⎛
⎞
2
2
2
0 J + 0 J + Mgy1 = M (v0 )cm
+ ⎜ MR 2 ⎟ (ω 0 )cm
+ 0 J ⇒ Mgy1 = M (v0 )cm
+ MR 2 0 2cm
2
2⎝ 3
2
3
R
⎠
5
(v ) 2
5
5 (5.0 m/s) 2
2
⇒ gy1 = (v0 )cm
⇒ y1 = 6 0 cm =
= 2.126 m
6
6 9.8 m/s 2
g
The distance traveled along the incline is
y1
2.126 m
=
= 4.3 m
sin 30°
0.5
Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom
of the incline is approximately 10 mph.
s=
12.75. Model:
Visualize:
Solve:
The masses are particles.
(a) The moment of inertia of the barbell is
2
I = I M + I m = Mx 2 + m ( L − x )
The rotational kinetic energy is therefore
K rot =
(
)
1 2 1
2
Iω = Mx 2 + m ( L − x ) ω 2
2
2
To find x such that K rot is a minimum, set
dK rot
1
= 0 = ( 2 Mx − 2m ( L − x ) )ω 2
2
dx
(b) This is the center of mass location measured from the mass M.
⇒x=
m
L
m+M
12.76. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational
potential energy is transformed into rotational kinetic energy as the disk is released.
Visualize:
We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the
center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is
coincident with the origin of the coordinate system.
Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus
τ = (mg ) R = (5.0 kg)(9.8 m/s 2 )(0.30 m) = 14.7 N m
The moment of inertia about the disk’s edge is obtained using the parallel-axis theorem:
1
3
⎛ 3⎞
I = I cm + mR 2 = mR 2 + mR 2 = mR 2 = ⎜ ⎟ (5.0 kg)(0.30 m)2 = 0.675 kg m 2
2
2
⎝ 2⎠
14.7 N m
τ
⇒α = =
= 22 rad/s 2
I 0.675 kg m 2
(b) The energy conservation equation K f + U gf = K i = U gi is
1 2
1
1
Iω1 + mgy1 = Iω 02 + mgy0 ⇒ Iω12 + 0 J = 0 J + mgR
2
2
2
ω1 =
2mgR
2(5.0 kg)(9.8 m/s 2 )(0.30 m)
=
= 6.6 rad/s
I
0.675 kg m 2
Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below
the axle is reasonable.
12.77.
Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational
torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass
into rotational kinetic energy.
Visualize:
We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of
mass is a distance R above the origin, but it is a distance R below the origin in the final position.
Solve: (a) Applying the parallel-axis theorem, I edge = I cm + mR 2 = mR 2 + mR 2 = 2mR 2 . Using this expression in
the energy conservation equation K f + U gf = K i + U gi yields:
1
1
I edgeω12 + mgy1 = I edgeω 02 + mgy0
2
2
1
2g
(2mR 2 )ω12 − mgR = 0 J + mgR ⇒ ω1 =
2
R
(b) The speed of the lowest point on the hoop is
2g
(2 R ) = 8 gR
R
Note that the speed of the lowest point on the loop involves a distance of 2R instead of R.
v = (ω1 )(2 R) =
Assess:
12.78.
Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod.
The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod’s
center of mass into rotational kinetic energy.
Visualize:
We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance
1
L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational potential
2
energy.
Solve: (a) The energy conservation equation for the rod K f + U gf = K i + U gi is
1 2
1
Iω1 + mgy1 = Iω 02 + mgy0
2
2
1⎛1 2⎞ 2
⎜ mL ⎟ω1 + 0 J = 0 J + mg ( L /2) ⇒ ω1 = 3 g / L
2⎝3
⎠
(b) The speed at the tip of the rod is vtip = (ω1 ) L = 3 gL .
12.79.
Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is
vertical and the sphere solid.
Visualize: Please refer to Figure P12.79. The sphere rotates because the string wrapped around the rod exerts a torque τ.
Solve: The torque exerted by the string on the rod is τ = Tr.
From the parallel-axis theorem, the moment of inertia of the sphere about the rod’s axis is
2
MR 2 13
⎛R⎞ 2
I off center = I cm + M ⎜ ⎟ = MR 2 +
=
MR 2
4
20
⎝2⎠ 5
From Newton’s second law,
α=
τ
I
=
Tr
20Tr
=
(13MR 2 /20) 13MR 2
12.80.
Model: The pulley is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the floor. The pulley rotates about its center.
Solve: Using kinematics for the physics book (mass = m1 ),
1
1
y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2
0 m = 1.0 m + 0 m + a (0.71 s − 0 s) 2 ⇒ a = −3.967 m/s 2
2
2
Since the torque acting on the pulley is τ = −TR = Iα , we have
I =−
TR
α
=−
TR
TR 2
=−
a /R
a
We can compare the measured value of I for the pulley with the theoretical value. We first must find the tension
T. From the free-body diagram, Newton’s second law of motion is T − m1 g = m1a. This means
T = m1 ( g + a ) = (1.0 kg)(9.80 m/s 2 − 3.967 m/s 2 ) = 5.833 N
With these values of T and a, we can now find I as:
I =−
TR 2
(5.833 N)(0.06 m) 2
=−
= 5.3 × 10−3 kg m 2
a
(−3.967 m/s 2 )
Let us now calculate the theoretical values of I:
hoop about center: I = MR 2 = (2.0 kg)(0.06 m) 2 = 7.2 × 10 −3 kg m 2
1
1
disk about center: I disk = MR 2 = (2.0 kg)(0.06 m) 2 = 3.6 × 10−3 kg m 2
2
2
Since I disk < I < I hoop , the mass of the disk is not uniformly distributed. The mass is concentrated near the rim.
12.81.
Model: The angular momentum of the satellite in the elliptical orbit is a constant.
Visualize:
Solve: (a) Because the gravitational
force is always along the same direction as the direction of the moment
G G G
arm vector, the torque τ = r × Fg is zero at all points on the orbit.
(b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is
G
perpendicular to r at points a and b, so β = 90° and L = mvr. Thus
⎛r ⎞
Lb = La ⇒ mvb rb = mva ra ⇒ vb = ⎜ a ⎟ va
⎝ rb ⎠
30,000 km
30,000 km
− 9000 km = 6000 km and rb =
+ 9000 km = 24,000 km
ra =
2
2
⎛ 6000 km ⎞
⇒ vb = ⎜
⎟ (8000 m/s) = 2000 m/s
⎝ 24,000 km ⎠
(c) Using the conservation of angular momentum Lc = La , we get
⎛r ⎞
mvc rc sin β c = mva ra ⇒ vc = ⎜ a ⎟ va /sin β c
⎝ rc ⎠
rc = (9000 km)2 + (12,000 km) 2 = 1.5 × 107 m
From the figure, we see that sin β c = 12,000 15,000 = 0.80. Thus
⎛ 6000 km ⎞ (8000 m/s)
vc = ⎜
= 4000 m/s
⎟
0.80
⎝ 15,000 km ⎠
12.82. Model: The clay balls are particles and undergo a totally inelastic collision. Linear momentum is
conserved during the collision.
Visualize:
Solve:
(a)
The
angle
is
measured counterclockwise
⎛1⎞
β1 = 180° + tan −1 (1) = 225°, and β 2 = tan −1 ⎜ ⎟ = 26.6°. So
⎝ 2⎠
L = L1 + L2 + m1r1v1 sin β1 + m2 r2v2 sin β 2
= ( 0.015 kg )
from
G
r
to
G
v.
( 2 m ) ( 2 m/s ) sin 225° + ( 0.025 kg ) ( 5 m ) ( 2.0 m/s ) sin 26.6°
= 0.020 kg m 2 /s.
Note that the signs of L1 and L2 agree with those determined by the right-hand rule.
(b) At the instant before the clay balls collide they are located at (1.5 m, 1.0 m). Here,
⎛ 2⎞
β 2 = tan −1 ⎜ ⎟ = 33.7°
β1 = 180° + β 2 = 146.3°
⎝ 3⎠
So
( (1.0 m) + (1.5 m) )( 2.0 m/s) sin 213.7
+ ( 0.025 kg ) ( (1.0 m ) + (1.5 m ) ) ( 2.0 m/s ) sin 33.7°
L = ( 0.015 kg )
2
2
2
2
= 0.020 kg m 2 /s
(c) The clay balls have a final speed v after the collision. Linear momentum is conserved.
p1i + p2i = p(1+ 2) f
( 0.015 kg )( 2.0 m/s ) + ( 0.025 kg )( −2.0 m/s ) = ( 0.015 kg + 0.025 kg ) v
⇒ v = −0.50 m/s.
The balls are moving to the left.
The angle β = β 2 from part (b). The angular momentum after the collision is
L = ( 0.040 kg )
Assess:
From
( (1.0 m) + (1.5 m) )(0.50 m/s )sin 33.7° = 0.020 kg m /s
2
2
G
Angular momentum is also conserved since τ net = 0.
2
geometry,
Model: For the (bullet + block + rod) system, angular momentum is conserved. After the bullet is
stuck in the block, the mechanical energy of the system is conserved. Assume the block is small.
Visualize:
12.83.
The origin of the coordinate system was placed at the center-of-mass of the block as it freely hangs from the bottom
of the rod.
Solve: The initial angular momentum of the system about the pivot is due only to the bullet:
Li = mb vb L = (0.010 kg)vb (1.0 m) = (0.010 kg m)vb
The angular momentum immediately after the bullet hits and sticks in the block is equal to Iω . The moment of
inertia is
1
⎛1
⎞
I = I rod + I block + I bullet = mR L2 + mB L2 + mb L2 = ⎜ mR + mB + mb ⎟ L2
3
⎝3
⎠
⎡1
⎤
= ⎢ (1.0 kg) + 2.0 kg + 0.010 kg ⎥ (1.0 m) 2 = 2.343 kg m 2
⎣3
⎦
Angular momentum is conserved in the collision:
Lf = (2.343 kg m 2 )ω = Li = (0.010 kg m)vb
We need to determine ω before we can find vb . To find ω we use the conservation of mechanical energy
equation K f + U gf = K i + U gi as the pendulum swings out, which is
1 2
Iω + 0 J
2
L
1
(0.010 kg + 2.0 kg)(9.8 m/s 2 ) L(1 − cos30°) + (1.0 kg)(9.8 m/s 2 ) (1 − cos30°) = (2.343 kg m 2 )ω 2
2
2
0 J + (mb + mB ) gyB + mR gyR =
The energy equation can be further simplified to
2.6390 kg m 2 /s 2 + 0.6565 kg m 2 /s 2 = (1.1715 kg m 2 )ω 2 ⇒ ω = 1.677 rad/s
Finally, we can use the conservation of angular momentum equation to obtain the speed of the bullet:
vb =
(2.343 kg m 2 )(1.677 rad/s)
= 3.9 × 102 m/s
0.010 kg m
Assess: A speed of 390 m/s for a bullet is reasonable.
12.84.
Model: For the (bullet + door) system, the angular momentum is conserved in the collision.
Visualize:
G
G
Solve: As the bullet hits the door, its velocity v is perpendicular to r . Thus the initial angular momentum
about the rotation axis, with r = L, is
Li = mBvB L = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m 2 /s
After the collision, with the bullet in the door, the moment of inertia about the hinges is
1
1
I = I door + I bullet = mD L2 + mB L2 = (10.0 kg)(1.0 m) 2 + (0.010 kg)(1.0 m) 2 = 3.343 kg m 2
3
3
Therefore,
Lf = Iω = (3.343 kg m 2 )ω .
Using
the
angular
Lf = Li (3.343 kg m )ω = 4.0 kg m /s and thus ω = 1.20 rad/s.
2
2
momentum
conservation
equation
12.85.
Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial
gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is
a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere.
Visualize:
The zero of gravitational potential energy is chosen at the bottom of the slope.
Solve: (a) The energy conservation equation for the sphere or hoop K f + U gf = K i + U gi is
1
1
1
1
I (ω1 ) 2 + m(v1 ) 2 + mgy1 = I (ω 0 ) 2 + m(v0 ) 2 + mgy0
2
2
2
2
For the sphere, this becomes
2
1⎛ 2
1
2 ⎞ (v1 )s
2
⎜ mR ⎟ 2 + m(v1 )s + 0 J = 0 J + 0 J + mghs
2⎝ 5
2
⎠ R
⇒
7
(v1 )s2 = gh ⇒ (v1 )s = 10 gh /7 = 10(9.8 m/s 2 )(0.30 m)/7 = 2.05 m/s
10
For the hoop, this becomes
1
(v ) 2 1
(mR 2 ) 1 2 h + m(v1 ) h2 + 0 J = 0 J + 0 J + mghhoop
2
R
2
(v ) 2
⇒ hhoop = 1 h
g
For the hoop to have the same velocity as that of the sphere,
hhoop =
(v1 )s2 (2.05 m/s) 2
=
= 42.9 cm
g
9.8 m/s 2
The hoop should be released from a height of 43 cm.
(b) As we see in part (a), the speed of a hoop at the bottom depends only on the starting height and not on the
mass or radius. So the answer is No.
12.86.
Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from
above and stick on the turntable, the turntable slows down due to increased rotational inertia of the
(turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular
momentum of the system is conserved.
Visualize: The initial moment of inertia is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 and the final
moment of inertia is
I 2 = I1 + 2mR 2 = 0.010 kg m 2 + 2(0.500 kg) × (0.10 m) 2 = 0.010 kg m 2 + 0.010 kg m 2 = 0.020 kg m2
Let ω1 and ω 2 be the initial and final angular velocities. Then
I ω (0.010 kg m 2 )(100 rpm)
Lf = Li ⇒ ω 2 I 2 = ω1I1 ⇒ ω 2 = 1 1 =
= 50 rpm
I2
0.020 kg m 2
12.87.
Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the
(turntable + block) system, no external torques act as the block moves outward towards the outer edge. Angular
momentum is thus conserved.
Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (0.2 kg)(0.2 m) 2 = 0.0040 kg m 2 . As the block
reaches the outer edge, the final moment of inertia is
I 2 = I1 + mB R 2 = 0.0040 kg m 2 + (0.020 kg)(0.20 m) 2
= 0.0040 kg m 2 + 0.0008 kg m 2 = 0.0048 kg m 2
Let ω1 and ω 2 be the initial and final angular velocities, then the conservation of angular momentum equation is
I ω (0.0040 kg m 2 )(60 rpm)
= 50 rpm
Lf = Li ⇒ ω 2 I 2 = ω1I1 ⇒ ω 2 = 1 1 =
I2
(0.0048 kg m 2 )
Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of
inertia is reasonable.
12.88.
Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in
the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps
on the merry-go-round. Angular momentum is thus conserved.
Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the
angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to
I finalω final .
Solve: John’s initial angular momentum is that of a particle: LJ = mJ vJ R sin β = mJ vJ R. The angle β = 90° since
John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is
⎛1
⎞
I finalω final = Ldisk + LJ = ⎜ MR 2 ⎟ωi + mJ vJ R
⎝2
⎠
2π ⎛ rad ⎞
⎛1⎞
2
= ⎜ ⎟ (250 kg)(1.5 m) 2 (20 rpm) ⎜
⎟ + (30 kg)(5.0 m/s)(1.5 m) = 814 kg m /s
60 ⎝ rpm ⎠
⎝ 2⎠
⇒ ω final =
814 kg m 2 /s
I final
1
1
I final = I disk + I J = MR 2 + mJ R 2 = (250 kg)(1.5 m) 2 + (30 kg)(1.5 m)2 = 349 kg m 2
2
2
814 kg m 2 /s
ω final =
= 2.33 rad/s = 22 rpm
349 kg m 2
12.89. Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating
about its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum
of the car and track is conserved.
Visualize:
Solve:
The toy car’s steady speed of 0.75 m/s relative to the track means that
vc − vt = 0.75 m/s ⇒ v c = vt + 0.75 m/s,
where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum
implies that
Li = Lf
0 = I cω c + I tω t = ( mr 2 )ω c + ( Mr 2 )ω t = mω c + M ω t
The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track.
v
v
Since ω c = c , ω t = t , we have
r
r
0 = mvc + Mvt
⇒ m ( vt + 0.75 m/s ) + Mvt = 0
⇒ vt = −
M
( 0.200 kg )
( 0.75 m/s ) = −
( 0.75 m/s ) = −0.125 m/s
m+M
( 0.200 kg + 1.0 kg )
The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the
track is
v ( 0.125 m/s )
ωt = t =
= 0.417 rad/s clockwise.
r
0.30 m
In rpm,
⎛ rev ⎞⎛ 60 s ⎞
ω t = ( 0.417 rad/s ) ⎜
⎟⎜
⎟
⎝ 2π rad ⎠⎝ min ⎠
= 4.0 rpm
Assess: The speed of the track is less than that of the car because it is more massive.
12.90.
Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the
axis of the torso in the initial position and collapse into the torso in the final position.
Visualize:
Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is
that of a 66 cm long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis
theorem. In the final position, the moment of inertia is I 2 = 12 MR 2 . The equation for the conservation of angular
momentum Lf = Li can be written I 2ω 2 = I1ω1 ⇒ ω 2 = ( I1 / I 2 )ω1. Calculating I1 and I 2 ,
1
⎡1
⎤
I1 = M T R 2 + 2 ⎢ M A L2A + M A d 2 ⎥
2
12
⎣
⎦
1
2⎤
⎡1
2
= (40 kg)(0.10 m) + 2 ⎢ (2.5 kg)(0.66 m) 2 + (2.5 kg) ( 0.33 m + 0.10 m ) ⎥ = 1.306 kg m 2
2
⎣12
⎦
1
1
(1.306 kg m 2 )
(1.0 rev/s) = 5.8 rev /s
I 2 = MR 2 = (45 kg)(0.10 m) 2 = 0.225 kg m 2 ⇒ ω 2 =
2
2
(0.225 kg m 2 )
12.91.
Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop.
The mechanical energy of the marble is conserved.
Visualize:
Solve: The ball’s center of mass moves in a circle of radius R − r. The free-body diagram on the marble at its
highest position shows that Newton’s second law for the marble is
mv12
R−r
The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when
the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal
acceleration needed for the circular motion. For n → 0 N,
mg + n =
mg =
mv 2
⇒ v12 = g ( R − r )
(R − r)
Since rolling motion requires v12 = r 2ω12 , we have
ω12 r 2 = g ( R − r ) ⇒ ω12 =
g (R − r)
r2
The conservation of energy equation is
1
1
( K f + U gf ) top of loop = ( K i + U gi )initial ⇒ mv12 + Iω12 + mgy1 = mgy0 = mgh
2
2
Using the above expressions and I = 52 mr 2 the energy equation simplifies to
1
1⎛ 2⎞
⎛ g(R − r) ⎞
mg ( R − r ) + ⎜ ⎟ mr 2 ⎜
⎟ + mg 2( R − r ) = mgh ⇒ h = 2.7 ( R − r )
2
2
2⎝ 5⎠
⎝ r
⎠
12.92. Model: The Swiss cheese wedge is of uniform density—or at least uniform enough that its center of
mass is at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the
cheese will be treated as a particle, and the model of static friction will be used.
Visualize:
Solve: The angle at which the cheese starts sliding, θS , will be compared to the critical angle θ c for stability.
Use Newton’s second law with the free body diagram.
( Fnet ) x = 0 = fs − FG sinθ s
( Fnet ) y = 0 = n − FG cosθ s
With FG = mg , the y-direction equation gives n = mg cosθ . The cheese starts sliding when μ s is at its maximum
value. Combining that with the x-direction equation and fs = μ s n,
0 = μs ( mg cosθ s ) − mg sin θ s
⇒ θ s = tan −1 ( μ s ) = tan −1 ( 0.90 ) = 42°
The cheese will start sliding at an angle of 42°.
The center of mass of the cheese wedge can be found using the result of problem 12.53. There, the center of mass
of a triangle with the same proportions as the cheese wedge was found. So xcm is at the center of the cheese
wedge (by symmetry). The ycm can be found by proportional reasoning.
ycm
( 30 cm − 20 cm ) ⇒ y = 4.0 cm
=
cm
12 cm
30 cm
Note that here we have measured ycm from the base of the wedge.
Stability considerations require that the center of mass be no further than the left corner of the wedge. At the
critical angle geometry shown in the figure above, the right triangle formed by the wedge’s center of mass, lower
left corner, and center point of the base is a 45°-45°-90° triangle. So θ c = 45°.
Assess: The cheese will slide first as the incline reaches 42°. It would not topple until the angle reaches 45°. So
Emily is correct.
12.93. Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a
perfectly elastic collision in the absence of a net external torque. The rod is uniform.
Visualize: Please refer to Figure CP12.93.
Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be ω . Energy is
conserved, and angular momentum around the rod’s pivot point is conserved.
1
1
1
Ei = Ef ⇒ mv02 = mvf2 + I rodω 2
2
2
2
⎛d⎞
⎛d⎞
Li = Lf ⇒ mv0 ⎜ ⎟ = mvf ⎜ ⎟ + I rodω
⎝ 2⎠
⎝ 2⎠
This is two equations in the two unknowns vf and ω . From Table 12.2,
I rod =
1
1
1
Md 2 = ( 2m ) d 2 = md 2
12
12
6
From the angular momentum equation,
d
v0 = vf + ω
3
⇒ω =
3
( v0 − vf )
d
Substituting into the energy equation,
1
1
1⎛1
2
⎞⎛ 9 ⎞
mv0 2 = mvf 2 + ⎜ md 2 ⎟⎜ 2 ⎟ ( v0 − vf )
2
2
2⎝ 6
⎠⎝ d ⎠
3
2
v0 2 = vf 2 + ( v0 − vf )
2
6
1
0 = vf 2 − v0vf + v0 2
5
5
This is a quadratic equation in vf . The roots are
2
⎛v 2⎞
6
⎛6 ⎞
v0 ± ⎜ v0 ⎟ − 4 ⎜ 0 ⎟
5
⎝5 ⎠
2
⎝ 5 ⎠ 3
= v0 ± v0
vf =
2
5
5
⎧1
⎪ v
= ⎨5 0
⎪⎩ v0
1
The answer vf = v0 means the ice cube missed the rod. So vf = v0 to the right.
5
12.94. Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular
momentum is conserved in the collision.
Visualize:
Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as
the ball rises like a pendulum. The angular momentum conservation equation about the rod’s pivot point is
Li = Lf ⇒ mv0 r = ( I ball+rod )ω
1
L
= 15 cm. The rod and ball are a composite object. From Table 12.2, I rod = ML2 , so
12
2
1
L2 1
L2 ⎛
M⎞
I ball+rod = I ball + I rod = mr 2 + ML2 = m + ML2 = ⎜ m + ⎟
12
4 12
4⎝
3 ⎠
v
2v
If vf is the final velocity of the clay ball, ω = f = f since the ball sticks to the rod. Thus
r
L
2
mv0 L L ⎛
M ⎞⎛ 2v ⎞
= ⎜ m + ⎟⎜ f ⎟
2
4⎝
3 ⎠⎝ L ⎠
Note r =
mv0
( 0.010 kg )( 2.5 m/s ) = 0.714 m/s
=
M
( 0.075 kg )
m+
( 0.010 kg ) +
3
3
Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to
when the ball reaches the maximum height. Note that the center of mass of the rod does not change position.
1
Ei = Ef ⇒ ( I rod + ball )ω 2 = mgh
2
Thus
⇒ vf =
2
1 L2 ⎛
M ⎞⎛ 2vf ⎞
M⎞
2⎛
⎜ m + ⎟⎜
⎟ = mgh ⇒ vf ⎜ m + ⎟ = mgL (1 − cosθ )
2 4⎝
3 ⎠⎝ L ⎠
3 ⎠
⎝
⇒1−
vf2 ⎛
M⎞
⎜ m + ⎟ = cosθ
mgL ⎝
3 ⎠
Using the various values, cosθ = 0.393 ⇒ θ = 67°.
L
Assess: The clay ball rises h = (1 − cosθ ) = 9.1 cm. This is about 2/3 of the height of the pivot point, and is
2
reasonable.
12.95.
Model: Because no external torque acts on the star during gravitational collapse, its angular
momentum is conserved. Model the star as a solid rotating sphere.
Solve: (a) The equation for the conservation of angular momentum is
⎛2
⎞
⎛2
⎞
Li = Lf ⇒ I iωi = I f ω f ⇒ ⎜ mRi2 ⎟ωi = ⎜ mRf2 ⎟ω f
5
5
⎝
⎠
⎝
⎠
⇒ Rf = Ri
ωi
ωf
The angular velocity is inversely proportional to the period T. We can write
Rf = Ri
Tf
0.10 s
= ( 7.0 × 108 m )
= 1.3749 × 105 m = 137 km
Ti
2.592 × 106 s
(b) A point on the equator rotates with r = Rf . Its speed is
v=
2π Rf 2π (137,490 m)
=
= 8.6 × 106 m/s
0.10 s
T
Model: For the (turntable + bicycle wheel + professor) system the angular momentum is conserved because
the turntable is frictionless and no external torques act on the system.
Solve: (a) Nothing happens. The bicycle wheel already has an angular momentum and nothing changes for the wheel
when it is handed to the professor. So, nothing happens to the professor.
(b) The initial angular momentum is
12.96.
⎛ 180 × 2π rad ⎞
2
( Lwheel )i = Iω = (mR 2 )ω = (4.0 kg)(0.32 m) 2 ⎜
⎟ = 7.72 kg m /s
s ⎠
⎝ 60
When the wheel is turned upside down, the angular momentum of the wheel becomes
( Lwheel )f = −7.72 kg m 2 /s
Since the initial and final total angular momentum should be equal, the professor must acquire some angular momentum.
From the angular momentum conservation equation,
Lprof + ( Lwheel )f = ( Lwheel )i ⇒ Lprof = ( Lwheel )i − ( Lwheel )f = 7.72 kg m 2 /s − (−7.72 kg m 2 /s) = 15.44 kg m 2 /s
Let us now calculate the angular velocity of the turntable by using Lprof = I totalω , where I total = I turntable + I body + I arms +
I wheel in hand . We assume that the axis of the professor and turntable also become the axis of the arms and wheel (i.e., no
leaning).
1
1
I turntable = mT RT2 = (5.0 kg)(0.25 m) 2 = 0.156 kg m 2
2
2
1
2
I body = mbody R = (70 kg)(0.125 m) 2 = 0.547 kg m 2
2
1
⎛
⎞ 2
I arms = 2 ⎜ marms R 2 ⎟ = (2.5 kg)(0.45 m) 2 = 0.338 kg m 2
⎝3
⎠ 3
I wheel in hand = mwheel r 2 = (4.0 kg)(0.45 m) 2 = 0.810 kg m 2
Thus, I total = (0.156 + 0.547 + 0.338 + 0.810) kg m 2 = 1.851 kg m 2 . Using this value for I total , we can now find ω to be
ω=
Lprof
I total
=
15.44 kg m 2 /s
= 8.34 rad/s = 79.7 rpm
1.851 kg m 2
12-1
13.1.
Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between
your size and mass and that of either the sun or the earth, a human body can be treated as a particle.
GM s M y
GM e M y
and Fe on you =
Solve: Fs on you =
2
rs − e
re2
Dividing these two equations gives
2
2
⎛ M ⎞⎛ r ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 6.37 × 106 m ⎞
−4
= ⎜ s ⎟⎜ e ⎟ = ⎜
⎟⎜
⎟ = 6.00 × 10
Fe on y ⎝ M e ⎠⎝ rs − e ⎠ ⎝ 5.98 × 1024 kg ⎠⎝ 1.50 × 1011 m ⎠
Fs on y
13.2.
Model:
Assume the two lead balls are spherical masses.
Gm1m2 (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(10 kg)(0.100 kg)
=
= 6.7 × 10−9 N
Solve: (a) F1 on 2 = F2 on 1 =
r2
(0.10 m) 2
(b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is
6.67 × 10−9 N
= 6.81 × 10−9
(0.100 kg)(9.8 m/s 2 )
Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated
by 10 cm compared to the gravitational force on the 100 g ball.
13.3.
Model:
Model the sun (s), the moon (m), and the earth (e) as spherical masses.
GM s M m
GM e M m
Solve: Fs on m =
and Fe on m =
rs2− m
re2− m
Dividing the two equations and using the astronomical data from Table 13.2,
2
2
Fs on m ⎛ M s ⎞⎛ re − m ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 3.84 × 108 m ⎞
=⎜
⎟⎜
⎟ =⎜
⎟⎜
⎟ = 2.18
Fe on m ⎝ M e ⎠⎝ rs − m ⎠ ⎝ 5.98 × 1024 kg ⎠⎝ 1.50 × 1011 m ⎠
Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.
13.4.
Solve:
Fsphere on particle =
GM s M p
rs2− p
and
Fearth on particle =
GM e M p
re2
Dividing the two equations,
2
2
⎛ M ⎞ ⎛ r ⎞ ⎛ 5900 kg ⎞ ⎛ 6.37 × 106 m ⎞
−7
= ⎜ s ⎟⎜ e ⎟ = ⎜
⎟ = 1.60 × 10
⎟⎜
Fearth on particle ⎝ M e ⎠ ⎜⎝ rs − p ⎟⎠ ⎝ 5.98 × 1024 kg ⎠⎝ 0.50 m ⎠
Fsphere on particle
13.5.
Solve:
Model:
Model the woman (w) and the man (m) as spherical masses or particles.
GM w M m (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(50 kg)(70 kg)
Fw on m = Fm on w =
=
= 2.3 × 10−7 N
rm2 − w
(1.0 m) 2
13.6.
Model:
Visualize:
Model the earth (e) as a sphere.
The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re + rs = 6.37 × 106 m + 0.30 × 106 m = 6.67 × 106 m
away from the center of the earth.
GM e M s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.0 kg)
=
= 9.0 N
Solve: (a) Fe on s =
( Re + rs ) 2
(6.67 × 106 m) 2
(b) Because the sphere and the shuttle are in free fall with the same acceleration around the earth, there cannot be
any relative motion between them. That is why the sphere floats around inside the space shuttle.
13.7.
Solve:
Model:
Model the sun (s) as a spherical mass.
GM
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
= 274 m/s 2
(a) gsun surface = 2 s =
Rs
(6.96 × 108 m) 2
(b) gsun at earth =
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 5.90 × 10−3 m/s 2
rs2− e
(1.50 × 1011 m) 2
13.8.
Solve:
Model:
Model the moon (m) and Jupiter (J) as spherical masses.
GM m (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)
=
= 1.62 m/s 2
(a) g moon surface =
Rm2
(1.74 × 106 m) 2
(b) g Jupiter surface =
GM J (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
=
= 25.9 m/s 2
RJ2
(6.99 × 107 m) 2
13.9.
Model: Model the earth (e) as a spherical mass.
Visualize: The acceleration due to gravity at sea level is 9.83 m/s 2 (see Table 13.1) and Re = 6.37 × 106 m (see
Table 13.2).
Solve:
g observatory =
GM e
=
( Re + h) 2
GM e
⎛
h⎞
Re2 ⎜1 + ⎟
⎝ Re ⎠
2
=
g earth
⎛
h⎞
⎜1 + ⎟
⎝ Re ⎠
2
= (9.83 − 0.0075) m/s 2
Here g earth = GM e Re2 is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the
value 9.83 m/s 2 . Solving for h,
⎛ 9.83
⎞
h = ⎜⎜
− 1⎟⎟ Re = 2.43 km
⎝ 9.8225 ⎠
13.10. Model: Model the earth (e) as a spherical mass.
Solve:
Let the acceleration due to gravity be 3gsurface when the earth is shrunk to a radius of x. Then,
gsurface =
⇒3
GM e
Re2
and
3 gsurface =
GM e
x2
GM e GM e
R
= 2 ⇒ x = e = 0.577 Re
Re2
x
3
The earth’s radius would need to be 0.577 times its present value.
13.11.
Model:
Model Planet Z as a spherical mass.
GM Z
(6.67 × 10−11 N ⋅ m 2 /kg 2 ) M Z
⇒ 8.0 m/s 2 =
⇒ M Z = 3.0 × 1024 kg
Solve: (a) g Z surface =
2
RZ
(5.0 × 106 m) 2
(b) Let h be the height above the north pole. Thus,
g above N pole =
GM Z
=
( RZ + h) 2
GM Z
⎛
h ⎞
RZ2 ⎜1 +
⎟
⎝ RZ ⎠
=
2
g Z surface
⎛
h ⎞
⎜1 +
⎟
⎝ RZ ⎠
=
2
8.0 m/s 2
⎛ 10.0 × 106 m ⎞
⎜1 +
⎟
5.0 × 106 m ⎠
⎝
2
= 0.89 m/s 2
13.12.
Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as
an isolated system, so mechanical energy is conserved.
Visualize:
Solve: A height of 15 m is very small in comparison with the radius of earth or Mars. We can use flat-earth
gravitational potential energy to find the speed with which the astronaut can throw the ball. On earth, with
yi = 0 m and vf = 0 m/s, energy conservation is
1
2
mvf2 + mgyf = 12 mvi2 + mgyi ⇒ vi = 2 g e yf = 2(9.80 m/s 2)(15 m) = 17.1 m/s
Energy is also conserved on Mars, but the acceleration due to gravity is different.
GM M (6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.42 × 1023 kg)
g Mar’s surface =
=
= 3.77 m/s 2
RM2
(3.37 × 106 m) 2
On Mars, with yi = 0 m and vf = 0 m/s, energy conservation is
1
2
mvf2 + mgyf = 12 mvi2 + mgyi ⇒ yf =
vi2
(17.1 m/s) 2
=
= 39 m
2 g m 2(3.77 m/s 2)
13.13. Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter
form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is
called the escape speed, causes an object to stop only as the distance approaches infinity.
Visualize:
Solve:
The energy conservation equation K 2 + U 2 = K1 + U1 is
1
GM J mo 1
GM J mo
mo v22 −
= mo v12 −
r2
RJ
2
2
where RJ and M J are the radius and mass of Jupiter. Using the asymptotic condition v2 = 0 m/s as r2 → ∞,
GM J mo
1
2GM J
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg)
0 J = mo v12 −
⇒ v1 =
=
= 6.02 × 104 m/s
RJ
RJ
2
6.99 × 107 m
Thus, the escape velocity from Jupiter is 60.2 km/s.
13.14. Model: Model the earth (e) as a spherical mass. Compared to the earth’s size and mass, the rocket (r)
is modeled as a particle. This is an isolated system, so mechanical energy is conserved.
Visualize:
Solve:
The energy conservation equation K 2 + U 2 = K1 + U1 is
1
GM e mr 1
GM e mr
mr v22 −
= mr v12 −
r2
Re
2
2
In the present case, r2 → ∞, so
1
1
GM e mr
2GM e
mr v22 = mr v12 −
⇒ v22 = v12 −
2
2
Re
Re
⇒ v2 = (1.50 × 104 m/s) 2 −
= 9.99 × 103 m/s
2(6.67 × 10−11 N m 2 /kg 2 )(5.98 × 1024 kg)
6.37 × 106 m
13.15. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The
minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance
approaches infinity.
Visualize:
We denote by mp the mass of the probe. M S is the sun’s mass, and RS − p is the separation between the centers of
the sun and the probe.
Solve: The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM Smp 1
GM Smp
1
= mp v12 −
mp v22 −
2
( r2 )
2
RS− p
Using the condition v2 = 0 m/s asymptotically as r2 → ∞,
GM Smp
1
2GM S
2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
2
=
⇒ vescape =
=
= 4.21× 104 m/s
mp vescape
2
(1.50 × 1011 m)
RS − p
RS − p
13.16. Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated,
the mechanical energy of the object on both the planet and the earth is conserved.
Visualize:
Let us denote the mass of the planet by M p and that of the earth by M e . Your mass is m0 . Also, acceleration
due to gravity on the surface of the planet is g p and on the surface of the earth is g e . Rp and Re are the radii of
the planet and the earth, respectively.
Solve: (a) We are given that M P = 2M e and g p = 14 g e .
Since g p =
GM p
Rp2
and g e =
GM p
2
p
R
=
GM e
, we have
Re2
1 GM e 1 G ( M p /2)
=
⇒ Rp = 8Re = 8(6.37 × 106 m) = 1.80 × 107 m
4 Re2
4
Re2
(b) The conservation of energy equation K 2 + U 2 = K1 + U1 is
GM p mo 1
GM p mo
1
mov22 −
= mo v12 −
2
r2
2
Rp
Using v2 = 0 m/s as r2 → ∞, we have
GM p mo
1
2
=
mo vescape
2
Rp
⇒ vescape =
2GM p
Rp
=
2G (2M e )
4(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
= 9.41 km/s
Rp
1.80 × 107 m
13.17. Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle.
Visualize: The asteroid, having mass ma and velocity va , orbits the sun in a circle of radius ra . The asteroid’s
time period is Ta = 5.0 earth years = 1.5779 × 108 s.
Solve: The gravitational force between the sun (mass = M S ) and the asteroid provides the centripetal
acceleration required for circular motion.
2
1/ 3
⎛ GM sTa2 ⎞
GM s ma ma va2
GM s ⎛ 2π ra ⎞
=
⇒
=⎜
⎟ ⇒ ra = ⎜
⎟
2
2
ra
ra
ra
⎝ 4π ⎠
⎝ Ta ⎠
Substituting G = 6.67 × 10−11 N ⋅ m 2 /kg 2 , M s = 1.99 × 1030 kg, and the time period of the asteroid, we obtain
ra = 4.37 × 1011 m. The velocity of the asteroid in its orbit will therefore be
va =
2π ra (2π )(4.37 × 1011 m)
=
= 1.74 × 104 m/s
Ta
1.5779 × 108 s
13.18. Model: Model the sun (s) and the earth (e) as spherical masses.
Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs-e . The sun’s
and the earth’s masses are denoted by M s and me .
Solve: The gravitational force provides the centripetal acceleration required for circular motion.
GM s me meve2 me (2π rs − e ) 2
=
=
rs2− e
rs − e
rs − eTe2
⇒ Ms =
4π 2 rs3− e
4π 2 (1.50 × 1011 m)3
=
= 2.01 × 1030 kg
2
−11
GTe
(6.67 × 10 N ⋅ m 2 /kg 2 )(365 × 24 × 3600 s) 2
Assess: The tabulated value is 1.99 × 1030 kg. The slight difference can be ascribed to the fact that the earth’s
orbit isn’t exactly circular.
13.19.
Solve:
Model: Model the star (s) and the planet (p) as spherical masses.
A planet’s acceleration due to gravity is
gp =
⇒ Mp =
g p Rp2
G
=
GM p
Rp2
(12.2 m/s 2 )(9.0 × 106 m) 2
= 1.48 × 1025 kg
6.67 × 10−11 N ⋅ m 2 /kg 2
(b) A planet’s orbital period is
⎛ 4π 2 ⎞ 3
T2 =⎜
⎟r
⎝ GM s ⎠
4π 2 r 3
4π 2 (2.2 × 1011 m)3
⇒ Ms =
=
= 5.2 × 1030 kg
−11
2
GT
(6.67 × 10 N ⋅ m 2 /kg 2 )(402 × 24 × 3600 s)2
13.20. Model: Model the planet and satellites as spherical masses.
Visualize: Please refer to Figure EX13.20.
Solve: (a) The period of a satellite in a circular orbit is T = [(4π 2 /GM ) r 3 ]1/ 2 . This is independent of the
satellite’s mass, so we can find the ratio of the periods of two satellites a and b:
⎛r ⎞
Ta
= ⎜ a⎟
Tb
⎝ rb ⎠
3
Satellite 2 has r2 = r1 , so T2 = T1 = 250 min. Satellite 3 has r3 = (3/2) r1 , so T3 = (3/2)3/ 2 T1 = 459 min.
(b) The force on a satellite is F = GMm/r 2 . Thus the ratio of the forces on two satellites a and b is
2
Fa ⎛ rb ⎞ ⎛ ma ⎞
=⎜ ⎟ ⎜ ⎟
Fb ⎝ ra ⎠ ⎝ mb ⎠
Satellite 2 has r2 = r1 and m2 = 2m1 , so F2 = (1) 2(2) F1 = 20,000 N. Similarly, satellite 3 has r3 = (3/2) r1 and
m3 = m1 , so F3 = (2/3) 2 (1) F1 = 4440 N.
(c) The speed of a satellite in a circular orbit is v = (GM/r ) 2 , so its kinetic energy is K = 12 mv 2 = GMm/2r. Thus
the ratio of the kinetic energy of two satellites a and b is
K a ⎛ rb ⎞⎛ ma ⎞
= ⎜ ⎟⎜ ⎟
K b ⎝ ra ⎠⎝ mb ⎠
Satellite 3 has r3 = (3/2) r1 and m3 = m1 , so K1/K 3 = (3/2)(1/1) = 3/2 = 1.50.
13.21. Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle.
Visualize: The satellite, having mass ms and velocity vs , orbits the sun with a mass M S in a circle of radius
rs .
Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration
for circular motion. Newton’s second law is
GM Sms ms vs2
=
rs2
rs
Because vs = 2π rs / Ts where Ts is the period of the satellite, this equation simplifies to
GM S (2π rs ) 2
GM STs2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(24 × 3600 s) 2
3
=
⇒
r
=
=
⇒ rs = 2.9 × 109 m
s
rs2
Ts2 rs
4π 2
4π 2
13.22. Model: Model the earth (e) as a spherical mass and the shuttle (s) as a point particle.
Visualize: The shuttle, having mass ms and velocity vs , orbits the earth in a circle of radius rs . We will
denote the earth’s mass by M e .
Solve: The gravitational force between the earth and the shuttle provides the necessary centripetal acceleration
for circular motion. Newton’s second law is
GM e ms msvs2
GM e
GM e
=
⇒ vs2 =
⇒ vs =
rs2
rs
rs
rs
Because rs = Re + 350 km = 6.72 × 106 m,
vs =
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
= 7704 m/s
(6.72 × 106 m)
Ts =
2π rs 2π (6.72 × 106 m)
=
= 5580 s = 91.3 minutes
vs
7.704 × 103 m/s
13.23. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle.
Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs . The radius of the circular orbit
is denoted by rs and the mass of the earth by M e .
Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for
circular motion:
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
GM e ms ms vs2
GM
=
⇒ rs = 2 e =
= 1.32 × 107 m
2
(5500 m/s) 2
rs
rs
vs
⇒ Ts =
2π Rs (2π )(1.32 × 107 m)
=
= 1.51 × 104 s = 4.2 hr
(5500 m/s)
vs
13.24. Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle.
Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs
around Mars. Let us denote mass of Mars by M m .
Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for
circular motion:
1/ 3
⎛ GM mTs2 ⎞
GM m ms ms vs2 ms (2π rs ) 2
=
=
⇒
=
r
⎜
⎟
s
2
rs2
rs
rs (Ts ) 2
⎝ 4π
⎠
Using G = 6.67 × 10−11 N ⋅ m 2 /kg 2 ,
M m = 6.42 × 1023 kg, and Ts = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we
obtain rs = 2.05 × 107 m. Thus, altitude = rs − Rm = 1.72 × 107 m.
13.25.
Solve: We are given M 1 + M 2 = 150 kg which means M 1 = 150 kg − M 2 . We also have
GM 1M 2
( 0.20 m )
Thus,
2
= 8.00 × 10−6 N ⇒ M 1M 2 =
(8.00 × 10−6 N)(0.20 m) 2
= 4798 kg 2
6.67 × 10−11 N ⋅ m 2 /kg 2
(150 kg − M 2 ) M 2 = 4798 kg 2 or M 22 − (150 kg) M 2 + (4798 kg 2 ) = 0.
M 2 = 103.75 kg and 46.25 kg. The two masses are 104 kg and 46 kg.
Solving
this
equation
gives
13.26. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will
make an angle of θ with the vertical. The distance between the sphere centers is therefore going to be less than 1
m. The free-body diagram shows the forces acting on the lead sphere.
Solve: We can see from the diagram that the distance between the centers is d = 1.000 m − 2 L sin θ . Each
sphere is in static equilibrium, so Newton’s second law is
∑ F = F − T sinθ = 0 ⇒ T sinθ = F
x
grav
grav
∑ F = T cosθ − mg = 0 ⇒ T cosθ = mg
y
Dividing these two equations to eliminate the tension T yields
F
Gmm / d 2 Gm
sin θ
= tan θ = grav =
= 2
mg
mg
d g
cosθ
We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to
be enough to affect the value of Fgrav , the gravitational attraction between the two masses, so we’ll evaluate the
right side of this equation by using 1.00 m for d. This gives
tan θ =
(6.67 × 10−11 N m 2 /kg 2 ) (100 kg)
= 6.81 × 10−10 ⇒ θ = (3.90 × 10−8 )°
(1.00 m) 2 (9.80 m/s 2 )
This small angle causes the two spheres to move closer by 2 L sin θ = 1.4 × 10−7 m = 0.00000014 m.
Consequently, the distance between their centers is d = 0.99999986 m.
13.27. Visualize:
We placed the origin of the coordinate system on the 20 kg sphere (m1 ). The sphere (m2 ) with a mass of 10 kg is
20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses m1
and m2 . This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As
the gravitational force is directly proportional to the two masses and inversely proportional to the square of
distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the
left of m1 or to the right of m2 , will experience gravitational forces from m1 and m2 pointing in the same
direction, thus always leading to a nonzero force.
mm
m2 m
20
10
Solve: Fm1 on m = Fm2 on m ⇒ G 12 = G
⇒ 2 =
2
(0.20 − x )
(0.20 − x) 2
x
x
⇒ 10 x 2 − 8 x + 0.8 = 0 ⇒ x = 0.683 m and 0.117 m
The value x = 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2 . Thus,
the point ( x, y ) = (11.7 cm, 0 cm) is where a small mass is to be placed for a zero gravitational force.
13.28.
We placed the origin of the coordinate system on the 20.0 kg mass (m1 ) so that the 5.0 kg
Visualize:
mass (m3 ) is on the x-axis and the 10.0 kg mass (m2 ) is on the y-axis.
Solve:
(a) The forces acting on the 20 kg mass (m1 ) are
G
Gm1m2 ˆ (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(20.0 kg)(10.0 kg) ˆ
Fm2 on m1 =
j=
j = 3.335 × 10−7 ˆj N
r122
(0.20 m) 2
G
Gm1m3 ˆ (6.67 × 10−7 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg) ˆ
Fm3 on m1 =
i=
i = 6.67 × 10−7 iˆ N
r132
(0.10 m) 2
G
Fon m = 6.67 × 10−7 iˆ N + 3.335 × 10−7 ˆj N ⇒ Fon m = 7.46 × 10−7 N
1
1
⎛F
⎞
⎛ 6.67 × 10−7 N ⎞
θ = tan −1 ⎜ m3 on m1 ⎟ = tan −1 ⎜
⎟ = 63.4°
−7
⎜ Fm on m ⎟
⎝ 3.335 × 10 N ⎠
⎝ 2 1⎠
G
Thus the force is Fon m1 = ( 7.5 × 10−7 N, 63° cw from the y -axis ).
(b) The forces acting on the 5 kg mass (m3 ) are
G
G
Fm1 on m3 = − Fm3 on m1 = −6.67 × 10−7 iˆ N
Gm2 m3 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(10.0 kg)(5.0 kg)
=
= 6.67 × 10−8 N
r232
[(0.20) 2 + (0.10) 2 ] m
G
Fm2 on m3 = −(6.67 × 10−8 N)cos φ iˆ + (6.67 × 10−8 N)sin φ ˆj
Fm2 on m3 =
⎛ 10 cm ⎞ ˆ
⎛ 20 cm ⎞ ˆ
−8
= −(6.67 × 10−8 N) ⎜
⎟ i + (6.67 × 10 N) ⎜
⎟j
⎝ 22.36 cm ⎠
⎝ 22.36 cm ⎠
= −(2.983 × 10−8 N)iˆ + (5.966 × 10−8 N) ˆj
G
Fon m = −6.968 × 10−7 iˆ N + 5.966 × 10−8 ˆj N
3
Fon m3 = (−6.968 × 10−7 N) 2 + (5.966 × 10−8 N) 2 = 6.99 × 10−7 N
⎛ 6.968 × 10−7 N ⎞
⎟ = 85.1°
−8
⎝ 5.966 × 10 N ⎠
θ ′ = tan −1 ⎜
G
Thus Fon m3 = ( 7.0 × 10−7 N, 85° ccw from the y -axis ).
13.29. Visualize:
2
2
⎛ 5 ⎞
The angle θ = tan −1 ⎜ ⎟ = 14.04°. The distance r1 = r2 = ( 0.050 m ) + ( 0.200 m ) = 0.206 m. The
20
⎝ ⎠
forces on the 20.0 kg mass are
G
Mm
F1 = G 2 1 − sinθ iˆ + cosθ ˆj
r1
G
Mm
F2 = G 2 2 sin θ iˆ + cosθ ˆj
r2
Solve:
(
)
(
)
Note m1 = m2 and r1 = r2 . Thus the net force on the 20.0 kg mass is
G
G G
Mm
Fnet = F1 + F2 = 2G 2 1 cosθ ˆj
r1
=
2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 20.0 kg )( 5.0 kg ) cos14.04°
= 3.0 × 10−7 ˆj N
( 0.206 m )
2
ˆj
13.30. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the
pairs of masses.
U = U12 + U13 + U 23
= −G
m1m2
mm
mm
−G 1 3 −G 2 3
r12
r13
r23
With m1 = 20.0 kg, m2 = 10.0 kg, m3 = 5.0 kg, r12 = 0.20 m, r13 = 0.10 m, and r23 = 0.2236 m,
U = −1.48 × 10 −7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so
there are no vector calculations required.
13.31. Visualize:
Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the
pairs of masses.
U = U12 + U13 + U 23
= −G
With m1 = 20.0 kg, m2 = m3 = 10.0 kg, r12 = r13 =
m1m2
mm
mm
−G 1 3 −G 2 3
r12
r13
r23
( 0.050 m ) + ( 0.200 m ) = 0.206 m, and r23 = 0.100 m,
2
2
U = −1.96 × 10−7 J
Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so
there are no vector calculations required.
13.32. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle.
Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity ( g altitude ) is
10% of the surface value ( gsurface ).
Solve:
(a) Since g altitude = (0.10) gsurface , we have
GM e
GM
= (0.10) 2 e ⇒ ( Re + h) 2 = 10Re2
2
Re
( Re + h)
⇒ h = 2.162 Re ⇒ h = (2.162)(6.37 × 106 m) = 1.377 × 107 m= 1.38 × 107 m
(b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between
the earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting
with velocity vs ,
GM e ms
ms vs2
=
⇒ vs =
2
( Re + h)
( Re + h)
GM e
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
= 4.45 km/s
Re + h
(6.37 × 106 m + 1.377 × 107 m)
13.33. Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air
resistance. This is an isolated system, so mechanical energy is conserved.
Visualize:
Solve:
(a) The conservation of energy equation K 2 + U g2 = K1 + U g1 is
1
GM e mo 1
GM e mo
mov22 −
= mo v12 −
2
2
( Re + y1 )
Re
⎛ 1
1 ⎞
⇒ v2 = 2GM e ⎜ −
⎟
⎝ Re Re + y1 ⎠
1
1
⎛
⎞
= 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) ⎜
−
⎟ = 3.02 km/s
6
6
⎝ 6.37 × 10 m 6.87 × 10 m ⎠
(b) In the flat-earth approximation, U g = mgy. The energy conservation equation thus becomes
1
1
mov22 + mo gy2 = mov12 + mo gy1
2
2
⇒ v2 = v12 + 2 g ( y1 − y2 ) = 2(9.80 m/s 2 )(5.00 × 105 m − 0 m) = 3.13 km/s
(c) The percent error in the flat-earth calculation is
3130 m/s − 3020 m/s
≈ 3.6%
3020 m/s
13.34. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an
isolated system, so mechanical energy is conserved.
Visualize:
A pictorial representation of the before-and-after events is shown.
Solve: After using v2 = 0 m/s, the energy conservation equation K 2 + U 2 = K1 + U1 is
0 J−
GM e mp
GM e mp
1
= mpv12 −
Re + h 2
Re
The projectile mass cancels. Solving for h, we find
−1
⎡1
v2 ⎤
h = ⎢ − 1 ⎥ − Re
⎣ Re 2GM e ⎦
= 4.18 × 105 m = 418 km
13.35. Model: Model the planet (p) as a spherical mass and the projectile as a point mass. This is an isolated
system, so mechanical energy is conserved.
Visualize: The projectile of mass m was launched on the surface of the planet with an initial velocity v0 .
Solve:
(a) The energy conservation equation K1 + U1 = K 0 + U 0 is
1 2 GM p m 1 2 GM p m
mv1 −
= mv0 −
Rp + y1 2
Rp
2
−1
⎡1
v02 ⎤
6
⇒ y1 = ⎢ −
⎥ − Rp = 2.8 × 10 m
R
GM
2
⎥
p⎦
⎣⎢ p
(b) Using the energy conservation equation K1 + U1 = K 0 + U 0 with y1 = 1000 km = 1.000 × 106 m:
1 2 GM p m 1 2 GM p m
mv1 −
= mv0 −
2
Rp + y1 2
Rp
12
⎡
⎛ 1
1 ⎞⎤
⇒ v1 = ⎢v02 + 2GM p ⎜
− ⎟⎥
⎜
⎟
⎢⎣
⎝ Rp + y1 Rp ⎠ ⎥⎦
⎡
⎛
⎞⎤
1
1
= ⎢(5000 m/s) 2 + 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(2.6 × 1024 kg) ⎜
−
⎟⎥
6
6
6
⎝ (5.0 × 10 m + 1.000 × 10 m) 5.0 × 10 m ⎠ ⎦
⎣
= 3.7 km/s
12
13.36. Model: The object is a particle. The planet is a spherical mass.
Solve:
Conservation of mechanical energy of the object gives
−G
Mm 1 2
Mm
= mv − G
R+h 2
R
The object’s mass drops out. Solving for the speed as it hits the ground,
⎛ R+h−R⎞
1 ⎞
⎛1
v = 2GM ⎜ −
⎟⎟ =
⎟ = 2GM ⎜⎜
⎝ R R+h⎠
⎝ R ( R + h) ⎠
Assess:
Compare this to v = 2 gh =
2GMh
R ( R + h)
2GMh
, which is the result if the potential U = mgh is used.
R2
13.37.
Model:
Visualize:
Solve:
Model the earth as a spherical mass and the meteoroids as point masses.
(a) The energy conservation equation K 2 + U 2 = K1 + U1 is
1/ 2
1
2
⎡
⎛ 1 1 ⎞⎤
GM e m 1 2 GM e m
= 2 mv1 −
⇒ v2 = ⎢ v12 + 2GM e ⎜ − ⎟ ⎥
mv −
Re
rm
⎝ Re rm ⎠ ⎦⎥
⎣⎢
2
2
= 1.13 × 104 m/s =11.3 km/s
The speed does not depend on the meteoroid’s mass.
(b) This part differs in that r2 = Re + 5000 km = 1.137 × 107 m. The shape of the meteoroid’s trajectory is not
important for using energy conservation. Thus
1
2
⎡
⎛
GM e m
GM e m
1
1 ⎞⎤
= 1 mv12 −
⇒ v2 = ⎢v12 + 2GM e ⎜
− ⎟⎥
mv −
Re + 5000 km 2
rm
R
r
5000
km
+
⎥
m ⎠⎦
⎝ e
⎣⎢
2
2
= 8.94 × 103 m/s = 8.94 km/s
1/ 2
13.38. Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated
system, so mechanical energy is conserved.
Visualize:
In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential
energy. In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both
kinetic energy and potential energy.
Solve: The conservation of energy equation K f + U f = K i + U i is
1 2 GMm GMm
mvf −
−
=0 J +0 J
rf1
rf2
2
⇒ vf =
Assess:
4GM
4(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 32,600 m/s
rf
(0.50 × 1012 m)
Note that the final velocity of 33 km/s does not depend on the mass of the comet.
13.39. Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated
system, so mechanical energy is conserved.
Visualize: The radius of the asteroid is M a and its mass is Ra .
Solve:
The conservation of energy equation K f + U f = K i + U i for the asteroid is
1 2 GM a m 1 2 GM a m
mvf −
= mvi −
Ra + r 2
Ra
2
The minimum speed for escape is the one that will cause you to stop only when the separation between you and
the asteroid becomes very large. Noting that vf → 0 m/s as r → ∞, we have
vi2 =
2GM a 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1014 kg)
=
⇒ vi = 2.58 m/s
Ra
(2.0 × 103 m)
That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on
the earth. The conservation of energy equation is
1
0 J − mvi2 = − mg (0.50 m) ⇒ vi = 2(9.8 m/s 2 )(0.50 m) = 3.13 m/s
2
This means you can escape from the asteroid.
13.40. Model: The projectile is a particle. The earth and moon are spherical masses.
Solve: The projectile is attracted to both the moon and earth. Its final velocity and potential energy are zero.
Since the projectile is fired from the far side of the moon, its initial distance from the center of the earth is the
earth-moon distance Re-m plus the radius of the moon Rm . Let the projectile have mass m. The conservation of
mechanical energy equation is
1 2
M em
M m
mvescape − G
−G m = 0 J +0 J
+
2
R
R
Rm
( e-m m )
⎛ Me
M ⎞
2
⇒ vescape
= 2G ⎜
+ m⎟
+
R
R
Rm ⎠
m
⎝ e-m
From
Table
13.2,
Re-m = 3.84 × 108 m, Rm = 1.74 × 106 m, M e = 5.98 × 1024 kg,
and
M m = 7.36 × 1022 kg.
vescape = 2.78 km/s.
Assess:
The escape velocity does not depend on the mass of the object which is trying to escape.
So
13.41. Model: The earth and sun are spherical masses. The earth is in a circular orbit around the sun. The
projectile is a particle.
Visualize:
Solve: At the earth’s distance from the sun, to escape the sun’s gravitational pull a projectile must have speed
vescape . The energy conservation equation for the projectile is
1 2
M m
mvescape − G 2 = 0 J
2
Re − s
⇒ vescape = 2G
2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg )
Ms
=
= 4.207 × 104 m/s
Re − s
(1.50 ×1011 m )
The earth’s speed in its orbit is found by considering its period T = 1 year × 365.25 × 24 × 3600 = 3.156 × 107 s.
ve =
11
2π Re −s 2π (1.50 × 10 m )
=
= 2.987 × 104 m/s
T
3.156 × 107 s
The projectile’s total speed is the sum of its launch speed and the earth’s speed.
vescape = ve + vlaunch
⇒ vlaunch = vescape − ve = 1.22 × 104 m/s
Assess: The projectile need only attain a launch speed of 12.2 km/s if launched in the direction of earth’s
motion. This is about a factor of 3.5 times less than what is required from rest.
13.42. Model:
Visualize:
Gravity is a conservative force, so we can use conservation of energy.
The planets will be pulled together by gravity and each will have speed v2 as they crash and the separation
between their centers will be 2R.
Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential
and kinetic energy. Thus,
⎡
1
1
GMM ⎤ ⎡
GMM ⎤
2
2
⎢ K 2 + U 2 = Mv2 + Mv2 −
⎥ = ⎢ K1 + U1 = 0 J −
⎥
2
2
r2 ⎦ ⎣
r1 ⎦
⎣
⎛ 1 1⎞
⇒ v2 = GM ⎜ − ⎟
⎝ r2 r1 ⎠
Because the planet is “Jupiter-size,” we’ll use M = M Jupiter = 1.9 × 1027 kg and r2 = 2 RJupiter = 1.4 × 108 m. The
crash speed of each planet is v2 = 3.0 × 104 m/s.
Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant
acceleration motion. The formulas from constant-acceleration kinematics do not work for problems such as this.
13.43. Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also
use the law of conservation of momentum for our system.
Visualize:
Solve:
The conservation of momentum equation pfx = pix is
M (vfx )1 + 2 M (vfx ) 2 = 0 kg m/s ⇒ (vfx )1 = −2(vfx ) 2
The equation for mechanical energy conservation K f + U f = K i + U i is
1
1
G ( M )(2M )
G ( M )(2M )
1
0.8GM
M (vfx )12 + (2M )(vfx ) 22 −
=−
⇒ [2(vfx ) 2 ]2 + (vfx ) 22 =
2
2
2R
10 R
2
R
GM
GM
⇒ (vfx ) 2 = −0.516
⇒ (vfx )1 = −2(vfx ) 2 = 1.032
R
R
The heavier asteroid has a speed of 0.516(GM/R)1/ 2 and the lighter one a speed of 1.032(GM/R)1/ 2 .
13.44.
Solve:
Model: Model the distant planet (P) as a spherical mass.
The acceleration at the surface of the planet and at the altitude h are
gsurface =
GM P
R2
and g altitude =
1
MP
1 GM P
=
g surface ⇒ G
2
2
( R + h)
2 R2
⇒ ( R + h) 2 = 2 R 2 ⇒ R + h = 2 R ⇒ h = ( 2 − 1) R = 0.414 R
That is, the starship is orbiting at an altitude of 0.414R.
13.45. Model: The stars are spherical masses.
Visualize:
Solve: The starts are identical, so their final speeds vf are the same. They collide when their centers are 2R
apart. From energy conservation,
⎛
⎞
⎛ M2 ⎞
M2
⎛1
2⎞
3 ⎜ −G
⎟ = 3 ⎜ Mvf ⎟ − 3 ⎜ G
⎟
9
(5.0 × 10 m) ⎠
⎝2
⎠ ⎝ 2R ⎠
⎝
1
⎛ 1
⎞
vf2 = 2GM ⎜
−
⎟
9
⎝ 2 R 5.0 × 10 m ⎠
⇒ vf = 3.71 × 105 m/s
13.46. Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated
system, so mechanical energy is conserved.
Visualize: The initial position of the lunar lander (mass = m1 ) is at a distance r1 = Rm + 50 km from the center
of the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is
r2 = Rm + 300 km.
Solve:
The external work done by the thrusters is
Wext = ΔEmech = 12 ΔU g
where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
r1 = Rm + 50 km to the final orbit rf = Rm + 300 km. Thus
1 ⎛ −GM m m −GM m m ⎞ GM m m ⎛ 1 1 ⎞
−
Wext = ⎜
⎟=
⎜ − ⎟
2⎝
2 ⎝ ri rf ⎠
rf
ri
⎠
(6.67 × 10−11 N m 2 / kg 2 )(7.36 × 1022 kg)(4000 kg) ⎛
1
1
⎞
=
−
⎜
⎟
6
6
×
×
2
1.79
10
m
2.04
10
m
⎝
⎠
= 6.72 × 108 J
13.47.
Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an
isolated system, so the mechanical energy is conserved.
Visualize: The space shuttle (mass = ms ) is at a distance of r1 = Re + 250 km.
Solve:
The external work done by the thrusters is
Wext = ΔEmech = 12 ΔU g
where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at
r1 = Re + 250 km to the final orbit rf = Re + 610 km. Thus
1 ⎛ −GM e m −GM e m ⎞ GM e m ⎛ 1 1 ⎞
−
Wext = ⎜
⎟=
⎜ − ⎟
2 ⎝ rf
2 ⎝ ri rf ⎠
ri
⎠
(6.67 × 10−11 N m 2 / kg 2 )(5.98 × 1024 kg)(75,000 kg) ⎛
1
1
⎞
=
−
⎜
⎟
6
6
2
⎝ 6.62 × 10 m 6.98 × 10 m ⎠
= 1.17 × 1011 J
This much energy must be supplied by burning the on-board fuel.
13.48.
Solve:
(a) Using Equation 13.31 for a satellite in a circular orbit,
⎛
⎞
1
1
1
1
ΔEmech = ΔU g = (−GM e m) ⎜
−
⎟
R
R
2
2
300
km
500
km
+
+
e
⎝ e
⎠
=
−1
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(5.00 × 104 kg)
2
1
1
⎛
⎞
10
×⎜
−
⎟ = −4.35 × 10 J
6
5
6
5
⎝ 6.37 × 10 m + 3 × 10 m 6.37 × 10 m + 5 × 10 m ⎠
The negative sign indicates that 4.35 × 1010 J of energy is lost.
(b)
The shuttle would fire retro-rockets to lose enough energy to go into an elliptical orbit. When halfway around, at
rmin , the shuttle would fire retro-rockets again to lose the rest of the energy to stay at rmin .
13.49. Model: Planet Physics is a spherical mass. The cruise ship is in a circular orbit.
Solve:
(a) At the surface, the free-fall acceleration is g = G
M
. From kinematics,
R2
y f = yi + v0 Δt − 2 g ( Δt ) ⇒ 0 m = 0 m + (11 m/s )( 2.5 s ) − 2 g ( 2.5 s )
2
2
⇒ g = 2.20 m/s 2
The period of the cruise ship’s orbit is 230 × 60 = 13,800 s. For the circular orbit of the cruise ship,
⎛ 4π 2 ⎞
⎛ R2 ⎞
⎛1⎞
T2
3
T2 =⎜
R
=
⎟ ( 2R ) ⇒
⎜
⎟ = R⎜ ⎟
2
32π
⎝g⎠
⎝ GM ⎠
⎝ GM ⎠
( 2.20 m/s ) (13,800 s ) = 1.327 × 10 m.
2
⇒R=
R2
g = 5.8 × 1022 kg.
G
(b) From part (a), R = 1.33 × 106 m.
The mass is thus M =
2
6
32π 2
13.50. Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated
system, so mechanical energy is conserved.
Visualize:
The orbital radius of the satellite is
r = Ra + h = 8,800 m + 5,000 m = 13,800 m
Solve:
(a) The speed of a satellite in a circular orbit is
12
GM ⎡ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) ⎤
=⎢
v=
⎥
r
(13,800 m)
⎣
⎦
= 7.0 m/s
(b) The minimum launch speed for escape (vi ) will cause the satellite to stop asymptotically (vf = 0 m/s) as
rf → ∞. Using the energy conservation equation K 2 + U 2 = K1 + U1 , we get
1
GM a ms 1
GM a ms
1 2
GM a
ms vf2 −
= msvi2 −
⇒ 0 J − 0 J = vescape
−
2
2
2
rf
Ra
Ra
⇒ vescape =
2GM a
2(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg)
=
= 12.3 m/s
Ra
8800 m
13.51. Model: Model the moon as a spherical mass and the satellite as a point mass.
Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own
axis. This time happens to be 27.3 days.
Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration
necessary for circular motion around the moon. Therefore,
GM m m
⎛ 2π ⎞
= mrω 2 = mr ⎜
⎟
r2
⎝ T ⎠
⇒ r3 =
2
GM mT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(27.3 × 24 × 3600 s) 2
=
4π 2
4π 2
⇒ r = 8.84 × 107 m
Since r = Rm + h, then h = r − Rm = 8.84 × 107 m − 1.74 × 106 m = 8.67 × 107 m.
13.52.
Model: Model the earth as a spherical mass and the satellite as a point mass.
Visualize: The satellite is directly over a point on the equator once every two days. Thus, T = 2Te = 2 × 24 ×
3600 s = 1.728 × 105 s.
Solve: A satellite’s period is
⎛ 4π 2 ⎞ 3
T2 = ⎜
⎟r
⎝ GM e ⎠
GM eT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.728 × 105 ) 2
⇒ r3 =
=
4π 2
4π 2
7
⇒ r = 6.71 × 10 m
Assess:
The radius of the orbit is larger than the geosynchronous orbit.
13.53.
Solve:
(a) Taking the logarithm of both sides of v p = Cu q gives
[log(v p ) = p log v] = [log(Cu q ) = log C + q log u ] ⇒ log v =
q
log C
log u +
p
p
But x = log u and y = log v, so x and y are related by
⎛q⎞
log C
y = ⎜ ⎟x +
p
p
⎝ ⎠
(b) The previous result shows there is a linear relationship between x and y, hence there is a linear relationship
between log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u
will be a straight line.
(c) The slope of the straight line represented by the equation y = (q / p ) x + log C / p is q/p. Thus, the slope of the
log v-versus-log u graph will be q/p.
(d) From Newton’s theory, the period T and radius r of an orbit around the sun are related by
⎛ 4π 2 ⎞ 3
T2 =⎜
⎟r
⎝ GM ⎠
This equation is of the form T p = Cr q , with p = 2, q = 3, and C = 4π 2 /GM . If the theory is correct, we expect
a graph of log T-versus-log r to be a straight line with slope q/p = 3/2 = 1.500. The experimental measurements
of actual planets yield a straight line graph whose slope is 1.500 to four significant figures. Note that the graph
has nothing to do with theory—it is simply a graph of measured values. But the fact that the shape and slope of
the graph agree precisely with the prediction of Newton’s theory is strong evidence for its correctness.
(e) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating
these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is M = M sun .
⎛ 4π 2 ⎞
1
1
4π 2
1
log C = log ⎜
= 10−18.528 = 18.528
⎟ = −9.264 ⇒
GM sun
2
2
10
⎝ GM sun ⎠
⇒ M sun =
4π 2
⋅ 1018.528 = 1.996 × 1030 kg
G
The tabulated value, to three significant figures, is M sum = 1.99 × 1030 kg. We have used the orbits of the planets
to “weigh the sun!”
13.54.
Solve:
Visualize: Please refer to Figure P13.54.
The gravitational force on one of the masses is due to the star and the other planet. Thus
2
G
Mm Gmm mv 2 m ⎛ 2π r ⎞
GM Gm 4π 2 r 2
+
=
=
⇒
+
=
⎜
⎟
r2
(2r ) 2
r
r⎝ T ⎠
r
4r
T2
12
⎡ 4π 2 r 3
⎤
G⎛
m ⎞ 4π 2 r 2
1
T
⇒
=
⎢
⎥
⎜M + ⎟ =
2
r⎝
4⎠
T
⎣ G ( M + m /4) ⎦
13.55.
Solve:
(a) Dividing the circumference of the orbit by the period,
2π Rs 2π (1.0 × 104 m)
=
= 6.3 × 104 m/s
1.0 s
T
(b) Using the formula for the acceleration at the surface,
v=
gsurface =
GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
=
= 1.33 × 1012 m/s 2
Rs2
(1 × 104 m) 2
(c) The mass of an object on the earth will be the same as its mass on the star. The gravitational force is
( FG )star = mg surface = 1.33 × 1012 N
(d) The radius of the orbit of the satellite is r = 1 × 104 m + 1.0 × 103 m = 1.1 × 104 m. The period is
T2 =
4π 2 r 3
4π 2 (1.1 × 104 m)3
=
⇒ T = 6.29 × 10−4 s
−11
GM s (6.67 × 10 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
This means there are 1589 revolutions per second or 9.5 × 104 orbits per minute.
(e) Applying Equation 13.25 for a geosynchronous orbit,
r3 =
GM s 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(1.0 s)
⇒ r = 1.50 × 106 m
T =
4π 2
4π 2
13.56.
Solve:
Model: Assume the solar system is a point particle.
(a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means
r = 25,000 light years = 2500(3 × 108 )(365)(24)(3600) m = 2.36 × 1020 m
T=
2π r (2π )(2.36 × 1020 m)
=
= 6.64 × 1015 s = 2.05 × 108 years
v
(2.30 × 105 m/s)
5.0 × 109 years
= 24 orbits.
2.05 × 108 years
(c) Applying Newton’s second law yields
GM g center mss mss v 2
v 2 r (2.30 × 105 m/s) 2 (2.36 × 1020 m)
=
⇒
M
=
=
= 1.87 × 1041 kg
g
center
r2
r
G
6.67 × 10−11 N ⋅ m 2 /kg 2
(b) The number of orbits =
(d) The number of stars in the center of the galaxy is
1.87 × 1041 kg
= 9.4 × 1010
1.99 × 1030 kg
13.57. Model: Assume the two stars are spherical masses.
Visualize: The gravitational force between the two stars provides the centripetal acceleration required for
circular motion about the center of mass.
Solve:
Newton’s second law is
2
1
⎛ GT 2 M ⎞ 3
GMM
⎛ 2π ⎞
= MRω 2 = MR ⎜
Fgravitation =
⎟ ⇒ R=⎜
2
2 ⎟
(2 R )
⎝ T ⎠
⎝ 16π ⎠
Using T = 90 days = 90 × 24 × 3600 s and M = 2M sun = 3.98 × 1030 kg, we get R = 4.667 × 1010 m. Thus the star
separation is 2 R = 9.33 × 1010 m.
13.58. Model:
Visualize:
Assume the three stars are spherical masses.
The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three
stars. The gravitational force between any two stars is the same. On a given star the two forces from the other
stars
make
an
angle
of 60°.
Solve: The value of r can be found as follows:
L /2
L
1.0 × 1012 m
= cos30° ⇒ r =
=
= 0.577 × 1012 m
r
2cos30°
2cos30°
The gravitational force between any two stars is
Fg =
GM 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2
=
= 2.64 × 1026 N
L2
(1.0 × 1012 m) 2
The component of this force toward the center is
Fc = Fg cos30° = (2.64 × 10 26 N)cos30° = 2.29 × 1026 N
The net force on a star toward the center is twice this force, and that force equals MRω 2 . This means
⎛ 2π ⎞
2 × 2.29 × 1026 N = MRω 2 = MR ⎜
⎟
⎝ T ⎠
⇒T =
2
4π 2 MR
4π 2 (1.99 × 1030 kg)(0.577 × 1012 m)
=
= 3.15 × 108 s = 10 years
26
4.58 × 10 N
4.58 × 1026 N
13.59. Model:
Visualize:
Angular momentum is conserved for a particle following a trajectory of any shape.
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the
G
G
velocity tangent to the trajectory and β is the angle between r and v .
Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no
tangential force (the only force being the radial force), angular momentum must be conserved:
mPluto v1rmin = mPlutov2 rmax
⇒ v2 = v1 (rmin / rmax ) = (6.12 × 103 m/s)(4.43 × 1012 m/7.30 × 1012 m) = 3.71 km/s
13.60. Model:
Visualize:
Angular momentum is conserved for a particle following a trajectory of any shape.
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the velocity
G
G
tangent to the trajectory, and β is the angle between r and v .
Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no
tangential force (the only force being the radial force), angular momentum must be conserved:
mMercuryv1rmin = mMercury v2 rmax
⇒ rmin = rmax (v2 / v1 ) =
(6.99 × 1010 m)(38.8 km/s)
= 4.60 × 1010 m
59.0 km/s
13.61.
Model:
Visualize:
Solve:
For the sun + comet system, the mechanical energy is conserved.
The conservation of energy equation K f + U f = K i + U i is
1
GM s M c 1
GM s M c
M cv22 −
= M cv12 −
r2
r1
2
2
Using G = 6.67 × 10−11 Nm 2 /kg 2 , M s = 1.99 × 1030 kg, r1 = 8.79 × 1010 m, r2 = 4.50 × 1012 m, and v1 = 54.6 km/s,
we get v2 = 4.49 km/s.
13.62. Model:
Visualize:
Solve:
Model the planet (p) as a spherical mass and the spaceship (s) as a point mass.
(a) For the circular motion of the spaceship around the planet,
GM p ms
2
0
r
=
GM p
mv02
⇒ v0 =
r0
r0
Immediately after the rockets were fired v1 = v0 /2 and r1 = r0 . Therefore,
v1 =
1 GM p
2
r0
(b) The spaceship’s maximum distance is rmax = r0 . Its minimum distance occurs at the other end of the ellipse.
The energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is,
GM p ms 1
GM p ms
1
ms v12 −
= ms v22 −
2
r1
2
r2
Since the angular momentum at these two ends is conserved, we have
mv1r1 = mv2 r2 ⇒ v2 = v1 (r1 / r2 )
With this expression for v2 , the energy equation simplifies to
GM p
1 2 GM p 1 2
v1 −
= v1 (r1 / r2 ) 2 −
2
r1
2
r2
Using r1 = r0 and v1 = v0 /2 =
1 GM p
,
2
r0
1 ⎛ 1 GM p ⎞ GM p 1 ⎛ 1 GM p ⎞ r02 GM p
1 1
r
1
= ⎜
⇒
− = 0 −
⎜
⎟−
⎟ −
2 ⎝ 4 r0 ⎠
r0
2 ⎝ 4 r0 ⎠ r22
r2
8r0 r0 8r22 r2
⇒
⎛ 7 ⎞ 2
7
r
1
r0
+ 02 − = 0 ⇒ ⎜
⎟ r2 − r2 + = 0
8r0 8r2 r2
8
⎝ 8r0 ⎠
The solutions are r2 = r0 (the initial distance) and r2 = r0 / 7. Thus the minimum distance is rmin = r0 / 7.
1
2
13.63. Solve: (a) The satellite is in a circular orbit if K = − Ug. Calculate K and U g :
2
1
1
K = mv 2 = m ( 5.5 × 103 m/s ) = 1.51 × 107 m.
2
2
( 5.98 × 1024 kg ) m
M em
U g = −G
= − ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )
1.10 × 107 m
r
7
= −3.63 × 10 m
1
The quantity − U g = 1.81× 107 m ≠ K , so no, the satellite is not in a circular orbit.
2
(b) The orbit is bound if the total mechanical energy is negative.
Emech = K + U g = 1.51 × 107 m − 3.63 × 107 m < 0.
Yes, the orbit is bound.
Assess: Even without knowing the mass of the satellite we could answer questions about its orbit.
13.64. Visualize:
Solve:
We choose two equal time intervals tb − ta and tc − tb . A constant velocity and equal time intervals
means that xb − xa = xc − xb . The area swept from t = 0 s to t = ta is Rxa /2 and the area swept from
t = 0 s to t = tb is Rxb /2. Thus, the area swept between t = ta and t = tb is R( xb − xa ) /2. In the same way, the area
swept between t = tb and t = t is R ( xc − xb )/2. Since xc − xb = xb − xa , the area swept during time (tb − ta ) is
the same as the area swept during an equal time (tc − tb ). Kepler’s second law is obeyed.
13.65. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on
the surface of the earth?
(b)
(c) The distance is 6.21 × 107 m. This is 1.06RSaturn
13.66.
Solve:
(a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the
orbit?
(b)
(c) The radius of the orbit is
r=
GM E
6.67 × 10−11 N ⋅ m 2 /kg 2 (5.98 × 1024 kg)
=
= 1.00 × 108 m
(vpayload ) 2
(1997 m/s) 2
13.67. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moon’s
radius. At what speed does it impact the moon’s surface?
(b)
(c) The speed is v2 = 1680 m/s.
13.68.
Solve:
(a) A 2.0 × 1030 kg star and a 4.0 × 1030 kg star, each 1.0 × 109 m in diameter, are at rest
1.0 × 1012 m apart. What are their speeds as they crash together?
(b)
(c) The first equation is the conservation of momentum. It can be used in the conservation of energy equation to
give vf1 = 596 km/s and vf2 = −298 km/s. That is, the speeds of the two stars are 6.0 × 102 km/s and
3.0 × 102 km/s.
13.69. Solve: (a) Kepler’s third law for circular orbits is
⎛ 4π 2 ⎞ 3
4π 2 32
T2 =⎜
r
⎟r ⇒ T =
GM
⎝ GM ⎠
Letting a =
4π 2
3
, the first satellite obeys T = ar 2 . For the second satellite, which orbits the same mass,
GM
3
⎛ Δr ⎞ 2
T + ΔT = a ( r + Δr ) = ar ⎜1 +
⎟
r ⎠
⎝
3
2
Since
3
2
Δr
n
<< 1, we can use the approximation (1 ± x ) ≈ 1 ± nx.
r
Thus
3 ⎛
3 Δr ⎞
T + ΔT ≈ ar 2 ⎜ 1 +
⎟
⎝ 2 r ⎠
3
Subtracting the equation T = ar 2 for the first satellite from this,
3 3 ⎛ Δr ⎞
ΔT = ar 2 ⎜ ⎟
2
⎝ r ⎠
Dividing this by the equation for the first satellite,
ΔT 3 Δr
=
T
2 r
(b) The satellites orbit the earth. The fractional difference in their periods is
ΔT 3 1 km
=
= 2.24 × 10−4
T
2 6700 km
After
1
= 4467 periods they will meet again. For the inner satellite,
2.24 × 10−4
T=
3
4π 2
6.700 × 106 m ) 2
(
24
G ( 5.98 × 10 kg )
= 5456 s = 1.52 hrs
So the satellites will meet again in 4467 × 1.52 hrs = 6770 hrs = 282 days.
Assess: A communications satellite has an orbital period of around 1.5 h. The surprising length of time
between the two satellites meeting is due to the small differences in their periods.
13.70. Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the
satellite’s distance from the earth is very small compared to the earth’s (and satellite’s) distance to the sun.
Visualize:
Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational
force toward the earth. This net force is responsible for circular motion around the sun. We want to chose the
distance d to make the period T match the period Te with which the earth orbits the sun. The earth’s orbital
period is given by T e2 = (4π 2 / GM s ) Re3 . Thus
2
Fnet =
GM s m GM e m
mv 2 m ⎛ 2π r ⎞
4π 2
GM
−
=
=
=
=
= mr 3 s
ma
mr
⎜
⎟
centripetal
2
2
2
r
d
r
r ⎝ Te ⎠
Te
Re
Using r = Re − d and canceling the Gm term gives
Ms
M
M
− 2e = 3s ( Re − d )
2
( Re − d )
d
Re
This equation can’t be solved exactly, but we can make use of the fact that d Re to use the binomial
approximation. Factor the Re out of the expressions ( Re − d ) to get
Ms
M
M
− e = 2s (1 − d / Re )
Re2 (1 − d / Re ) 2 d 2
Re
If we think of d/Re = x 1, we can simplify the first term by using (1 ± x )n ≈ 1 ± nx. Here n = −2, so we get
Ms
M
M
M
M
(1 − (−2)d / Re ) − 2e = 2s (1 − d / Re ) ⇒ 2e = 3 2s
Re2
d
Re
d
Re
Thus d = ( M e /3M s )1/3 Re = 1.50 × 109 m.
Assess:
d/Re = 0.010, so our assumption that d/Re 1 is justified.
13.71. Model: Model the earth as a spherical mass and the shuttle and payload as point masses. We’ll
assume mpayload mshuttle .
Visualize:
Solve: (a) The payload, in steady state, is undergoing uniform circular motion. This means that the net force is
directed toward the center of the earth. There is no tangential force component, since such a force would cause
the payload to speed up or slow down and the motion wouldn’t be uniform. Since the net force is due to gravity
and tension, and the gravitational force is radial, the tension force cannot have any tangential component. Thus
the rope is radially outward at angle 0°.
(b) To move with the shuttle, the payload’s period Tp in an orbit of radius rp must exactly match the period Ts
of the shuttle in an orbit of radius rs . The shuttle’s period around the earth is given by T s2 = (4π 2 /GM e ) rs3 ,
where we’ve used the assumption mp ms to infer that the rope’s tension will be too small to have any
influence on the shuttle’s motion. Newton’s second law for the payload’s motion is
( Fnet ) r =
GM e mp
rp 2
− T = mp ar =
mp v 2
rp
2
m ⎛ 2π rp ⎞
4π 2
= p⎜
= mp rp 2
⎟
rp ⎜⎝ Tp ⎟⎠
Tp
Using Tp = Ts and the above expression for Ts , this becomes
GM e mp
r 2p
⇒T =
3
− T = mp rp
GM e GM e mp r p
=
rs3
r 2p rs3
3
3
GM e mp ⎡ ⎛ rp ⎞ ⎤ GM e mp ⎡ ⎛ Re + 290 km ⎞ ⎤
⎢
⎥
⎢
1
1
−
=
−
⎜ ⎟
⎜
⎟ ⎥ = 4.04 N
r 2p ⎢ ⎝ rs ⎠ ⎥
r 2p ⎢ ⎝ Re + 300 km ⎠ ⎥
⎣
⎦
⎣
⎦
Assess: The fact that the tension is so small justifies our assumption that it will have no effect on the motion of
the shuttle.
13.72. Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a
spherical mass. Momentum is conserved during the inelastic collision of the two satellites.
Visualize:
Solve:
For the given orbit, r0 = Re + 1× 106 m = 7.37 × 106 m. The speed of a satellite in this orbit is
GM e
= 7357 m/s
r0
v0 =
The two satellites collide, stick together, and move with velocity v1. The equation for momentum conservation
for the perfectly inelastic collision is
(400 kg + 100 kg)v1 = (400 kg)(7357 m/s) − (100 kg)(7357 m/s)
⇒ v1 = 4414 m/s
The new satellite’s radius immediately after the collision is still r1 = r0 = 7.37 × 106 m. Now it is moving in an
elliptical orbit. We need to determine if the minimum distance r2 is larger or smaller than the earth’s radius
Re = 6.37 × 106 m.
The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is
L = mrv sin β (see Equation 13.26) and is conserved in a trajectory of any shape. The angle β is 90° when
v1 = 4414 m/s and when the satellite is at its closest approach to the earth. From the conservation of angular
momentum, we have
r1v1 = r2v2 = (7.37 × 106 )(4414 m/s) = 3.253 × 1010 m 2 /s
⇒ r2 =
3.253 × 1010 m 2 /s
v2
Using the conservation of energy equation at positions 1 and 2,
1
GM e (500 kg) 1
GM e (500 kg)
(500 kg)(4414 m/s) 2 −
= (500 kg)v22 −
2
7.37 × 106 m 2
r2
Using the above expression for r2 , we can simplify the energy equation to
v22 − (2.452 × 104 m/s)v2 + (8.876 × 107 m 2 /s 2 ) = 0 m 2 /s 2
⇒ v2 = 20,107 m/s and 4414 m/s
A velocity of 20,107 m/s for v2 yields
r2 =
3.253 × 1010 m 2 /s
= 1.62 × 106 m
20,107 m/s
Since r2 < Re = 6.37 × 106 m, the combined mass of the two satellites will crash into the earth.
13.73.
Model: The stars are spherical masses. They each rotate about the system’s center of mass.
Visualize:
Solve: (a) The stars rotate about the system’s center of mass with the same period: T1 = T2 = T . We can locate
the center of mass by letting the origin be at the smaller-mass star. Then
r1 = rcm =
(2.0 × 1030 kg)(0 m) + (6.0 × 1030 kg)(2.0 × 1012 m)
= 1.5 × 1012 m
2.0 × 1030 kg + 6.0 × 1030 kg
Mass m2 undergoes uniform circular motion with radius r2 = 0.5 × 1012 m due to the gravitational force of mass
m1 at distance R = 2.0 × 1012 m. The gravitational force is responsible for the centripetal acceleration, so
2
Fgrav =
1/ 2
⎡ 4π 2
⎤
r2 R 2 ⎥
⇒ T =⎢
⎣ Gm1
⎦
Gm1m2
m2v22 m2 ⎛ 2π r2 ⎞ 4π 2 m2 r2
m
a
=
=
=
2
centripetal
⎜
⎟ =
R2
r2
r2 ⎝ T ⎠
T2
1/ 2
⎡ 4π 2 (0.5 × 1012 m)(0.5 × 1012 m)2 ⎤
=⎢
⎥
−11
2
2
30
⎣ (6.67 × 10 N m /kg ) (2.0 × 10 kg) ⎦
(b) The speed of each star is v = (2π r )/T . Thus
2π r1 2π (1.5 × 1012 m)
=
= 12.3 km/s
T
7.693 × 108 s
2π r2 2π (0.5 × 1012 m)
v2 =
=
= 4.1 km/s
T
7.693 × 108 s
v1 =
= 7.693 × 108 s = 24 years
13.74. Model: The moon is a spherical mass. The moon lander is originally in a circular orbit.
Visualize:
Solve: Energy and momentum are conserved between points 1 and 2 in the elliptical orbit. Also, at both points
β = 90°, so L = mrv sin β = mrv. Let h = 1000 km. The original speed of the lander is
v0 =
GM m
v
r R +h
= 1338 m/s. Conservation of angular momentum requires mr1v1 = mr2v2 ⇒ 2 = 1 = m
. The
Rm + h
v1 r2
Rm
energy conservation equation is
1
M m 1
M m
mv12 − G m = mv22 − G m
Rm + h 2
Rm
2
⎛ R +h⎞
Substituting v2 = ⎜ m
⎟ v1 ,
⎝ Rm ⎠
2
1 2
Mm
1 ⎛ R +h⎞
Mm
v1 − G
= v12 ⎜ m
⎟ −G
2
Rm + h 2 ⎝ Rm ⎠
Rm
⎛ ⎛ R +h⎞ ⎞
⎛ 1
1 ⎞
v12 ⎜1 − ⎜ m
−
⎟ ⎟ = 2GM m ⎜
⎟
⎜ ⎝ Rm ⎠ ⎟
⎝ Rm + h Rm ⎠
⎝
⎠
2
2GM m h
1
2GM m ⎛ Rm ⎞
6
2 2
=
⎜
⎟ = 1.392 × 10 m /s
Rm ( Rm + h ) ⎛ R + h ⎞ 2
Rm + h ) ⎝ 2 Rm + h ⎠
(
m
⎜
⎟ −1
⎝ Rm ⎠
v1 = 1180 m/s
v12 =
The fractional change in speed required to just graze the moon at point 2 is
1338 m/s − 1180 m/s
= 11.8%
1338 m/s
Assess:
A reduction in speed by almost 12% is reasonable.
13.75. Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated
system, so mechanical energy is conserved. Also, the angular momentum of the satellite is conserved.
Visualize: Please refer to Figure CP13.75.
Solve: (a) Angular momentum is L = mrv sin β . The angle β = 90° at points 1 and 2, so conservation of
angular momentum requires
⎛r ⎞
mr1v1′ = mr2v2′ ⇒ v1′ = ⎜ 2 ⎟ v′2
⎝ r1 ⎠
The energy conservation equation is
⎛ 1 1⎞
GMm 1
GMm
1
m(v′2 ) 2 −
= m(v1′ ) 2 −
⇒ (v′2 ) 2 − (v1′ ) 2 = 2GM ⎜ − ⎟
r2
r1
2
2
⎝ r2 r1 ⎠
Using the angular momentum result for v1′ gives
2
⎛r ⎞
⎛r −r ⎞
⎡ r2 − r2 ⎤
⎛r −r ⎞
(v2′ ) 2 − (v2′ ) 2 ⎜ 2 ⎟ = 2GM ⎜ 1 2 ⎟ ⇒ (v2′ ) 2 ⎢ 1 2 2 ⎥ = 2GM ⎜ 1 2 ⎟
r
r
r
r
⎝ 1⎠
⎝ 12 ⎠
⎣ 1 ⎦
⎝ r1r2 ⎠
v2′ =
2GM ( r1 / r2 )
r1 + r2
⎛r ⎞
2GM (r2 / r1 )
and v1′ = ⎜ 2 ⎟ v2′ =
r
r1 + r2
⎝ 1⎠
(b) For the circular orbit,
v1 =
GM
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
= 7730 m/s
r1
(6.37 × 106 m + 3 × 105 m)
For the elliptical orbit,
r1 = Re + 300 km = 6.37 × 106 m + 3 × 105 m = 6.67 × 106 m
r2 = Re + 35,900 km = 6.37 × 106 m + 3.59 × 107 m = 4.23 × 107 m
v1′ =
2GM ( r2 / r1 )
⇒ v1′ = 10,160 m/s
( r1 + r2 )
1
1
(c) From the work-kinetic energy theorem, W = ΔK = mv1′2 − mv12 = 2.17 × 1010 J
2
2
(d) v2′ =
2GM (r1 / r2 )
r1 + r2
Using the same values of r1 and r2 as in (b), v′2 = 1600 m/s. For the circular orbit,
v2 =
GM
= 3070 m/s
r2
1
1
W = mv22 − mv′22 = 3.43 × 109 J
2
2
(f) The total work done is 2.513 × 1010 J. This is the same as in Example 13.6, but here we’ve learned how the
work has to be divided between the two burns.
(e)
13.76.
Model: The rod is thin and uniform.
Visualize:
Solve: (a) The rod is not spherical so must be divided into thin sections each dr wide and having mass dm.
Since the rod is uniform,
dm dr
M
=
⇒ dm =
dr
M
L
L
The width of the rod is small enough so that all of dm is distance r away from m. The gravitational potential
energy of dm and m is
m dm
= −G
dU = −G
r
⎛M ⎞
m ⎜ ⎟ dr
⎝ L⎠
r
The total potential energy of the rod and mass m is found by adding the contributions dU from every point along
the rod in an integral:
L
L⎞
⎛
x+
x+ ⎟
GMm 2 dr
GMm ⎜
2
U = ∫ dU = −
=−
ln ⎜
⎟
L
L ∫L r
L
⎜ x− ⎟
x−
2
2⎠
⎝
Note r is increasing with the limits chosen as they are.
(b) The force on m when at x is
⎛
⎞
1 ⎟
dU GMm d ⎛ ⎛
L⎞
L ⎞ ⎞ GMm ⎜ 1
⎛
=
−
F =−
⎜
⎟
⎜ ln ⎜ x + ⎟ − ln ⎜ x − ⎟ ⎟ =
2⎠
2 ⎠⎠
dx
L dx ⎝ ⎝
L ⎜ x+ L x− L⎟
⎝
⎝
2
2⎠
L
4
⎛
⎞
= −GMm ⎜ 2
, x≥
2 ⎟
2
⎝ 4x − L ⎠
Assess: The direction of the force is towards the –x direction, as expected. The force magnitude approaches ∞
1
as the mass m approaches the end of the rod, but goes to zero like 2 as x gets large. This is expected since from
x
far away the rod looks like a point mass.
13.77.
Model: The ring is uniform and is so thin that every point on it is the same distance r from m.
Visualize:
Solve: (a) We must determine the gravitational potential (dU) between m and an arbitrary part of the ring dm, then add
using an integral all the contributions to U. Since the ring is uniform,
dm
dl
M
=
⇒ dm =
dl
M 2π r
2π r
The distance from m to dm is r = x 2 + R 2 . The gravitational potential between m and dm is
dU = −G
m dm
mM
= −G
r
2π r
dl
x + R2
2
The total gravitational potential is
U = ∫ dU = −
GmM
∫ dl
2π r x 2 + R 2 ring
Note that x and R do not change for any location of dm. The integral is just the length of the ring.
∫ dl = 2π R
ring
Thus
U =−
GmM
x2 + R2
(b) The force on m when at x is
(
)
−1
−3
dU
d
⎛ 1⎞
= GmM
x 2 + R 2 ) 2 = GmM ⎜ − ⎟ ( x 2 + R 2 ) 2 ( 2 x )
(
dx
dx
⎝ 2⎠
x
= −GmM
3
2
( x + R2 ) 2
Fx = −
Thus the magnitude of the force is
x
.
( x 2 + R 2 )3 2
Assess: The force is zero at the center of the ring. Elsewhere its direction is toward the origin. As x gets large, the force
1
decreases like 2 . This is expected since from far away the ring looks like a point mass.
x
F = GmM
13-1
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency,
hence
T=
1
1
=
= 2.27 × 10−3 s = 2.27 ms
f 440 Hz
14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion.
Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark.
33 s
(a)
T=
= 3.3 s oscillation = 3.3 s
10 oscillations
1
1
(b)
f = =
= 0.303 Hz ≈ 0.30 Hz
T 3.3 s
ω = 2π f = 2π ( 0.303 Hz ) = 1.90 rad s
(c)
(d) The oscillation from one side to the other is equal to 60 cm − 10 cm = 50 cm = 0.50 m. Thus, the amplitude is
A = 12 ( 0.50 m ) = 0.25 m.
(e) The maximum speed is
⎛ 2π ⎞
vmax = ω A = ⎜
⎟ A = (1.90 rad s )( 0.25 m ) = 0.48 m s
⎝ T ⎠
14.3. Model: The air-track glider attached to a spring is in simple harmonic motion.
Visualize: The position of the glider can be represented as x (t ) = A cos ω t.
Solve: The glider is pulled to the right and released from rest at t = 0 s. It then oscillates with a period
T = 2.0 s and a maximum speed vmax = 40 cm s = 0.40 m s.
v
2π
2π
0.40 m s
=
= π rad s ⇒ A = max =
= 0.127 m = 12.7 cm
T
2.0 s
ω
π rad s
(b) The glider’s position at t = 0.25 s is
(a) vmax = ω A and ω =
x0.25 s = ( 0.127 m ) cos ⎡⎣(π rad s )( 0.25 s ) ⎤⎦ = 0.090 m = 9.0 cm
14.4. Model: The oscillation is the result of simple harmonic motion.
Visualize: Please refer to Figure EX14.4.
Solve: (a) The amplitude A = 10 cm.
(b) The time to complete one cycle is the period, hence T = 2.0 s and
f =
(c)
The
position
of
an
object
1
1
=
= 0.50 Hz
T 2.0 s
undergoing
simple
harmonic
motion
is
At t = 0 s, x0 = −5 cm, thus
−5 cm = (10 cm ) cos ⎣⎡ω ( 0 s ) + φ0 ⎦⎤
⇒ cos φ0 =
1
2π
−5 cm
⎛ 1⎞
rad or ± 120°
= − ⇒ φ0 = cos −1 ⎜ − ⎟ = ±
10 cm
2
3
⎝ 2⎠
Since the oscillation is originally moving to the left, φ0 = +120°.
x ( t ) = A cos (ω t + φ0 ) .
14.5. Model: The oscillation is the result of simple harmonic motion.
Visualize: Please refer to Figure EX14.5.
Solve: (a) The amplitude A = 20 cm.
(b) The period T = 4.0 s, thus
f =
1
1
=
= 0.25 Hz
T 4.0 s
(c) The position of an object undergoing simple harmonic motion is
x ( t ) = A cos (ω t + φ0 ) . At
t = 0 s, x0 = 10 cm. Thus,
π
⎛ 10 cm ⎞
−1 ⎛ 1 ⎞
10 cm = ( 20 cm ) cos φ0 ⇒ φ0 = cos −1 ⎜
⎟ = cos ⎜ ⎟ = ± rad = ±60°
3
⎝ 20 cm ⎠
⎝ 2⎠
Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus
must have a phase constant between π and 2π radians. Therefore, φ0 = −
π
3
rad = −60°.
14.6. Visualize: The phase constant 32 π has a plus sign, which implies that the object undergoing simple
harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the
left.
Solve: The position of the object is
x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) = ( 4.0 cm ) cos ⎡⎣( 4π rad s ) t + 32 π rad ⎤⎦
The amplitude is A = 4 cm and the period is T = 1 f = 0.50 s. A phase constant φ0 = 2π 3 rad =120° (second
quadrant) means that x starts at − 12 A and is moving to the left (getting more negative).
Assess:
We can see from the graph that the object starts out moving to the left.
14.7. Visualize: A phase constant of −
π
implies that the object that undergoes simple harmonic motion is
2
in the lower half of the circular motion diagram. That is, the object is moving to the right.
Solve: The position of the object is given by the equation
⎡⎛ π
⎤
⎞ π
x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) = ( 8.0 cm ) cos ⎢⎜ rad s ⎟ t − rad ⎥
2
2
⎠
⎣⎝
⎦
The amplitude is A = 8.0 cm and the period is T = 1 f = 4.0 s. With φ0 = −π 2 rad, x starts at 0 cm and is
moving to the right (getting more positive).
Assess:
As we see from the graph, the object starts out moving to the right.
14.8. Solve: The position of the object is given by the equation
x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 )
We can find the phase constant φ0 from the initial condition:
0 cm = ( 4.0 cm ) cos φ0 ⇒ cos φ0 = 0 ⇒ φ0 = cos −1 ( 0 ) = ± 12 π rad
Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence,
φ0 = − 12 π rad. The final result, with f = 4.0 Hz, is
x ( t ) = ( 4.0 cm ) cos ⎡⎣( 8.0π rad s ) t − 12 π rad ⎤⎦
14.9. Solve: The position of the object is given by the equation
x ( t ) = A cos (ω t + φ0 )
The amplitude A = 8.0 cm. The angular frequency ω = 2π f = 2π ( 0.50 Hz ) = π rad/s. Since at t = 0 it has its
most negative velocity, it must be at the equilibrium point x = 0 cm and moving to the left, so φ0 =
x ( t ) = ( 8.0 cm ) cos[(π rad/s)t +
π
2
rad]
π
2
. Thus
14.10. Model: The air-track glider is in simple harmonic motion.
Solve:
(a) We can find the phase constant from the initial conditions for position and velocity:
x0 = A cos φ0
v0 x = −ω A sin φ0
Dividing the second by the first, we see that
sin φ0
v
= tan φ0 = − 0 x
cos φ0
ω x0
The glider starts to the left ( x0 = −5.00 cm) and is moving to the right (v0 x = +36.3 cm/s). With a period of
1.5 s =
3
2
s, the angular frequency is ω = 2π /T = 34 π rad/s. Thus
⎛
⎞ 1
36.3 cm/s
2
⎟ = 3 π rad (60°) or – 3 π rad (–120°)
(4
/3
rad/s)(
5.00
cm)
π
−
⎝
⎠
φ0 = tan −1 ⎜ −
The tangent function repeats every 180°, so there are always two possible values when evaluating the arctan
function. We can distinguish between them because an object with a negative position but moving to the right is
in the third quadrant of the corresponding circular motion. Thus φ0 = − 23 π rad, or −120°.
(b) At time t, the phase is φ = ω t + φ0 = ( 34 π rad/s) t − 32 π rad. This gives φ = − 23 π rad, 0 rad, 32 π rad, and
4
3
π rad at, respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.
14.11. Model: The block attached to the spring is in simple harmonic motion.
Solve:
The period of an object attached to a spring is
T = 2π
m
= T0 = 2.0 s
k
where m is the mass and k is the spring constant.
(a) For mass = 2m,
T = 2π
2m
=
k
( 2 )T = 2.8 s
0
(b) For mass 12 m,
T = 2π
1
2
m
= T0
k
2 = 1.41 s
(c) The period is independent of amplitude. Thus T = T0 = 2.0 s
(d) For a spring constant = 2k ,
T = 2π
m
= T0
2k
2 = 1.41 s
14.12. Model: The air-track glider attached to a spring is in simple harmonic motion.
Solve:
Experimentally, the period is T = (12.0 s ) (10 oscillations ) = 1.20 s. Using the formula for the period,
2
T = 2π
2
m
⎛ 2π ⎞
⎛ 2π ⎞
⇒k =⎜
⎟ m=⎜
⎟ ( 0.200 kg ) = 5.48 N m
k
⎝ T ⎠
⎝ 1.20 s ⎠
14.13. Model: The mass attached to the spring oscillates in simple harmonic motion.
Solve: (a) The period T = 1 f = 1 2.0 Hz = 0.50 s.
(b) The angular frequency ω = 2π f = 2π (2.0 Hz) = 4π rad/s.
(c) Using energy conservation
1
2
kA2 = 12 kx02 + 12 mv02x
Using x0 = 5.0 cm, v0 x = −30 cm/s and k = mω 2 = (0.200 kg)(4π rad/s) 2 , we get A = 5.54 cm.
(d) To calculate the phase constant φ0 ,
A cos φ0 = x0 = 5.0 cm
⎛ 5.0 cm ⎞
⇒ φ0 = cos −1 ⎜
⎟ = 0.45 rad
⎝ 5.54 cm ⎠
(e) The maximum speed is vmax = ω A = ( 4π rad s )( 5.54 cm ) = 70 cm s.
(f) The maximum acceleration is
amax = ω 2 A = ω (ω A ) = ( 4π rad s )( 70 cm s ) = 8.8 m s 2
2
(g) The total energy is E = 12 mvmax
= 12 ( 0.200 kg )( 0.70 m s ) = 0.049 J.
2
(h) The position at t = 0.40 s is
x0.4 s = ( 5.54 cm ) cos ⎡⎣( 4π rad s )( 0.40 s ) + 0.45 rad ⎤⎦ = +3.8 cm
14.14. Model: The oscillating mass is in simple harmonic motion.
Solve: (a) The amplitude A = 2.0 cm.
(b) The period is calculated as follows:
ω=
2π
2π
= 10 rad s ⇒ T =
= 0.63 s
T
10 rad s
(c) The spring constant is calculated as follows:
ω=
k
2
⇒ k = mω 2 = ( 0.050 kg )(10 rad s ) = 5.0 N m
m
(d) The phase constant φ0 = − 14 π rad.
(e) The initial conditions are obtained from the equations
x ( t ) = ( 2.0 cm ) cos (10t − 14 π ) and vx ( t ) = − ( 20.0 cm s ) sin (10t − 14 π )
At t = 0 s, these equations become
x0 = ( 2.0 cm ) cos ( − 14 π ) = 1.41 cm and v0 x = − ( 20 cm s ) sin ( − 14 π ) = 14.1 cm s
In other words, the mass is at +1.41 cm and moving to the right with a velocity of 14.1 cm/s.
(f) The maximum speed is vmax = Aω = ( 2.0 cm )(10 rad s ) = 20 cm s.
(g) The total energy E = 12 kA2 = 12 ( 5.0 N m )( 0.020 m ) = 1.00 × 10−3 J.
2
(h) At t = 0.41 s, the velocity is
v0 x = − ( 20 cm s ) sin ⎡⎣(10 rad s )( 0.40 s ) − 14 π ⎤⎦ = 1.46 cm s
14.15. Model: The block attached to the spring is in simple harmonic motion.
Visualize:
Solve:
(a) The conservation of mechanical energy equation K f + U sf = K i + U si is
1
2
mv12 + 12 k ( Δx ) = 12 mv02 + 0 J ⇒ 0 J + 12 kA2 = 12 mv02 + 0 J
⇒ A=
2
m
1.0 kg
v0 =
( 0.40 m s ) = 0.10 m = 10.0 cm
k
16 N m
(b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation
K 2 + U s2 = K i + U si is
2
1 2 1 ⎛ A⎞ 1 2
1
1
1⎛1
1⎛1
⎞ 1
⎞ 3⎛ 1
⎞
mv2 + k ⎜ ⎟ = mv0 + 0 J ⇒ mv22 = mv02 − ⎜ kA2 ⎟ = mv02 − ⎜ mv02 ⎟ = ⎜ mv02 ⎟
2
2 ⎝2⎠ 2
2
2
4⎝ 2
2
4
2
4
2
⎠
⎝
⎠
⎝
⎠
⇒ v2 =
The velocity is 35 cm/s.
3
3
v0 =
( 0.40 m s ) = 0.346 m s
4
4
14.16. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve:
(a) At equilibrium, Newton’s first law applied to the physics book is
( F ) − mg = 0 N ⇒ −k Δy − mg = 0 N
⇒ k = − mg Δy = − ( 0.500 kg ) ( 9.8 m s ) ( −0.20 m ) = 24.5 N m
sp y
2
(b) To calculate the period:
ω=
k
24.5 N m
2π
2π rad
=
= 7.0 rad s and T =
=
= 0.90 s
m
0.500 kg
ω 7.0 rad s
(c) The maximum speed is
vmax = Aω = ( 0.10 m )( 7.0 rad s ) = 0.70 m s
Maximum speed occurs as the book passes through the equilibrium position.
14.17. Model: The vertical oscillations constitute simple harmonic motion.
Solve: To find the oscillation frequency using ω = 2π f = k m , we first need to find the spring constant k. In
equilibrium, the weight mg of the block and the spring force k ΔL are equal and opposite. That is,
mg = k ΔL ⇒ k = mg ΔL . The frequency of oscillation f is thus given as
f =
1
2π
k
1
=
m 2π
mg ΔL
1
=
2π
m
g
1
=
ΔL 2π
9.8 m s 2
= 3.5 Hz
0.020 m
14.18. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve:
The period and angular frequency are
T=
20 s
2π
2π
= 0.6667 s and ω =
=
= 9.425 rad s
30 oscillations
T
0.6667 s
(a) The mass can be found as follows:
ω=
k
k
15 N/m
⇒m= 2 =
= 0.169 kg
2
m
ω
( 9.425 rad s )
(b) The maximum speed vmax = ω A = ( 9.425 rad s )( 0.060 m ) = 0.57 m/s.
14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve:
The period of the pendulum is
T0 = 2π
L0
= 4.0 s
g
(a) The period is independent of the mass and depends only on the length. Thus T = T0 = 4.0 s.
(b) For a new length L = 2 L0 ,
T = 2π
2 L0
= 2T0 = 5.7 s
g
T = 2π
L0 2
1
T0 = 2.8 s
=
g
2
(c) For a new length L = L0 /2,
(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s.
14.20. Model: The pendulum undergoes simple harmonic motion.
Solve: (a) The amplitude is 0.10 rad.
(b) The frequency of oscillations is
f =
ω
5
=
Hz = 0.796 Hz
2π 2π
(c) The phase constant φ = π rad.
(d) The length can be obtained from the period:
2
2
ω = 2π f =
⎛
⎞
⎛ 1 ⎞
g
1
2
⇒L=⎜
⎟⎟ ( 9.8 m s ) = 0.392 m
⎟ g = ⎜⎜
L
2
0.796
Hz
π
(
)
⎝ 2π f ⎠
⎝
⎠
(e) At t = 0 s, θ 0 = ( 0.10 rad ) cos (π ) = −0.10 rad. To find the initial condition for the angular velocity we take
the derivative of the angular position:
θ ( t ) = ( 0.10 rad ) cos ( 5t + π ) ⇒
At t = 0 s, ( dθ dt )0 = ( −0.50 rad ) sin (π ) = 0 rad s.
dθ ( t )
= − ( 0.10 rad )( 5 ) sin ( 5t + π )
dt
(f) At t = 2.0 s, θ 2.0 = ( 0.10 rad ) cos ( 5 ( 2.0 s ) + π ) = 0.084 rad.
14.21. Model: Assume the small-angle approximation so there is simple harmonic motion.
Solve:
The period is T = 12 s 10 oscillations = 1.20 s and is given by the formula
2
T = 2π
2
L
⎛ T ⎞
⎛ 1.20 s ⎞
2
⇒L=⎜
⎟ g =⎜
⎟ ( 9.8 m s ) = 36 cm
g
⎝ 2π ⎠
⎝ 2π ⎠
14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve:
(a) On the earth the period is
Tearth = 2π
L
1.0 m
= 2π
= 2.0 s
g
9.80 m s 2
(b) On Venus the acceleration due to gravity is
g Venus =
GM Venus ( 6.67 × 10
=
2
RVenus
⇒ TVenus = 2π
−11
N ⋅ m 2 kg 2 )( 4.88 × 1024 kg )
( 6.06 ×106 m )
2
L
1.0 m
= 2π
= 2.1 s
g Venus
8.86 m s 2
= 8.86 m s 2
14.23. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes
simple harmonic motion.
Solve: Using the formula g = GM R 2 , the periods of the pendulums on the moon and on the earth are
Tearth = 2π
L
L R2
L R2
= 2π earth earth and Tmoon = 2π moon moon
g
GM earth
GM moon
Because Tearth = Tmoon ,
2
2π
2
⎛M
⎞⎛ R ⎞
Learth Rearth
L R2
= 2π moon moon ⇒ Lmoon = ⎜ moon ⎟⎜ earth ⎟ Learth
GM earth
GM moon
⎝ M earth ⎠⎝ Rmoon ⎠
2
⎛ 7.36 × 1022 kg ⎞⎛ 6.37 × 106 m ⎞
=⎜
⎟⎜
⎟ ( 2.0 m ) = 33 cm
24
6
⎝ 5.98 × 10 kg ⎠⎝ 1.74 × 10 m ⎠
14.24. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion.
Solve:
The time periods of the pendulums on the earth and on Mars are
Tearth = 2π
L
g earth
and TMars = 2π
L
g Mars
Dividing these two equations,
Tearth
=
TMars
2
2
⎛T ⎞
g Mars
⎛ 1.50 s ⎞
2
⇒ g Mars = g earth ⎜ earth ⎟ = ( 9.8 m s 2 ) ⎜
⎟ = 3.67 m s
g earth
⎝ 2.45 s ⎠
⎝ TMars ⎠
14.25. Visualize: Please refer to Figure Ex14.25.
Solve: The mass of the wrench can be obtained from the length that it stretches the spring. From Equation 14.41,
ΔL =
mg
k ΔL 360 N/m ( 0.030 m )
⇒m=
=
= 1.10 kg
k
g
9.8 m/s 2
When swinging on a hook the wrench is a physical pendulum. From Equation 14.52,
2π f =
mgl
2π
mgl
⇒
=
I
T
I
From the figure, l = 0.14 m. Thus
2
2
⎛ T ⎞
⎛ 0.90 s ⎞
−2
2
2
I =⎜
⎟ mgl = ⎜
⎟ (1.10 kg ) ( 9.8 m/s ) ( 0.14 m ) = 3.1 × 10 kg m
⎝ 2π ⎠
⎝ 2π ⎠
14.26. Model: The spider is in simple harmonic motion.
Solve:
Your tapping is a driving frequency. Largest amplitude at f ext = 1.0 Hz means that this is the resonance
frequency, so f 0 = f ext = 1.0 Hz. That is, the spider’s natural frequency of oscillation f 0 is 1.0 Hz and
ω 0 = 2π f 0 = 2π rad s. We have
ω0 =
k
2
⇒ k = mω 02 = ( 0.0020 kg )( 2π rad s ) = 0.079 N m
m
14.27. Model: The motion is a damped oscillation.
Solve:
The amplitude of the oscillation at time t is given by Equation 14.58: A ( t ) = A0e − t 2τ , where τ = m b is
the time constant. Using x = 0.368 A and t = 10.0 s, we get
0.368 A = Ae−10.0 s 2τ ⇒ ln ( 0.368 ) =
−10 s
10.0 s
⇒τ = −
= 5.00 s
2τ
2ln ( 0.368 )
14.28. Model: The motion is a damped oscillation.
The position of a damped oscillator is x ( t ) = Ae− ( t 2τ ) cos (ω t + φ0 ) . The frequency is 1.0 Hz and the
Solve:
damping time constant τ is 4.0 s. Let us assume φ0 = 0 rad and A = 1 with arbitrary units. Thus,
x ( t ) = e − t (8.0 s ) cos ⎣⎡ 2π (1.0 Hz ) t ⎦⎤ ⇒ x ( t ) = e −0.125 t cos ( 2π t )
where t is in s. Values of x(t) at selected values of t are displayed in the following table:
t(s)
x(t)
t(s)
x(t)
t(s)
x(t)
0
1
2.00
0.779
6.00
0.472
0.25
0
2.50
−0.732
6.50
− 0.444
0.50
−0.939
3.00
0.687
7.00
0.417
0.75
0
3.50
−0.646
7.50
− 0.392
1.00
0.882
4.00
0.607
8.00
0.368
1.25
0
4.50
−0.570
8.50
− 0.346
1.50
−0.829
5.00
0.535
9.00
0.325
1.75
0
5.50
−0.503
9.50
− 0.305
10.00
0.286
14.29. Model: The pendulum is a damped oscillator.
Solve:
The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:
T = 2π
4 ( 3600 s )
L
15.0 m
= 2π
= 7.773 s ⇒ N osc =
= 1853
2
g
9.8 m s
7.773 s
The amplitude of the pendulum as a function of time is A ( t ) = Ae − bt / 2 m . The exponent of this expression can be
calculated to be
−
bt
( 0.010 kg s )( 4 × 3600 s ) = −0.6545
=−
2m
2 (110 kg )
We have A ( t ) = (1.50 m ) e −0.6545 = 0.780 m.
14.30. Model: The vertical oscillations are damped and follow simple harmonic motion.
Solve: The position of the ball is given by x ( t ) = Ae − ( t 2τ ) cos (ω t + φ0 ) . The amplitude A ( t ) = Ae − ( t 2τ ) is a
function of time. The angular frequency is
ω=
k
=
m
(15.0 N m ) = 5.477 rad s ⇒ T = 2π = 1.147 s
0.500 kg
ω
Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 × 1.147 s =
34.41 s, we have
3.0 cm = ( 6.0 cm ) e− (34.414 s 2τ ) ⇒ 0.50 = e − ( 34.41 s 2τ ) ⇒ ln ( 0.50 ) =
−34.41 s
⇒ τ = 25 s
2τ
14.31. Visualize: Please refer to Figure P14.31.
Solve:
The position and the velocity of a particle in simple harmonic motion are
x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) = −vmax sin (ω t + φ0 )
(a) At t = 0 s, the equation for x yields
( −5.0 cm ) = (10.0 cm ) cos (φ0 ) ⇒ φ0 = cos−1 ( −0.5) = ± 23 π rad
Because the particle is moving to the left at t = 0 s, it is in the upper half of the circular motion diagram, and the
phase constant is between 0 and π radians. Thus, φ0 = 23 π rad.
(b) The period is 4.0 s. At t = 0 s,
⎛ 2π ⎞ ⎛ 2π ⎞
v0 x = − Aω sin φ0 = − (10.0 cm ) ⎜
⎟ sin ⎜
⎟ = −13.6 cm/s
⎝ T ⎠ ⎝ 3 ⎠
(c) The maximum speed is
⎛ 2π ⎞
vmax = ω A = ⎜
⎟ (10.0 cm ) = 15.7 cm s
⎝ 4.0 s ⎠
Assess: The negative velocity at t = 0 s is consistent with the position-versus-time graph and the positive sign
of the phase constant.
14.32. Visualize: Please refer to Figure P14.32.
Solve:
The position and the velocity of a particle in simple harmonic motion are
x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) = −vmax sin (ω t + φ0 )
From the graph, T = 12 s and the angular frequency is
ω=
2π
2π π
=
=
rad s
T 12 s 6
(a) Because vmax = Aω = 60 cm s, we have
A=
60 cm s
ω
=
60 cm s
= 115 cm
π 6 rad s
(b) At t = 0 s,
v0x = − Aω sin φ0 = −30 cm/s ⇒ − ( 60 cm s ) sin φ0 = −30 cm/s
⇒ φ0 = sin −1 ( 0.5 rad ) = 16 π rad ( 30° ) or 56 π rad (150° )
Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant
of the circular motion diagram. Thus φ0 = 56 π rad.
(c) At t = 0 s, x0 = (115 cm ) cos ( 56 π rad ) = −100 cm.
14.33. Model: The vertical mass/spring systems are in simple harmonic motion.
Visualize: Please refer to Figure P14.33.
Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to the
maximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz.
(b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the
first time is at t = 1.5 s. The time period of system B is thus 6.0 s.
(c) Spring/mass A undergoes three oscillations in 12 s, giving it a period TA = 4.0 s. Spring/mass B undergoes 2
oscillations in 12 s, giving it a period TB = 6.0 s. We have
TA = 2π
⎛ m ⎞⎛ k ⎞ 4.0 s 2
mA
mB
T
⇒ A = ⎜ A ⎟⎜ B ⎟ =
=
and TB = 2π
kA
kB
TB
⎝ mB ⎠⎝ kA ⎠ 6.0 s 3
If mA = mB , then
kB 4
k
9
= ⇒ A = = 2.25
kA 9
kB 4
14.34. Solve: The object’s position as a function of time is x ( t ) = A cos (ω t + φ0 ) . Letting x = 0 m at t = 0 s,
gives
0 = A cos φ0 ⇒ φ0 = ± 12 π
Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase
constant between −π and 0 radians. Thus, φ0 = − 12 π and
x ( t ) = A cos (ω t − 12 π ) ⇒ x ( t ) = A sin ω t = ( 0.10 m ) sin ( 12 π t )
where we have used A = 0.10 m and
ω=
2π 2π rad π
rad s
=
=
T
4.0 s
2
Let us now find t where x = 0.60 m:
2
⎛π ⎞
⎛ 0.060 m ⎞
0.060 m = ( 0.10 m ) sin ⎜ t ⎟ ⇒ t = sin −1 ⎜
⎟ = 0.41 s
π
⎝2 ⎠
⎝ 0.10 m ⎠
Assess:
The answer is reasonable because it is approximately 18 of the period.
14.35. Model: The block attached to the spring is in simple harmonic motion.
Visualize:
The position and the velocity of the block are given by the equations
x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 )
Solve:
To graph x(t) we need to determine ω , φ0 , and A. These quantities will be found by using the initial
(t = 0 s) conditions on x(t) and vx (t ). The period is
T = 2π
m
1.0 kg
2π 2π rad
= 2π
= 1.405 s ⇒ ω =
=
= 4.472 rad s
k
T 1.405 s
20 N m
At t = 0 s, x0 = A cos φ0 and v0 x = − Aω sin φ0 . Dividing these equations,
tan φ0 = −
v0 x
( −1.0 m s )
=−
= 1.1181 ⇒ φ0 = 0.841 rad
ω x0
( 4.472 rad s )( 0.20 m )
From the initial conditions,
2
⎛v ⎞
A = x + ⎜ 0x ⎟ =
⎝ω ⎠
2
0
2
⎛ −1.0 m s ⎞
( 0.20 m ) + ⎜
⎟ = 0.300 m
⎝ 4.472 rad s ⎠
2
The position-versus-time graph can now be plotted using the equation
x ( t ) = ( 0.300 m ) cos ⎡⎣( 4.472 rad s ) t + 0.841 rad ⎤⎦
14.36. Model: The astronaut attached to the spring is in simple harmonic motion.
Visualize: Please refer to Figure P14.36.
Solve: (a) From the graph, T = 3.0 s, so we have
2
T = 2π
2
m
⎛ T ⎞
⎛ 3.0 s ⎞
⇒m=⎜
⎟ k =⎜
⎟ ( 240 N m ) = 55 kg
k
⎝ 2π ⎠
⎝ 2π ⎠
(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A = 12 ( 0.80 m ) = 0.40 m, φ0 =
0 rad, and
ω=
2π
2π
=
= 2.1 rad s
T
3.0 s
The equation for the position of the astronaut is
x ( t ) = A cos ω t + 1.0 m = ( 0.4 m ) cos ⎡⎣( 2.1 rad s ) t ⎤⎦ + 1.0 m
⇒ 1.2 m= ( 0.4 m ) cos ⎡⎣( 2.1 rad s ) t ⎤⎦ + 1.0 m ⇒ cos ⎡⎣( 2.1 rad s ) t ⎤⎦ = 0.5 ⇒ t = 0.50 s
The equation for the velocity of the astronaut is
vx ( t ) = − Aω sin (ω t )
⇒ v0.5 s = − ( 0.4 m )( 2.1 rad s ) sin ⎡⎣( 2.1 rad s )( 0.50 s ) ⎤⎦ = −0.73 m s
Thus her speed is 0.73 m/s.
14.37. Model: The particle is in simple harmonic motion.
Solve:
The equation for the velocity of the particle is
vx ( t ) = − ( 25 cm )(10 rad s ) sin (10 t )
Substituting into K = 2U gives
2
2
1 2
⎛1
⎞ 1
mvx ( t ) = 2 ⎜ kx 2 ( t ) ⎟ ⇒ m ⎣⎡ − ( 250 cm s ) sin (10 t ) ⎦⎤ = k ⎣⎡( 25 cm ) cos (10 t ) ⎦⎤
2
2
2
⎝
⎠
sin 2 (10 t )
⎛ k ⎞ ( 25 cm )
⎛ 1 ⎞ 2
= 2⎜ ⎟
= 2ω 2 ⎜
⎟s
2
cos 2 (10 t )
⎝ m ⎠ ( 250 cm s )
⎝ 100 ⎠
2
⇒
1
2⎛ 1 ⎞ 2
−1
2.0 = 0.096 s
⇒ tan 2 (10 t ) = 2 (10 rad s ) ⎜
⎟ s = 2.0 ⇒ t = tan
10
⎝ 100 ⎠
14.38. Model: The spring undergoes simple harmonic motion.
Solve:
(a) Total energy is E = 12 kA2 . When the displacement is x = 12 A, the potential energy is
U = 12 kx 2 = 12 k ( 12 A) 2 = 14 ( 12 kA2 ) = 14 E ⇒ K = E − U = 34 E
One quarter of the energy is potential and three-quarters is kinetic.
(b) To have U = 12 E requires
U = 12 kx 2 = 12 E = 12 ( 12 kA2 ) ⇒ x =
A
2
14.39. Solve: Average speed is vavg = Δx/Δt. During half a period (Δt = 12 T ), the particle moves from
x = − A to x = + A( Δx = 2 A). Thus
vavg =
Δx 2 A 4 A
4A
2
2
π
=
=
=
= (ω A) = vmax ⇒ vmax = vavg
Δt T /2 T
2π /ω π
π
2
14.40. Model: The ball attached to a spring is in simple harmonic motion.
Solve:
(a) Let t = 0 s be the instant when x0 = −5.0 cm and v0 = 20 cm/s. The oscillation frequency is
k
2.5 N m
=
= 5.0 rad / s
m
0.100 kg
ω=
Using Equation 14.27, the amplitude of the oscillation is
2
⎛v ⎞
A = x02 + ⎜ 0 ⎟ =
⎝ω ⎠
2
⎛ 20 cm / s ⎞
2
( −5.0 cm ) + ⎜
⎟ = 6.4 cm
⎝ 5.0 rad / s ⎠
(b) The maximum acceleration is amax = ω 2 A = 160 cm s 2 .
(c) For an oscillator, the acceleration is most positive ( a = amax ) when the displacement is most negative
( x = − xmax = − A) . So the acceleration is maximum when x = −6.4 cm.
(d) We can use the conservation of energy between x0 = −5.0 cm and x1 = 3.0 cm:
1
2
mv02 + 12 kx02 = 12 mv12 + 12 kx12 ⇒ v1 = v02 +
k 2
( x0 − x12 ) = 0.283 m s
m
The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units
of m, m/s, and kg.
14.41. Model: The block on a spring is in simple harmonic motion.
Solve:
(a) The position of the block is given by x ( t ) = A cos (ω t + φ0 ) . Because x ( t ) = A at t = 0 s, we have
φ0 = 0 rad, and the position equation becomes x ( t ) = A cos ω t. At t = 0.685 s, 3.00 cm = A cos ( 0.685ω ) and at
t = 0.886 s, −3.00 cm = A cos ( 0.886ω ) . These two equations give
cos ( 0.685ω ) = − cos ( 0.886ω ) = cos (π − 0.886ω )
⇒ 0.685ω = π − 0.886ω ⇒ ω = 2.00 rad s
(b) Substituting into the position equation,
3.00 cm = A cos ( ( 2.00 rad s )( 0.685 s ) ) = A cos (1.370 ) = 0.20 A ⇒ A =
3.00 cm
= 15.0 cm
0.20
14.42. Model: The oscillator is in simple harmonic motion. Energy is conserved.
Solve:
The energy conservation equation E1 = E2 is
1
2
mv12 + 12 kx12 = 12 mv22 + 12 kx22
1
1
1
1
2
2
2
2
( 0.30 kg )( 0.954 m s ) + k ( 0.030 m ) = ( 0.30 kg )( 0.714 m s ) + k ( 0.060 m )
2
2
2
2
⇒ k = 44.48 N m
The total energy of the oscillator is
1
1
1
1
2
2
Etotal = mv12 + kx12 = ( 0.30 kg )( 0.954 m s ) + ( 44.48 N m )( 0.030 m ) = 0.1565 J
2
2
2
2
2
,
Because Etotal = 12 mvmax
0.1565 J =
Assess:
1
2
⇒ v max = 1.02 m s
( 0.300 kg ) vmax
2
A maximum speed of 1.02 m/s is reasonable.
14.43. Model: The transducer undergoes simple harmonic motion.
Solve:
Newton’s second law for the transducer is
Frestoring = ma max ⇒ 40,000 N = ( 0.10 × 10−3 kg ) amax ⇒ amax = 4.0 × 108 m s 2
Because amax = ω 2 A,
A=
amax
ω
2
=
4.0 × 108 m s 2
⎡ 2π (1.0 × 10 Hz ) ⎤
⎣
⎦
6
2
= 1.01 × 10−5 m = 10.1 μ m
(b) The maximum velocity is
vmax = ω A = 2π (1.0 × 106 Hz )(1.01 × 10−5 m ) = 64 m s
14.44. Model: The block attached to the spring is in simple harmonic motion.
Solve:
(a) The frequency is
f =
1
2π
k
1
=
m 2π
2000 N m
= 3.183 Hz
5.0 kg
The frequency is 3.2 Hz.
(b) From energy conservation,
2
2
⎛v ⎞
⎛ 1.0 m/s ⎞
A = x02 + ⎜ 0 ⎟ = (0.050 m) 2 + ⎜
⎟ = 0.0707 m
ω
⎝ ⎠
⎝ 2π ⋅ 3.183 Hz ⎠
The amplitude is 7.1 cm.
(c) The total mechanical energy is
2
E = 12 kA2 = 12 ( 2000 N m )( 0.0707 m ) = 5.0 J
14.45. Model: The tips of the tuning fork are in simple harmonic oscillation.
Solve:
(a) The maximum speed is related to the amplitude.
vmax = ω A = 2π fA = 2π ( 440 Hz ) ( 5.0 × 10−4 m ) = 1.38 m/s
(b) The acceleration of the flea can not be greater than that allowed by the maximum force with which it can hold
on. From Newton’s second law, the maximum acceleration that the flea can withstand is
aflea =
F 1.0 × 10−3 N
=
= 100 m/s 2
m 10 × 10−6 kg
The maximum acceleration at the tip of the prong is
amax = ω 2 A = ( 2π f ) A = ( 2π ( 440 Hz ) ) ( 5.0 × 10−4 m ) = 382 m/s 2
2
The flea will not be able to hold on to the tuning fork.
2
14.46. Model: The block undergoes simple harmonic motion.
Visualize:
Solve:
(a) The frequency of oscillation is
f =
1
2π
k
1
=
m 2π
10 N/m
= 1.125 Hz
0.20 kg
The frequency is 1.13 Hz.
(b) Using conservation of energy, 12 mv12 + 12 kx12 = 12 mv02 + 12 kx02 , we find
m 2 2
0.20 kg
(v0 − v1 ) = (−0.20 m) 2 +
((1.00 m/s) 2 − (0.50 m/s) 2 )
k
10 N/m
= 0.2345 m or 23 cm
x1 = x02 +
(c) At time t, the displacement is x = A cos(ω t + φ0 ). The angular frequency is ω = 2π f = 7.071 rad/s. The
amplitude is
2
2
⎛v ⎞
⎛ 1.00 m/s ⎞
A = x02 + ⎜ 0 ⎟ = (−0.20 m) 2 + ⎜
⎟ = 0.245 m
ω
⎝ ⎠
⎝ 7.071 rad/s ⎠
The phase constant is
⎛x ⎞
⎝ A⎠
⎛ −0.200 m ⎞
⎟ = ±2.526 rad or ± 145°
⎝ 0.245 m ⎠
φ0 = cos −1 ⎜ 0 ⎟ = cos −1 ⎜
A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the
corresponding circular motion is in the third quadrant, so φ0 = −2.526 rad. Thus at t = 1.0 s,
x = (0.245 m)cos ( (7.071 rad/s)(1.0 s) − 2.526 rad ) = −0.0409 m = −4.09 cm
The block is 4.1 cm below the equilibrium point.
14.47. Model: The mass is in simple harmonic motion.
Visualize:
The high point of the oscillation is at the point of release. This conclusion is based on energy conservation.
Gravitational potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches
the spring, then the elastic potential energy is converted 100% back into gravitational potential energy as the
mass rises, bringing the mass back to exactly its starting height. The total displacement of the oscillation—high
point to low point—is 20 cm. Because the oscillations are symmetrical about the equilibrium point, we can
deduce that the equilibrium point of the spring is 10 cm below the point where the mass is released. The mass
oscillates about this equilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm
above and 10 cm below the equilibrium point.
Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp = FG = mg . At the
equilibrium point, the spring is stretched by Δy = 10 cm = 0.10 m. Hooke’s law is Fsp = k Δy, so the equilibrium
condition is
k
g 9.8 m s 2
⎡⎣ Fsp = k Δy ⎤⎦ = [ FG = mg ] ⇒ =
=
= 98 s −2
m Δy
0.10 m
The ratio k m is all we need to find the oscillation frequency:
f =
1
2π
k
1
98 s −2 = 1.58 Hz
=
m 2π
14.48. Model: The spring is ideal, so the apples undergo SHM.
Solve: The spring constant of the scale can be found by considering how far the pan goes down when the
apples are added.
ΔL =
20 N
mg
mg
⇒k =
=
= 222 N/m
k
ΔL 0.090 m
The frequency of oscillation is
f =
1
2π
k
1
=
m 2π
222 N/m
= 1.66 Hz
( 20 N 9.8 m/s 2 )
Assess: An oscillation of fewer than twice per second is reasonable.
14.49. Model: The compact car is in simple harmonic motion.
Solve:
(a) The mass on each spring is (1200 kg ) 4 = 300 kg. The spring constant can be calculated as follows:
ω2 =
2
k
2
⇒ k = mω 2 = m ( 2π f ) = ( 300 kg ) ⎡⎣ 2π ( 2.0 Hz ) ⎤⎦ = 4.74 × 104 N m
m
The spring constant is 4.7 × 104 N/m.
(b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That
is, m = 300 kg + 70 kg = 370 kg. Thus,
f =
ω
1
=
2π 2π
k
1
=
m 2π
4.74 × 104 N m
= 1.80 Hz
370 kg
Assess: A small frequency change from the additional mass is reasonable because frequency is inversely
proportional to the square root of the mass.
14.50. Model: Hooke’s law for the spring. The spring’s compression and decompression constitutes simple
harmonic motion.
Visualize:
Solve: (a) The spring’s compression or decompression is one-half of the oscillation cycle. This means the
contact time is Δt = 12 T , where T is the period. The period is calculated as follows:
ω=
50 N m
1 2π
2π
k
=
= 10 rad s ⇒ T = =
=
= 0.628 s
0.500 kg
ω 10 rad s
m
f
⇒ Δt =
T
= 0.31 s
2
(b) There is no change in contact time, because period of oscillation is independent of the amplitude or the
maximum speed.
14.51. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will
also apply the model of static friction between the two blocks.
Visualize:
Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2 . The
model of static friction gives the maximum force of static friction as
( fs )max = μs n = μs ( m1g ) = m1amax ⇒ amax = μs g
Using μ s = 0.5, amax = μS g = ( 0.5) ( 9.8 m s 2 ) = 4.9 m s 2 . That is, the two blocks will ride together if the
maximum acceleration of the system is equal to or less than amax . We can calculate the maximum value of A as
follows:
amax = ω 2 Amax =
2
a ( m + m2 ) ( 4.9 m s ) (1.0 kg + 5.0 kg )
k
Amax ⇒ Amax = max 1
=
= 0.59 m
m1 + m2
k
50 N m
14.52. Model: Assume simple harmonic motion for the two-block system without the upper block slipping.
We will also use the model of static friction between the two blocks.
Visualize:
Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2 . The two
blocks ride together as long as the static friction doesn’t exceed its maximum possible value. The model of static
friction gives the maximum force of static friction as
( fs )max = μs n = μs ( m1g ) = m1amax ⇒ amax = μs g
2
⇒ μs =
2
amax ω 2 Amax ⎛ 2π ⎞ ⎛ Amax ⎞ ⎛ 2π ⎞ ⎛ 0.40 m ⎞
=
=⎜
= 0.72
⎟=⎜
⎟ ⎜
⎟ ⎜
2 ⎟
g
g
⎝ T ⎠ ⎝ g ⎠ ⎝ 1.5 s ⎠ ⎝ 9.8 m s ⎠
Assess: Because the period is given, we did not need to use the block masses or the spring constant in our
calculation.
14.53. Model: The DNA and cantilever undergo SHM.
Visualize: Please refer to figure P14.53.
Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of
oscillation without the DNA is
k
ω1 = 1
3M
With the DNA, the frequency of oscillation is
ω2 =
1
3
k
M +m
where m is the mass of the DNA.
Divide the two equations, and express ω 2 = ω1 − Δω , where Δω = 2πΔf = 2π ( 50 Hz ) .
f1
ω1 f1
=
=
=
ω 2 f 2 f1 − Δf
Thus
k
( 13 M )
=
k
( 13 M + m)
1
3
M +m
1
M
3
f12 ( 13 M ) = ( f1 − Δf ) ( 13 M + m )
2
( f − ( f − Δf ) ) = M ⎛
m= M
1
3
2
1
2
1
( f1 − Δf )
2
⎞
− 1⎟
⎜⎜
2
⎟
⎝ ( f1 − Δf )
⎠
1
3
f12
−2
Δf ⎞
−2
⎛ Δf ⎞
−2 ⎛
Since Δf f1 , (50 Hz 12MHz), ( f1 − Δf ) = f1−2 ⎜ 1 −
⎟ ≈ f1 ⎜ 1 + 2 f ⎟ . Thus
f
1 ⎠
1 ⎠
⎝
⎝
⎛
⎞
Δf
Δf
m = 13 M ⎜1 + 2
− 1⎟ = 23 M
f
f1
1
⎝
⎠
The mass of the cantilever
M = (2300 kg/m3 )(4000 × 10−9 m)(100 × 10−9 m) = 3.68 × 10−16 kg
Thus the mass of the DNA molecule is
2
⎛ 50 Hz ⎞
−21
m = ( 3.68 × 10−16 kg ) ⎜
⎟ = 1.02 × 10 kg
6
3
⎝ 12 × 10 Hz ⎠
Assess: The mass of the DNA molecule is about 6.2 × 105 atomic mass units, which is reasonable for such a large
molecule.
14.54. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple
harmonic motion.
Visualize:
Solve:
(a) Using the formula for the period of a pendulum,
2
T = 2π
2
L
⎛ T ⎞
2 ⎛ 5.5 s ⎞
⇒ L = g⎜
⎟ = ( 9.8 m s ) ⎜
⎟ = 7.5 m
g
2
π
⎝
⎠
⎝ 2π ⎠
(b) The conservation of mechanical energy equation K 0 + U g0 = K1 + U g1 for the swinging lamp is
1
2
2
mv02 + mgy0 = 12 mv12 + mgy1 ⇒ 0 J + mgh = 12 mvmax
+0 J
⇒ vmax = 2 gh = 2 g ( L − L cos3° )
= 2 ( 9.8 m s 2 ) ( 7.5 m )(1 − cos3° ) = 0.45 m s
14.55. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there
is simple harmonic motion.
Solve: The angle θ made by the string with the vertical as a function of time is
θ ( t ) = θ max cos (ω t + φ0 )
The pendulum starts from maximum displacement, thus φ0 = 0. Thus, θ ( t ) = θ max cos ω t. To find the time t when
the pendulum reaches 4.0° on the opposite side:
( −4.0° ) = (8.0° ) cos ω t ⇒ ω t = cos −1 ( −0.5) = 2.094 rad
Using the formula for the angular frequency,
ω=
g
9.8 m s 2
2.0944 rad
2.094 rad
=
= 3.130 rad s ⇒ t =
=
= 0.669 s
L
1.0 m
3.130 rad s
ω
The time t = 0.67 s.
Assess: Because T = 2π ω = 2.0 s, a value of 0.67 s for the pendulum to cover a little less than half the
oscillation is reasonable.
14.56. Model: Assume a small angle oscillation of the pendulum so that it has simple harmonic motion.
Solve:
(a) At the equator, the period of the pendulum is
Tequator = 2π
1.000 m
= 2.009 s
9.78 m s 2
Tpole = 2π
1.000 m
= 2.004 s
9.83 m s 2
The time for 100 oscillations is 200.9 s.
(b) At the north pole, the period is
The time for 100 oscillations is 200.4 s.
(c) The difference between the two answers is 0.5 s, and this difference is quite measurable with a hand-operated
stopwatch.
(d) The period on the top of the mountain is 2.010 s. The acceleration due to gravity can be calculated by
rearranging the formula for the period:
2
2
⎛ 2π ⎞
⎛ 2π ⎞
2
g mountain = L ⎜
⎟ = (1.000 m ) ⎜
⎟ = 9.772 m s
T
2.010
s
⎝
⎠
mountain
⎝
⎠
Assess:
This last result is reasonable because g decreases with altitude.
14.57. Model: The mass is a particle and the string is massless.
Solve:
Equation 14.52 is
ω=
Mgl
I
The moment of inertia of the mass on a string is I = Ml 2 , where l is the length of the string. Thus
ω=
Mgl
=
Ml 2
g
l
This is Equation 14.49 with L = l.
Assess: Equation 14.49 is really a specific case of the more general physical pendulum described by
Equation 14.52.
14.58. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a
particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are
small.
Visualize:
Solve:
The moment of inertia of the composite pendulum formed by the rod and clay ball is
1
I rod + ball = I rod + I ball = mrod L2 + mball L2
3
1
2
2
= ( 0.200 kg )( 0.15 m ) + ( 0.020 kg )( 0.15 m ) = 1.95 × 10−3 kg m 2
3
The center of mass of the rod and ball is located at a distance from the pivot point of
ycm =
( 0.200 kg ) ⎛⎜
0.15 ⎞
m ⎟ + ( 0.020 kg )( 0.15 m )
⎝ 2
⎠
= 8.18 × 10−2 m
( 0.200 kg + 0.020 kg )
The frequency of oscillation of a physical pendulum is
1
2π
Mgl
1
=
I
2π
The period of oscillation T =
1
= 0.66 s.
f
f =
( 0.220 kg ) ( 9.8 m/s 2 )(8.18 × 10−2 m )
1.95 × 10−3 kg m 2
= 1.51 Hz
14.59. Model: The circular hoop can be modeled as a cylindrical hoop and its moment of inertia about the
point of rotation found with the parallel-axis theorem.
Visualize: Please refer to Figure P14.59.
Solve: Using the parallel-axis theorem, the moment of inertia of the cylindrical hoop about the rotation point is
I = MR 2 + MR 2 = 2MR 2
The frequency of small oscillations is given by Equation 14.52.
f =
1
2π
Mgl
I
The center of mass of the hoop is its center, so l = R. Thus
f =
1
2π
MgR
1
=
2
2 MR
2π
g
2R
14.60. Model: The motion is a damped oscillation.
Solve:
The position of the air-track glider is x ( t ) = Ae − ( t 2τ ) cos (ω t + φ0 ) , where τ = m b and
k
b2
−
m 4m 2
ω=
Using A = 0.20 m, φ0 = 0 rad, and b = 0.015 kg/s,
4.0 N m ( 0.015 kg s )
−
= 16 − 9 × 10−4 rad s = 4.0 rad s
0.250 kg 4 ( 0.250 kg )2
2
ω=
Thus the period is
T=
2π
ω
=
2π rad
= 1.57 s
4.0 rad s
The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e −1 A at a time given by
1
m
A = Ae − ( t 2τ ) ⇒ t = 2τ = 2 = 33.3 s
e
b
The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21.
14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion.
Visualize:
Let us denote the 250 g and 500 g masses as m1 and m2 , which have initial velocities vi1 and vi2 . After m1
collides with and sticks to m2 , the two masses move together with velocity vf .
Solve:
The momentum conservation equation pf = pi for the completely inelastic collision is ( m1 + m2 ) vf =
m1vi1 + m2vi2 . Substituting the given values,
( 0.750 kg ) vf = ( 0.250 kg )(1.20 m s ) + ( 0.500 kg )( 0 m s ) ⇒ vf = 0.400 m s
We now use the conservation of mechanical energy equation:
( K + U s )compressed = ( K + U s )equilibrium ⇒ 0 J + 12 kA2 = 12 ( m1 + m2 ) vf2 + 0 J
⇒ A=
m1 + m2
0.750 kg
vf =
( 0.400 m s ) = 0.110 m
k
10 N m
The period is
T = 2π
m1 + m2
0.750 kg
= 2π
= 1.72 s
k
10 N m
14.62. Model: The block attached to the spring is oscillating in simple harmonic motion.
Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude
and/or the maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz.
(b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of
mechanical energy equation as follows:
1
2
⇒ v0 =
2
kA2 = 12 mvmax
= 12 mv02
k
A = ω A = ( 2π f ) A = ( 2π )( 2.0 Hz )( 0.02 m ) = 0.25 m s
m
The blow to the block provides an impulse that changes the velocity of the block:
( −20 N ) (1.0 × 10
J x = Fx Δt = Δp = mvf − mv0
−3
s ) = ( 0.200 kg ) vf − ( 0.200 kg )( 0.25 m s ) ⇒ vf = 0.150 m s
Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to Aω . Thus,
A=
Assess:
0.150 m s
ω
=
0.150 m s
= 0.012 m = 1.19 cm
2π ( 2.0 Hz )
Because vf is positive, the block continues to move to the right even after the blow.
14.63. Model: The pendulum falls, then undergoes small-amplitude oscillations in simple harmonic motion.
Visualize:
We placed the origin of the coordinate system at the bottom of the arc.
Solve: We need to find the length of the pendulum. The conservation of mechanical energy equation for the
:
pendulum’s fall is ( K + U g ) = ( K + U g )
top
1
2
bottom
mv + mgy0 = mv12 + mgy1 ⇒ 0 J + mg ( 2 L ) = 12 m ( 5.0 m s ) + 0 J
2
0
2
1
2
1 ( 5.0 m s )
= 0.6377 m
g
4
Using L = 0.6377 m, we can find the frequency f as
2
⇒L=
f =
1
2π
g
1
=
L 2π
9.8 m s 2
= 0.62 Hz
0.6377 m
14.64. Model: Assume the small-angle approximation.
Visualize:
Solve:
The tension in the two strings pulls downward at angle θ . Thus Newton’s second law is
∑ Fy = −2T sinθ = ma y
From the geometry of the figure we can see that
sin θ =
y
L + y2
2
If the oscillation is small, then y L and we can approximate sin θ ≈ y / L. Since y/L is tanθ , this
approximation is equivalent to the small-angle approximation sinθ ≈ tanθ if θ 1 rad. With this
approximation, Newton’s second law becomes
2T
d2y
d2y
2T
−2T sin θ ≈ −
=−
y = ma y = m 2 ⇒
y
2
L
dt
dt
mL
This is the equation of motion for simple harmonic motion (see Equations 14.33 and 14.47). The constants
2T/mL are equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation
frequency is
f =
1
2π
2T
mL
14.65. Visualize: Please refer to Figure P14.65.
Solve:
The potential energy curve of a simple harmonic oscillator is described by U = 12 k ( Δx ) , where
2
Δx = x − x0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is
x0 = 0.13 nm. We can find the bond’s spring constant by reading the value of the potential energy U at a
displacement Δx and using the potential energy formula to calculate k.
x (nm)
0.11
Δx (nm)
U (J)
k (N/m)
0.02
0.8 × 10−19 J
400
0.03
0.10
0.04
0.09
1.9 × 10
−19
J
422
3.4 × 10−19 J
425
The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the
spring constant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be
f =
1
2π
k
1
416 N m
=
= 7.9 × 1013 Hz
m 2π 1.67 × 10−27 kg
14.66. Model:
Visualize:
Solve:
is
Assume that the size of the ice cube is much less than R and that θ is a small angle.
The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction
− ( FG ) sin θ = ma = mRα = mR
d 2θ
d 2θ
⇒
−
mg
sin
=
mR
θ
dt 2
dt 2
where α is the angular acceleration. With the small-angle approximation sin θ ≈ θ , this becomes
d 2θ
g
= − θ = −ω 2θ
dt 2
R
This is the equation of motion of an object in simple harmonic motion with a period of
T=
2π
ω
= 2π
R
g
14.67.
Visualize:
Solve:
(a) Newton’s second law applied to the penny along the y-axis is
Fnet = n − mg = ma y
G
Fnet is upward at the bottom of the cycle (positive a y ), so n > mg . The speed is maximum when passing
G
through equilibrium, but a y = 0 so n = mg . The critical point is the highest point. Fnet points down and a y is
negative. If a y becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the
surface.
(b) When the penny loses contact (n = 0), the equation for Newton’s law becomes amax = g . For simple
harmonic motion,
g
9.8 m s 2
=
= 15.65 rad
A
0.040 m
ω 15.65 rad s
⇒ f =
=
= 2.5 Hz
2π
2π
amax = Aω 2 ⇒ ω =
14.68. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve:
At the equilibrium position, the net force on mass m on Planet X is:
Fnet = k ΔL − mg X = 0 N ⇒
k gX
=
m ΔL
For simple harmonic motion k m = ω 2 , thus
2
ω2 =
2
⎛ 2π
⎞
gX
g X 2π
⎛ 2π ⎞
2
⇒ω =
=
⇒ gX = ⎜
⎟ ( 0.312 m ) = 5.86 m s
⎟ ΔL = ⎜
14.5
s
10
ΔL
ΔL T
⎝ T ⎠
⎝
⎠
14.69. Model: The doll’s head is in simple harmonic motion and is damped.
Solve:
(a) The oscillation frequency is
f =
1
2π
k
2
2
2
⇒ k = m ( 2π f ) = ( 0.015 kg )( 2π ) ( 4.0 Hz ) = 9.475 N m
m
The spring constant is 9.5 N/m.
(b) The maximum speed is
vmax = ω A =
k
9.475 N m
A=
( 0.020 m ) = 0.50 m s
m
0.015 kg
(c) Using A ( t ) = A0e − bt / 2 m , we get
( 0.5 cm ) = ( 2.0 cm ) e−b(4.0 s)/(2×0.015 kg) ⇒ 0.25 = e− (133.3 s/kg)b
⇒ − (133.33 s kg ) b = ln 0.25 ⇒ b = 0.0104 kg / s
14.70. Model: The oscillator is in simple harmonic motion.
Solve:
(a) The maximum displacement at time t of a damped oscillator is
xmax ( t ) = Ae −t 2τ ⇒ −
⎛ x (t ) ⎞
t
= ln ⎜ max ⎟
2τ
⎝ A ⎠
Using xmax = 0.98 A at t = 0.50 s, we can find the time constant τ to be
τ =−
0.50 s
= 12.375 s
2ln ( 0.98 )
25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be
xmax, 12.5 s = (10 cm ) e−12.5 s ( 2)(12.375 s ) = 6.0 cm
(b) The energy of a damped oscillator decays more rapidly than the amplitude: E (t ) = E0e−1/ τ . When the energy
is 60% of its initial value, E (t )/E0 = 0.60. We can find the time this occurs as follows:
⎛ E (t ) ⎞
⎛ E (t ) ⎞
t
− = ln ⎜
⎟ ⇒ t = −τ ln ⎜
⎟ = − (12.375 s ) ln ( 0.60 ) = 6.3 s
τ
⎝ E0 ⎠
⎝ E0 ⎠
14.71. Model: The oscillator is in simple harmonic motion.
Solve:
The maximum displacement, or amplitude, of a damped oscillator decreases as xmax (t ) = Ae− t 2τ , where
τ is the time constant. We know xmax A = 0.60 at t = 50 s, so we can find τ as follows:
−
⎛ x (t ) ⎞
t
50 s
= ln ⎜ max ⎟ ⇒ τ = −
= 48.9 s
2τ
2ln ( 0.60 )
⎝ A ⎠
Now we can find the time t30 at which xmax A = 0.30 :
⎛ x (t ) ⎞
t30 = −2τ ln ⎜ max ⎟ = −2 ( 48.9 s ) ln ( 0.30 ) = 118 s
⎝ A ⎠
The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillation
frequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, which
allows the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz.
Then the number of oscillations before the spring decays to 30% of its initial amplitude is
N = f ⋅ t30 = ( 2 oscillations s ) ⋅ (118 s ) = 236 oscillations
14.72. Solve: The solution of the equation
d 2 x b dx k
+
+ x=0
dt 2 m dt m
is x ( t ) = Ae − bt 2 m cos (ω t + φ0 ) . The first and second derivatives of x(t) are
dx
Ab −bt / 2 m
e
=−
cos (ω t + φ0 ) − Aω e − bt / 2 m sin (ω t + φ0 )
dt
2m
⎤
⎞
d 2 x ⎡⎛ Ab 2
ω Ab
= ⎢⎜
− Aω 2 ⎟ cos (ω t + φ0 ) +
sin (ω t + φ0 ) ⎥ e − bt / 2 m
dt 2 ⎣⎝ 4m 2
m
⎠
⎦
Substituting these expressions into the differential equation, the terms involving sin(ω t + φ0 ) cancel and we
obtain the simplified result
⎛ −b 2
k⎞
2
⎜ 2 − ω + ⎟ cos (ω t + φ0 ) = 0
m⎠
⎝ 4m
Because cos (ω t + φ0 ) is not equal to zero in general,
k
k
b2
−b 2
−ω2 + = 0 ⇒ ω =
−
2
4m
m
m 4m 2
14.73. Model: The two springs obey Hooke’s law.
Visualize:
Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring
forces—one pushing, the other pulling—are directed to the left and have negative values:
( Fnet ) x = ( Fsp 1 ) x + ( Fsp 2 ) x = −k1x − k2 x = − ( k1 + k2 ) x = −keff x
where keff = k1 + k2 is the effective spring constant. This means the oscillatory motion of the block under the
influence of the two springs will be the same as if the block were attached to a single spring with spring constant
keff . The frequency of the blocks, therefore, is
f =
1
2π
keff
1
=
2π
m
k1 + k2
k1
k
=
+ 22 =
2
4π m 4π m
m
f12 + f 22
14.74. Model: The two springs obey Hooke’s law. Assume massless springs.
Visualize:
Solve:
Each spring is shown separately. Note that Δx = Δx1 + Δx2 .
Only spring 2 touches the mass, so the net force on the mass is Fm = F2 on m . Newton’s third law tells us
that F2 on m = Fm on 2 and that F2 on 1 = F1 on 2 . From Fnet = ma, the net force on a massless spring is zero. Thus
Fw on 1 = F2 on 1 = k1Δx1 and Fm on 2 = F1 on 2 = k2 Δx2 . Combining these pieces of information,
Fm = k1Δx1 = k2 Δx2
The net displacement of the mass is Δx = Δx1 + Δx2 , so
Δx = Δx1 + Δx2 =
Fm Fm ⎛ 1 1 ⎞
k +k
+
= ⎜ + ⎟ Fm = 1 2 Fm
k1 k2 ⎝ k1 k2 ⎠
k1k2
Turning this around, the net force on the mass is
Fm =
k1k2
kk
Δx = keff Δx where keff = 1 2
k1 + k2
k1 + k2
keff , the proportionality constant between the force on the mass and the mass’s displacement, is the effective
spring constant. Thus the mass’s angular frequency of oscillation is
ω=
keff
1 k1k2
=
m
m k1 + k2
Using ω12 = k1 / m and ω 22 = k2 / m for the angular frequencies of either spring acting alone on m, we have
ω=
( k1 / m) ( k2 / m)
ω12ω 22
=
ω12 + ω 22
( k1 / m) + (k2 / m)
Since the actual frequency f is simply a multiple of ω , this same relationship holds for f:
f =
f 12 f 22
f 12 + f 22
14.75. Model: The blocks undergo simple harmonic motion.
Visualize:
The length of the stretched spring due to a block of mass m is ΔL1. In the case of the two block system, the spring
is further stretched by an amount ΔL2 .
Solve: The equilibrium equations from Newton’s second law for the single block and double block systems are
( ΔL1 ) k = mg and ( ΔL1 + ΔL2 ) k = ( 2m ) g
Using ΔL2 = 5.0 cm, and subtracting these two equations, gives us
( ΔL1 + ΔL2 ) k − ΔL1k = ( 2m ) g − mg ⇒ ( 0.05 m ) k = mg
⇒
k 9.8 m/s 2
=
= 196 s –2
m 0.05 m
With both blocks attached, giving total mass 2m, the angular frequency of oscillation is
ω=
k
1k
1
196 s –2 = 9.90 rad/s
=
=
2m
2m
2
Thus the oscillation frequency is f = ω /2π = 1.58 Hz.
14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic
motion.
Visualize:
Solve:
(a) The equation for conservation of energy after the collision is
k
1 2 1
2500 N m
kA = ( mb + mB ) vf2 ⇒ vf =
A=
( 0.10 m ) = 5.0 m s
mb + mB
2
2
1.010 kg
The momentum conservation equation for the perfectly inelastic collision pafter = pbefore is
( mb + mB ) vf = mbvb + mBvB
(1.010 kg )( 5.0 m s ) = ( 0.010 kg ) vb + (1.00 kg )( 0 m s ) ⇒ vb = 5.0 × 102 m s
(b) No. The oscillation frequency k ( mb + mB ) depends on the masses but not on the speeds.
14.77. Model: The block undergoes SHM after sticking to the spring. Energy is conserved throughout the
motion.
Visualize:
It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched
upward.
Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit
on the spring at rest. The spring is compressed by ΔL at this point. Balancing the forces requires k ΔL = mg . The
angular frequency is w2 = k/m = g/ΔL, so we can find the oscillation frequency by finding ΔL. The block hits
the spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have
been transformed into the spring’s elastic energy. Equate the energies at these points:
K1 + U1g = U 3s + U 3g ⇒ 12 mv12 + mg ΔL = 12 k (ΔL + A) 2 + mg (− A)
We’ve used y1 = ΔL as the block hits and y3 = − A at the bottom. The spring has been compressed by
Δy = ΔL + A. Speed v1 is the speed after falling distance h, which from free-fall kinematics is v12 = 2 gh.
Substitute this expression for v12 and mg/ΔL for k, giving
mgh + mg ΔL =
mg
(ΔL + A) 2 + mg (− A)
2(ΔL)
The mg term cancels, and the equation can be rearranged into the quadratic equation
(ΔL) 2 + 2h(ΔL) − A2 = 0
The positive solution is
ΔL = h 2 + A2 − h = (0.030 m) 2 + (0.100 m) 2 − 0.030 m = 0.0744 m
Now that ΔL is known, we can find
ω=
g
9.80 m/s 2
ω
=
= 11.48 rad/s ⇒ f =
= 1.83 Hz
0.0744 m
2π
ΔL
14.78. Model: Model the bungee cord as a spring. The motion is damped SHM.
Visualize:
Solve:
(a) For light damping, the oscillation period is
2
2
m
⎛ 2π ⎞
⎛ 2π ⎞
⇒ k = m⎜
⎟ = (75 kg) ⎜
⎟ = 185 N/m
k
⎝ T ⎠
⎝ 4.0 s ⎠
T = 2π
(b) The maximum speed is vmax = ω A = (2π /4.0 s)(11.0 m) = 17.3 m/s.
(c) Jose oscillates about the equilibrium position at which he would hang at rest. Balancing the forces,
ΔL = mg / k = (75 kg)(9.80 m/s 2 )(185 N/m) = 3.97 m. Jose’s lowest point is 11.0 m below this point, so the
bungee cord is stretched by Δymax = ΔL + A = 14.97 m. Choose this lowest point as y = 0. Because Jose is
instantaneously at rest at this point, his energy is entirely the elastic potential energy of the stretched bungee
cord. Initially, his energy was entirely gravitational potential energy. Equating his initial energy to his energy at
the lowest point,
U lowest point = U highest point ⇒ 12 k (Δymax ) 2 = mgh
h=
k (Δymax ) 2 (185 N/m)(14.97 m) 2
=
= 28.2 m
2mg
2(75 kg)(9.80 m/s 2)
Jose jumped 28 m above the lowest point.
(d) The amplitude decreases due to damping as A(t ) = Ae −bt/2m . At the time when the amplitude has decreased
from 11.0 m to 2.0 m,
2.0 m
2m ⎛ 2 ⎞
2(75 kg)
= e − bt / 2 m ⇒ t = −
ln ⎜ ⎟ = −
(−1.705) = 42.6 s
11.0 m
b ⎝ 11 ⎠
6.0 kg/s
With a period of 4.0 s, the number of oscillations is N osc = (42.6 s)/(4.0 s) = 10.7 oscillations.
14.79. Model: The vertical movement of the car is simple harmonic motion.
Visualize:
The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the road
provide a periodic external force to the car’s suspension system, and a resonance will occur when the “bump
frequency” f ext matches the car’s natural oscillation frequency f 0 .
Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps
are spaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period:
T=
Δx
3.0 m
=
= 0.60 s
v 5.0 m s
The external frequency due to the bumps is thus f ext = 1 T = 1.667 Hz. This matches the car’s natural frequency
f 0 , which is the frequency the car oscillates up and down if you push the car down and release it. This is enough
information to deduce the spring constant of the car’s suspension:
1.667 Hz =
1
2π
N
k
2
⇒ k = m ( 2π f ext ) = 131,600
m
m
where we used m = mtotal = mcar + 2mpassenger = 1200 kg. When at rest, the car is in static equilibrium with
Fnet = 0 N. The downward weight mtotal g of the car and passengers is balanced by the upward spring force k Δy
of the suspension. Thus the compression Δy of the suspension is
Δy =
mtotal g
k
Initially mtotal = mcar + 2mpassenger = 1200 kg, causing an initial compression Δyi = 0.0894 m = 8.94 cm. When three
additional passengers get in, the mass increases to mtotal = mcar + 5mpassenger = 1500 kg. The final compression is
Δyf =
0.1117 m = 11.17 cm. Thus the three new passengers cause the suspension to “sag” by
11.17 cm − 8.94 cm = 2.23 cm.
14.80. Model: The rod is thin and uniform.
Visualize: Please refer to Figure CP14.80.
G
Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations, Fs
remains horizontal. The net torque around the pivot point is
⎛ L⎞
⎝ ⎠
τ net = Iα = − Fs L cosθ − FG ⎜ ⎟ sinθ
2
With α =
d 2θ
1
, FG = mg , Fs = k Δx = kL sin θ , and I = mL2 ,
dt 2
3
d 2θ
3k
3g
sin θ
= − sin θ cosθ −
dt 2
m
2L
1
We can use sin θ cosθ = sin 2θ . For small angles, sinθ ≈ θ and sin 2θ ≈ 2θ . So
2
d 2θ
⎛ 3k 3 g ⎞
= −⎜ +
⎟θ
dt 2
⎝ m 2L ⎠
This is the same as Equations 14.33 and 14.47 with
ω=
3k 3 g
+
m 2L
The frequency of oscillation is thus
f =
1
2π
2
3 ( 3.0 N/m ) 3 ( 9.8 m/s )
+
= 1.73 Hz
( 0.200 kg ) 2 ( 0.20 m )
1
= 0.58 s.
f
Assess: Fewer than two oscillations per second is reasonable. The rod’s angle from the vertical must be small
enough that sin 2θ ≈ 2θ . This is more restrictive than other examples, which only require that sin θ ≈ θ .
The period T =
14-1
15.1. Solve: The density of the liquid is
ρ=
Assess:
m 0.240 kg
0.240 kg
=
=
= 960 kg m3
V
250 mL 250 × 10−3 × 10−3 m3
The liquid’s density is near that of water (1000 kg/m3 ) and is a reasonable number.
15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B.
That is, VA = VB . Using the definition of mass density ρ = m V , the above relationship becomes
mA
ρA
=
mB
ρB
⇒
mHe
ρ He
=
7000 mHe
ρB
⇒ ρ B = ( 7000 ) ρ He = ( 7000 ) ( 0.18 kg m3 ) = 1260 kg m3
Referring to Table 15.1, we find that the liquid is glycerine.
15.3. Model: The density of water is 1000 kg/m3 .
Visualize:
Solve:
Volume of water in the swimming pool is
V = 6 m × 12 m × 3 m − 12 ( 6 m × 12 m × 2 m ) = 144 m3
The mass of water in the swimming pool is
m = ρV = (1000 kg m3 )(144 m3 ) = 1.44 × 105 kg
15.4. Model: The densities of gasoline and water are given in Table 15.1.
Solve:
(a) The total mass is
mtotal = mgasoline + mwater = 0.050 kg + 0.050 kg = 0.100 kg
The total volume is
Vtotal = Vgasoline + Vwater =
mgasoline
ρ gasoline
⇒ ρ avg =
+
mwater
ρ water
=
0.050 kg
0.050 kg
+
= 1.24 × 10−4 m3
680 kg m3 1000 kg m3
mtotal
0.100 kg
=
= 8.1× 102 kg m3
Vtotal 1.24 × 10−4 m3
(b) The average density is calculated as follows:
mtotal = mgasoline + mwater = ρ waterVwater + ρ gasolineVgasoline
⇒ ρ avg =
Assess:
ρ waterVwater + ρgasolineVgasoline
Vwater + Vgasoline
( 50 cm )(1000 kg/m + 680 kg/m ) = 8.4 × 10 kg/m
3
=
3
3
2
100 cm
3
The above average densities are between those of gasoline and water, and are reasonable.
3
15.5. Model: The density of sea water is 1030 kg/m3 .
Solve:
The pressure below sea level can be found from Equation 15.6 as follows:
p = p0 + ρ gd = 1.013 × 105 Pa + (1030 kg m3 )( 9.80 m s 2 )(1.1 × 104 m )
= 1.013 × 105 Pa + 1.1103 × 108 Pa = 1.1113 × 108 Pa = 1.10 × 103 atm
where we have used the conversion 1 atm = 1.013 × 105 Pa.
Assess: The pressure deep in the ocean is very large.
15.6. Model: The density of water is 1000 kg/m3 and the density of ethyl alcohol is 790 kg/m3 .
Solve:
(a) The volume of water that has the same mass as 8.0 m 3 of ethyl alcohol is
Vwater =
mwater
ρ water
=
malcohol
ρ water
=
ρ alcoholValcohol ⎛ 790 kg/m3 ⎞
8.0 m3 ) = 6.3 m3
=⎜
3 ⎟(
ρ water
⎝ 1000 kg/m ⎠
(b) The pressure at the bottom of the cubic tank is p = p0 + ρ water gd :
p = 1.013 × 105 Pa + (1000 kg/m3 )( 9.80 m s 2 ) ( 6.3) = 1.19 × 105 Pa
13
where we have used the relation d = (Vwater ) .
13
15.7.
Visualize:
Solve:
The pressure at the bottom of the vat is p = p0 + ρ gd = 1.3 atm. Substituting into this equation gives
1.013 × 105 Pa + ρ ( 9.8 m s 2 ) ( 2.0 m ) = (1.3) (1.013 × 105 ) Pa ⇒ ρ = 1550.5 kg m3
The mass of the liquid in the vat is
m = ρV = ρπ ( 0.50 m ) d = (1550.5 kg m3 )π ( 0.50 m ) ( 2.0 m ) = 2.4 × 103 kg
2
2
15.8. Model:
Visualize:
The density of oil ρoil = 900 kg m 3 and the density of water ρ water = 1000 kg m3 .
Solve: The pressure at the bottom of the oil layer is p1 = p0 + ρ oil gd1 , and the pressure at the bottom of the
water layer is
p2 = p1 + ρ water gd 2 = p0 + ρoil gd1 + ρ water gd 2
⇒ p2 = (1.013 × 10 Pa ) + ( 900 kg m 3 )( 9.80 m s 2 ) ( 0.50 m ) + (1000 kg m3 )( 9.80 m s 2 ) (1.20 m ) = 1.18 × 105 Pa
5
Assess:
A pressure of 1.18 × 105 Pa = 1.16 atm is reasonable.
15.9. Model: The density of seawater ρseawater = 1030 kg m 2 .
Visualize:
The pressure outside the submarine’s window is pout = p0 + ρseawater gd , where d is the maximum safe
Solve:
depth for the window to withstand a force F. This force is F A = pout − pin , where A is the area of the window.
With pin = p0 , we simplify the pressure equation to
pout − p0 =
Assess:
F
F
= ρseawater gd ⇒ d =
A
Aρseawater g
d=
1.0 × 106 N
π ( 0.10 m ) (1030 kg m 2 )( 9.8 m s 2 )
2
A force of 1.0 × 106 N corresponds to a pressure of
ρ=
A depth of 3 km is therefore reasonable.
F 1.0 × 106 N
=
= 314 atm
A π ( 0.10 m )2
= 3.2 km
15.10.
Visualize:
We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the
cover and the forces cancel. Thus, only the 10 cm diameter opening inside the seal is relevant, not the 20 cm
diameter of the cover.
Solve: Within the 10 cm diameter area where the pressures differ,
Fto left = patmos A
Fto right = pgas A
where A = π r 2 = 7.85 × 10−3 m 2 is the area of the opening. The difference between the forces is
Fto left − Fto right = ( patmos − pgas ) A = (101,300 Pa − 20,000 Pa ) ( 7.85 × 10−3 m 2 ) = 0.64 kN
Normally, the rubber seal exerts a 0.64 kN force to the right to balance the air pressure force. To pull the cover
off, an external force must pull to the right with a force ≥ 0.64 kN.
15.11. Model: The density of water is ρ = 1000 kg m3 .
Visualize: Please refer to Figure 15.17.
Solve: From the figure and the equation for hydrostatic pressure, we have
p0 + ρ gh = patmos
Using p0 = 0 atm, and patmos = 1.013 × 105 Pa, we get
0 Pa + (1000 kg m3 )( 9.8 m s 2 ) h = 1.013 × 105 Pa ⇒ h = 10.3 m
Assess:
This large value of h is due to water having a much smaller density than mercury.
15.12. Model: Assume that the oil is incompressible. Its density is 900 kg/m3 .
Visualize: Please refer to Figure 15.19. Because the liquid is incompressible, the volume displaced in the left
cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder.
Solve: Equating the two volumes,
2
2
⎛r ⎞
⎛ 0.04 m ⎞
A1d1 = A2 d 2 ⇒ (π r12 ) d1 = (π r22 ) d 2 ⇒ d1 = ⎜ 2 ⎟ d 2 = ⎜
⎟ ( 0.20 m ) = 3.2 m
r
⎝ 0.01 m ⎠
⎝ 1⎠
15.13.
Visualize:
Solve: By sucking on the top and reducing the pressure, the straw is essentially a barometer. Atmospheric
pressure pushes the liquid up the straw. The length of the longest straw can be obtained from the
formula p = p0 + ρ gd . If one could reduce the mouth pressure to zero, the length of the straw would be
1.013 × 105 Pa = 0 Pa + (1000 kg m 3 )( 9.8 m s 2 ) d ⇒ d = 10.3 m
15.14. Model:
Visualize:
The vacuum cleaner can create zero pressure.
Solve: The gravitational force on the dog is balanced by the force resulting from the pressure difference
between the atmosphere and the vacuum ( phose = 0) in the hose. The force applied by the hose is
F = ( patmos − phose ) A = patmos A = mg
(10 kg ) ( 9.8 m/s2 )
= 9.7 × 10−4 m 2
1.013 × 105 Pa
2
A
⎛d⎞
= 0.035 m = 3.5 cm.
Since A = π ⎜ ⎟ , the diameter of the hose is d = 2
π
⎝2⎠
Assess: The dog will have a terrible skin rash. It’s a good thing the miscreant holding the vacuum does not
have a younger sibling.
⇒ A=
15.15. Model: The buoyant force on the sphere is given by Archimedes’ principle.
Visualize:
Solve:
The sphere is in static equilibrium because it is neutrally buoyant. That is,
∑F = F − F = 0 N ⇒ ρV g − m g = 0 N
y
B
G
l l
s
The sphere displaces a volume of liquid equal to its own volume, Vl = Vs , so
ρl =
ms
m
0.0893 kg
= 4 s3 =
= 7.9 × 102 kg m3
3
4
Vs 3 π rs
π
0.030
m
(
)
3
A density of 790 kg/m3 in Table 15.1 identifies the liquid as ethyl alcohol.
Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.
15.16. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Vcyl is the volume of the cylinder and Vw is the volume of the water displaced by the cylinder. Note that the
volume displaced is only from the part of the cylinder that is immersed in water.
Solve: The cylinder is in static equilibrium, so FB = FG . The buoyant force is the weight ρ wVw g of the
displaced water. Thus
FB = ρ wVw g = FG = mg = ρcylVcyl g ⇒ ρ wVw = ρ cylVcyl ⇒ ρ cyl = ρ w
⇒ ρ cyl = (1000 kg m3 )
Assess:
A ( 0.040 m )
= 6.7 × 102 kg m3
A ( 0.060 m )
ρcyl < ρ w for a cylinder floating in water is an expected result.
Vw
Vcyl
15.17. Model: The buoyant force on the sphere is given by Archimedes’ principle.
Visualize:
Solve:
The sphere is in static equilibrium. The free body diagram on the sphere shows that
1
4
∑F = F −T − F = 0 N ⇒ F = T + F = 3 F + F = 3 F
y
⇒ ρ wVsphere g =
B
G
B
G
G
G
G
4
3
3
ρsphereVsphere g ⇒ ρsphere = ρ w = (1000 kg m3 ) = 750 kg m3
3
4
4
15.18. Model: The buoyant force on the rock is given by Archimedes’ principle.
Visualize:
Solve:
Because the rock is in static equilibrium, Newton’s first law is
Fnet = T + FB − ( FG )rock = 0 N
⎞
1
1
ρ
⎛1
⎞
⎛
⎞
⎛
⎞⎛ m g ⎞ ⎛
⇒ T = ρ rockVrock g − ρ water ⎜ Vrock ⎟ g = ⎜ ρ rock − ρ water ⎟Vrock g = ⎜ ρ rock − ρ water ⎟ ⎜ rock ⎟ = ⎜ 1 − water ⎟ mrock g
2
2
⎝2
⎠
⎝
⎠
⎝
⎠ ⎝ ρ rock ⎠ ⎝ 2ρ rock ⎠
Using ρ rock = 4800 kg m3 and mrock = 5.0 kg, we get T = 44 N.
15.19. Model: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of
aluminum and ethyl alcohol are ρ Al = 2700 kg m3 and ρ ethyl alcohol = 790 kg m3 .
Visualize:
The buoyant force FB and the tension due to the string act vertically up, and the gravitational force on the
aluminum block acts vertically down. The block is submerged, so the volume of displaced fluid equals VAl , the
volume of the block.
Solve: The aluminum block is in static equilibrium, so
∑F = F +T − F = 0 N ⇒ ρ V g +T − ρ V g = 0 N ⇒T =V g(ρ − ρ )
y
B
G
T = (100 × 10
−6
f
Al
Al Al
Al
Al
f
m )( 9.80 m s )( 2700 kg m − 790 kg m ) = 1.87 N
3
2
3
3
where we have used the conversion 100 cm3 = 100 × (10−2 m ) = 10−4 m3 .
3
Assess: The gravitational force on the aluminum block is ρ AlVAl g = 2.65 N. A similar order of magnitude for T
is reasonable.
15.20. Model: The buoyant force on the steel cylinder is given by Archimedes’ principle.
Visualize:
The length of the cylinder above the surface of mercury is d.
Solve: The cylinder is in static equilibrium with FB = FG . Thus
FB = ρ HgVHg g = FG = mg = ρ cylVcyl g ⇒ ρ HgVHg = ρcylVcyl ⇒ ρ Hg A ( 0.20 m − d ) = ρ cyl A ( 0.20 m )
⇒ d = 0.20 m −
ρ cyl
⎛
7900 kg m3 ⎞
= 0.084 m = 8.4 cm
( 0.20 m ) = ( 0.20 m ) ⎜1 −
3 ⎟
ρ Hg
⎝ 13,600 kg m ⎠
That is, the length of the cylinder above the surface of the mercury is 8.4 cm.
15.21. Model: The buoyant force is determined by Archimedes’ principle. Ignore any compression the air in
the beach ball may undergo as a result of submersion.
Solve: The mass of the beach ball is negligible, so the force needed to push it below the water is equal to the
buoyant force.
3⎞
⎛4
⎞
⎛4
FB = pw ⎜ π R 3 ⎟ g = (1000 kg/m3 ) ⎜ π ( 0.30 m ) ⎟ ( 9.8 m/s 2 ) = 1.11 kN
⎝3
⎠
⎝3
⎠
Assess: It would take a 113 kg (250 lb) person to push the ball below the water. Two people together could do
it. This seems about right.
15.22. Model: The buoyant force on the sphere is given by Archimedes’ principle.
Visualize:
Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the
buoyant force FB must be equal to the sum of the gravitational force on the Styrofoam sphere and the attached
mass. The volume of displaced water equals the volume of the sphere, so
FB = ρ waterVwater g = (1000 kg m3 ) 43π ( 0.25 m ) ( 9.80 m s 2 ) = 641.4 N
3
( FG )Styrofoam = ρStyrofoamVStyrofoam g = (150 kg m3 ) ⎡⎣ 34 π ( 0.25 m ) ⎤⎦ ( 9.80 m s 2 ) = 96.2 N
3
Because ( FG )Styrofoam + mg = FB ,
m=
The mass is 56 kg.
FB − ( FG )Styrofoam
g
=
641.4 N − 96.2 N
= 55.6 kg
9.80 m s 2
15.23. Model: Treat the water as an ideal fluid. The pipe is a flow tube, so the equation of continuity applies.
Solve:
The volume flow rate is
Q=
300 L 300 × 10−3 m3
=
= 1.0 × 10−3 m3 s
5.0 min
5.0 × 60 s
Using the definition Q = vA, we get
v=
Q 1.0 × 10−3 m3 s
=
= 3.2 m s
A π ( 0.010 m )2
15.24. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity
applies.
Visualize:
Note that A1 , A2 , and A3 and v1 , v2 , and v3 are the cross-sectional areas and the speeds in the first, second, and
third segments of the pipe.
Solve: (a) The equation of continuity is
A1v1 = A2v2 = A3v3 ⇒ π r12v1 = π r22v2 = π r32v3 ⇒ r12v1 = r22v2 = r32v3
⇒ ( 0.0050 m ) ( 4.0 m s ) = ( 0.010 m ) v2 = ( 0.0025 m ) v3
2
2
2
⎛ 0.0050 m ⎞
⇒ v2 = ⎜
⎟ ( 4.0 m s ) = 1.00 m s
⎝ 0.010 m ⎠
2
2
⎛ 0.0050 m ⎞
v3 = ⎜
⎟ ( 4.0 m s ) = 16.0 m s
⎝ 0.0025 m ⎠
(b) The volume flow rate through the pipe is
Q = A1v1 = π ( 0.0050 m ) ( 4.0 m s ) = 3.1 × 10−4 m3 s
2
15.25. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation.
Visualize:
Solve:
The equation of continuity is v0 A0 = v1 A1 , where A0 = L2 and A1 = π ( 12 L ) . The above equation simplifies
2
to
2
⎛ L⎞
⎛ 4⎞
v0 L2 = v1π ⎜ ⎟ ⇒ v1 = ⎜ ⎟ v0 = 1.27v0
2
⎝ ⎠
⎝π ⎠
15.26. Model: Treat the oil as an ideal fluid obeying Bernoulli’s equation. Consider the path connecting point 1
in the lower pipe with point 2 in the upper pipe a streamline.
Visualize: Please refer to Figure EX15.26.
Solve: Bernoulli’s equation is
p2 + 12 ρ v22 + ρ gy2 = p1 + 12 ρ v12 + ρ gy1 ⇒ p2 = p1 + 12 ρ ( v12 − v22 ) + ρ g ( y1 − y2 )
Using p1 = 200 kPa = 2.00 × 105 Pa, ρ = 900 kg m3 , y2 − y1 = 10.0 m, v1 = 2.0 m s, and v2 = 3.0 m s, we
get p2 = 1.096 × 105 Pa = 110 kPa.
15.27. Model: Turning the tuning screws on a guitar string creates tensile stress in the string.
Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the crosssectional area of the string. From the definition of Young’s modulus,
Y=
T/A
T ⎛ L⎞
⇒ ΔL = ⎜ ⎟
ΔL / L
A⎝Y ⎠
Using T = 2000 N, L = 0.80 m, A = π (0.00050 m)2 , and Y = 20 × 1010 N/m 2 (from Table 15.3), we obtain
ΔL = 0.010 m = 1.02 cm.
Assess: 1.02 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.
15.28. Model: The dangling mountain climber creates tensile stress in the rope.
Solve:
Young’s modulus for the rope is
Y=
F / A stress
=
ΔL / L strain
The tensile stress is
( 70 kg ) ( 9.8 m/s2 )
= 8.734 × 106 Pa
2
π ( 0.0050 m )
and the strain is 0.080 m 50 m = 0.00160. Dividing the two quantities yields Y = 5.5 × 109 N m 2 .
15.29. Model: The hanging mass creates tensile stress in the wire.
Solve: The force (F) pulling on the wire, which is simply the gravitational force (mg) on the hanging mass,
produces tensile stress given by F/A, where A is the cross-sectional area of the wire. From the definition of Young’s
modulus, we have
(π r 2 )Y ΔL = π ( 2.50 × 10−4 m ) ( 20 × 1010 N/m2 )(1.0 × 10−3 m ) = 2.0 kg
mg / A
⇒m=
ΔL / L
gL
( 9.80 m/s 2 ) ( 2.0 m )
2
Y=
15.30. Model: The load supported by a concrete column creates compressive stress in the concrete column.
Solve: The gravitational force on the load produces tensile stress given by F/A, where A is the cross-sectional
area of the concrete column and F equals the gravitational force on the load. From the definition of Young’s
modulus,
Y=
Assess:
2
F/A
3.0 m
⎛ F ⎞⎛ L ⎞ ⎛ 200,000 kg × 9.8 m/s ⎞ ⎛
⎞
⎟⎜
⇒ ΔL = ⎜ ⎟⎜ ⎟ = ⎜
= 1.0 mm
2
10
2 ⎟
⎜
⎟
ΔL / L
×
3
10
N/m
⎝ A ⎠⎝ Y ⎠ ⎝
⎝
⎠
π ( 0.25 m )
⎠
A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.
15.31. Model: Water is almost incompressible and it applies a volume stress.
Solve:
(a) The pressure at a depth of 5000 m in the ocean is
p = p0 + ρsea water g ( 5000 m ) = 1.013 × 105 Pa + (1030 kg m3 )( 9.8 m s 2 ) ( 5000 m ) = 5.057 × 107 Pa
(b) Using the bulk modulus of water,
ΔV
p
5.057 × 107 Pa
=− =−
= −0.025
V
B
0.2 × 1010 Pa
(c) The volume of a mass of water decreases from V to 0.975V. Thus the water’s density increases from ρ to
ρ /0.975. The new density is
ρ5000m =
1030 kg/m3
= 1056 kg/m3
0.975
15.32. Solve: The pressure p at depth d in a fluid is p = p0 + ρ gd . Using 1.29 kg/m3 for the density of air,
pbottom = ptop + ρ air gd ⇒ pbottom − ptop = 202 Pa = 1.99 × 10−3 atm
Assuming pbottom = 1 atm,
pbottom − ptop
pbottom
=
1.99 × 10−3 atm
= 0.2%
1 atm
15.33. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume
that the atmospheric pressure in the room is 1 atm.
Visualize: Please refer to Figure P15.33.
2
Solve: (a) The flat end of each cylinder has an area A = π r 2 = π ( 0.30 m ) = 0.283 m 2 . The force on each end is
thus
Fatm = p0 A = (1.013 × 105 Pa )( 0.283 m 2 ) = 2.86 × 104 N
The force on each end is 2.9 × 104 N.
(b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is
∑F = F
y
atm
− ( FG )players = 0 N ⇒ ( FG )players = Fatm = 2.86 × 104 N
⇒ number of players =
2.86 × 104 N
= 29.2
(100 kg ) ( 9.8 m s 2 )
That is, 30 players are needed to pull the two cylinders apart.
15.34. Model: Assume that the oil is incompressible and its density is 900 kg/m3 .
Visualize: Please refer to Figure P15.34.
Solve: (a) The pressure at depth d in a fluid is p = p0 + ρ gd . Here, pressure p0 at the top of the fluid is due
both to the atmosphere and to the gravitational force on the piston. That is, p0 = patm + ( FG ) p / A. At point A,
pA = patm +
( FG )P
A
+ ρ g (1.00 m − 0.30 m )
(10 kg ) ( 9.8 m s 2 )
+ ( 900 kg m3 )( 9.8 m s 2 ) ( 0.70 m ) = 185,460 Pa
2
π ( 0.02 m )
2
⇒ FA = pA A = (185,460 Pa )π ( 0.10 m ) = 5.8 kN
= 1.013 × 105 Pa +
(b) In the same way,
pB = patm +
Assess:
( FG )P
A
+ ρ g (1.30 m ) = 190,752 Pa ⇒ FB = 6.0 kN
FB is larger than FA , because pB is larger than pA .
15.35. Model: The tire flattens until the pressure force against the ground balances the upward normal force
of the ground on the tire.
Solve: The area of the tire in contact with the road is A = ( 0.15 m )( 0.13 m ) = 0.0195 m 2 . The normal force on
each tire is
FG (1500 kg ) ( 9.8 m s )
=
= 3675 N
4
4
2
n=
Thus, the pressure inside each tire is
pinside =
n
3675 N
14.7 psi
=
= 188,500 Pa = 1.86 atm ×
= 27 psi
A 0.0195 m 2
1 atm
15.36.
Visualize:
Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg
above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 × 105 Pa,
the minimum pressure is 100 mm = 1.333 × 10 4 Pa. The excess pressure in the fluid is due to force F pushing on the
internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to
the syringe is
F = fluid pressure × area of plunger = (1.333 × 104 Pa ) ⎡π ( 0.0030 m ) ⎤ = 0.38 N
⎣
⎦
2
(b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the
needle. Thus,
v=
Q 2.0 × 10−6 m3 2.0 s
=
= 20 m s
A π ( 0.125 × 10−3 m )2
Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also
note that the syringe is not drawn to scale.
15.37. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa = 101,300 N/m 2 means that the
weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square
meter is m = (101,300 N)/g = (101,300 N)/(9.80 m/s 2 ) = 10,340 kg per m 2 . Multiplying by the earth’s surface
area will give the total mass. Using Re = 6.27 × 106 m for the earth’s radius, the total mass of the atmosphere is
M air = Aearth m = (4π Re2 )m = 4π (6.37 × 106 m) 2 (10,340 kg/m 2 ) = 5.27 × 1018 kg
15.38. Visualize: Let d be the atmosphere’s thickness, p the atmospheric pressure on the earth’s surface, and
p0
(= 0 atm) the pressure beyond the earth’s atmosphere.
Solve:
The pressure at a depth d in a fluid is p = p0 + ρ gd . This equation becomes
1 atm = 0 atm + ρair gd ⇒ d =
1 atm
1.013 × 105 Pa
=
= 7.95 km
ρair g (1.3 kg/m3 )( 9.8 m/s 2 )
15.39. Solve: (a) We can measure the atmosphere’s pressure by measuring the height of the liquid column in a
barometer, because patmos = ρ gh. In the case of the water barometer, the height of the column at a pressure of 1 atm
is
h=
patmos
1.013 × 105 Pa
=
= 10.337 m
ρ water g (1000 kg/m3 )( 9.8 m/s 2 )
Because the pressure of the atmosphere can vary by 5 percent, the height of the barometer must be at least be
1.05 greater than this amount. That is, hmin = 10.85 m.
(b) Using the conversion rate 1 atm = 29.92 inches of Hg, we have
29.55 inches of Hg =
29.55
× 1 atm = 0.9876 atm
29.92
The height of the water in your barometer will be
h=
patmos
0.9876 × 1.013 × 105 Pa
=
= 10.21 m
ρ water g (1000 kg/m3 )( 9.8 m/s 2 )
15.40. Model: Oil is incompressible and has a density 900 kg/m3 .
Visualize: Please refer to Figure P15.40.
Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is
pA = p0 + ρ oil g (1.00 m − 0.50 m ) = 101,300 Pa + ( 900 kg m3 )( 9.8 m s 2 ) ( 0.50 m ) = 106 kPa
(b) The pressure difference between A and B is
pB − pA = ( p0 + ρ gd B ) − ( p0 + ρ gd A ) = ρ g (d B − d A ) = (900 kg/m3 )(9.8 m/s 2 )(0.50 m) = 4.4 kPa
Pressure depends only on depth, and C is the same depth as B. Thus pC − pA = 4.4 kPa also, even though C isn’t
directly under A.
15.41. Model: Assume that oil is incompressible and its density is 900 kg/m3 .
Visualize: Please refer to Figure P15.41.
Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level.
Equation 15.11, therefore, simplifies to
Fleft piston Fright piston
( FG )student ( FG )elephant
=
⇒
=
2
2
Aleft piston Aright piston
π ( rstudent ) π ( relephant )
⎛ (F )
⎞
⇒ rstudent = ⎜ G student ⎟ ( relephant ) =
⎜ ( FG )elephant ⎟
⎝
⎠
( 70 kg ) g 1.0 m = 0.2415 m
(
)
(1200 kg ) g
The diameter of the piston the student is standing on is therefore 2 × 0.2415 m = 0.48 m.
(b) From Equation 15.13, we see that an additional force ΔF is required to increase the elephant’s elevation
through a distance d 2 . That is,
ΔF = ρ g ( Aleft piston + Aright piston ) d 2
2
2
⇒ ( 70 kg ) ( 9.8 m s 2 ) = ( 900 kg m 3 )( 9.8 m s 2 )π ⎡( 0.2415 m ) + (1.0 m ) ⎤ d 2
⎣
⎦
⇒ d 2 = 0.0234 m
The elephant moves 2.3 cm.
15.42. Model: Assume that the oil is incompressible and its density is 900 kg/m3 .
Visualize:
Solve:
The pressures p1 and p2 are equal. Thus,
p0 +
F1
F
F F
= p0 + 2 + ρ gh ⇒ 1 = 2 + ρ gh
A1
A2
A1 A2
With F1 − m1 g , F2 = 4m2 g , A1 = π r12 , and A2 = π r2 2 , we have
12
⎞
m1 g 4m2 g
⎛ 4m g ⎞ ⎛ m g
=
+ ρ gh ⇒ r2 = ⎜ 2 ⎟ ⎜ 1 2 − ρ gh ⎟
r
π r12 π r22
π
π
⎝
⎠ ⎝ 1
⎠
Using
m1 = 55 kg, m2 = 110 kg, r1 = 0.08 m, ρ = 900 kg/m3 ,
and
−1 2
h = 1.0 m,
the
r2 = 0.276 m. The diameter is 55 cm.
Assess: Both pistons are too small to hold the people as shown, but the ideas are correct.
calculation
yields
15.43. Model: Assume that the oil is incompressible.
Visualize:
Solve:
When the force F1 balances the gravitational force mg, the pressure p2 is related to the pressure p1 as
p1 = p2 + ρ gh
When F1 is increased to F1′, the weight is raised higher through a distance d 2 and the left piston is lowered
through a distance d1. The pressures p′2 and p1′ are now related through
p1′ = p′2 + ρ g ( h + d1 + d 2 )
Subtracting these two equations,
p1′ − p1 = ( p2′ − p2 ) + ρ g ( d1 + d 2 )
Because p1 = F1 A1 , p1′ = F1′ A1 , and p2 = p′2 = mg A2 , the above equation simplifies to
( F1′ − F1 ) A1 = ρ g ( d1 + d 2 ) ⇒ ΔF = ρ g ( A1d1 + A1d 2 )
Since the oil is incompressible, A1d1 = A2 d 2 . The equation for ΔF thus becomes
ΔF = ρ g ( A2 d 2 + A1d 2 ) = ρ g ( A1 + A2 ) d 2
15.44. Model: Water and mercury are incompressible and immiscible liquids.
Visualize:
The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because
points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be
equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption
of static equilibrium.
Solve: The pressure at point 1 is due to water of depth d w = 10 cm:
p1 = patmos + ρ w gd w
Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the
right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth d Hg = 2h :
p2 = patmos + ρ Hg gd Hg = patmos + 2 ρ Hg gh
Equating p1 and p2 gives
patmos + ρ w gd w = patmos + 2 ρ Hg gh ⇒ h =
1 ρw
1 1000 kg m3
dw =
(10 cm ) = 3.7 mm
2 ρ Hg
2 13,600 kg m3
The mercury in the right arm rises 3.7 mm above its initial level.
15.45. Model: Glycerin and ethyl alcohol are incompressible and do not mix.
Visualize:
Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h
from its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each
other and the fluids are in static equilibrium, so the pressures at these two points must be equal:
p1 = p2 ⇒ p0 + ρ eth gd eth = p0 + ρ gly gd gly ⇒ ρ eth g (20 cm) = ρ gly g (2h)
h=
1 ρ eth
1 790 kg/m3
(20 cm) =
(20 cm) = 6.27 cm
2 ρgly
2 1260 kg/m3
You can see from the figure that the difference between the top surfaces of the fluids is
Δy = 20 cm − 2h = 20 cm − 2(6.27 cm) = 7.46 cm ≈ 7.5 cm
15.46. Model: The water is in hydrostatic equilibrium.
Visualize: Please refer to figure P15.46.
Solve: (a) Can 2 has moved down with respect to Can 1 since the water level in Can 2 has risen. Since the total
volume of water stays constant, the water level in Can 1 has fallen by the same amount. The water level is
equalized in the two cans at the middle of the height change, so the change in height of the water is half the
relative change in height of the cans. Can 2 has moved relative to Can 1 (6.5 cm − 5.0 cm) × 2 = 3.0 cm down.
(b) The water level in Can 1 has fallen by the same amount. The new level is 5.0 − 1.5 cm = 3.5 cm
Assess: The two cans are an inexpensive method of measuring relative changes in height.
15.47.
Visualize:
The figure shows a dam with water height d. We chose a coordinate system with the origin at the bottom of the
dam. The horizontal slice has height dy, width w, and area dA = wdy. The slice is at the depth of d − y.
Solve: (a) The water exerts a small force dF = pdA = pwdy on this small piece of the dam, where
p = ρ g (d − y ) is the pressure at depth d − y. Altogether, the force on this small horizontal slice at position y is
dF = ρ gw(d − y ) dy. Note that a force from atmospheric pressure is not included. This is because atmospheric
pressure exerts a force on both sides of the dam. The total force of the water on the dam is found by adding up all
the small forces dF for the small slices dy between y = 0 m and y = d . This summation is expressed as the
integral
d
d
d
1 ⎞
1
⎛
Ftotal = ∫ dF = ∫ dF = ρ gw∫ ( d − y ) dy = ρ gw ⎜ yd − y 2 ⎟ = ρ gwd 2
2
⎝
⎠0 2
all slices
0
0
(b) The total force is
Ftotal = 12 (1000 kg m3 )( 9.8 m s 2 ) (100 m )( 60 m ) = 1.76 × 109 N
2
15.48.
Visualize:
The figure shows an aquarium tank with water at a height d. We chose a coordinate system with the origin at the
bottom of the tank. The slice has a height dy, length l, and area dA = ldy. It is at depth d − y.
Solve: (a) Pressure in excess of atmospheric pressure of the water on the bottom is pbottom = ρ water gd . Atmospheric
pressure is ignored because it exerts an equal force on both sides of the bottom. Therefore, the force of the water on
the bottom is
Fbottom = pbottom ( lw ) = ( ρ water gd )( lw ) = (1000 kg m3 )( 9.8 m s 2 ) ( 0.40 m )(1.0 m )( 0.35 m ) = 1.37 kN
(b) The water exerts a small force dF = pdA = pldy on a small piece of the window, where p = ρ g (d − y ) is
the excess pressure at a depth of d − y. The force on this small horizontal slice at position y is
dF = ρ gl ( d − y ) dy. The total force on the front window is
d
Ftotal = ∫ dF = ∫ ρ water gL ( d − y ) dy = ρ water gl ( yd − 12 y 2 ) =
d
0
0
=
1
2
1
ρ water gld 2
2
(1000 kg m )( 9.8 m s ) (1.00 m )( 0.40 m ) = 0.78 kN
3
2
2
15.49.
Visualize:
The figure shows a small column of air of thickness dz, of cross-sectional area A = 1 m 2 , and of density ρ ( z ).
The column is at a height z above the surface of the earth.
Solve: (a) The atmospheric pressure at sea level is 1.013 × 105 Pa. That is, the weight of the air column with a
1 m 2 cross section is 1.013 × 105 N. Consider the weight of a 1 m 2 slice of thickness dz at a height z. This slice has
volume dV = Adz = (1 m 2 ) dz, so its weight is dw = ( ρ dV ) g = ρ g (1 m 2 )dz = ρ 0e− z/z0 g (1 m 2 )dz. The total weight
of the 1 m 2 column is found by adding all the dw. Integrating from z = 0 to z = ∞,
∞
w = ∫ ρ 0 g (1 m 2 )e− z z0 dz
0
= ( − ρ 0 g (1 m 2 ) z0 ) ⎡⎣e− z z0 ⎤⎦
∞
0
= ρ 0 g (1 m ) z0
2
Because w = 101,300 N = ρ 0 g (1 m 2 ) z0 ,
z0 =
101,300 N
= 8.08 × 103 m
(1.28 kg/m3 )( 9.8 m/s2 ) (1.0 m2 )
(b) Using the density at sea level from Table 15.1,
ρ = (1.28 kg m3 ) e − z /(8.08×10 m) = (1.28 kg m3 ) e −1600 m/(8.08×10 m) = 1.05 kg m3
3
This is 82% of ρ 0 .
3
15.50. Model: The buoyant force on the ceramic statue is given by Archimedes’ principle.
Visualize:
Solve: The gravitational force on the statue is the 28.4 N registered on the scale in air. In water, the
gravitational force on the statue is balanced by the sum of the buoyant force FB and the spring’s force on the
statue. That is,
( FG )statue = FB + Fspring on statue ⇒ 28.4 N = ρ wVstatue g + 17.0 N ⇒ Vstatue =
⇒ ρstatue =
11.4 N mstatue
=
g ρw
ρstatue
( mstatue g ) ρ w = ( 28.4 N ) (1000 kg m3 ) = 2.49 × 103 kg m3
(11.4 N )
(11.4 N )
15.51. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the
gravitational force on the cylinder. The volume of displaced liquid is Ah, so
FB = ρ liq ( Ah) g = FG
Force F pushes the cylinder down distance x, so the submerged length is h + x and the volume of displaced
liquid is A(h + x ). The cylinder is again in equilibrium, but now the buoyant force balances both the
gravitational force and force F. Thus
FB = ρ liq ( A(h + x)) g = FG + F
Since ρ liq ( Ah) g = FG , we’re left with
F = ρ liq Agx
(b) The amount of work dW done by force F to push the cylinder from x to x + dx is dW = Fdx = ( ρ liq Agx)dx.
To push the cylinder from xi = 0 m to xf = 10 cm = 0.10 m requires work
xf
xf
xi
xi
W = ∫ F dx = ρliq Ag ∫ x dx = 12 ρ liq Ag ( xi2 − xf2 )
= 12 (1000 kg/m3 )π (0.020 m) 2 (9.8 m/s 2 )(0.10 m) 2 = 0.62 J
15.52. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Let d1 be the length of the cylinder in the less-dense liquid with density ρ1 , and d 2 be the length of the cylinder
in the more-dense liquid with density p2 .
Solve: The cylinder is in static equilibrium, so
∑ F = F + F − F = 0 N ⇒ ρ ( Ad ) g + ρ ( Ad ) g = ρ A ( d + d ) g
y
B1
B2
G
1
1
2
2
1
ρ
ρ
⇒ 1 d1 + d 2 = ( d1 + d 2 )
ρ2
ρ2
Since l = d1 + d 2 ⇒ d1 = l − d 2 , we can simplify the above equation to obtain
⎛ ρ − ρ1 ⎞
d2 = ⎜
⎟l
⎝ ρ 2 − ρ1 ⎠
That is, the fraction of the cylinder in the more dense liquid is f = ( ρ − ρ1 ) ( ρ 2 − ρ1 ) .
Assess:
As expected f = 0 when ρ is equal to ρ1 , and f = 1 when ρ = ρ 2 .
2
15.53. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
Solve:
The tube is in static equilibrium, so
∑F = F − (F )
y
B
G tube
⇒ ρ liquid =
Assess:
− ( FG )Pb = 0 N ⇒ ρ liquid A ( 0.25 m ) g = ( 0.030 kg ) g + ( 0.250 kg ) g
( 0.280 kg )
( 0.280 kg ) =
= 8.9 × 102 kg m 3
A ( 0.25 m ) π ( 0.020 m )2 ( 0.25 m )
This is a reasonable value for a liquid.
15.54. Model: Archimedes’ Principle determines the buoyant force.
Visualize:
Solve: The plastic hemisphere will hold the most weight when its rim is at the surface of the water. The
buoyant force balances the gravitational force on the bowl and rock.
∑F = F − (F )
y
B
G rock
− ( FG )bowl = 0 N
Thus
3⎞
⎛ 1 ⎞⎛ 4
mg = pwVbowl g − m bowl g = (1000 kg/m3 ) ⎜ ⎟⎜ π ( 0.040 m ) ⎟ ( 9.8 m/s 2 ) − ( 0.021 kg ) ( 9.8 m/s 2 ) ⇒ m = 0.155 kg
2
3
⎝ ⎠⎝
⎠
Assess:
Putting a rock as big as 155 g in an 8 cm diameter bowl before it sinks is reasonable.
15.55. Model: The buoyant force is determined by Archimedes’ principle. The spring is ideal.
Visualize:
Solve: The spring is stretched by the same amount that the cylinder is submerged. The buoyant force and
spring force balance the gravitational force on the cylinder.
∑ F = F + F − mg = 0 N
y
B
S
⇒ pw Ayg + ky = mg
y=
(1.0 kg ) ( 9.8 m/s 2 )
mg
=
pw Ag + k (1000 kg/m3 )π ( 0.025 m )2 ( 9.8 m/s 2 ) + 35 N/m
= 0.181 m = 18.1 cm
Assess: This is difficult to assess because we don’t know the height h of the cylinder and can’t calculate it
without the density of the metal material.
15.56. Model: The balloon displaces air, so the air exerts an upward buoyant force on the balloon. The
buoyant force on the balloon is given by Archimedes’ principle.
Visualize:
FB is the buoyant force on the balloon, ( FG ) b is the gravitational force on the balloon, ( FG ) He is the
gravitational force on the helium gas, and FG is the gravitational force on the mass tied to the balloon.
Solve:
The volume of the balloon Vballoon = 34 π ( 0.10 m ) = 4.1888 × 10−3 m3 . We have
3
FB = ρ air gVballoon = (1.28 kg m3 )( 9.8 m s 2 )( 4.188 × 10−3 m3 ) = 0.05254 N
( FG )He = ρ He gVballoon = ( 0.18 kg m3 )( 9.8 m s 2 )( 4.188 × 10−3 m3 ) = 0.007387 N
( FG )b = ( 0.0010 kg ) g = 0.0098 N
For the balloon to be in static equilibrium,
FB = ( FG )He + ( FG )b + FG = 0.05254 N=0.01719 N + mg
⇒ m = 0.0036 kg = 3.6 g
15.57. Model: The buoyant force on the can is given by Archimedes’ principle.
Visualize:
The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is
A.
Solve: The can is in static equilibrium, so
∑F = F − (F )
y
B
G can
− ( FG ) water = 0 N ⇒ ρ water A ( L − d ) g = ( 0.020 kg ) g + mwater g
The mass of the water in the can is
355 × 10
⎛V ⎞
mwater = ρ water ⎜ can ⎟ = (1000 kg m3 )
2
2
⎝
⎠
Because
−6
m3
= 0.1775 kg
⇒ ρ water A ( L − d ) = 0.020 kg + 0.1775 kg = 0.1975 kg ⇒ d − L = −
0.1975 kg
= 0.0654 m
ρ water A
Vcan = π ( 0.031 m ) L = 355 × 10−6 m3 ,
this
2
L = 0.1176 m.
Using
value
of
L,
we
get
d = 0.0522 m ≈ 5.2 cm.
Assess: d L = 5.22 cm 11.76 cm = 0.444, thus 44.4% of the length of the can is above the water surface. This
is reasonable.
15.58. Model: The buoyant force on the boat is given by Archimedes’ principle.
Visualize:
The minimum height of the boat that will enable the boat to float in perfectly calm water is h.
Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of
displaced water is the volume of the boat. Archimedes’ principle in equation form is FB = ρ wVboat g . For the boat
to float FB = ( FG )boat . Let us first calculate the gravitational force on the boat:
( FG )boat = ( FG )bottom + 2 ( FG )side 1 + 2 ( FG )side 2 , where
( FG )bottom = ρsteelVbottom g = ( 7900 kg m3 )( 5 × 10 × 0.02 m3 ) g = 7900g N
( FG )side 1 = ρsteelVside 1g = ( 7900 kg m3 )( 5 × h × 0.005 m3 ) g = 197.5gh N
( FG )side 2 = ρsteelVside 2 g = ( 7900 kg m3 )(10 × h × 0.005 m3 ) g = 395gh
⇒ ( FG )boat = ⎡⎣7900g + 2 (197.5gh ) + 2 ( 395gh ) ⎤⎦ N = ( 7900+1185h ) g N
Going back to the Archimedes’ equation and remembering that h is in meters, we obtain
ρ wVboat g = ( 7900 + 1185h ) g N ⇒ (1000 ) (10 × 5 × ( h + 0.020 ) ) = 7900 + 1185h
⇒ h = 14.1 cm
15.59. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. There is a streamline
connecting point 1 in the wider pipe with point 2 in the narrower pipe.
Visualize:
Solve:
Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2
For no height change y1 = y2 . The flow rate is given as 5.0 L/s, which means
Q = v1 A1 = v2 A2 = 5.0 × 10−3 m3 s ⇒ v1 =
5.0 × 10−3 m3 s
π ( 0.05 m )
2
= 0.6366 m s
2
⇒ v2 = v1
A1
⎛ 0.050 m ⎞
= ( 0.6366 m s ) ⎜
⎟ = 2.546 m s
A2
⎝ 0.025 m ⎠
Bernoulli’s equation now simplifies to
p1 + 12 ρ v12 = p2 + 12 ρ v22
⇒ p1 = p2 + 12 ρ ( v22 − v12 ) = 50 × 103 Pa + 12 (1000 kg m3 ) ⎡( 2.546 m s ) − ( 0.6366 m s ) ⎤ = 53 kPa
⎣
⎦
2
Assess:
Reducing the pipe size reduces the pressure because it makes v2 > v1.
2
15.60. Model: The two pipes are identical.
Visualize:
Solve:
The water speed is the same in both pipes. The flow rate
Q = 3.0 × 106 L/min = 2 ( vA )
( 3.0 ×10 )(10 ) ⎛⎜⎝ 601 ⎞⎟⎠ m /s
6
Q
⇒v=
=
2A
−3
2π (1.5 m )
3
2
= 3.5 m/s
15.61. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins in the
bigger size pipe and ends at the exit of the narrower pipe.
Visualize: Please see Figure P15.61. Let point 1 be beneath the standing column and point 2 be where the
water exits the pipe.
Solve: (a) The pressure of the water as it exits into the air is p2 = patmos .
(b) Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ p1 − p2 = 12 ρ ( v22 − v12 ) + ρ g ( y2 − y1 )
From the continuity equation,
v1 A1 = v2 A2 = ( 4 m s ) ( 5 × 10−4 m 2 ) ⇒ v1 (10 × 10−4 m 2 ) = 20 × 10−4 m3 s ⇒ v1 = 2.0 m s
Substituting into Bernoulli’s equation,
2
2
p1 − p2 = p1 − patmos = 12 (1000 kg m3 ) ⎡( 4.0 m s ) − ( 2.0 m s ) ⎤ + (1000 kg m 3 ) ( 9.8 m s )( 4.0 m )
⎣
⎦
= 6000 Pa + 39,200 Pa = 45 kPa
But p1 − p2 = ρ gh, where h is the height of the standing water column. Thus
h=
45 × 103 Pa
= 4.6 m
(1000 kg/m3 )( 9.8 m/s2 )
15.62. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins at the
faucet and continues down the stream.
Visualize:
The pressure at point 1 is p1 and the pressure at point 2 is p2 . Both p1 and p2 are atmospheric pressure. The
velocity and the area at point 1 are v1 and A1 and they are v2 and A2 at point 2. Let the distance of point 2 below
point 1 be d.
Solve: The flow rate
Q = v1 A =
3
2.0 × 10−4 m
3
2.0 × 1000 × 10−6 m3
2 = 1.0 m/s
= 2.0 × 10−4 m ⇒ v1 =
2
s
10 s
π ( 0.0080 m )
Bernoulli’s equation at points 1 and 2 is
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ ρ gd = 12 ρ ( v22 − v12 )
From the continuity equation,
v1 A1 = v2 A2 ⇒ (1.0 m s )π ( 8.0 × 10−3 m ) = v2π ( 5.0 × 10−3 m ) ⇒ v2 = 2.56 m s
2
2
Going back to Bernoulli’s equation, we have
2
2
gd = 12 ⎡( 2.56 m s ) − (1.0 m s ) ⎤ ⇒ d = 0.283 m ≈ 28 cm
⎣
⎦
15.63. Model: Treat the air as an ideal fluid obeying Bernoulli’s equation.
Solve: (a) The pressure above the roof is lower due to the higher velocity of the air.
(b) Bernoulli’s equation, with yinside ≈ youtside , is
2
pinside = poutside + 12 ρair v 2 ⇒ Δp =
1
1
⎛ 130 × 1000 m ⎞
ρ air v 2 = (1.28 kg m3 ) ⎜
⎟ = 835 Pa
2
2
⎝ 3600 s ⎠
The pressure difference is 0.83 kPa
(c) The force on the roof is ( Δp ) A = ( 835 Pa )( 6.0 m × 15.0 m ) = 7.5 × 104 N. The roof will blow up, because
pressure inside the house is greater than pressure on the top of the roof.
15.64. Model: The ideal fluid obeys Bernoulli’s equation.
Visualize: Please refer to Figure P15.64. There is a streamline connecting point 1 in the wider pipe on the left
with point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are v1 and v2 and the crosssectional area of the pipes at these points are A1 and A2 . Points 1 and 2 are at the same height, so y1 = y2 .
Solve:
The volume flow rate is Q = A1v1 = A2v2 = 1200 × 10−6 m3 s. Thus
v2 =
1200 × 10−6 m3 s
= 95.49 m s
π (0.0020 m) 2
v1 =
1200 × 10−6 m
3
π ( 0.010 m )
2
s = 3.82 m s
Now we can use Bernoulli’s equation to connect points 1 and 2:
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2
2
2
⇒ p1 − p2 = 12 ρ ( v22 − v12 ) + ρ g ( y2 − y1 ) = 12 (1.28 kg m3 ) ⎡( 95.49 m s ) − ( 3.82 m s ) ⎤ + 0 Pa = 5.83 kPa
⎣
⎦
Because the pressure above the mercury surface in the right tube is p2 and in the left tube is p1 , the difference
in the pressures p1 and p2 is ρ Hg gh. That is,
p1 − p2 = 5.83 kPa = ρ Hg gh ⇒ h =
5.83 × 103 Pa
= 4.4 cm
(13,600 kg m3 )(9.8 m s 2 )
15.65. Model: The ideal fluid (that is, air) obeys Bernoulli’s equation.
Visualize: Please refer to Figure P15.65. There is a streamline connecting points 1 and 2. The air speeds at
points 1 and 2 are v1 and v2 , and the cross-sectional areas of the pipes at these points are A1 and A2 . Points 1
and 2 are at the same height, so y1 = y2 .
Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by
ρ Hg g ( 0.10 m ) = (13,600 kg m3 )( 9.8 m s 2 ) ( 0.10 m ) = 13,328 Pa ⇒ p2 = p1 + 13,328 Pa
Using Bernoulli’s equation,
p1 + 12 ρair v12 + ρ air gy1 = p2 + 12 ρ air v22 + ρ air gy2 ⇒ p2 − p1 = 12 ρair ( v12 − v22 )
⇒ v12 − v22 =
2 ( p2 − p1 )
ρ air
=
2 (13,328 Pa )
(1.28 kg/m )
3
= 20,825 m 2 s 2
From the continuity equation, we can obtain another equation connecting v1 and v2 :
π ( 0.005 m )
A
A1v1 = A2v2 ⇒ v1 = 2 v2 =
v = 25 v2
2 2
A1
π ( 0.001 m )
2
Substituting v1 = 25v2 in the Bernoulli equation, we get
( 25 v2 ) − v22 = 20,825 m 2 s 2 ⇒ v2 = 5.78 m s
2
Thus v2 = 5.8 m/s and v1 = 25v2 = 144 m/s.
(b) The volume flow rate A2v2 = π ( 0.0050 m ) ( 5.78 m s ) = 4.5 × 10−3 m3 s.
2
15.66. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from
the top of the water to the hole.
Visualize: Please refer to Figure P15.66. The top of the water (at y1 = h) and the hole (at y2 = y ) are at
atmospheric pressure. The speed of the water at the top is zero because the tank is kept filled.
Solve: (a) Bernoulli’s equation connecting the two points is
0 + ρ gh = 12 ρ v 2 + ρ gy ⇒ v = 2 g ( h − y )
(b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic
equations. For the y-motion, using t0 = 0 s, we have
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = y + ( 0 m s ) t1 + 12 ( − g ) t12 ⇒ t1 = 2 y g
2
For the x-motion,
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ x = 0 m + vt1 + 0 m ⇒ x = v 2 y g
2
(c) Combining the results of (a) and (b), we obtain
x = 2g ( h − y ) 2 y g = 4 y ( h − y )
To find the maximum range relative to the vertical height,
dx
1
h
=0⇒
⎡⎣ 4 ( h − y ) − 4 y ⎤⎦ = 0 ⇒ y =
dy
2
4y(h − y)
With y = 12 h, the maximum range is
h⎞
⎛ h ⎞⎛
xmax = 4 ⎜ ⎟⎜ h − ⎟ = h
2⎠
⎝ 2 ⎠⎝
15.67. Model: Treat water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline connecting
the top of the tank with the hole.
Visualize: Please refer to Figure P15.67. We placed the origin of the coordinate system at the bottom of the
tank so that the top of the tank (point 1) is at a height of h + 1.0 m and the hole (point 2) is at a height h. Both
points 1 and 2 are at atmospheric pressure.
Solve: (a) Bernoulli’s equation connecting points 1 and 2 is
p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2
⇒ patmos + 12 ρ v12 + ρ g ( h + 1.0 m ) = patmos + 12 ρ v22 + ρ gh
⇒ v22 − v12 = 2g (1.0 m ) = 19.6 m 2 s 2
Using the continuity equation A1v1 = A2v2 ,
π ( 2.0 × 10 m )
⎛A ⎞
v2
v1 = ⎜ 2 ⎟ v2 =
v2 =
2
250,000
A
π (1.0 m )
⎝ 1⎠
−3
2
Because v1 v2 , we can simply put v1 ≈ 0 m/s. Bernoulli’s equation thus simplifies to
v22 = 19.6 m 2 s 2 ⇒ v2 = 4.43 m s
Therefore, the volume flow rate through the hole is
Q = A2v2 = π ( 2.0 × 10−3 m ) ( 4.43 m s ) = 5.56 × 10−5 m3 s = 3.3 L min
2
(b) The rate at which the water level will drop is
v2
4.43 m s
v1 =
=
= 1.77 × 10−2 mm s = 1.06 mm min
250,000 250,000
Assess: Because the hole through which water flows out of the tank has a diameter of only 4.0 mm, a drop in
the water level at the rate of 1.06 mm/min is reasonable.
15.68. Model: The aquarium creates tensile stress.
Solve:
Weight of the aquarium is
FG = mg = ρ waterVg = (1000 kg m3 )(10 m3 )( 9.8 m s 2 ) = 9.8 × 104 N
where we have used the conversion 1 L = 10−3 m 3 . The weight supported by each wood post is 14 (9.8 × 104 N) =
2.45 × 104 N. The cross-sectional area of each post is A = ( 0.040 m ) = 1.6 × 10−3 m 2 . Young’s modulus for the
2
wood is
F/A
FL
=
ΔL / L AΔL
4
×
2.45
10
N
0.80
m
(
)
(
)
FL
⇒ ΔL =
=
= 1.23 × 10−3 m = 1.23 mm
2
10
−3
AY (1.6 × 10 m )(1× 10 N/m 2 )
Y = 1 × 1010 N m 2 =
Assess:
A compression of 1.23 mm due to a weight of 2.45 × 104 N is reasonable.
15.69. Model: The water pressure applies a volume stress to the sphere.
Solve:
The volume change is ΔV = −1.0 × 10−3 V . The volume stress is
F
ΔV
−1.0 × 10−3V
= p = −B
= − ( 7 × 1010 N m 2 )
= 7.0 × 107 Pa
A
V
V
Using p = p0 + ρ gh, we get
7.0 × 107 Pa = 1.013 × 105 Pa + (1030 kg m3 )( 9.8 m s 2 ) d ⇒ d = 6.9 km
Assess: A pressure of 7.0 × 107 Pa = 690 atm causes only a volume change of 0.1%. This is reasonable
because liquids and solids are nearly incompressible.
15.70. Model: Pressure applies a volume stress to water in the cylinder.
Solve:
The volume strain of water due to the pressure applied is
p
ΔV
2 × 106 Pa
=− =−
= −1.0 × 10−3
V
B
0.20 × 1010 Pa
⇒ ΔV = V ′ − V = − (1.0 × 10−3 )(1.30 m 3 ) = −1.30 × 10−3 m3 = −1.3 L
As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. The volume
of water that comes out is 1.30 L.
15.71. Model: Air is an ideal gas and obeys Boyle’s law.
Visualize: Please refer to Figure CP15.71. h is the length of the air column when the mercury fills the cylinder
to the top. A is the cross-sectional area of the cylinder.
Solve: For the column of air, Boyle’s law is p0V0 = p1V1 , where p0 and V0 are the pressure and volume before
any mercury is poured, and p1 and V1 are the pressure and volume when mercury fills the cylinder above the air.
Using p1 = p0 + ρ Hg g (1.0 m − h ) , Boyle’s law becomes
p0V0 = ⎡⎣ p0 + ρ Hg g (1.0m − h ) ⎤⎦ V1 ⇒ p0 A (1.0 m ) = ⎡⎣ p0 + ρ Hg g (1.0m − h ) ⎤⎦ Ah
p0 (1.0m − h ) = ρ Hg g (1.0m − h ) h ⇒ h =
p0
ρ Hg g
=
1.013 × 105 Pa
= 0.76 m = 76 cm
(13,600 kg/m3 )( 9.8 m/s 2 )
15.72. Model: The buoyant force on the cone is given by Archimedes’ principle.
Visualize:
Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know
it, but it isn’t essential. The volume is clearly the area of the base multiplied by the height multiplied by some
constant. That is, the cone shown above has V = aAd , where a is some constant. But the radius of the base is
r = d tan α , where α is the angle of the apex of the cone, and A = π r 2 , making A proportional to d 2 . Thus the
volume of a cone of height d is V = cd 3 , where c is a constant. Because the cone is floating in static equilibrium,
we must have FB = FG . The cone’s density is ρ 0 , so the gravitational force on it is FG = ρ0Vg = ρ0cl 3 g. The
buoyant force is the gravitational force on the displaced fluid. The volume of displaced fluid is the full volume of
the cone minus the volume of the cone of height h above the water, or Vdisp = cl 3 = ch3 . Thus
FB = ρf Vdisp g = ρ f c(l 3 − h3 ) g , and the equilibrium condition is
1/ 3
⎛ h3 ⎞
h ⎛ ρ ⎞
FB = FG ⇒ ρ f c(l 3 − h3 ) g = ρ 0cl 3 g ⇒ ρ f ⎜1 − 3 ⎟ = ρ 0 ⇒ = ⎜1 − 0 ⎟
l ⎠
l ⎝ ρf ⎠
⎝
15.73. Model: The grinding wheel is a uniform disk. We will use the model of kinetic friction and
hydrostatics.
Visualize: Please refer to figure CP15.73.
Solve: This is a three-part problem. First find the desired angular acceleration, then use that to find the force
applied by each brake pad, then finally the needed oil pressure. The angular acceleration required to stop the
wheel is found using rotational kinematics.
1
Δω ωf − ωi 0 rad/s − 900 × 2π × 60 rad/s
α=
=
=
= −18.85 rad/s 2
5.0 s
Δt
Δt
Each brake pad applies a frictional force f k = μ k n to the wheel. The normal force is equal to the force applied by
the piston by Newton’s third law. Rotational dynamics can be used to find the magnitude of the force. f k is
applied 12 cm from the rotation axis on both sides of the disk.
τ net = Iα
⎛1
⎞
−2 f k ( 0.12 m ) = ⎜ MR 2 ⎟α
⎝2
⎠
(15 kg )( 0.13 m ) ( −18.85 rad/s )
MR 2α
⇒n=
=
= 16.6 N
−4 μ k ( 0.12 m )
−4 ( 0.60 )( 0.12 m )
2
2
The oil pressure required to generate this much force at each brake pad is
F
16.6 N
p= =
= 53 kPa
A π ( 0.010 m )2
relative to atmospheric pressure.
Assess: The required oil pressure is about half an atmosphere, which is quite reasonable.
15.74. Model: The buoyant force on the cylinder is given by Archimedes’ principle.
Visualize:
A is cross-sectional area of the cylinder.
Solve: (a) The gravitational force on the cylinder is FG = ρ0 ( Al ) g and the weight of the displaced fluid is
FB = ρf ( Ah ) g. In static equilibrium, FB = FG and we can write
ρf Ahg = ρ0 Alg ⇒ h = ( ρ0 ρf ) l
(b) Now the volume of the displaced liquid is A(h − y ). Applying Newton’s second law in the y-direction,
∑ F = − F + F = − ρ Alg + ρ A ( h − y ) g
y
G
Using ρf Ahg = ρ0 Alg from part (a), we find
B
0
f
( Fnet ) y = − ρf Agy
(c) The result in part (b) is F = −ky, where k = ρ f Ag . This is Hooke’s law.
(d) Since the cylinder’s equation of motion is determined by Hooke’s law, the angular frequency for the resulting
simple harmonic motion is ω = k m , and the period is
T=
2π
ω
= T = 2π
m
m
ρ0 Al
h
= 2π
= 2π
= 2π
k
ρf Ag
ρf Ag
g
where we have used the expression for h from part (a).
(e) The oscillation period for the 100-m-tall iceberg ( pice = 917 kg/m3 ) in sea water is
T = 2π
ρ 0l
h
= 2π
= 2π
ρf g
g
( 917 kg/m ) (100 m ) = 18.9 s
(1030 kg/m )( 9.8 m/s )
3
3
2
15.75. Model: A streamline connects every point on the surface of the liquid to a point in the drain. The
drain diameter is much smaller than the tank diameter (r R ).
Visualize:
Solve: The pressures at the surface and drain (points 1 and 2) are equal to one atmosphere. When the liquid is a
depth y, Bernoulli’s principle comparing points 1 and 2 is
1
1
p1 + ρ v12 + ρ gy = p2 + ρ v22 + ρ g ( 0 )
2
2
⇒ v22 = v12 + 2 gy
The flow rate through the drain is the same as through a horizontal layer in the tank:
A
Q = v1 A1 = v2 A2 ⇒ v1 = v2 2
A1
Thus the velocity of the liquid through the drain is
2
⎛ A ⎞
2 gy
v22 = ⎜ v2 2 ⎟ + 2 gy ⇒ v2 =
π r2
⎝ A1 ⎠
1−
π R2
Since r R, v2 ≈ 2 gy , and the flow rate through the drain is
Q ≈ π r 2 2 gy
The flow rate gives the volume of liquid flowing out per unit time. The inverse gives the time needed for a unit
volume of liquid to flow out. For the volume of liquid to decrease by dV requires a time
2
− R 2 dy
dV − (π R dy )
=
=
dt =
Q π r 2 2 gy
2 gr 2 y
Note that dV < 0 implies the volume of liquid is decreasing. Integrating both sides from the initial condition
(t = 0, y = d ) to the final condition (t , y = 0) yields
0
R2
2 R 2 1 2 0 R 2 2d
− 12
y
dy
y
=
−
= 2
d
r
g
2 gr 2 ∫d
2 gr 2
Assess: The time for the tank to drain depends on the ratio of the cross-sectional areas of the tank to drain,
which makes sense, as well as on the strength of the free-fall acceleration. Note that if g were larger (say we were
on Jupiter) the time to drain would be shorter, while if the tank were taller (larger d) the time would be longer.
t=−
15.76.
Visualize:
The column of air has a cross-sectional area A. We chose the origin of the coordinate system at z = 0 m and denote the
vertical axis as the z-axis.
Solve: (a) The pressure at height z is pz . The pressure at a height z + dz is
pz + dz = pz −
weight of air column of height dz
ρ Adzg
= pz −
= pz − ρ gdz
A
A
(b) The change in pressure is dp = pz + dz − pz = − ρ gdz. The sign is minus because pressure decreases as z increases.
(c) From the ideal gas law,
p
ρ
=
p0
ρ0
⇒ρ=
p ρ0
p0
The result for part (b) thus becomes dp = − ( p ρ 0 p0 ) gdz.
(d) To find the pressure p at a height z, we rewrite the result of part (c) as
⎛ρ ⎞
dp
= − ⎜ 0 g ⎟ dz
p
⎝ p0 ⎠
Integrating both sides
p
⎛ p⎞
dp
ρ0 g z
ρg
dz ⇒ ln ⎜ ⎟ = − 0 z ⇒ p = p0e − ρ0 gz p0 = p0e − z z0
=
−
∫p p
∫
p
p
p0
0 0
⎝ 0⎠
0
where z0 = p0 ρ 0 g .
(e) The scale height is z0 =
(1.013 × 105 Pa ) = 8076 m.
p0
=
ρ0 g (1.28 kg m3 )( 9.8 m s 2 )
(f) A table showing values of p = p0e− z z0 = e − z (8076 m ) atm at selected values of z follows.
z (m)
0
1000
2000
4000
6000
8000
10,000
12,000
14,000
15,000
p (atm)
1
0.884
0.781
0.609
0.476
0.371
0.290
0.226
0.177
0.156
15-1
Fluids and Elasticity
15-2
16.1. Model: Recall the density of water is 1000 kg/m3.
Solve: The mass of lead mPb = ρ PbVPb = (11,300 kg m3 )( 2.0 m3 ) = 22,600 kg . For water to have the same
mass its volume must be
Vwater =
Assess:
mwater
ρ water
=
22,600 kg
= 22.6 m3
1000 kg m3
Since the lead is 11.3 times as dense we expect the water to take 11.3 times the volume.
16.2. Model: Assume the nucleus is spherical.
Solve:
The volume of the uranium nucleus is
V = 34 π R 3 = 34 π ( 7.5 × 10−15 m ) = 1.767 × 10−42 m3
3
The density of the uranium nucleus is
ρ nucleus =
Assess:
mnucleus
4.0 × 10−25 kg
=
= 2.3 × 1017 kg m3
Vnucleus 1.767 × 10−42 m 3
This density is extremely large compared to the typical density of materials.
16.3. Solve: The volume of the aluminum cube is 10−3 m3 and its mass is
mAl = ρ AlVAl = ( 2700 kg m3 )(1.0 × 10−3 m3 ) = 2.7 kg
The volume of the copper sphere with this mass is
VCu =
m
4π
2.7 kg
3
= 3.027 × 10−4 m3
( rCu ) = Cu =
3
ρ Cu 8920 kg m3
1
⎡ 3 ( 3.027 × 10−4 m3 ) ⎤ 3
⎥ = 0.042 m
⇒ rCu = ⎢
4π
⎢⎣
⎥⎦
The diameter of the copper sphere is 0.0833 m = 8.33 cm.
Assess: The diameter of the sphere is a little less than the length of the cube, and this is reasonable considering
the density of copper is greater than the density of aluminum.
16.4. Model: The volume of a hollow sphere is
V=
Solve:
4p 3
( rout − rin3 )
3
We are given m = 0.690 kg, rout = 0.050 m, and we know that for aluminum r = 2700 kg/m3 .
Solve the above equation for rin .
V
3
rin = 3 rout
−4
3p
m/r
3
= 3 rout
− 4
p
3
= 3 0.050 m3 −
0.690 kg/2700 kg/m3
4
p
3
= 0.040 m
So the inner diameter is 8.0 cm.
Assess: We are happy that the inner diameter is less than the outer diameter, and in a reasonable range.
16.5. Solve: The volume of the aluminum cube V = 8.0 × 10−6 m3 and its mass is
M = ρV = (2700 kg/m3 )(8.0 × 10−6 m3 ) = 0.0216 kg = 21.6 g
One mole of aluminum (27Al) has a mass of 27 g. The number of atoms is
⎛ 6.02 × 1023 atoms ⎞⎛ 1 mol ⎞
23
N =⎜
⎟⎜
⎟ ( 21.6 g ) = 4.8 × 10 atoms
1
mol
27
g
⎠
⎝
⎠⎝
Assess:
This is almost one mole of atoms, which is a reasonable value.
16.6. Solve: The volume of the copper cube is 8.0 × 10−6 m3 and its mass is
M = ρV = (8920 kg/m3 )(8.0 × 10−6 m3 ) = 0.07136 kg = 71.36 g
Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the
cube is
⎛ 1 mol ⎞
n=⎜
⎟ ( 71.36 g ) = 1.1 mol
⎝ 64 g ⎠
Assess:
This answer is in the same ballpark as the previous exercise.
16.7. Solve: (a) The number density is defined as N V , where N is the number of particles occupying a
volume V. Because Al has a mass density of 2700 kg/m3, a volume of 1 m3 has a mass of 2700 kg. We also know
that the molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is
⎛ 1 mol ⎞
5
n = ( 2700 kg ) ⎜
⎟ = 1.00 × 10 mol
⎝ 0.027 kg ⎠
The number of Al atoms in 1.00 × 105 mols is
N = nN A = (1.00 × 105 mol )( 6.02 × 1023 atoms mol ) = 6.02 × 1028 atoms
Thus, the number density is
N 6.02 × 1028 atoms
=
= 6.02 × 1028 atoms m3
V
1 m3
(b) Pb has a mass of 11,300 kg in a volume of 1 m3. Since the atomic mass number of Pb is 207, the number of
moles in 11,300 kg is
⎛ 1 mole ⎞
n = (11,300 kg ) ⎜
⎟
⎝ 0.207 kg ⎠
The number of Pb atoms is thus N = nNA, and hence the number density is
atoms
N nN A ⎛ 11,300 kg ⎞ ⎛
23 atoms ⎞ (1 mol )
=
=⎜
= 3.28 ×1028
⎟⎜ 6.02 ×10
⎟
3
V
V
mol ⎠ 1 m
m3
⎝ 0.207 kg ⎠ ⎝
Assess:
We expected to get very large numbers like this.
16.8. Solve: The mass density is ρ = M/V . The mass M of the sample is related to the number of atoms N
and the mass m of each atom by M = Nm. Combining these, the atomic mass is
m=
M ρV
ρ
1750 kg/m3
=
=
=
= 3.986 × 10−26 kg/atom
N
N
N / V 4.39 × 1028 atoms/m3
the atomic mass in m = A u, where A is the atomic mass number. Thus
A=
m 3.986 × 10−26 kg
=
= 24
1 u 1.661 × 10−27 kg
The element’s atomic mass number is 24.
Assess: This is a reasonable answer for an isotope of neon (although neon is a gas at normal temperatures),
sodium, or magnesium.
4
16.9 Model: Assume the gold is shaped into a solid sphere of volume V = π r 3 .
3
Visualize: We want to know D = 2r. Because the atomic mass number of gold is 197,
1.0 mol of gold has a mass of 197 g or 0.197 kg. Table 16.1 gives ρ = 19,300 kg/m3 .
Solve:
4
3 M
3
0.197 kg
=3
= 0.0135 m = 1.35 cm
V = π r3 = M / ρ ⇒ r = 3
3
4π ρ
4π 19,300 kg/m3
D = 2r = 2(1.35 cm) = 2.70 cm.
Assess: This is about the right size for a chunk that contains one mole of material.
16.10. Solve: The mass of mercury is
⎛ 10−6 m3 ⎞
= 0.136 kg = 136 g
M = ρV = (13,600 kg m3 )(10 cm3 ) ⎜
3 ⎟
⎝ 1 cm ⎠
and the number of moles is
n=
M
0.136 g
=
= 0.6766 mol
M mol 201 g mol
The mass of aluminum with 0.6766 mol of Al is
⎛ 27 g ⎞
M = ( 0.6766 mol ) M mol = ( 0.6766 mol ) ⎜
⎟ = 18.27 g = 0.01827 kg
⎝ mol ⎠
This mass M of aluminum corresponds to a volume of
V=
M
ρ
=
0.01827 kg
= 6.8 × 10−6 m3 = 6.8 cm3
2700 kg m3
Assess: We expected an answer in the same order of magnitude. The size of atoms doesn’t vary as much as the
density of atoms from element to element.
16.11. Solve: The lowest temperature is
TF = 95 TC + 32° ⇒ −127°F = 95 TC + 32° ⇒ TC = −88°C ⇒ Tk = ( −88.3 + 273) K = 185 K
In the same way, the highest temperature is
136°F = 95 TC + 32° ⇒ TC = 58°C = 331 K
Assess:
On the absolute scale the highest recorded temperature is not quite twice the lowest.
16.12. Solve: Let TF = TC = T :
TF = 95 TC + 32° ⇒ T = 95 T + 32° ⇒ T = −40°
That is, the Fahrenheit and the Celsius scales give the same numerical value at −40° .
Assess: It is usually unnecessary to specify the scale when the temperature is reported as − 40° .
16.13. Model: A temperature scale is a linear scale.
Solve: (a) We need a conversion formula for °C to °Z, analogous to the conversion of °C to °F. Since
temperature scales are linear, TC = aTZ + b, where a and b are constants to be determined. We know the boiling
point of liquid nitrogen is 0°Z and –196°C. Similarly, the melting point of iron is 1000°Z and 1538°C. Thus
−196 = 0a + b
1538 = 1000a + b
From the first, b = −196°. Then from the second, a = (1538 + 196)/1000 = 1734/1000. Thus the conversion is
TC = (1734/1000)TZ − 196°. Since the boiling point of water is TC = 100°C, its temperature in °Z is
⎛ 1000 ⎞
TZ = ⎜
⎟ (100° + 196°) = 171°Z
⎝ 1734 ⎠
(b) A temperature TZ = 500°Z is
⎛ 1734 ⎞
TC = ⎜
⎟ 500° − 196° = 671°C = 944 K
⎝ 1000 ⎠
16.14. Solve: (a) The triple point of water is T = 0.01°C and p = 0.006 atm. Thus
TF = 95 TC + 32° = 95 ( 0.01) + 32 = 32.02°F
p = 0.006 atm ×
1.013 × 105 Pa
= 608 Pa
1 atm
(b) The triple point of carbon dioxide is T = −56°C and p = 5 atm. Thus
TF = 95 TC + 32° = 95 ( −56° ) + 32° = −68.8°F
⎛ 1.013 × 105 Pa ⎞
5
p = 5.0 atm = ( 5.0 atm ) ⎜
⎟ = 5.06 × 10 Pa
1
atm
⎝
⎠
16.15. Model: Treat the gas in the container as an ideal gas.
Solve:
From the ideal-gas law pV = nRT , the pressure of the gas is
p=
Assess:
nRT ( 3.0 mol )( 8.31 J mol K ) ⎡⎣( 273 − 120 ) K ⎤⎦
=
= 1.9 × 106 Pa = 19 atm
V
( 2.0 × 10−3 m3 )
19 atm is a high pressure, but not unreasonable.
16.16. Model: Treat the nitrogen gas in the closed cylinder as an ideal gas.
Solve: (a) The density before and after the compression are ρ before = m1 V1 and ρ after = m2 V2 . Noting that
m1 = m2 and V2 = 12 V1 ,
ρ after m V1
=
= 2 ⇒ ρ after = 2 ρ before
ρ before V2 m
The mass density has changed by a factor of 2.
(b) The number of atoms in the gas is unchanged, implying that the number of moles in the gas remains the
same; hence the number density is unchanged.
16.17. Model: Treat the gas in the sealed container as an ideal gas.
Solve:
(a) From the ideal gas law equation pV = nRT , the volume V of the container is
V=
nRT ( 2.0 mol )( 8.31 J mol K ) ⎡⎣( 273 + 30 ) K ⎤⎦
=
= 0.050 m3
p
1.013 × 105 Pa
Note that pressure must be in Pa in the ideal-gas law.
(b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is
p1V p2V
T
( 273 + 130 ) K = 1.3 atm
=
⇒ p2 = p1 2 = (1.0 atm )
T1
T2
T1
( 273 + 30 ) K
Note that gas-law calculations must use T in kelvins.
16.18. Model: Treat the gas as an ideal gas in the sealed container.
Solve:
(a) For p1 = p0 , the before-and-after relationship of an ideal gas in a sealed container is
V1 V0
T
473 K
= ⇒ V1 = V0 1 = V0
= 1.27V0
T1 T0
T0
373 K
where T0 = (273 + 100) K and T1 = (273 + 200) K.
(b) When the Kelvin temperature is doubled, T1 = 2T0 = 2(373 K) = 746 K and the above equation becomes
T
746 K
V1 = V0 1 = V0
= 2V0
T0
373 K
Assess: When we use the Kelvin scale we expect the volume to double when the temperature doubles (if the
pressure is kept constant).
16.19. Model: Treat the air in the compressed-air tank as an ideal gas.
Solve:
(a) From the ideal-gas law pV = nRT , the number of moles n is
pV p (π r h )
=
=
RT
RT
2
n=
⎛ 1.013 × 105 Pa ⎞ ⎡
2
⎟ ⎣π ( 0.075 m ) ( 0.50 m ) ⎤⎦
1
atm
⎝
⎠
= 55.1 mol ≈ 55 mol
( 8.31 J mol K ) ⎣⎡( 273 + 20 ) K ⎦⎤
(150 atm ) ⎜
(b) At STP, the ideal-gas law yields
V=
nRT ( 55.1 mol )( 8.31 J mol K )( 273 K )
=
= 1.234 m3 ≈ 1.2 m3
p
1.013 × 105 Pa
Assess: Because the volume of the compressed air tank is (π r 2 )h = 8.8357 × 10−3 m3 , the volume at STP is 140
times the volume of the tank.
16.20. Model: Treat the oxygen gas in the cylinder as an ideal gas.
Solve:
(a) The number of moles of oxygen is
n=
M
50 g
=
= 1.563 mol ≈ 1.6 mol
M mol 32 g mol
(b) The number of molecules is
N = nN A = (1.563 mol ) ( 6.02 × 1023 mol−1 ) = 9.41× 1023 ≈ 9.4 × 1023
(c) The volume of the cylinder V = π r 2 L = π ( 0.10 m ) ( 0.40 m ) = 1.257 × 10−2 m3. Thus,
2
N
9.41 × 1023
=
= 7.5 × 1025 m −3
V 1.257 × 10−2 m3
(d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be
p=
nRT (1.563 mol )( 8.31 J mol K )( 293 K )
=
= 303 kPa
1.257 × 10−2 m3
V
where we used T = 20°C = 293 K. But a pressure gauge reads gauge pressure:
pg = p − 1 atm = 303 kPa − 101 kPa = 202 kPa ≈ 200 kPa
16.21. Model: Treat the helium gas in the sealed cylinder as an ideal gas.
2
Solve: The volume of the cylinder is V = π r 2 h = π ( 0.05 m ) ( 0.30 m ) = 2.356 × 10−3 m3. The gauge pressure of
the gas is 120 psi ×
1 atm 1.013 × 105 Pa
×
= 8.269 ×105 Pa, so the absolute pressure of the gas is
14.7 psi
1 atm
8.269 ×105 Pa + 1.013 × 105 Pa = 9.282 × 105 Pa. The temperature of the gas is T = (273 + 20) K = 293 K. The
number of moles of the gas in the cylinder is
pV ( 9.282 × 10 Pa )( 2.356 × 10 m )
=
= 0.898 mol
RT
( 8.31 J mol K )( 293 K )
5
n=
−3
3
(a) The number of atoms is
N = nN A = ( 0.898 mol ) ( 6.02 × 1023 mol−1 ) = 5.41× 1023 atoms ≈ 5.4 × 1023 atoms
(b) The mass of the helium is
M = nM mol = ( 0.898 mol )( 4 g mol ) = 3.59 g = 3.59 × 10−3 kg ≈ 3.6 × 10−3 kg
(c) The number density is
N 5.41 × 1023 atoms
=
= 2.3 × 1026 atoms m3
V
2.356 × 10−3 m 3
(d) The mass density is
ρ=
M
3.59 × 10−3 kg
=
= 1.5 kg m3
V 2.356 × 10−3 m3
16.22. Model: The gas is assumed to be ideal and it expands isothermally.
Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2 = T1.
(b) The before-and-after relationship of an ideal gas under isothermal conditions is
⎛V ⎞ p
p1V1 p2V2
V
=
⇒ p2 = p1 1 = p1 ⎜ 1 ⎟ = 1
T1
T1
V2
⎝ 2V1 ⎠ 2
16.23. Model: The gas is assumed to be ideal.
Solve: (a) Isochoric means the volume stays unchanged. That is V2 = V1.
(b) The before-and-after relationship of an ideal gas under isochoric conditions is
⎛1 ⎞
⎜ p1 ⎟ T
p1V1 p2V2
p2
=
⇒ T2 = T1
= T1 ⎜ 3 ⎟ = 1
T1
T1
p1
⎜⎜ p1 ⎟⎟ 3
⎝
⎠
16.24. Model: The gas is assumed to be ideal, and since the container is rigid V2 = V1.
Solve:
Convert both temperatures to the Kelvin scale.
p1V1 p2V2
T
⎛ 283 K ⎞
=
⇒ p2 = p1 2 = 3 atm ⎜
⎟ = 3.1 atm
T1
T1
T1
⎝ 275 K ⎠
Assess:
bit.
On the absolute scale the temperature only went up a little bit, so we expect the pressure to rise a little
16.25. Model: The rigid sphere’s volume does not change, so this is an isochoric process. The air is
assumed to be an ideal gas.
Solve: (a) When the valve is closed, the air inside is at p1 = 1 atm and T1 = 100°C. The before-and-after
relationship of an ideal gas in the closed sphere (constant volume) is
⎛T ⎞
p1V p2V
( 273 + 0 ) K = 0.73 atm
=
⇒ p2 = p1 ⎜ 2 ⎟ = (1.0 atm )
T1
T2
( 273 + 100 ) K
⎝ T1 ⎠
(b) Dry ice is CO2. From Figure 16.4, we can see that the solid-gas transition line gives a temperature of −78°C
when p = 1 atm. Cooling the sphere to –78°C gives
⎛T ⎞
( 273 − 78) K = 0.52 atm
p2 = p1 ⎜ 2 ⎟ = (1.0 atm )
373 K
⎝ T1 ⎠
16.26. Model: Assume the gas is an ideal gas.
Visualize: The pressure in the gas must exert exactly enough upward force to counteract the gravitational force
on the piston; this is true at both temperatures ( p2 = p1 = p ).
The initial volume of the cylinder is V1 = pr 2 h1 = p ( 0.12m ) ( 0.84m ) = 0.038m 3 . The initial temperature is
2
T1 = 303°C + 273 = 576K.
Solve: (a)
F mg ( 20 kg ) ( 9.80m/s )
=
=
= 4333Pa < 4300Pa
2
A pr 2
p ( 0.12m )
2
p=
(b) Knowing the initial temperature, pressure, and volume allows us to compute the number of moles of gas
(which will stay constant).
n=
( 4333Pa ) ( 0.038m3 )
pV1
=
= 0.0344mol < 0.034mol
RT1 ( 8.31J/mol ⋅ K )( 576K )
Now apply the n just found and the new temperature (T2 = 15°C + 273 = 288K) to find V2 .
V2 =
nRT2 ( 0.0344 mol )( 8.31 J/mol ⋅ K )( 288 K )
=
= 0.019 m3
p
4333 Pa
Solve V2 = pr 2 h2 for h2 .
h2 =
0.019m3
V2
=
= 0.42m
pr 2 p ( 0.12m )2
The new height of the piston is 42 cm.
Assess:
Due to a coincidence in the data we could have used the shortcut that T2 = 12 T1 on the absolute scale to
deduce that V2 = 12 V1 1 h2 = 12 h1 .
16.27. Model: In an isochoric process, the volume of the container stays unchanged. Argon gas in the
container is assumed to be an ideal gas.
Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and
T1 = 20°C = 293 K. This produces a pressure
p1 =
nRT ( 0.1 mol )( 8.31 J mol K )( 293 K )
=
= 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa
V1
50 × 10−6 m3
An ideal gas process has p2V2 /T2 = p1V1 / T1. Isochoric heating to a final temperature T2 = 300°C = 573 K has
V2 = V1 , so the final pressure is
p2 =
V1 T2
573
p1 = 1 ×
× 4870 kPa = 9520 kPa ≈ 9500 kPa
V2 T1
293
Note that it is essential to express temperatures in kelvins.
(b)
Assess:
All isochoric processes will be a straight vertical line on a pV diagram.
16.28. Model: The isobaric heating means that the pressure of the argon gas stays unchanged.
Solve:
(a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and
T1 = 20°C = 293 K. This produces a pressure
p1 =
nRT1 ( 0.1 mol )( 8.31 J mol K )( 293 K )
=
= 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa
V1
50 × 10−6 m3
An ideal gas process has p2V2 /T2 = p1V1/T1. Isobaric heating to a final temperature T2 = 300°C = 573 K has
p2 = p1 , so the final volume is
V2 =
p1 T2
573
V1 = 1 ×
× 50 cm3 = 97.8 cm3 ≈ 98 cm3
p2 T1
293
(b)
Assess:
All isobaric processes will be a straight horizontal line on a pV diagram.
16.29. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is
assumed to be an ideal gas.
Solve: (a) The container has only argon inside with n = 0.1 mol,
V1 = 50 cm3 = 50 × 10−6 m3 , and
T1 = 20°C = 293 K. This produces a pressure
p1 =
nRT1 (0.1 mol)(8.31 J/mol K) (293 K)
=
= 4.87 ×106 Pa = 12.02 atm ≈ 12 atm
50 ×10−6 Pa
V1
An ideal-gas process obeys p2V2 /T2 = p1V1/T1. Isothermal expansion to V2 = 200 cm3 gives a final pressure
p2 =
(b)
T2 V1
200
p1 = 1 ×
× 12.02 atm = 48 atm
50
T1 V2
16.30. Model: Assume the gas to be an ideal gas.
Solve: (a) Because the volume stays unchanged, the process is isochoric.
(b) The ideal-gas law p1V1 = nRT1 gives
T1 =
5
−6
3
p1V1 ( 3 × 1.013 × 10 Pa )(100 × 10 m )
=
= 914 K = 641°C
nR
( 0.0040 mol )( 8.31 J mol K )
The final temperature T2 is calculated as follows for an isochoric process:
p1 p2
p
⎛ 1 atm ⎞
=
⇒ T2 = T1 2 = ( 914 K ) ⎜
⎟ = 305 K = 32°C
p1
T1 T2
⎝ 3 atm ⎠
Assess:
All straight vertical lines on a pV diagram represent isochoric processes.
16.31. Model: Assume that the gas is an ideal gas.
Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal.
(b) From the ideal-gas law,
T1 =
5
−6
3
p1V1 ( 3 × 1.013 × 10 Pa )(100 × 10 m )
=
= 914 K = 641°C
nR
( 0.0040 mol )( 8.31 J mol K )
T2 is also 914 K, because the process is isothermal.
(c) The before-and-after relationship of an ideal gas under isothermal conditions is
p1V1 = p2V2 ⇒ V2 = V1
p1
⎛ 3 atm ⎞
3
= (100 cm3 ) ⎜
⎟ = 300 cm
p2
⎝ 1 atm ⎠
16.32. Model: Assume that the gas is ideal.
Solve: (a) Because the process is at a constant pressure, it is isobaric.
(b) For an ideal gas at constant pressure,
V2 V1
V
100 cm3
=
⇒ T2 = T1 2 = ⎡⎣( 273 + 900 ) K ⎤⎦
= 391 K = 118°C
V1
300 cm3
T2 T1
(c) Using the ideal-gas law p2V2 = nRT2 ,
p2V2 ( 3 × 1.013 × 10 Pa )(100 × 10 m )
=
= 9.35 × 10−3 mol
RT2
( 8.31 J mol K )( 391 K )
5
n=
−6
3
Assess: All straight horizontal lines on a pV diagram represent isobaric processes.
16.33. Visualize:
Solve: Suppose we have a 1 m × 1 m × 1 m block of copper of mass M containing N atoms. The atoms are
spaced a distance L apart along all three axes of the cube. There are Nx atoms along the x-edge of the cube, Ny
atoms along the y-edge, and Nz atoms along the z-edge. The total number of atoms is N = Nx Ny Nz . If L is
expressed in meters, then the number of atoms along the x-edge is N x = (1 m)/L. Thus,
13
N=
⎛ 1 m3 ⎞
1 m 1 m 1 m 1 m3
×
×
= 3 ⇒L=⎜
⎟
L
L
L
L
⎝ N ⎠
This relates the spacing between atoms to the number of atoms in a 1-meter cube. The mass of the large cube of
copper is
M = ρ CuV = ( 8920 kg m 3 )(1 m3 ) = 8920 kg
But M = mN , where m = 64 u = 64 × (1.661 × 10−27 kg ) is the mass of an individual copper atom. Thus,
N=
M
8920 kg
=
= 8.39 × 1028 atoms
m 64 × (1.661 × 10−27 kg )
13
⎛ 1 m3 ⎞
⇒ L=⎜
= 2.28 × 10−10 m = 0.228 nm
28 ⎟
8.39
10
×
⎝
⎠
Assess:
This is a reasonable interatomic spacing in a crystal lattice.
16.34. Solve: The volume of the cube associated with each atom is
Vatom = ( 0.227 × 10−9 m ) = 1.1697 × 10−29 m3
3
The volume of a mole of atoms is
V = Vatom N A = (1.1697 × 10−29 m3 )( 6.02 × 1023 mol−1 ) = 7.0416 × 10−6 m3 mol
Thus, the mass of a mole of atoms is
M mol = V ρ = ( 7.0416 m3 mol )( 7950 kg m3 ) = 0.056 kg/mol = 56 g/mol
The atomic mass number of the element is 56.
Assess: This is likely the most common isotope of iron.
1.0 g
.
1.0 cm3
There are 10 protons in each water molecule. 1 mol of water molecules has a mass of 18 g.
16.35. Model: Assume the density of the liquid water is
Visualize:
Solve:
23
⎛ 1.0 g ⎞ ⎛ 1 mol ⎞ ⎛ 6.02 × 10 molecules ⎞ ⎛ 10 protons ⎞
26
1.0L = 1000cm3 ⎜
⎜
⎟⎜
⎜
⎟
⎟
⎟ = 3.3 × 10 protons
3
1 mol
⎝ 1.0 cm ⎠ ⎝ 18 g ⎠⎝
⎠ ⎝ 1 molecule ⎠
Assess: This is an unimaginably large number, but reasonable considering the data.
16.36. Model: Air is an ideal gas. The pressure at sea level is 1 atm.
Solve:
From the ideal gas law pV = NkBT , the number density is
1.013 ×105 Pa
N
p
=
=
= 2.5 × 1025 molecules / m3
V kBT (1.38 ×10−23 J/K)(293 K)
Assess:
Each cubic meter of air holds an unimaginably large number of molecules.
16.37. Model: Assume the gas in the solar corona is an ideal gas.
Solve:
The number density of particles in the solar corona is N V . Using the ideal-gas equation,
pV = NkBT ⇒
N
p
=
V kBT
N
( 0.03 Pa )
=
= 1.1 × 1015 particles m3
−23
V (1.38 × 10 J K )( 2 × 106 K )
Assess:
This answer is a lot smaller than the one in the previous problem.
16.38. Model: Assume the gas in the evacuated volume is an ideal gas.
Solve:
The number density of particles is N V . Using the ideal-gas equation,
pV = NkBT ⇒
N
p
=
V kBT
The pressure is
p = (1 × 10−10 mm of Hg ) ×
⇒
1 atm
1.013 × 105 Pa
×
= 1.33 × 10−8 Pa
760 mm of Hg
1 atm
N
1.33 × 10−8 Pa
=
= 3.3 × 1012 m −3 =3.3 × 106 cm −3
V (1.38 × 10−23 J/K)(293 K)
Assess: Even the best vacuum we can achieve in the laboratory contains 3 million molecules per cubic
centimeter.
16.39. Model: Assume that the gas in the vacuum chamber is an ideal gas.
Solve:
(a) The fraction is
pvacuum chamber 1.0 × 10−10 mm of Hg
=
= 1.3 × 10−13
patmosphere
760 mm of Hg
(b) The volume of the chamber V = π ( 0.20 m ) ( 0.30 m ) = 0.03770 m3 . From the ideal-gas equation
2
pV = NkBT , the number of molecules of gas in the chamber is
N=
−13
5
3
pV (1.32 × 10 )(1.013 × 10 Pa )( 0.03770 m )
=
= 1.2 × 1011 molecules
−23
kBT
1.38
×
10
J
K
293
K
)
(
)(
16.40 Model: Assume the nebula gas is ideal.
Visualize: Use pV = NkBT . We are given N / V = 100 atoms/cm3 = 1× 108 atoms/m3 .
Solve:
p=
NkBT
= (1.0 × 108 atoms/m3 )(1.38 × 10−23 J/K)(7500 K) = 1.0 ×10−11 Pa = 1.0 ×10−16 atm
V
Assess: This is much lower pressure than the best vacuum we can achieve in a laboratory,
but it is a higher pressure than in non-nebula space.
16.41. Model: Assume the air is pure N 2 , with a molar mass of 28g/mol.
Visualize:
We will use pV = nRT both before and after. Our intermediate goal is n1 − n2 . We are given
T2 = T1 = T = 20°C + 273 = 293 K and V2 = V1 = V = pr 2C = p(0.011 m 2 )(2.0 m) = 7.60 × 1024 m3 .
Solve:
We convert the gauge pressures to absolute pressures with
p = pg + 1 atm = pg + 14.7 psi.
p1 = 110 psi + 14.7 psi = 124.7 psi = 860 kPa. p2 = 80 psi + 14.7 psi = 94.7 psi = 653 kPa.
n1 − n2 =
p1V1 p2V2
−
RT1 RT2
V
( p1 − p2 )
RT
7.60 × 10−4 m3
=
(860 kPa − 653 kPa)
(8.31 J/mol ⋅ K)(293 K)
= 0.0646 mol
=
Thus 0.0646 mol of N2 was lost; this is
⎛ 28g ⎞
0.0646mol ⎜
⎟ = 1.8g
⎝ 1mol ⎠
Assess:
The result seems to be a reasonable number.
16.42. Model: The carbon dioxide in the cube is an ideal gas.
Solve:
Using the ideal gas equation and n = M/M mol ,
pV = nRT ⇒ V =
nRT
MRT
=
p
pM mol
The molar mass of CO2 is 44 g/mol or 0.044 kg/mol. Thus,
V=
(10,000kg )(8.31 J mol K )( 273 K ) = 5090 m3
(1.013 × 105 Pa ) ( 0.044 kg mol )
The length of the cube is L = (V )1/3 = 17.2 m.
Assess: This is sobering when you multiply it by the number of people in the industrialized world. Good thing
plants take up CO2 in large quantities.
16.43. Model: Assume the gas is an ideal gas.
Solve:
The before-and-after relationship of an ideal gas is
1
p 3V
p1V1 p2V2
p V
⇒ T2 = T1 2 2 = ( 298 K ) 2 1 1 = 447 K = 174°C
=
p1 V1
T1
T2
p1 V1
16.44. Model: Assume the evaporated water is an ideal gas with a molar mass of 18 g/mol. Assume the
pressure is 1 atm = 101.3 kPa.
Visualize: We are given T = 35°C + 273 = 308 K. n = 1000 g (1 mol/ 18 g ) = 55.6 mol.
Solve:
(a)
pV = nRT 1V =
nRT ( 55.6 mol )( 8.31 J/mol ⋅ K )( 308 K )
=
= 1.4 m3
p
101.3 kPa
(b) In the liquid state r = 1000 kg/m3 .
V=
m
1.0 kg
=
= 0.0010 m3
r 1000 kg/m3
The factor by which the volume of the evaporated water is larger than the liquid water is
1.4 m3
= 1400
0.0010 m3
Assess:
Gases really do take up a lot more volume than the equivalent mass of a liquid.
16.45. Model: Assume that the steam (as water vapor) is an ideal gas.
Solve:
The volume of the liquid water is
V=
m
ρ
=
nM mol
ρ
5
−6
3
⎛ pV ⎞ M mol 20 (1.013 × 10 Pa )(10,000 × 10 m ) ( 0.018 kg mol )
=
=⎜
⎟
⎝ RT ⎠ ρ
(8.31 J mol K )( 473 K ) (1000 kg m3 )
= 9.28 × 10−5 m 3 = 92.8 cm3 ≈ 93 cm3
Assess:
The liquid takes a lot smaller volume than the same number of atoms as a gas.
16.46. Model: Assume that the steam is an ideal gas.
Solve:
V=
(a) The volume of water is
M
ρ
=
nM mol
ρ
pV M mol 50 (1.013 × 10 Pa )( 5.0 m ) ( 0.018 kg mol )
=
= 0.0815 m3 = 81.5 L ≈ 82 L
RT ρ
( 8.31 J mol K )( 673 K ) (1000 kg m3 )
5
=
3
(b) Using the before-and-after relationship of an ideal gas,
⎛ ( 273 + 150 ) K ⎞ ⎛ 50 atm ⎞
p2V2 p1V1
T p
3
3
3
=
⇒ V2 = 2 1 V1 = ⎜
⎟⎜
⎟ ( 5.0 m ) = 78.6 m ≈ 79 m
T2
T1
T1 p2
673
K
2.0
atm
⎝
⎠
⎝
⎠
16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant.
Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using
the before-and-after relationship of an ideal gas for an isochoric (constant volume) process,
p1 p2
T
⎛ 273 + 45 ⎞
=
⇒ p2 = 2 p1 = ⎜
⎟ ( 44.7 psi ) = 49.4 psi
T1
T1 T2
⎝ 273 + 15 ⎠
Your tire gauge will read a gauge pressure pg = 49.4 psi − 14.7 psi = 34.7 psi. ≈ 35 psi.
16.48. Model: The air is assumed to be an ideal gas.
Solve:
At 20°C and 1 atm pressure, the number of moles in the container is
n1 =
5
−3
3
p1V1 (1.013 × 10 Pa )(10 m )
=
= 0.0416 mol
RT1
( 8.31 J mol K )( 293 K )
At 100°C and 1 atm pressure, the number of moles is
n2 =
5
−3
3
p2V2 (1.013 × 10 Pa )(10 m )
=
= 0.0327 mol
RT2
( 8.31 J mol K )( 373 K )
When heated, the pressure will rise as the number of moles remains n1. When opened, the pressure drops to 1 atm
as gas escapes. Thus, the number of moles of air that escape as the container is opened is
n1 − n2 = 0.0416 mol − 0.0327 mol = 0.0089 mol.
16.49. Model: The gas’s pressure does not change, so this is an isobaric process.
Solve:
The triple point of water is 0.01°C or 273.16 K, so T1 = 273.16 K. Because the pressure is a constant,
V1 V2
V
⎛ 1638 mL ⎞
⇒ T2 = T1 2 = ( 273.16 K ) ⎜
=
⎟ = 447.44 K = 174.3°C
V1
T1 T2
⎝ 1000 mL ⎠
16.50. Model: Assume the gas in the manometer is an ideal gas.
Solve:
In the ice-water mixture the pressure is
p1 = patoms + ρ Hg g (0.120 m)
= 1.013 × 105 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.120 m) = 1.173 × 105 Pa
In the freezer the pressure is
p2 = patm + ρ Hg g (0.030 m)
= 1.013 × 105 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.030 m) = 1.053 × 105 Pa
Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total
volume of the gas cell. This means the volume of the chamber can be considered constant. Hence,
⎛p ⎞
p1 p2
1.053 × 105 Pa
=
⇒ T2 = T1 ⎜ 2 ⎟ = ( 273 K )
= 245 K= − 28°C
T1 T2
1.173 × 105 Pa
⎝ p1 ⎠
Assess:
This is a reasonable temperature for an industrial freezer.
16.51. Model: The air in the closed section of the U-tube is an ideal gas.
Visualize: The length of the tube is l = 1.0 m and its cross-sectional area is A.
Solve: Initially, the pressure of the air in the tube is p1 = patmos and its volume is V1 = Al. After the mercury is
poured in, compressing the air, the air-pressure force supports the weight of the mercury. Thus the compressed
pressure equals the pressure at the bottom of the column: p2 = patmos + ρ gL. The volume of the compressed air is
V2 = A(l − L). Because the mercury is poured in slowly, we will assume that the gas remains in thermal
equilibrium with the surrounding air, so T2 = T1. In an isothermal process, pressure and volume are related by
p1V1 = patmos Al = p2V2 = ( patmos + ρ gL) A(l − L)
Canceling the A, multiplying through, and solving for L gives
L=l−
patmos
101,300 Pa
= 1.00 m −
= 0.240 m = 24.0 cm
(13,600 kg/m3 )(9.8 m/s 2 )
ρg
16.52. Model: Assume that the air bubble is always in thermal equilibrium with the surrounding water, and
the air in the bubble is an ideal gas.
Solve: The pressure inside the bubble matches the pressure of the surrounding water. At 50 m deep, the
pressure is
p1 = p0 + ρ water gd = 1.013 × 105 Pa + (1000 kg/m3 )(9.8 m/s 2 )(50 m) = 5.913 × 105 Pa
At the lake’s surface, p2 = p0 = 1.013 × 105 Pa. Using the before-and-after relationship of an ideal gas,
p2V2 p1V1
p T
=
⇒ V2 = V1 1 2
T2
T1
p2 T1
⇒
5
4π 3 4π
3 ⎛ 5.913 × 10 Pa ⎞ ⎛ 293 K ⎞
r2 =
( 0.005 m ) ⎜
⎟⎜
⎟ ⇒ r2 = 0.0091 m
5
3
3
⎝ 1.013 × 10 Pa ⎠ ⎝ 283 K ⎠
The diameter of the bubble is 2r2 = 0.0182 m = 1.82 cm.
Assess: The bubble’s diameter just about doubled; this seems reasonable.
16.53. Model: Assume that the compressed air in the cylinder is an ideal gas. The volume of the air in the
cylinder is a constant.
Solve: Using the before-and-after relationship of an ideal gas,
p2V2 p1V1
T V
⎛ 1223 K ⎞ V1
=
⇒ p2 = p1 2 1 = ( 25 atm ) ⎜
⎟ = 104 atm
T2
T1
T1 V2
⎝ 293 K ⎠ V1
where we have converted to the Kelvin temperature scale. Because the pressure does not exceed 110 atm, the
compressed air cylinder does not blow.
16.54. Model: Assume that the gas is an ideal gas.
Solve:
Assess: For the isothermal process, the pressure must be halved as the volume doubles. This is because p1 is
proportional to 1/V1 for isothermal processes.
16.55. Model: Assume that the gas is an ideal gas.
Solve:
Assess: For the isothermal process, the pressure must double as the volume is halved. This is because p is
proportional to 1/V for isothermal processes.
16.56. Model: Assume that the helium gas is an ideal gas.
Visualize: Process 1 → 2 is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric.
Solve: The number of moles of helium is
n=
M
8.0 g
=
= 2.0 mol
M mol 4 g mol
Using the ideal-gas equation,
V1 =
nRT1 ( 2.0 mol )( 8.31 J mol K ) ⎡⎣( 273 + 37 ) K ⎤⎦
=
= 0.0254 m3 ≈ 0.025 m3
p1
2 (1.013 × 105 Pa )
For the isochoric process V2 = V1 , and
p1 p2
T
⎛ 657 + 273 ⎞
=
⇒ p2 = p1 2 = ( 2 atm ) ⎜
⎟ = 6.0 atm
T1
T1 T2
⎝ 37 + 273 ⎠
For the isothermal process, the equation p3V3 = p2V2 is
V3 = V2
p2
⎛ 6 atm ⎞
3
3
= ( 0.0254 m3 ) ⎜
⎟ = 0.0762 m ≈ 0.076 m
p3
⎝ 2 atm ⎠
For the isothermal process, T3 = T2 = 657°C.
16.57. Model: Assume the nitrogen gas is an ideal gas.
Solve:
(a) The number of moles of nitrogen is
n=
M
1g
⎛ 1 ⎞
=
= ⎜ ⎟ mol
M mol 28 g mol ⎝ 28 ⎠
Using the ideal-gas equation,
p1 =
nRT1 (1 28 mol )( 8.31 J mol K )( 298 K )
=
= 8.84 × 105 Pa = 884 kPa ≈ 880 kPa
V1
(100 × 10−6 m3 )
(b) For the process from state 1 to state 3:
⎛ 1.5 p1 ⎞⎛ 50 cm3 ⎞
p1V1 p3V3
p V
=
⇒ T3 = T1 3 3 = ( 298 K ) ⎜
= 223.5 K ≈ −49°C
⎟⎜
3 ⎟
T1
T3
p1 V1
⎝ p1 ⎠ ⎝ 100 cm ⎠
For the process from state 3 to state 2:
⎛ p ⎞⎛ V ⎞
⎛ 2.0 p1 ⎞⎛ 100 cm3 ⎞
p2V2 p3V3
=
⇒ T2 = T3 ⎜ 2 ⎟⎜ 2 ⎟ = ( 223.5 K ) ⎜
= 596 K = 323°C
⎟⎜
3 ⎟
T2
T3
⎝ 1.5 p1 ⎠ ⎝ 50 cm ⎠
⎝ p3 ⎠⎝ V3 ⎠
For the process from state 1 to state 4:
⎛ 1.5 p1 ⎞⎛ 150 cm3 ⎞
p4V4 p1V1
p V
⇒ T4 = T1 4 4 = ( 298 K ) ⎜
= 670.5 K ≈ 398°C
=
⎟⎜
3 ⎟
p1 V1
T4
T1
⎝ p1 ⎠ ⎝ 100 cm ⎠
16.58. Model: The gas is an ideal gas.
Solve:
(a) Using the ideal-gas equation,
T1 =
(1.0 × 105 Pa )( 2.0 m3 ) = 301 K
p1V1
=
nR ( 80 mol )( 8.31 J mol K )
Because points 1 and 2 lie on the isotherm, T2 = T1 = 301 K. The temperature of the isothermal process is 301 K.
(b) The straight-line process 1 → 2 can be represented by the equation
p = (3 − V ) × 105
where V is in m3 and p is in Pa. We can use the ideal gas law to find that the temperature along the line varies as
T=
pV
105
= (3V − V 2 ) ×
nR
nR
We can maximize T by setting the derivative dT/dV to zero:
dT
105
3
2
)×
= (3 − 2Vmax
= 0 ⇒ Vmax = m3 = 1.50 m3
dV
2
nR
At this volume, the pressure is pmax = 1.5 × 105 Pa and the temperature is
Tmax =
pmaxVmax (1.50 × 105 Pa)(1.5 m3 )
=
= 338 K
nR
(80 mol)(8.31 J/mol K)
16.59. Model: Assume the gas is an ideal gas.
Solve:
units:
(a) We can find the temperatures directly from the ideal-gas law after we convert all quantities to SI
T1 =
p1V1 ( 3.0 atm × 101,300 Pa atm ) (1000 cm × 10
=
nR
( 0.10 mol )(8.31 J mol K )
T2 =
p2V2 (1.0 atm × 101,300 Pa atm ) ( 3000 cm × 10
=
nR
( 0.10 mol )(8.31 J mol K )
3
3
−6
−6
m3 cm3 )
m3 cm3 )
= 366 K = 93°C
= 366 K = 93°C
(b) T2 = T1 , so this is an isothermal process.
(c) A constant volume process has V3 = V2 . Because p1 = 3 p2 , restoring the pressure to its original value means
that p3 = 3p2. From the ideal-gas law,
⎛ p ⎞⎛ V ⎞
p3V3 p2V2
=
⇒ T3 = ⎜ 3 ⎟⎜ 3 ⎟ T2 = 3 × 1 × T2 = 3 × 366 K = 1098 K = 825°C
T3
T2
⎝ p2 ⎠⎝ V2 ⎠
16.60. Model: Assume the gas is an ideal gas.
Solve:
(a) Using the ideal-gas law,
T1 =
5
−6
3
p1V1 (1.013 × 10 Pa )(100 × 10 m )
=
= 244 K = −29°C
nR ( 5.0 × 10−3 mol ) ( 8.31 J mol K )
(b) Using the before and after relationship of an ideal gas,
3
p1V1 p2V2
T V
⎛ 2926 K ⎞ ⎛ 100 cm ⎞
=
⇒ p2 = p1 2 1 = (1 atm ) ⎜
= 4.0 atm
⎟⎜
3 ⎟
T1
T2
T1 V2
⎝ 244 K ⎠ ⎝ 300 cm ⎠
(c) Using the before and after relationship of an ideal gas,
p3V3 p2V2
p T
⎛ 4 atm ⎞⎛ 2438 K ⎞
3
3
⇒ V3 = 2 3 V2 = ⎜
=
⎟⎜
⎟ ( 300 cm ) = 500 cm
p3 T2
T3
T2
⎝ 2 atm ⎠⎝ 2926 K ⎠
16.61. Model: We assume the oxygen gas is ideal.
Visualize: From the figure we glean p2 = 3 p1 and V2 = 3V1 . We are given T1 = 20°C + 273 = 293K.
Solve: Use the before-and-after version of the ideal-gas law.
p1V1 p2V2
=
T1
T2
T2 =
Assess:
( 3 p1 )( 3V1 ) T = 9T = 9 293K = 2637 K = 2364°C
p2V2
T1 =
(
)
1
1
p1V1
p1V1
This is a hot temperature—higher than the melting point of many elements.
16.62. Model: Assume CO2 gas is an ideal gas.
Solve:
(a) The molar mass for CO2 is M mol = 44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes
0.227 mol of gas at 0°C. With V1 = 10,000 cm3 = 0.010 m3 and T1 = 0°C = 273 K, the pressure is
p1 =
nRT1 ( 0.2273 mol )( 8.31 J mol K )( 273 K )
=
= 5.156 × 104 Pa = 0.509 atm ≈ 0.50 atm
V1
0.010 m3
(b) From the isothermal compression,
p2V2 = p1V1 ⇒ V2 = V1
p1
⎛ 0.509 atm ⎞
−3
3
3
= ( 0.010 m3 ) ⎜
⎟ = 1.70 × 10 m = 1700 cm
p2
⎝ 3.0 atm ⎠
From the isobaric compression,
T3 = T2
(c)
⎛ 1000 cm 3 ⎞
V3
= ( 273 K ) ⎜
= 161 K = −112°C
3 ⎟
V2
⎝ 1700 cm ⎠
16.63. Model: The gas in the container is assumed to be an ideal gas.
Solve:
(a) The gas starts at pressure p1 = 2.0 atm, temperature T1 = 127°C = (127 + 273) K = 400 K and
volume V1. It is first compressed at a constant temperature T2 = T1 until V2 = 12 V1 and the pressure is p2. It is then
further compressed at constant pressure p3 = p2 until V3 = 12 V2 . From the ideal-gas law,
p2V2 p1V1
V T
V
=
⇒ p2 = p1 1 2 = ( 2.0 atm ) 1 1 × 1 = 4.0 atm
T2
T1
V2 T1
V
2 1
Note that T2 = T1 = 400 K. Using the ideal-gas law once again,
1
V
p3V3 p2V2
V p
⇒ T3 = T2 3 3 = ( 400 K ) 2 2 × 1 = 200 K = −73°C
=
V2 p2
V2
T3
T2
The final pressure and temperature are 4.0 atm and –73°C.
(b)
16.64. Model: Assume that the nitrogen gas is an ideal gas.
Solve:
(a) The molar mass of N2 gas is 28 g/mol. The number of moles is n = (5 g)/(28 g / mol) = 0.1786 mol.
The initial conditions are p1 = 3.0 atm and T1 = 293 K. We use the ideal gas law to find the initial volume as
follows:
V1 =
nRT1 ( 0.1786 mol )( 8.31 J mol K )( 293 K )
=
= 1.430 × 10−3 m3 = 1430 cm 3 ≈ 1400 cm3
p1
3.0 atm × 101,300 Pa atm
An isobaric expansion until the volume triples results in V2 = 3V1 = 4290 cm3 .
(b) After the expansion,
p2V2 p1V1
p V
=
⇒ T2 = 2 2 T1 = 1 × 3 × T1 = 3T1 = 879 K = 606°C
T2
T1
p1 V1
(c) A constant volume decrease at V3 = V2 = 4290 cm3 back to T3 = T1 = 13 T2 results in the following:
p3V3 p2V2
T V
1
1
=
⇒ p3 = 3 2 p2 = × 1 × p2 = × 3.0 atm = 1.0 atm
T3
T2
T2 V3
3
3
(d) An isothermal compression at T4 = T3 back to the initial volume V4 = V1 = 13 V3 results in the following:
p4V4 p3V3
T V
1
=
⇒ p4 = 4 3 p3 = 1 × 1 × p3 = 3 × 1.0 atm = 3.0 atm
T4
T3
T3 V4
3
(e)
16.65. Solve: (a) A gas is compressed isothermally from a volume 300 cm3 at 2 atm to a volume of 100 cm3.
What is the final pressure?
(b)
(c) The final pressure is p2 = 6 atm.
16.66. Solve: (a) A gas at 400°C and 500 kPa is cooled at a constant volume to a pressure of 200 kPa. What
is the final temperature in C°?
(b)
(c) The final temperature is T2 = −3.8°C.
16.67. Solve: (a) A gas expands at constant pressure from 200 cm3 at 50°C until the temperature is 400°C.
What is the final volume?
(b)
(c) The final volume is V2 = 417 cm3 .
16.68. Solve: (a) 0.12 g of neon gas at 2.0 atm and 100 cm3 expands isobarically to twice its initial volume.
What is the final temperature of the gas?
(b)
(c) The number of moles of neon is
n=
M
0.12 g
=
= 0.006 mol
M mol
20 g
The initial temperature is
T1 =
5
−4
3
p1V1 2 (1.013 × 10 Pa )(1.0 × 10 m )
=
= 406 K
nR
( 0.006 mol )(8.31 J mol K )
The final temperature is
T2 =
p2 V2
p 2V1
T1 =
( 406 K ) = 812 K
p1 V1
p V1
16.69. Model: Assume that the compressed air is an ideal gas.
Solve:
(a) Because the piston is floating in equilibrium,
Fnet = ( p1 − patoms ) A − w = 0 N
where the piston’s cross-sectional area A = π (r 2 ) = π (0.050 m) 2 = 7.854 ×10−3 m 2 and the piston’s weight
w = (50 kg)(9.8) = 490 N. Thus,
p1 =
w
490 N
+ patmos =
+ 1.013 × 105 Pa=1.637 × 105 Pa
A
7.854 × 10−3 m 2
Using the ideal-gas equation p1V1 = nRT1 ,
(1.637 × 105 Pa)Ah1 = (0.12 mol)(8.31 J/mol K)[(273 + 30) K]
With the value of A given above, this equation yields h1 = 0.235 m = 23.5 cm.
(b) When the temperature is increased from T1 = 303 K to T2 = (303 + 100) K = 403 K, the volume changes
from V1 = Ah1 to V2 = Ah2 at a constant pressure p2 = p1. From the before-and-after relationship of the ideal gas:
p1 ( Ah1 ) p2 ( Ah2 )
=
T1
T2
⇒ h2 =
p1 T2
⎛ 403 K ⎞
h1 = (1) ⎜
⎟ ( 0.235 m ) = 0.313 m = 31.3 cm
p2 T1
⎝ 303 K ⎠
Thus, the piston moves h2 − h1 = 7.8 cm.
16.70. Model: The air in the diving bell is an ideal gas.
Visualize:
Solve:
(a) Initially p1 = p0 (atmospheric pressure), V1 = AL, and T1 = 293 K. When the diving bell is
submerged to d = 100 m at the bottom edge, the water comes up height h inside. The volume is V2 = A( L − h)
and the temperature is T2 = 283 K. Like a barometer, the pressure at points a and b must be the same. Thus
p2 + ρ gh = p0 + ρ gd , or p2 = p0 + ρ g (d − h). Using the before and after relationship of an ideal gas,
p1V1 p2V2
p AL [ p0 + ρ g (d − h)] A( L − h)
=
⇒ 0
=
T1
T2
293 K
283 K
Multiplying this out gives the following quadratic equation for h:
⎛
⎝
ρ gh 2 − [ p0 + ρ g (d + L)]h + ⎜1 −
283 ⎞
⎟ p0 L + ρ gLd = 0
293 ⎠
Inserting the known values (using ρ = 1030 kg/m3 for seawater) and dividing by ρ g gives
h 2 − 113.04h + 301.03 = 0 ⇒ h = 110 m or 2.7 m
The first solution is not physically meaningful, so the water rises to height h = 2.7 m.
(b) To expel all the water, the air pressure inside the bell must be increased to match the water pressure at the
bottom edge of the bell, d = 100 m. The necessary pressure is
p = p0 + ρ gd = 101,300 Pa + (1030 kg/m3 )(9.8 m/s 2 )(100 m) = 1111 kPa = 10.96 atm ≈ 11 atm
16.71. Model: Assume the trapped air to be an ideal gas.
Visualize:
Initially, as the pipe is touched to the water surface and the gas inside is thus closed off from the air, the pressure
p1 = patmos = 1 atm and the volume is V1 = L1 A, where A is the cross-sectional area of the pipe. By pushing the
pipe in slowly, the gas temperature in the pipe remains the same as the water temperature. Thus, this is an
isothermal compression of the gas with T2 = T1.
Solve: From the ideal-gas law,
p2V2 = p1V1 ⇒ p2 L2 A = p1L1 A ⇒ p2 L2 = patmos L1 ⇒ p2 = patmos ( L1 L2 )
As the pipe is pushed down, the increasing water pressure pushes water up into the pipe, compressing the air. In
equilibrium, the pressure at points a and b, along a horizontal line, must be equal. (This is like the barometer. If
the pressures at a and b weren’t equal, the pressure difference would cause the liquid level in the pipe to move up
or down.) The pressure at point a is just the gas pressure inside the pipe: pa = p2 . The pressure at point b is the
pressure at depth L2 in water: pb = patmos + ρ gL2 . Equating these gives
p2 = patmos = ρ gL2
Substituting the expression for p2 from the ideal-gas equation above, the pressure equation becomes
patmos L1
= patmos + ρ gL2 ⇒ ρ gL22 + patmos L2 − patmos L1 = 0
L2
This is a quadratic equation for L2 with solutions
L2 =
− patmos ±
( patmos ) + 4 ρ gpatmos L1
2
2ρ g
Length has to be a positive quantity, so the one physically acceptable solution is
L2 =
−101,300 Pa +
(101,300 Pa ) + 4 (1000 kg m3 )( 9.8 m s 2 ) (101,300 Pa )( 3.0 m )
2
2 (1000 kg m3 )( 9.8 m s 2 )
= 2.4 m
16.72. Model: The gas in the cylinder is assumed to be an ideal gas.
Solve:
The gas at T1 = 20°C = 293 K has a pressure p1 = 1 atm and a volume V1 = AL0 . (The pressure has to be
1 atm to balance the external force of the air on the piston.) At a temperature T2 = 100°C = 373 K, its volume
becomes V2 = A( L0 + Δx) and its pressure increases to
p2 = p1 +
F
k Δx
= p1 +
A
A
where F = k Δx is the spring force on the piston. Using the before-and-after relationship of an ideal-gas equation,
p1V1 p2V2
p L A ( p + k Δx A )( L0 + Δx ) A
=
⇒ 1 0 = 1
T1
T2
T1
T2
⇒
T2 ( p1 + k Δx A )( L0 + Δx ) ⎛ k Δx ⎞ ⎛ Δx ⎞
=
= ⎜1 +
⎟
⎟ ⎜1 +
T1
p1L0
p1 A ⎠⎝
L0 ⎠
⎝
L0 can be obtained from the ideal-gas law as follows:
L0 =
V1 ⎛ nRT1 ⎞ 1 ( 0.004 mol )( 8.31 J mol K )( 293 K )
=⎜
= 0.0961 m = 9.61 cm
⎟ =
A ⎝ p1 ⎠ A
(1.013 ×105 Pa )(10 ×10−4 m2 )
Substituting this value of L0 and the values for p1, T1, T2, and k, the above equation can be simplified to
154.02Δx 2 + 25.210Δx − 0.2730 = 0 ⇒ Δx = 0.0102 m = 1.02 cm
The spring is compressed by 1.0 cm.
16.73. Model: The gas in containers A and B is assumed to be an ideal gas.
Solve:
(a) The number of moles of gas in containers A and B can be expressed as follows:
( 5.0 ×10 Pa ) ( 4VA ) = 5000 VA
pAVA (1.0 × 10 Pa )VA
V
pV
=
= 333.3 A nB = B B =
RTA
R ( 300 K )
R
RTB
R ( 400 K )
R
5
nA =
5
where nA and nB are expressed in moles. Let n′A and n′B be the number of moles after the valve is opened. Since
the total number of moles is the same,
V
nA + nB = 5333.3 A = nA′ + nB′
R
The pressure is now the same in both containers:
p′A =
⇒ nA′ = nB′
n′A RTA
n′ RT
= p′B = B B
VA
VB
TB VA
⎛ 400 K ⎞ VA
= n′B ⎜
⇒ 3n′A = n′B
⎟
TA VB
⎝ 300 K ⎠ 4VA
Solving the above equations,
V
V
V
5333.3 A = n′A + 3n′A = 4n′A ⇒ nA′ = 1333.3 A ⇒ n′B = 4000 A
R
R
R
The new pressure is
pA′ =
n′A RTA 1333.3 ⎛ VA ⎞
5
=
⎜ ⎟ RTA = 4.0 × 10 Pa
VA
VA ⎝ R ⎠
pB′ =
nB′ RTB 4000 ⎛ VA ⎞
5
=
⎜ ⎟ RTB = 4.0 × 10 Pa
VB
VB ⎝ R ⎠
Gas flows from B to A until the pressure is 4.0 × 105 Pa.
(b) The change is irreversible because opening the valve is like breaking a membrane. It is not a quasi-static
process.
16.74 Model: Assume the gas in the left chamber is ideal.
Visualize: For the piston to be in equilibrium ( Fx ) net = 0. The initial volume is V0 = L0 A.
Solve:
(a) The force to the right on the piston by the gas in the left chamber is simply p0 A. . The
magnitude of the force to the left on the piston by the spring is k (ΔL).
p A
( Fx ) net = p0 A − k (ΔL) ⇒ ΔL = 0
k
(b) When the piston is moved x to the right the spring is compressed a total of ΔL + x and the
pressure in the left chamber changes to p1 according to the ideal gas law. When the temperature
doesn’t change the ideal gas equation reduces to p0V0 = p1V1 , where V0 = L0 A and V1 = ( L0 + x) A .
Therefore
pL
p1 = 0 0
( L0 + x)
Now, similar to part (a) above,
( Fx ) net = p1 A − k (ΔL + x)
Substitute in ΔL from part (a) and p1 from above.
( Fx ) net =
p0 L0
⎛p A
⎞
A− k ⎜ 0 + x⎟
( L0 + x)
⎝ k
⎠
Now it is time to apply the binomial approximation: ( L0 + x) −1 ≈ 1/ L0 (1 − x / L0 ) for x << L0 .
( Fx ) net = ( p0 L0 ) (1/ L0 (1 − x / L0 ) ) A − p0 A − kx
= p0 A − p0 Ax / L0 − p0 A − kx
= − ( p0 A / L0 + k ) x
This is clearly a linear restoring force of the form of Hooke’s law where the usual k is replaced
with ( p0 A / L0 + k ) . That is, the gas in the left chamber is increasing the k of the system (by
p0 A / L0 ) and making it act like a stiffer spring.
(c) Since we have a Hooke’s law for which we know the modified k, we simply replace k in the
period equation for harmonic oscillators with the modified constant.
M
T = 2π
p0 A / L0 + k
Assess: This is simple harmonic motion when the approximation x << L0 is valid.
16-1
17.1. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve.
Visualize: The gas is compressing, so we expect the work to be positive.
Solve: The work done on the gas is
W = − ∫ p dV = − ( area under the pV curve )
(
)
= − − ( 200 cm3 ) ( 200 kPa ) = ( 200 × 10−6 m3 )( 2.0 × 105 Pa ) = 40 J
Assess: The area under the curve is negative because the integration direction is to the left. Thus, the
environment does positive work on the gas to compress it.
17.2. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve.
Visualize: The gas is expanding, so we expect the work to be negative.
Solve: The area under the pV curve is the area of the rectangle and triangle. We have
( 200 × 10 m )( 200 × 10 Pa ) + ( 200 × 10 m )( 200 × 10 Pa ) = 60 J
−6
3
3
1
2
−6
3
Thus, the work done on the gas is W = −60 J.
Assess: The environment does negative work on the gas as it expands.
3
17.3. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve.
Solve:
The work done on gas in an isobaric process is
W = − pΔV = − p (Vf − Vi )
Substituting into this equation,
80 J = − ( 200 × 103 Pa ) (V1 − 3V1 ) ⇒ Vi = 2.0 × 10−4 m3 = 200 cm3
Assess:
The work done to compress a gas is positive.
17.4. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes.
Solve:
(a) Since the pressure ( pi = pf = p ) is constant the work done is
Won gas = − pΔV = − p(Vf − Vi ) = −
nRTi
(Vf − V i )
Vi
= −( 0.10 mol )( 8.31 J mol K )( 573 K )
(1000 cm − 2000 cm ) = 240 J
3
3
2000 cm3
(b) For compression at a constant temperature,
Won gas = −nRT ln (Vf Vi )
⎛ 1000 × 10−6 m3 ⎞
= − ( 0.10 mol )( 8.31 J mol K )( 573 K ) ln ⎜
= 330 J
−6
3 ⎟
⎝ 2000 × 10 m ⎠
(c) For the isobaric case,
p=
nRTi
= 2.38 × 105 Pa
Vi
For the isothermal case, pi = 2.38 × 105 Pa and the final pressure is
pf =
nRTf
= 4.76 × 105 Pa
Vf
17.5.
Visualize:
Solve:
Because W = − ∫ p dV and this is an isochoric process, W = 0 J. The final point is on a higher isotherm
than the initial point, so Tf > Ti . Heat energy is thus transferred into the gas (Q > 0) and the thermal energy of
the gas increases ( Eth f > Eth i ) as the temperature increases.
17.6.
Visualize:
Solve:
Because this is an isobaric process W = − ∫ pdV = − p (Vf − Vi ) . Since Vf is smaller than Vi, W is positive.
That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, Tf < Ti . In other
words, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be larger
than the work done.
17.7.
Visualize:
Solve:
Because the process is isothermal, ΔEth = Eth f − Eth i = 0 J. According to the first law of
thermodynamics, ΔEth = W + Q. This can only be satisfied if W = −Q. W is positive because the gas is
compressing, hence Q is negative. That is, heat energy is removed from the gas.
17.8.
Visualize:
Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the
environment. That is, Q = 0 J. According to the first law of thermodynamics, the work done on a gas in an
adiabatic process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas
is compressed.
17.9. Solve: The first law of thermodynamics is
ΔEth = W + Q ⇒ −200 J = 500 J + Q = Q ⇒ Q = −700 J
The negative sign means a transfer of energy from the system to the environment.
Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger in
magnitude than W so that Eth f < Eth i .
17.10. Solve: This is an isobaric process. W > 0 because the gas is compressed. This transfers energy into
the system. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics is
ΔEth = W + Q = − pΔV + Q = −(4.0 × 105 Pa)(200 − 600) × 10−6 m3 − 100 J = 60 J
Thermal energy increases by 60 J.
17.11. Model: The removal of heat from the ice reduces its thermal energy and its temperature.
Solve:
The heat needed to change an object’s temperature is Q = McΔT. The mass of the ice cube is
M = ρiceV = (920 kg/m3 )(0.06 × 0.06 × 0.06)m3 = 0.1987 kg
The specific heat of ice from Table 17.2 is cice = 2090 J/kg K, so
Q = (0.199 kg)(2090 J/kg K)(243 K − 273 K) = −12,460 J ≈ 12,000 J
Thus, the energy removed from the ice block is 12,000 J.
Assess: The negative sign with Q means loss of energy.
17.12. Model: The spinning paddle wheel does work and changes the water’s thermal energy and its
temperature.
Solve: (a) The temperature change is ΔT = Tf − Ti = 25°C − 21°C = 4 K. The mass of the water is
M = (200 × 10−6 m3 )(1000 kg/m3 ) = 0.20 kg
The work done is
W = ΔEth = Mcwater ΔT = (0.20 kg)(4190 J/kg K)(4 K) = 3350 J ≈ 3400 J
(b) Q = 0. No energy is transferred between the system and the environment because of a difference in
temperature.
17.13. Model: Heating the mercury at its boiling point changes its thermal energy without a change in
temperature.
Solve: The mass of the mercury is M = 20 g = 2.0 × 10−2 kg, the specific heat cmercury = 140 J/kg K, the boiling
point Tb = 357°C, and the heat of vaporization Lv = 2.96 × 105 J/kg. The heat required for the mercury to change
to the vapor phase is the sum of two steps. The first step is
Q1 = Mcmercury ΔT = (2.0 × 10−2 kg)(140 J/kg K)(357°C − 20°C) = 940 J
The second step is
Q2 = MLV = (2.0 ×10−2 kg)(2.96 × 105 J/kg) = 5920 J
The total heat needed is 6860 J.
17.14. Model: Heating the mercury changes its thermal energy and its temperature.
Solve:
(a) The heat needed to change the mercury’s temperature is
Q = McHg ΔT ⇒ ΔT =
Q
100 J
=
= 35.7 K ≈ 35°C
McHg ( 0.020 kg )(140 J kg K )
(b) The amount of heat required to raise the temperature of the same amount of water by the same number of
degrees is
Q = Mcwater ΔT = (0.020 kg)(4190 J/kg K)(35.7 K) = 3000 J
Assess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat
for mercury. This explains why Qwater > Qmercury.
17.15. Model: Changing ethyl alcohol at 20°C to solid ethyl alcohol at its melting point requires two steps:
lowering its temperature from 20°C to −114°C, then changing the ethyl alcohol to its solid phase at −114°C.
Solve: The change in temperature is −114°C − 20°C = −134°C = −134 K. The mass is
M = ρV = ( 789 kg/m3 )( 200 × 10−6 m3 ) = 0.1578 kg
The heat needed for the two steps is
Q1 = Mcalcohol ΔT = (0.1578 kg)(2400 J/kg K)(−134 K) = −5.07 × 104 J
Q2 = − MLf = −(0.1578 kg)(1.09 × 105 J/kg) = −1.72 × 104 J
The total heat required is
Q = Q1 + Q2 = −6.79 × 104 J ≈ −6.8 × 104 J
Thus, the minimum amount of energy that must be removed is 6.8 × 104 J.
Assess: The negative sign with Q indicates that 6.8 × 104 J will be removed from the system.
17.16. Model: Changing solid lead at 20°C to liquid lead at its melting point (Tm = 328°C) requires two
steps: raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm.
Solve: The equation for the total heat is
Q = Q1 + Q2 ⇒ 1000 J = Mclead (Tf − Ti ) + MLf
⇒ 1000 J = M (128 J/kg K)(328 − 20) K + M (0.25 × 105 J/kg) ⇒ M =
The maximum mass of lead you can melt with 1000 J of heat is 15.5 g.
1000 J
= 15.5 g
( 64,424 J/kg)
17.17. Model: We have a thermal interaction between the copper pellets and the water.
Solve:
The conservation of energy equation Qc + Ww = 0 is
Mc cc (Tf − 300°C) + Mw cw (Tf − 20°C) = 0 J
Solving this equation for the final temperature Tf gives
Tf =
=
Mc cc (300°C) + Mw cw (20°C)
Mc cc + Mw cw
(0.030 kg)(385 J/kg K) (300°C) + (0.10 kg)(4190 J/kg K)(20°C)
= 28°C
(0.030 kg)(385 J/kg K) + (0.10 kg)(4190 J/kg K)
The final temperature of the water and the copper is 28°C.
17.18. Model: We have a thermal interaction between the copper block and water.
Solve:
The conservation of energy equation Qcopper + Qwater = 0 J is
M copper ccopper (Tf − Ti copper ) + M water cwater (Tf − Ti water ) = 0 J
Both the copper and the water reach the common final temperature Tf = 25.5°C. Thus
M copper (385 J/kg K)(25.5°C − 300°C) + (1.00 × 10−3 m3 )(1000 kg/m3 )(4190 J/kg K)(25.5°C − 20°C) = 0 J
⇒ M copper = 0.218 kg
17.19. Model: We have a thermal interaction between the thermometer and the water.
Solve:
The conservation of energy equation Qthermo + Qwater = 0 J is
M thermocthermo (Tf − (Ti ) thermo ) + M water cwater (Tf − (Ti ) water ) = 0 J
The thermometer slightly cools the water until both have the same final temperature Tf = 71.2°C. Thus
(0.050 kg)(750 J/kg K)(71.2°C − 20.0°C) + (200 × 10−6 m 3 )(1000 kg/m 3 )(4190 J/kg K)(71.2° C − Ti water )
= 1920 J + 838 (J/K)(71.2°C − Ti water ) = 0 J
⇒ Ti water = 73.5°C
Assess:
The thermometer reads 71.2°C for a real temperature of 73.5°C. This is reasonable.
17.20. Model: We have a thermal interaction between the aluminum pan and the water.
Solve:
The conservation of energy equation QAl + Qwater = 0 J is
M AlcAl (Tf − Ti A1 ) + M water cwater (Tf − Ti water )
The pan and water reach a common final temperature Tf = 24.0°C
(0.750 kg)(900 J/kg K)(24.0°C − Ti Al ) + (10.0 × 10 −3 m 3 )(1000 kg/m 3 )(4190 J/kg K)(24.0°C − 20.0°C )
= (675.0 J/K)(24.0°C − Ti Al ) + 167,600 J = 0 J
⇒ Ti Al = 272°C = [(272)(9/5) + 32]°F = 522°F
17.21. Model: We have a thermal interaction between the metal sphere and the mercury.
Solve:
The conservation of energy equation Qmetal + QHg = 0 J is
M metalcmetal (Tf − Ti metal ) + M Hg cHg (Tf − Ti Hg ) = 0 J
The metal and mercury reach a common final temperature Tf = 99.0°C. Thus
(0.500 kg)cmetal (99°C − 300°C) + (300 × 10−6 m3 )(13,600 kg/m3 )(140 J/kg K)(99°C − 20°C) = 0 J
We find that cmetal = 449 J kg K. The metal is iron.
17.22. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on the
Kelvin scale is the same as the change in temperature on the Celsius scale.
Solve: (a) The atomic mass number of argon is 40. That is, M mol = 40 g/mol. The number of moles of argon
gas in the container is
n=
M
1.0 g
=
= 0.025 mol
M mol 40 g mol
The amount of heat is
Q = nCV ΔT = (0.025 mol)(12.5 J/mol K)(100°C) = 31.25 J ≈ 31 J
(b) For the isobaric process Q = nCP ΔT becomes
31.25 J = (0.025 mol)(20.8 J/mol K) ΔT ⇒ ΔT = 60°C
17.23. Model: The heating processes are isobaric and isochoric. O2 is a diatomic ideal gas.
Solve:
(a) The number of moles of oxygen is
n=
M
1.0 g
=
= 0.03125 mol
M mol 32 g mol
For the isobaric process,
Q = nCp ΔT = (0.03125 mol)(29.2 J/mol K)(100°C) = 91.2 J ≈ 91 J
(b) For the isochoric process,
Q = nCV ΔT = 91.2 J = (0.03125 mol)(20.9 J/mol K) ΔT ⇒ ΔT = 140°C
17.24. Model: The heating is an isochoric process.
Solve:
The number of moles of helium is
n=
M
2.0 g
=
= 0.50 mol
M mol 4 g mol
For the isochoric processes,
QHe = nCV ΔT = ( 0.50 mol )(12.5 J mol K ) ΔT
⎛
⎞
M
QO2 = nCV ΔT = ⎜
⎟ ( 20.9 J mol K ) ΔT
32
g
mol
⎝
⎠
Because QHe = QO2 ,
⎛
⎞
M
⎟ ( 20.9 J mol K ) ⇒ M = 9.6 g
32
g
mol
⎝
⎠
( 0.50 mol )(12.5 J mol K ) = ⎜
17.25. Model: The gas is an ideal gas that is subjected to an adiabatic process.
Solve:
(a) For an adiabatic process,
γ
γ
pf Vfγ = pV
i i ⇒
pf ⎛ Vi ⎞
ln(2.5)
γ
= ⎜ ⎟ ⇒ 2.5 = ( 2.0 ) ⇒ γ =
= 1.32
pi ⎝ Vf ⎠
ln(2.0)
(b) Equation 17.39 for an adiabatic process is
γ −1
⇒
Tf Vfγ −1 = TV
i i
Tf ⎛ Vi ⎞
=⎜ ⎟
Ti ⎝ Vf ⎠
γ −1
= ( 2.0 )
1.32 −1
= ( 2.0 )
0.32
= 1.25
17.26. Model: We assume the gas is an ideal gas and γ = 1.40 for a diatomic gas.
Solve:
Using the ideal-gas law,
Vi =
nRTi ( 0.10 mol )( 8.31 J mol K )( 423 K )
=
= 1.157 × 10−3 m3
pi
( 3 × 1.013 × 105 Pa )
For an adiabatic process,
γ
γ
pV
i i = pf Vf
1γ
⎛p ⎞
⇒ Vf = Vi ⎜ i ⎟
⎝ pf ⎠
1 1.40
⎛ pi ⎞
= (1.157 × 10 m ) ⎜
⎟
⎝ 0.5 pi ⎠
−3
3
= 1.9 × 10−3 m3
To find the final temperature, we use the ideal-gas law once again as follows:
Tf = Ti
⎛ 0.5 pi ⎞⎛ 1.90 × 10−3 m3 ⎞
pf Vf
= ( 423 K ) ⎜
= 346.9 K ≈ 74°C
⎟⎜
−3
3 ⎟
pi Vi
⎝ pi ⎠ ⎝ 1.157 × 10 m ⎠
17.27. Model: The O2 gas has γ = 1.40 and is an ideal gas.
Solve:
(a) For an adiabatic process, pV γ remains a constant. That is,
γ
1.40
⎛ Vi ⎞
⎛ Vi ⎞
pV
⎟ = ( 3.0 atm ) ⎜
⎟
i i = pf Vf ⇒ pf = pi ⎜
V
⎝ f⎠
⎝ 2Vi ⎠
γ
γ
1.40
⎛1⎞
= ( 3.0 atm ) ⎜ ⎟
⎝2⎠
= 1.14 atm ≈ 1.1 atm
(b) Using the ideal-gas law, the final temperature of the gas is calculated as follows:
pV
pV
p V
⎛ 1.14 atm ⎞ ⎛ 2Vi ⎞
i i
= f f ⇒ Tf = Ti f f = ( 423 K ) ⎜
⎟ = 321.5 K ≈ 48°C
⎟⎜
pi Vi
Ti
Tf
⎝ 3.0 atm ⎠ ⎝ Vi ⎠
17.28. Visualize: We are asked for the heat-loss rate which is given by Equation 17.48:
Q
A
= k DT
Dt
L
We are given A = 10 m ×14 m = 140 m 2 , L = 0.12 m, and DT = 22°C − 5°C = 17°C = 17 K. We look up the
thermal conductivity of concrete in Table 17.5: k = 0.8W/m ⋅ K.
Solve:
⎛ 140 m 2 ⎞
Q
A
= k DT = ( 0.8W/m ⋅ K ) ⎜
⎟ (17 K ) = 16 kW
L
Dt
⎝ 0.12 m ⎠
Assess: The answer is in a reasonable range. The heat loss could be reduced with thicker concrete or adding a
layer of a different material.
17.29. Visualize: To determine the material we will solve for k in Equation 17.48:
Q
A
= k DT
Dt
L
We are given L = 0.20m and DT = 100 K. We compute A = pr 2 = p ( 0.010m ) = 3.14 × 1024 m 2 .
2
We also convert the heat conduction to watts.
⎛ 1h ⎞
4.5 ×104 J/h ⎜
⎟ = 12.5 W
⎝ 3600 s ⎠
Solve:
Solve the equation for k .
⎛
⎞⎛ 1 ⎞
0.20 m
⎛ Q ⎞L 1
k =⎜ ⎟
= (12.5 W ) ⎜
⎟ = 80 W/m ⋅ K
24
2 ⎟⎜
3
14
10
m
.
×
⎝ Dt ⎠ A DT
⎝
⎠⎝ 100K ⎠
Look this up in Table 17.5; the value corresponds to iron.
Assess: We are grateful that our answer was one of the entries in the table.
17.30. Model: Assume the lead sphere is an ideal radiator with e = 1 . Also assume that the highest
temperature the solid lead sphere can have is the melting temperature of lead.
Visualize: Use Equation 17.49. First look up the melting temperature of lead in Table 17.3:
Tm = 328°C = 601 K. Then compute the surface area of the sphere: A = 4pR 2 = 4p(0.050 m) 2 = 0.0314 m 2 *
Solve:
Q
= esAT 4 = (1)(5.6731028 W/m 2?K 4 )(0.0314 m 2 )(601 K) 4 = 230 W
Dt
Assess:
If the sphere were larger it could radiate more power without melting.
17.31. Model: We will ignore the bottom of the head and model it with just the cylindrical sides and top, all
covered in skin. As instructed, assume an emissivity of e = 0.95 .
Visualize: The
area
of
the
2
2
A = Aside + Atop = 2prh + pr = 2p(0.10 m)(0.20 m) + p(0.10 m) = 0.157 m 2 .
cylinder
Solve: Use Equation 17.50.
Qnet
= esA(T 4 − T04 ) = (0.95)((5.6731028 W/m 2?K 4 )(0.157 m 2 ))((308 K) 4 − (278 K) 4 ) = 26 W
Dt
Assess:
This is a significant amount of the heat lost by the body. Wearing a hat can help a lot.
is
17.32. Model: There are various steps to the problem. We must (1) raise the the temperature of the ice to
0°C, (2) melt the ice to liquid water at 0°C, (3) raise the water temperature to 100°C, (4) boil the water to
produce steam at 100°C , and (5) raise the temperature of the steam to 200°C.
Solve: The heat needed for each step is
Q1 = MciceDTice = ( 0.0050 kg )( 2090 J/kg?K )( 20 K ) = 209 J < 210 J
Q2 = MLf = ( 0.0050 kg ) ( 3.333105 J/kg ) = 1665 J < 1670 J
Q3 = Mcwater DTwater = ( 0.0050 kg )( 4190 J/kg?K )(100 K ) = 2095 J < 2100 J
Q4 = MLv = ( 0.0050 kg ) ( 22.63105 J/kg ) = 11,300 J
Q5 = Mcsteam DTsteam = ( 0.0050 kg )( 2009 J/kg?K )(100 K ) = 1005 J < 1000 J
The total heat is
Q = Q1 + Q2 + Q3 + Q4 + Q5 = 16,300 J
Assess: It is interesting to note which step required the most heat: The boiling to change the liquid to the gas is
by far the largest contributor to the total.
17.33. Solve: The area of the garden pond is A = π ( 2.5 m ) = 19.635 m 2 and its volume is V = A ( 0.30 m ) =
2
5.891 m 3 . The mass of water in the pond is
M = ρV = (1000 kg m3 )(19.635 m3 ) = 5891 kg
The water absorbs all the solar power which is
( 400 W m )(19.635 m ) = 7854 W
2
2
This power is used to raise the temperature of the water. That is,
Q = ( 7854 W ) Δt = Mcwater ΔT = ( 5891 kg )( 4190 J kg K )(10 K ) ⇒ Δt = 31,425 s ≈ 8.7 h
17.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture.
We assume the starting temperature of the bowling ball to be 0°C.
Solve: The potential energy of the bowling ball is
U g = M ball gh = (11 kg ) ( 9.8 m s 2 ) h = (107.8 kg m s 2 ) h
This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is,
(107.8 kg m s2 ) h = ΔEth = M w Lf ⇒ h =
( 0.005 kg ) ( 3.33 × 105 J kg )
(107.8 kg m s )
2
= 15.4 m
17.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and
raises the temperature of the kettle to 100°C.
Solve: The amount of heat energy from the electric stove’s output in 3 minutes is
Q = ( 2000 J s )( 3 × 60 s ) = 3.6 × 105 J
This heat energy heats the kettle and brings the water to a boil. Thus,
Q = M water cwater ΔT + M kettle ckettle ΔT
Substituting the given values into this equation,
3.6 × 105 J = M water ( 4190 J kg K )(100°C − 20°C ) + ( 0.750 kg )( 449 J kg K )(100°C − 20°C )
⇒ M water = 0.994 kg
The volume of water in the kettle is
V=
Assess:
M water
ρ water
=
0.994 kg
= 0.994 × 10−3 m3 = 994 cm3 ≈ 990 cm3
1000 kg m3
1 L = 103 cm3, so V ≈ 1 L. This is a reasonable volume of water.
17.36. Model: Each car’s kinetic energy is transformed into thermal energy.
Solve:
For each car,
K = 12 Mv 2 = ΔEth = Mccar ΔT ⇒ ΔT =
v2
2ccar
Assume ccar = ciron. The speed of the car is
80 × 1000 m
( 22.22 m s ) = 0.55°C
= 22.22 m s ⇒ ΔT =
3600 s
2 ( 449 J kg K )
2
v = 80 km hr =
Assess:
Notice the answer is independent of the car’s mass.
17.37. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol.
Solve: The aluminum, copper, and alcohol form a closed system, so Q = QAl + QCu + Qeth = 0 J. The mass of the
alcohol is
M eth = ρV = ( 790 kg m3 )( 50 × 10−6 m3 ) = 0.0395 kg
Expressed in terms of specific heats and using the fact that ΔT = Tf – Ti, the Q = 0 J condition is
M AlcAl ΔTAl + M Cu cCu ΔTCu + M eth ceth ΔTeth = 0 J
Substituting into this expression,
( 0.010 kg )( 900 J kg K )( 298 K − 473 K ) + ( 0.020 kg )( 385 J kg K )( 298 K − T )
+ ( 0.0395 kg )( 2400 J kg K )( 298 K − 288 K ) = −1575 J + ( 7.7 J K )( 298 − T ) + 948 J = 0 J
⇒ T = 216.6 K = −56.4°C ≈ −56°C
17.38. Model: There are two interacting systems: aluminum and ice. The system comes to thermal
equilibrium in four steps: (1) the ice temperature increases from −10°C to 0°C, (2) the ice becomes water at 0°C,
(3) the water temperature increases from 0°C to 20°C, and (4) the cup temperature decreases from 70°C to 20°C.
Solve: The aluminum and ice form a closed system, so Q = Q1 + Q2 + Q3 + Q4 = 0 J. These quantities are
Q1 = M icecice ΔT = ( 0.100 kg )( 2090 J kg K )(10 K ) = 2090 J
Q2 = M ice Lf = ( 0.100 kg ) ( 3.33 × 105 J kg ) = 33,300 J
Q3 = M icecwater ΔT = ( 0.100 kg )( 4190 J kg K )( 20 K ) = 8380 J
Q4 = MAl cAl ΔT = MAl ( 900 J kg K )( −50 K ) = − ( 45,000 J kg ) MAl
The Q = 0 J equation now becomes
43,770 J – (45,000 J/kg)MAl = 0 J
The solution to this is MAl = 0.973 kg.
17.39. Model: There are three interacting systems: metal, aluminum, and water.
Solve: The metal, aluminum container, and water form a closed system, so Qm + QAl + Qw = 0 J, where Qm is
the heat transferred to the metal sample. This equation can be written:
MmcmΔTm + MAlcAlΔTAl + MwcwΔT = 0 J
Substituting in the given values,
( 0.512 kg ) cm ( 351 K − 288 K ) + ( 0.100 kg )( 900 J kg K )( 351 K − 371 K )
+ ( 0.325 kg )( 4190 J kg K )( 351 K − 371 K ) = 0 J ⇒ cm = 900 J kg K
From Table 17.2, we see that this is the specific heat of aluminum.
17.40. Solve:
Equation 17.22,
For a monatomic gas, the molar specific heat at constant volume is CV = 12.5 J mol K . From
12.5 J mol K =
The gas is therefore neon.
M mol
625 J kg K ⇒ M mol = 20 g mol
1000 g kg
17.41. Model: Heating the water raises its thermal energy and its temperature.
Solve: A 5.0 kW heater has power P = 5000 W. That is, it supplies heat energy at the rate 5000 J/s. The heat
supplied in time ∆t is Q = 5000∆t J. The temperature increase is ∆TC = (5/9)∆TF = (5/9)(75°) = 41.67°C. Thus
Q = 5000Δt J = Mcw ΔT = (150 kg)(4190 J/kg K)(41.67°C) ⇒ Δt = 5283 s ≈ 87 min
Assess:
A time of ≈1.5 hours to heat 40 gallons of water is reasonable.
17.42. Model: Heating the material increases its thermal energy.
Visualize: Heat raises the temperature of the substance from –40°C to –20°C, at which temperature a solid to
liquid phase change occurs. From –20°C, heat raises the liquid’s temperature up to 40°C. Boiling occurs at 40°C
where all of the liquid is converted into the vapor phase.
Solve: (a) In the solid phase,
⎛ ΔQ ⎞ 1 ⎛ 20 kJ ⎞ ⎛ 1 ⎞
ΔQ = McΔT ⇒ c = ⎜
⎟ = 2000 J kg K
⎟ =⎜
⎟⎜
⎝ ΔT ⎠ M ⎝ 20 K ⎠ ⎝ 0.50 kg ⎠
(b) In the liquid phase,
c=
1 ⎛ ΔQ ⎞ ⎛ 1 ⎞ ⎛ 80 kJ ⎞
⎟⎜
⎜
⎟=⎜
⎟ = 2667 J kg K
M ⎝ ΔT ⎠ ⎝ 0.50 kg ⎠ ⎝ 60 K ⎠
(c) The melting point Tm = −20°C and the boiling point Tb = +40°C .
(d) The heat of fusion is
Lf =
Q 20,000 J
=
= 4.0 × 104 J kg
M
0.50 kg
Lv =
Q 60,000 J
=
= 1.2 × 105 J kg
M 0.50 kg
The heat of vaporization is
17.43. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogen’s temperature from
20°C to −196°C, and then liquefying it at −196°C. Assume the cooling occurs at a constant pressure of 1 atm.
Solve: The mass of 1.0 L of liquid nitrogen is M = ρV = ( 810 kg m3 )(10−3 m3 ) = 0.810 kg . This mass corresponds to
n=
M
810 g
=
= 28.9 mols
M mol 28 g mol
At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is
Q = MLv + nCP ΔT
= − ( 0.810 kg ) (1.99 × 105 J kg ) + ( 28.9 mols )( 29.1 J mol K )( 77 K − 293 K ) = −3.4 × 105 J
17.44. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee
temperature from 90°C to 60°C requires four steps: (1) raise the temperature of ice from −20°C to 0°C, (2)
change ice at 0°C to water at 0°C, (c) raise the water temperature from 0°C to 60°C, and (4) lower the coffee
temperature from 90°C to 60°C.
Solve: For the closed coffee-ice system,
Q = Qice + Qcoffee = ( Q1 + Q2 + Q3 ) + ( Q4 ) = 0 J
Q1 = M icecice ΔT = M ice ( 2090 J kg K )( 20 K ) = M ice ( 41,800 J kg )
Q2 = M ice Lf = M ice ( 330,000 J kg )
Q3 = M icecwater ΔT = M ice ( 4190 J kg K )( 60 K ) = M ice ( 251,400 J kg )
Q4 = M coffeeccoffee ΔT = ( 300 × 10−6 m3 )(1000 kg m3 ) ( 4190 J kg K )( −30 K ) = −37,000 J
The Q = 0 J equation thus becomes
M ice ( 41,800 + 330,000 + 251,400 ) J kg − 37,710 J = 0 J ⇒ M ice = 0.061 kg = 61 g
Assess:
61 g is the mass of approximately 1 ice cube.
17.45. Model: We have three interacting systems: the aluminum, the air, and the firecracker. The energy
released by the firecracker raises the temperature of the aluminum and the air. We will assume that air is
predominantly N2. Assume that the surrounding insulation is excellent.
Solve: For the closed firecracker + air + aluminum system, energy conservation requires that
Q = Qfirecracker + QAl + Qair = 0 J
QAl = mAlcAlΔT = (2.0 kg)(900 J/kg K)(3 K) = 5400 J
⎛ pV ⎞
Qair = nCV ΔT = ⎜
⎟ CV ΔT
⎝ RT ⎠
(1.01×10 Pa )( 20 ×10 m ) ( 20.8 J/mol K )( 3 K )
5
=
−3
3
(8.31 J/mol K )( 298 K )
= 51 J
Thus,
Qfirecracker = −QAl − Qair
= –5450 J
That is, ≈ 5500 J of energy are released on explosion of the firecracker.
Assess: The negative sign with Qfirecracker means that the firecracker has lost energy.
17.46. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by
the nuclear reactor is used to raise the water temperature.
Solve: For the closed reactor-water system, energy conservation per second requires
Q = Qreactor + Qwater = 0 J
The heat from the reactor in Δt = 1 s is Qreactor = −2000 MJ = −2.0 × 109 J and the heat absorbed by the water is
Qwater = mwater cwater ΔT = mwater ( 4190 J kg K )(12 K )
⇒ −2.0 × 109 J + mwater ( 4190 J kg K )(12 K ) = 0 J ⇒ mwater = 3.98 × 104 kg
Each second, 3.98 × 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per
minute is
3.98 × 104
kg 60 s
1m3
1L
×
×
×
= 2.4 × 106 L/min
s min 1000 kg 10−3 m3
17.47. Model: We have two interacting systems: the water and the gas. For the closed system comprised of
water and gas to come to equilibrium, heat is transferred from one interacting system to the other.
Solve: Energy conservation requires that
Qair + Qwater = 0 J ⇒ ngasCV (Tf – Ti gas) + mwaterc(Tf – Ti water) = 0 J
Using the ideal-gas law,
Ti gas =
pgasVgas
ngas R
(10 ×1.013 × 10 Pa )( 4000 ×10 m ) =1219 K
5
=
−6
3
( 0.40 mol )(8.31 J mol K )
The energy conservation equation with Ti water = 293 K becomes
( 0.40 mol )(12.5 J mol K )(T − 1219 K ) + ( 20 × 10−3 kg ) ( 4190 J kg K )(T − 293 K ) = 0 J ⇒ Tf = 345 K
We can now use the ideal-gas equation to find the final gas pressure.
pV
pV
T
⎛ 345 K ⎞
i i
= f f ⇒ pf = pi f = ⎜
⎟ (10 atm ) = 2.8 atm
Ti ⎝ 1219 K ⎠
Ti
Tf
17.48. Model: These are isothermal and isobaric ideal-gas processes.
Solve:
(a) The work done at constant temperature is
nRT
dV = − nRT ( ln Vf − ln Vi ) = − nRT ln (Vf Vi )
V
= − ( 2.0 mol )( 8.31 J mol K )( 303 K ) ln ( 13 ) = 5500 J
W = −∫
Vf
Vi
pdV = − ∫
Vf
Vi
(b) The work done at constant pressure is
Vf
⎛V
⎞ 2
W = − ∫ p dV = − p (Vf − Vi ) = − p ⎜ i − Vi ⎟ = pVi
Vi
⎝3
⎠ 3
2
2
= nRT = ( 2.0 mol )( 8.31 J mol K )( 303 K ) = 3400 J
3
3
(c) For an isothermal process in which Vf = 13 Vi , the pressure changes to pf = 3 pi = 4.5 atm.
17.49. Model: This is an isothermal process. The work done is positive for a compression.
Solve:
For an isothermal process,
W = − nRT ln (Vf Vi )
For the first process,
W = 500 J = − nRT ln ( 12 ) ⇒ nRT = 721.35 J
For the second process,
W = − nRT ln ( 101 ) ⇒ W = − ( 721.35 J ) ln ( 101 ) = 1660 J
17.50.
Visualize:
Solve:
(a) The gas exerts a force on the piston of magnitude
Fgas on piston = pgas A = ( 3 atm × 101,300 Pa atm ) ⎡π ( 0.080 m ) ⎤ = 6100 N
⎣
⎦
2
This force is directed toward the right.
(b) The piston is in static equilibrium, so the environment must exert a force on the piston of equal magnitude
Fenviron on piston = 6100 N but in the opposite direction, toward the left.
(c) The work done by the environment is
G
G
G
Wenviron = Fenviron on piston ⋅ Δr = − Fenviron on piston Δx = − ( 6110 N )( 0.10 m ) = −610 J
The work is negative because the force and the displacement are in opposite directions.
(d) The work done by the gas is
G
G
Wgas = Fgas on piston ⋅ Δr = + Fgas on piston Δx = ( 6110 N )( 0.10 m ) = 610 J
This work is positive because the force and the displacement are in the same direction.
(e) The first law of thermodynamics is W + Q = ΔEth where W is Wenviron. So here W = −610 J and we find
Q = ΔEth − W = (196 J ) − ( −610 J ) = 806 J
Thus, 806 J of heat is added to the gas.
17.51. Model: This is an isobaric process.
Visualize:
Solve: (a) The initial conditions are p1 = 10 atm = 1.013 × 106 Pa, T1 = 50°C = 323 K, V1 = πr2L1 = π(0.050 m)2
(0.20 m) = 1.57 × 10−3 m3. The gas is heated at a constant pressure, so heat and temperature change are related by
Q = nCPΔT. From the ideal gas law, the number of moles of gas is
p1V1 (1.013 × 10 Pa )(1.57 × 10 m )
=
= 0.593 mol
RT1
(8.31 J mol K )( 323 K )
6
n=
−3
3
The temperature change due to the addition of Q = 2500 J of heat is thus
ΔT =
Q
2500 J
=
= 203 K
nCP ( 0.593 mol )( 20.8 J mol K )
The final temperature is T2 = T1 + ΔT = 526 K = 253°C.
(b) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal gas relationship for an
isobaric process is
V2 V1
L
L
T
526 K
= ⇒ 2 = 1 ⇒ L2 = 2 L1 =
( 20 cm ) = 33 cm
T2 T1
T2 T1
T1
323 K
17.52. Model: The process in part (a) is isochoric and the process in part (b) is isobaric.
Solve: (a) Initially V1 = (0.20 m)3 = 0.0080 m3 = 8.0 L and T1 = 293 K. Helium has an atomic mass number A =
4, so 3 g of helium is n = M/Mmol = 0.75 mole of helium. We can find the initial pressure from the ideal-gas law:
p1 =
nRT1 ( 0.75 mol )( 8.31 J mol K )( 293 K )
=
= 228 kPa = 2.25 atm
V1
0.0080 m3
Heating the gas will raise its temperature. A constant volume process has Q = nCVΔT, so
ΔT =
Q
1000 J
=
= 107 K
nCV ( 0.75 mol )(12.5 J mol K )
This raises the final temperature to T2 = T1 + ΔT = 400 K. Because the process is isochoric,
p2 p1
T
400 K
= ⇒ p2 = 2 p1 =
( 2.25 atm ) = 3.1 atm
T2 T1
T1
293 K
(b) The initial conditions are the same as part a, but now Q = nCPΔT. Thus,
ΔT =
1000 J
Q
=
= 64.1 K
nCP ( 0.75 mol )( 20.8 J mol K )
Now the final temperature is T2 = T1 + ΔT = 357 K. Because the process is isobaric,
V2 V1
T
357 K
= ⇒ V2 = 2 V1 =
( 0.0080 m3 ) = 0.0097 m3 = 9.7 L
T2 T1
T1
293 K
(c)
17.53. Model: Assume the air and the nitrogen are ideal gases. For the piston to be in equilibrium the forces
on it must sum to zero. Also assume that the temperature doesn’t change in the second part.
2
Solve: The area is A = pr 2 = p ( 0.040 m ) = 5.031023 m 2 . V1 = Ah1 = ( 5.031023 m 2 ) ( 0.26 m ) = 0.0013 m3 .
Fnet = pin A − mg − pabove A = 0 N
(a)
mg + pabove A ( 5.1 kg ) ( 9.8 m/s ) + (100 kPa ) ( 5.0310
=
A
5.031023 m 2
2
pin =
23
m2 )
= 110 kPa
(b) Since T1 = T2 , then p1V1 = p2V2 where p1 is pin from above, and p2 is the inside pressure after the new
weight is added. Compute p2 exactly as above replacing 5.1 kg with 5.1 kg + 3.5kg = 8.6kg.
mg + pabove A ( 8.6 kg ) ( 9.8 m/s ) + (100 kPa ) ( 5.0310
=
5.031023 m 2
A
2
p2 =
V2 =
23
m2 )
= 117 kPa
3
p1V1 (110 kPa ) ( 0.0013 m )
=
= 1.2231023 m3
p2
117 kPa
h2 =
V2 1.2231023 m 3
=
= 0.2447 m < 24 cm
A 5.031023 m 2
Assess: The result seems to be a reasonable number; we expected the piston to be a little lower than before due
to the increased mass on it.
17.54. Model: This is an isothermal process.
Solve: (a) The final temperature is T2 = T1 because the process is isothermal.
(b) The work done on the gas is
V2
V2
V1
V1
W = − ∫ pdV = − ∫
nRT1
V
dV = − nRT1 ln 2 = −nRT1ln 2
V
V1
(c) From the first law of thermodynamics ΔEth = W + Q = 0 J because ΔT = 0 K. Thus, the heat energy transferred
to the gas is Q = −W = nRT1 ln 2 .
17.55. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process.
Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa and T1 = 293 K. Nitrogen has a molar mass Mmol
=
28 g/mol, so 5 g of nitrogen gas has n = M/Mmol = 0.1786 mol. From this, we can find the initial volume:
V1 =
nRT1 ( 0.1786 mol )( 8.31 J mol K )( 293 K )
=
= 1.430 × 10−3 m3 ≈ 1400 cm 3
p1
304,000 Pa
The volume triples, so V2 = 3V1 = 4300 cm3. The expansion is isobaric (p 2 = p1 = 3.0 atm), so
V2 V1
V
= ⇒ T2 = 2 T1 = ( 3) 293 K = 879 K = 606°C
T2 T1
V1
(b) The process is isobaric, so
Q = nCP ΔT = ( 0.1786 mol )( 29.1 J mol K )( 879 K − 293 K ) = 3000 J
(c) The pressure is decreased at constant volume (V3 = V2 = 4290 cm3) until the original temperature is reached
(T3 = T1 = 293 K). For an isochoric process,
p3 p2
T
293 K
=
⇒ p3 = 3 p2 =
( 3.0 atm ) = 1.0 atm
T3 T2
T2
879 K
(d) The process is isochoric, so
Q = nCV ΔT = ( 0.1786 mol )( 20.8 J mol K )( 293 K − 879 K ) = −2200 J
So, 2200 J of heat was removed to decrease the pressure.
(e)
17.56. Model: The gas is an ideal gas.
Visualize: In the figure, call the upper left corner (on process A) 2 and the lower right corner (on process B) 3.
Solve: The change in thermal energy is the same for any gas process that has the same ∆T. Processes A and B
have the same ∆T, since they start and end at the same points, so (∆Eth)A = (∆Eth)B. The first law is then
(∆Eth)A = QA + WA = (∆Eth)B = QB + WB ⇒ QA – QB = WB – WA
In process B, work W = –p∆V = –pi(2Vi – Vi) = –piVi is done during the isobaric process i → 3. No work is done
during the isochoric process 3 → f. Thus WB = –piVi. Similarly, no work is done during the isochoric process i →
2 of process A, but W = –p∆V = –2pi(2Vi – Vi) = –2piVi is done during the isobaric process 2→ f. Thus WA = –
2piVi. Combining these,
QA – QB = WB – WA = –piVi – (–2piVi) = piVi
17.57. Model: The two processes are isochoric and isobaric.
Solve:
Process A is isochoric which means
Tf Ti = pf pi ⇒ Tf = Ti ( pf pi ) = Ti (1 atm 3 atm ) = 13 Ti
From the ideal-gas equation,
( 3 ×1.013 ×10 Pa )( 2000 × 10 m ) = 731.4 K ⇒ T = 1 T = 243.8 K
pV
i i
=
f
3 i
nR
( 0.10 mol )(8.31 J mol K )
−6
5
Ti =
3
⇒ Tf − Ti = −487.6 K
Thus, the heat required for process A is
QA = nCV ΔT = ( 0.10 mol )( 20.8 J mol K )( −487.6 K ) = −1000 J
Process B is isobaric which means
Tf Vf = Ti Vi ⇒ Tf = Ti (Vf Vi ) = Ti ( 3000 cm3 1000 cm3 ) = 3Ti
From the ideal-gas equation,
( 2 × 1.013 × 10 Pa )(1000 × 10 m ) = 243.8 K
pV
i i
=
nR
( 0.10 mol )( 8.31 J mol K )
5
Ti =
−6
3
⇒ Tf = 3Ti = 731.4 K ⇒ Tf − Ti = 487.6 K
Thus, heat required for process B is
QB = nCP ΔT = ( 0.10 mol )( 29.1 J mol K )( 487.6 K ) = 1400 J
Assess:
Heat is transferred out of the gas in process A, but transferred into the gas in process B.
17.58. Model: We have an adiabatic and an isothermal process.
Solve: For the adiabatic process, no heat is added or removed. That is Q = 0 J.
Isothermal processes occur at a fixed temperature, so ∆T = 0 K. Thus ∆Eth = 0 J, and the first law of
thermodynamics gives
Q = −W = nRT ln (Vf Vi )
The temperature T can be obtained from the ideal-gas equation as follows:
pV
i i = nRT ⇒ T =
(1.013 ×105 Pa )( 3000 × 10−6 m3 ) = 366 K
pV
i i
=
nR
( 0.10 mol )(8.31 J mol K )
Substituting into the equation for Q we get
⎛ 1000 × 10−6 m3 ⎞
= −330 J
Q = ( 0.10 mol )( 8.31 J mol K )( 366 K ) ln ⎜
−6
3 ⎟
⎝ 3000 × 10 m ⎠
That is, 330 J of heat energy is removed from the gas.
17.59. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes.
Solve: (a) For the isochoric process, V2 = V1 = 800 × 10−6 m3, p1 = 4.0 atm, p2 = 2.0 atm. The temperature T1 of
the gas is obtained from the ideal-gas equation as:
T1 =
p1V1
= 390 K
nR
where n = 0.10 mol. T2 can be obtained from the ideal-gas equation as follows:
p1V1 p2V2
⎛ 2.0 atm ⎞
=
⇒ T2 = T1 ( p2 p1 ) = ( 390 K ) ⎜
⎟ = 195 K
T1
T2
⎝ 4.0 atm ⎠
The heat required for the process 1 → 2 is
Q = nCV (T2 − T1 ) = ( 0.10 mol )( 20.8 J/mol K )(195 K − 390 K ) = −406 J ≈ −410 J
Because of the negative sign, this is the amount of heat removed from the gas.
(b) For this isobaric process, p2 = p3 = 2.0 atm, V2 = 800 × 10−6 m3, and V3 = 1600 × 10−6 m3.
T3 = T2
⎛ 1600 × 10−6 m3 ⎞
V3
= T2 ⎜
= 2T2 = 390 K
−6
3 ⎟
V2
⎝ 800 × 10 m ⎠
Thus, the heat required for the process 2 → 3 is
Q = nCP (T3 − T2 ) = ( 0.10 mol )( 29.1 J mol K )(195 K ) = 567 J ≈ 570 J
This is heat transferred to the gas.
(c) The change in the thermal energy of the gas is
ΔEth = ( Q1→ 2 + Q2 →3 ) + (W1→ 2 + W2 →3 ) = −406 J + 567 J + 0 J +W2 →3 = 162 J − pΔV
= 162 J – (2.0 × 1.013 × 105 Pa)(1600 × 10−6 m3 – 800 × 10−6 m3) = 0 J
Assess:
This result was expected since T3 = T1.
17.60. Model: Assume that the gas is an ideal gas.
Visualize:
The volume of container A is a constant. On the other hand, heating container B causes the volume to change,
but the pressure remains the same.
Solve: (a) For the heating of the gas in container A, ΔTA = QA / nCV . Similarly, for the gas in container B,
ΔTB = QB / nCP . Because QA = QB and CP > CV, we see that ΔTA > ΔTB . The gases started at the same
temperature, so TA > TB.
(b)
(c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is,
pB = patmos +
wpiston
Apiston
= 1.013 × 105 Pa +
(10 kg ) ( 9.8 m s2 )
1.0 × 10−4 m 2
= 1.08 × 106 Pa
Container A has the same volume, temperature, and number of moles of gas as container B, so PA = PB =
1.08 × 106 Pa.
(d) The heating of container B is isobaric, so
Vf Vi
T
= ⇒ Vf = Vi f
Tf Ti
Ti
We have Ti = 293 K, and Tf can be obtained from
Q = PΔt = nCP (Tf − Ti )
The number of moles of gas is n = PV
i i / RTi = 0.355 mol. Thus
( 25 W )(15 s ) = ( 0.355 mol )( 20.8 J mol K )(Tf − 293 K )
⇒ Tf = 344 K ⇒ Vf = ( 8.0 × 10−4 m3 ) ( 344 K 293 K ) = 9.39 × 10−4 m3 = 939 cm3 ≈ 940 cm3
17.61. Model: Assume that the gas is an ideal gas. A diatomic gas has γ = 1.40.
Solve:
(a) For container A,
ViA =
nRTAi ( 0.10 mol )( 8.31 J mol K )( 300 K )
=
= 8.20 × 10−4 m3
pAi
3.0 × 1.013 × 105 Pa
For an isothermal process pAfVAf = pAiVAi. This means TAf = TAi = 300 K and
VAf = VAi ( pAi pAf ) = ( 8.20 × 10−4 m3 ) ( 3.0 atm 1.0 atm ) = 2.5 × 10−3 m3
The gas in container B starts with the same initial volume. For an adiabatic process,
1
1
⎛p ⎞ γ
⎛ 3.0 atm ⎞ 1.40
pBf VBf = pBiVBi ⇒ VBf = VBi ⎜ Bi ⎟ = ( 8.20 × 10−4 m3 ) ⎜
= 1.8 × 10−3 m3
⎟
⎝ 1.0 atm ⎠
⎝ pBf ⎠
γ
γ
The final temperature TBf can now be obtained by using the ideal-gas equation:
TBf = TiB
(b)
−3
3
pBf VBf
⎛ 1.0 atm ⎞ ⎛ 1.80 × 10 m ⎞
= 220 K
= ( 300 K ) ⎜
⎟⎜
−4
3 ⎟
pBi VBi
⎝ 3.0 atm ⎠ ⎝ 8.20 × 10 m ⎠
17.62. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes.
Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa, V1 = 100 cm3 = 1.0 × 10−4 m3, and T1 = 100°C =
373 K. The number of moles of gas is
n=
−4
3
p1V1 ( 304,000 Pa ) (1.0 × 10 m )
=
= 9.81 × 10−3 mol
RT1
( 8.31 J mol K ) (373 K)
At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so
V2 V1
V
= ⇒ T2 = 2 T1 = 3(373 K) = 1119 K
T2 T1
V1
The gas is heated to raise the temperature from T1 to T2. The amount of heat required is
Q = nCP ΔT = ( 9.81 × 10−3 mol ) ( 20.8 J mol K )(1119 K − 373 K ) = 152 J
This amount of heat is added during process 1 → 2.
(b) Point 3 returns to T3 = 100°C = 373 K. This is an isochoric process, so
Q = nCV ΔT = ( 9.81 × 10−3 mol ) (12.5 J mol K )( 373 K − 1119 K ) = −91.5 J
This amount of heat is removed during process 2 → 3.
17.63. Assume the gas to be an ideal gas.
Solve:
(a) The work done on the gas is the negative of the area under the p-versus-V graph, that is
W = −area under curve = −50.7 J
(b) The change in thermal energy is
ΔEth = nCV ΔT = nCV (Tf − Ti )
Using the ideal-gas law to calculate the initial and final temperatures,
( 4.0 ×1.013 × 10 Pa )(100 × 10 m ) = 325 K
pV
i i
=
nR
( 0.015 mol )(8.31 J mol K )
5
Ti =
Tf =
−6
3
5
−6
3
pfVf (1.013 × 10 Pa )( 300 × 10 m )
=
= 244 K
nR
( 0.015 mol )( 8.31 J mol K )
⇒ ΔEth = ( 0.015 mol )(12.5 J mol K )( 244 K − 325 K ) = −15.2 J ≈ −15 J
(c) From the first law of thermodynamics,
ΔEth = Q + W ⇒ Q = ΔEth − W = −15.2 J − ( −50.7 J ) = 35.5 J ≈ 36 J
That is, 36 J of heat energy is transferred to the gas.
17.64. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected
by the first law of thermodynamics.
Solve: (a) For point 1, V1 = 1000 cm3 = 1.0 × 10−3 m3, T1 = 133°C = 406 K, and the number of moles is
n=
⎛ 120 × 10−3 g ⎞
M
=⎜
⎟ = 0.030 mol
M mol ⎝ 4 g/mol ⎠
Thus, the pressure p1 is
p1 =
nRT1
= 1.012 × 105 Pa = 1.0 atm
V1
The process 1 → 2 is isochoric (V2 = V1) and p2 = 5p1 = 5.0 atm. Thus,
T2 = T1 ( p2 p1 ) = ( 406 K )( 5 ) = 2030 K = 1757°C
The process 2 → 3 is isothermal (T2 = T3), so
V3 = V2(p2/p3) = V2(p2/p1) = 5V2 = 5000 cm3
p (atm)
T (°C)
V (cm3)
1.0
5.0
1.0
133
1757
1757
1000
1000
5000
Point 1
Point 2
Point 3
(b) The work W1→ 2 = 0 J because it is an isochoric process. The work in process 2 → 3 can be found using
Equation 17.16 as follows:
W2 →3 = − nRT2 ln (V3 V2 ) = − ( 0.030 mol )( 8.31 J mol K )( 2030 K ) ln ( 5 ) = −815 J
The work in the isobaric process 3 → 1 is
W3→1 = − p (Vf − Vi ) = − (1.012 × 105 Pa )(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) = 405 J
(c) The heat transferred in process 1 → 2 is
Q1→ 2 = nCV ΔT = ( 0.030 mol )(12.5 J mol K )( 2030 K − 406 K ) = 609 J
The heat transferred in the isothermal process 2 → 3 is Q2 →3 = −W2 →3 = 815 J . The heat transferred in the
isobaric process 3 → 1 is
Q3→1 = nCP ΔT = ( 0.030 mol )( 20.8 J mol K )( 406 K − 2030 K ) = −1013 J
17.65. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.
Solve: The air admitted into the cylinder at T0 = 30°C = 303 K and p0 = 1 atm = 1.013 × 105 Pa has a volume V0
= 600 × 10−6 m3 and contains
pV
n = 0 0 = 0.024 mol
RT0
Using Equation 17.36 and the fact that Q = 0 J for an adiabatic process,
ΔEth = Q + W = nCV ΔT ⇒ W = nCVT
⇒ 400 J = (0.024 mol)(20.8 J/mol K)(Tf – 303 K) ⇒ Tf = 1100 K
For an adiabatic process Equation 17.40 is
1
1
⎛ T ⎞ γ −1
⎛ 303 K ⎞1.4 −1
= 2.39 × 10−5 m3 ≈ 24 cm3
TfVfγ −1 = T0V0γ −1 ⇒ Vf = V0 ⎜ 0 ⎟ = ( 6.0 × 10−4 m3 ) ⎜
⎟
T
⎝ 1100 K ⎠
⎝ f⎠
Assess:
Note that W is positive because the environment does work on the gas.
17.66. Model: γ is 1.40 for a diatomic gas and 1.67 for a monoatomic gas.
Solve:
(a) We will assume that air is a diatomic gas. For an adiabatic process,
γ −1
Tf Vfγ −1 = TV
i i
Thus
1
1
⎛ Vi ⎞ ⎛ Tf ⎞ γ −1 ⎛ 1123 K ⎞1.40 −1
= 26.4
⎜ ⎟=⎜ ⎟ =⎜
⎟
⎝ 303 K ⎠
⎝ Vf ⎠ ⎝ Ti ⎠
(b) For argon, a monatomic gas,
1
⎛ Vi ⎞ ⎛ 1123 K ⎞1.67 −1
= 7.07
⎜ ⎟=⎜
⎟
⎝ Vf ⎠ ⎝ 303 K ⎠
17.67. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.
Solve:
(a) Equation 17.40 for an adiabatic process is
1
γ −1
TfVf
= TV
i i
γ −1
V ⎛ T ⎞ γ −1
⇒ f =⎜ i ⎟
Vi ⎝ Tf ⎠
For the temperature to increase from Ti = 20°C = 293 K to Tf = 1000°C = 1273 K, the compression ratio will be
1
Vf ⎛ 293 K ⎞1.4 −1
V
1
= 39.3
=⎜
= 0.02542 ⇒ max =
⎟
Vi ⎝ 1273 K ⎠
Vmin 0.02542
(b) From the Equation 17.39,
γ
pf ⎛ Vi ⎞
1.4
pfVf = pV
= ⎜ ⎟ = ( 39.3) = 171
i i ⇒
pi ⎝ Vf ⎠
γ
γ
17.68. Model: The helium gas is assumed to be an ideal gas that is subjected to an isobaric process.
Solve:
(a) The number of moles in 2.0 g of helium is
n=
M
2.0 g
=
= 0.50 mol
M mol 4.0 g mol
At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume
Vi =
nRTi
= 0.0153 m3 = 15.3 L
pi
For an isobaric process (pf = pi) that doubles the volume Vf = 2Vi,
(Tf Ti ) = (Vf Vi ) = 2 ⇒ Tf = 2Ti = 2 ( 373 K ) = 746 K = 473°C
(b) The work done by the environment on the gas is
5
3
W = − pi (Vf − Vi ) = − pV
i i ( 2 − 1) = − (1.013 × 10 Pa )( 0.0153 m ) = −1550 J
(c) The heat input to the gas is
Q = nCP (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 746 K − 373 K ) = 3880 J ≈ 3900 J
(d) The change in the thermal energy of the gas is
ΔEth = Q + W = 3880 J − 1550 J = 2330 J ≈ 2300 J
(e)
Assess:
The internal energy can also be calculated as follows:
ΔEth = nCV ΔT = ( 0.5 mol )(12.5 J mol K )( 746 K − 373 K ) = 2330 J
This is the same result as we got in part (d).
17.69. Model: The helium gas is assumed to be an ideal gas that is subjected to an isothermal process.
Solve:
(a) The number of moles in 2.0 g of helium gas is
n=
M
2.0 g
=
= 0.50 mol
M mol 4.0 g mol
At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume
Vi =
nRTi
= 0.0153 m3 = 15.3 L
pi
For an isothermal process (Tf = Ti) that doubles the volume Vf = 2Vi,
1
pf Vf = pV
i i ⇒ pf = pi (Vi Vf ) = (1.0 atm ) ( 2 ) = 0.50 atm
(b) The work done by the environment on the gas is
W = − nRTi ln (Vf Vi ) = − ( 0.50 mol )( 8.31 J mol K )( 373 K ) ln ( 2 ) = −1074 J ≈ −1070 J
(c) Because ΔEth = Q + W = 0 J for an isothermal process, the heat input to the gas is Q = −W = 1074 J ≈ 1070 J .
(d) The change in internal energy ΔEth = 0 J.
(e)
17.70. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process.
Solve:
(a) The number of moles in 14.0 g of N2 gas is
n=
M
14.0 g
=
= 0.50 mol
M mol 28 g mol
At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume
Vi =
nRTi
= 0.0112 m3 = 11.2 L
pi
For an adiabatic process that compresses to a pressure pf = 20 atm, we can use Equation 17.39 and Equation
17.40 as follows:
TfVf
γ −1
= TV
i i
γ −1
⎛T ⎞ ⎛V ⎞
⇒⎜ f ⎟=⎜ i ⎟
⎝ Ti ⎠ ⎝ Vf ⎠
γ −1
1
⎛ pf ⎞ γ Vi
pf Vf = pV
⎟ =
i i ⇒ ⎜
Vf
⎝ pi ⎠
γ
γ
Combining the above two equations yields
γ −1
1.4 −1.0
(Tf Ti ) = ( pf pi ) γ ⇒ Tf = Ti ( 20 ) 1.4 = Ti ( 2.3535) = 643 K
(b) The work done on the gas is
W = ΔEth = nCV (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 643 K − 273 K ) = 3850 J ≈ 3800 J
(c) The heat input to the gas is Q = 0 J.
(d) From the above equation,
1
1
Vi ⎛ pf ⎞ γ
V
= ⎜ ⎟ = ( 20 )1.4 = 8.5 = max
Vf ⎝ pi ⎠
Vmin
(e)
17.71. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process.
Solve:
(a) The number of moles in 14.0 g of N2 gas is
n=
M
14.0 g
=
= 0.50 mol
M mol 28 g mol
At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume
Vi =
nRTi
= 0.0112 m3 = 11.2 L
pi
For an isochoric process (Vi = Vf),
Tf pf 20 atm
=
=
= 20 ⇒ Tf = 20 ( 273 K ) = 5460 K ≈ 5500 K
1 atm
Ti
pi
(b) The work done on the gas is W = − pΔV = 0 J.
(c) The heat input to the gas is
Q = nCV (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 5460 K − 273 K ) = 5.4 × 10 4 J
(d) The pressure ratio is
pmax pf 20 atm
=
=
= 20
pmin pi
1 atm
(e)
17.72. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchange
heat with its surroundings, the compression is an adiabatic process.
Solve: The initial pressure of air in the mountains behind Los Angeles is pi = 60 × 103 Pa at Ti = 273 K. The
pressure of this air when it is carried down to the elevation near sea level is pf = 100 × 103 Pa. The adiabatic
compression of a gas leads to an increase in temperature according to Equation 17.39 and Equation 17.40, which
are
γ −1
γ −1
TfVf
= TV
i i
γ −1
⎛T ⎞ ⎛V ⎞
⇒⎜ f ⎟=⎜ i ⎟
⎝ Ti ⎠ ⎝ Vf ⎠
1
⎛ pf ⎞ γ Vi
pfVf = pV
⎟ =
i i ⇒ ⎜
Vf
⎝ pi ⎠
γ
γ
Combining these two equations,
1.4 −1
(Tf Ti ) = ( pf pi )
γ −1
γ
⎛ 100 × 103 Pa ⎞ 1.4
⎛5⎞
⇒ Tf = Ti ⎜
= ( 273 K ) ⎜ ⎟
⎟
3
60
10
Pa
×
⎝3⎠
⎝
⎠
0.286
= 316 K = 43ºC = 109ºF
17.73. Model: Assume the collector has emissivity e = 1.0 . We will use Equation 17.50 and divide both
sides by A to get intensity in W/m 2 . At the equilibrium temperature the collector will absorb 800 W/m 2 on the
sun side and must radiate the same amount from the same side. T0 = 20°C = 293 K .
Solve:
Q/ Dt
= es (T 4 − T04 )
A
Solve this for T .
Q/ Dt 1
= T 4 − T04
A es
T=4
Q/ Dt 1
800 W/m 2
4
+ T04 = 4
+ ( 293K ) = 383K = 110°C
28
2
4
A es
1
0
5
67
10
W/m
K
.
.
×
⋅
( )(
)
Assess: 110°C is hot enough to boil water, so this seems feasible as a hot water heater, at least during the 5
hours a day the collector gets 800 W/m 2 . You can store the hot water in an insulated tank for use at other times
of the day.
17.74. Model: This is a problem about conduction of the heat from inside the box to outside. 100 W of heat
is generated by the light bulb inside the box, so in equilibrium that is how much must be conducted away through
the sides of the box.
Visualize: We are given L = 0.012 m , Q/ Dt = 100 W and the thermal conductivity of concrete
k = 0.8 W/m ⋅ K . We compute A = 6 ( 0.20m × 0.20 m ) = 0.24m 2 . TC = 20°C = 293K .
Solve:
Use Equation 17.48 for the rate of heat transfer by conduction.
Q
A
= k (TH − TC )
Dt
L
Solve for TH .
⎛Q⎞ L
= (TH − TC )
⎜ ⎟
⎝ Dt ⎠ kA
0.012 m
⎛Q⎞ L
TH = ⎜ ⎟ + TC = (100 W )
+ 293 K = 299 K = 26°C
D
t
kA
0
.
8
W/m
?K ) ( 0.24 m 2 )
⎝ ⎠
(
Assess:
We expected TH to be a few degrees hotter than TC , and so it is.
17.75. Model: Refer to Example 17.11. Assume the surface of the earth is an ideal radiator with e = 1 .
Visualize: Only half the earth faces the sun at a time, so the earth intercepts 1370 W/m 2 over a cross section of
pRe2 = p ( 6.37 × 106 m ) = 1.275 × 1014 m 2 . If the earth’s surface absorbs 70% of the incident power then the total
2
power absorbed is ( 0.70 ) (1370 W/m 2 )(1.275 × 1014 m 2 ) = 1.222 × 1017 W . This much power must also be
radiated away from the earth in equilibrium.
Solve: The power is radiated from the whole spherical surface of the earth: A = 4pRe2 . Use Equation 17.49 to
find the temperature of the earth.
⎡ Q/ Dt ⎤
⎥
T =⎢
2
⎢⎣ es ( 4pRe ) ⎥⎦
1/ 4
⎡
⎤
1.22231017 W
⎥
=⎢
2
⎢ (1) ( 5.6731028 W/m 2?K ) 4p ( 6.373106 m ) ⎥
⎣
⎦
1/ 4
= 255 K = 218°C
Assess: Without a moderate greenhouse effect the earth would be too cold for mammal life. On the other hand,
when the greenhouse effect gets out of hand, such as on Venus, it can be too hot for life.
17.76. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a
constant temperature of 77°C. How many moles of the gas are in the sample?
(b) The number of moles is
n=
50 J
= 0.0156 mol
− ( 8.31 J mol K )( 350 K ) ( ln 13 )
17.77. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15°C. If the mercury
temperature rises to 90°C, what was the initial temperature of the iron slug?
(b) The initial temperature was 217°C .
17.78. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What
is the compression ratio Vmax/Vmin?
1
(b) The ratio is Vmax Vmin = 101.4 = 5.18 .
17.79. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, and
adiabatic processes.
Visualize: Please refer to Figure CP17.79. The gas at point 1 has volume V1 = 1000 cm3 = 1.0 × 10−3 m3 and
pressure p1 = 3.0 atm. At point 2, V2 = 3000 cm3 = 3.0 × 10−3 m3 and p2 = 1.0 atm. These values mean that T2 =
T1, so process 1 → 2 is an isothermal process. The process 2 → 3 occurs at constant volume and is thus an
isochoric process. Finally, because temperature T3 is lower than T1 or T2, the process 3 → 1 is an adiabatic
process.
Solve: (a) The number of moles of gas is
n=
0.120 g
= 0.030 mols
4 g mol
The temperature T1 can be calculated to be
T1 =
5
−3
3
p1V1 ( 3.0 × 1.013 × 10 Pa )(1.0 × 10 m )
=
= 1219 K = 946°C
nR
( 0.030 mol )( 8.31 J mol K )
For the isothermal process 1 → 2, T2 = 1219 K. For the adiabatic process 3 → 1,
1.67
⎛ 1000 cm3 ⎞
p3 = p1 (V1 V3 ) = ( 3.0 atm ) ⎜
3 ⎟
⎝ 3000 cm ⎠
γ
Point
= 0.48 atm
p (atm)
T3 = T1 (V1 V3 )
γ −1
T (°C)
= (1219 K ) ( 13 )
0.67
= 583 K=310°C
V (cm3)
1
3.0
946
1000
2
1.0
946
3000
3
0.48
310
3000
Note that the values obtained above are consistent with the isochoric process 2 → 3, for which
p2 p3
=
⇒ p2 = (T2 T3 ) p3 = (1219 K 583 K )( 0.48 atm ) = 1 atm
T2 T3
(b) From Equation 17.15,
⎛V ⎞
W1→ 2 = − nRT1 ln ⎜ 2 ⎟ = − ( 0.030 mol )( 8.31 J mol K )(1219 K ) ln ( 3) = −334 J
⎝ V1 ⎠
The work done in the ischochoric process is W2 →3 = 0 J . The work done in the adiabatic process is
W3→1 = nCV (T1 − T3 ) = ( 0.030 mol )(12.5 J mol K )(1219 K − 583 K ) = 239 J
(c) For the process 1 → 2, ΔT = 0 K ⇒ ΔEth = 0 J ⇒ Q = −W = 334 J
For the process 2 → 3, W = 0 J,
ΔEth = Q = nCV (T3 − T2 ) = −239 J
For the process 3 → 1, Q = 0 J.
17.80. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving
molecules.
Visualize: Please refer to Figure P17.80.
Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the piston’s mass
must exactly balance the upward force of the gas, Fgas = pA where A = πr2 is the area of the face of the piston.
Since the upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus,
M piston g = pA ⇒ p =
M piston g
A
=
ρCuVpiston g
A
= ρ Cu gh = ( 8920 kg m3 )( 9.80 m s 2 ) ( 0.040 m ) = 3500 Pa
(b) The gas volume is V1 = π r 2 L = π (0.030) 2 (0.20 m) = 5.65 × 10−4 m3 . The number of moles is
n=
−4
3
p1V1 ( 3500 Pa ) ( 5.65 × 10 m )
=
= 8.12 × 10−4 mol
RT1
( 8.31 J mol K )( 293 K )
The number of molecules is
N = nNA = (8.12 × 10−4 mol)(6.02 × 1023 mol−1) = 4.9 × 1020
(c) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is
added, so the heating takes place at constant pressure with Q = nCPΔT. The temperature increase is
ΔT =
Q
2.0 J
=
= 85 K
nCP ( 8.12 × 10−4 mol ) ( 29.1 J mol K )
This raises the gas temperature to T2 = T1 + ΔT = 378 K = 105°C.
(d) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal-gas relationship for an
isobaric process is
V2 V1
L
L
T
⎛ 378 K ⎞
= ⇒ 2 = 1 ⇒ L2 = 2 L1 = ⎜
⎟ 20 cm = 25.8 cm
T2 T1
T2 T1
T1
⎝ 293 K ⎠
(e) The work done by the gas is Wgas = Fgas Δy . The force exerted on the piston by the gas is
Fgas = pA = pπ r 2 = 9.90 N
This force is applied through Δy = 5.8 cm = 0.058 m, so the work done is
Wgas = (9.90 N)(0.058 m) = 0.574 J ≈ 0.57 J
Thus, 0.57 J is the work done by the gas on the piston. The work done on the gas is 0.57 J.
17.81. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature
(20°C), and the liquid nitrogen. The boiling point of liquid nitrogen is –196°C = 77 K.
Solve: The piece of iron has mass Miron = 197 g = 0.197 kg and volume Viron = Miron/ρiron = (0.197 kg)/(7870
kg/m3) = 25 × 10−6 m3 = 25 cm3 = 25 mL. The heat lost by the iron is
Qiron = Mironciron∆T = (0.197 kg)(449 J/kg K)(77 K – 293 K) = –1.911 × 104 J
This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires
Qiron + QN2 = Qiron + MLf = 0 ⇒ M = −
Qiron
1.911 × 104 J
=
= 0.0960 kg
Lf
1.99 × 105 J/kg
The volume of liquid nitrogen boiled away is thus
Vboil =
M
ρ N2
=
0.0960 kg
= 1.19 × 10−4 m3 = 119 mL
810 kg/m3
Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of
iron excludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is V1 =
1475 mL. The pressure is p1 = 1.0 atm and the temperature is T1 = 77 K. Thus the number of moles of nitrogen
gas is
n1 =
p1V1 (101,300 Pa)(1.475 × 10−3 m3 )
=
= 0.234 mol
RT1
(8.31 J/mol K)(77 K)
119 mL of liquid boils away, so the gas volume increases to V2 = V1 + Vboil = 1475 mL + 119 mL = 1594 mL. The
temperature is still T2 = 77 K, but the number of moles of gas has been increased by the liquid that boiled. The
number of moles that boiled away is
96.0 g
nboil =
= 3.429 mol
28 mol/g
Thus the number of moles of nitrogen gas increases to n2 = n1 + nboil = 0.234 mol + 3.429 mol = 3.663 mol.
Consequently, the gas pressure increases to
p2 =
n2 RT2 (3.663 mol)(8.31 J/mol K)(77 K)
=
= 1.470 × 106 Pa = 14.5 atm ≈ 15 atm
V2
1.594 × 10−3 m3
Assess: Don’t try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading to serious
injuries.
17.82. Model: Ignore any radiation and assume all of the heat is transferred by conduction through the
compound rod from the hot end to the cold end.
Visualize: Look up the thermal conductivities of the two metals in Table 17.5: kCu = 400 W/m? k and
kFe = 80 W/m?K . TH = 373 K and TC = 273 K. Equation 17.48 will be applied to each rod and all of the heat
must go through both rods at the same rate.
⎛Q⎞
⎛Q⎞
⎜ ⎟ =⎜ ⎟
D
t
⎝ ⎠Cu ⎝ Dt ⎠ Fe
Solve:
Since A and L are the same for the two rods, then
kCu DTCu = kFeDTFe
Label the temperature at the joining point in the middle with a subscript M.
kCu (TH − TM ) = kFe (TM − TC )
Solve for TM .
kCuTH − kCuTM = kFeTM − kFeTC
kCuTH + kFeTC = kFeTM + kCuTM
( kFe + kCu )TM = kCuTH + kFeTC
TM =
TM =
kCuTH + kFeTC
kCu + kFe
(400 W/m ⋅ K)(373 K) + (80 W/m ⋅ K)(273 K)
= 356 K = 83°C
400 W/m ⋅ K + 80 W/m ⋅ K
Assess: Had we not arrived at an answer between 100°C and 0°C we would have been very worried.
Furthermore, because the conductivity of copper is greater than the conductivity of iron we expected the answer
to be above 50°C.
17.83. Model: Assume the gas is ideal. The process is not adiabatic, despite Equation 17.38, because the exponent
given in the problem is not equal to g . The gas is identified as diatomic which means that g = 1.40 , but the exponent is
explicitly given as 2, so while we are given pV 2 = constant, it is not true that pV g = constant. We will use the fact that
the gas is diatomic to deduce CV = 52 R = 20.8J/mol ⋅ k , although we won’t specifically need g = 1.40 .
We are given n = 0.020mol , Ti = 293K , Vi = 1500cm3 , and Vf = 500cm3 .
Visualize:
Solve:
(a) If we use the ideal-gas-law expression p = nRT/V in pV 2 = constant we get TV = constant .
Tf =
TV
(293 K)(1500 cm3 )
i i
=
= 879 K = 606°C
Vf
500 cm3
(b) The strategy to find Q will be to use the first law Q = DEth − W . We will use Eth = nCV DT and W = 2∫ pdV.
DEth = nCV DT = ( 0.020 mol )( 20.8 J/mol ⋅ k )( 879 K − 293 K ) = 243.8 J
To do the W = 2∫ pdV integral we need to know what the constant is in pV 2 = constant . Use the ideal gas law to
compute pi .
pi =
nRTi ( 0.020 mol )( 8.31 J/mol ⋅ K )( 293 K )
=
= 32,460 Pa
Vi
1.5 ×1023 m3
2
23
constant = pV
m3 ) = 0.0730 Pa ⋅ m 6
i i = ( 32,460Pa ) (1.5 × 10
2
Vf
W = − ∫ pdV = 2Ñ
Vi
constant
V2
0.00050 m3
dv
⎡1⎤
dV = 2( 0.0730 Pa ⋅ m ) Ñ
= ( 0.0730 Pa ⋅ m 6 ) ⎢ ⎥
= 97.4 J
0.0015 m3 V 2
⎣V ⎦ 0.0015 m3
6
0.00050 m3
Q = DEth − W = 243.8 J − 97.4 J = 146.4 J < 150 J
(c) We also need pf to complete the pV diagram. We could use the ideal-gas law again to compute pf , but let’s do it
another way.
2
2
pV
i i = pf Vf
2
2
⎛V ⎞
⎛ 1500 cm3 ⎞
pf = pi ⎜ i ⎟ = pi ⎜
= 9 pi = 292,200 Pa
3 ⎟
⎝ 500 cm ⎠
⎝ Vf ⎠
Assess: Both Q and W are positive because heat was added to the system and work was done on the system.
17-1
18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number
(N/V):
(1.013 × 105 Pa ) = 2.69 × 1025 m−3
N
p
=
=
V kBT (1.38 × 10−23 J K ) ( 273 K )
18.2. Solve: The volume of the nitrogen gas is 1.0 m3 and its temperature is 20°C or 293 K. The number of
gas molecules can be found as
N = nN A =
(1.013 × 105 Pa )(1.0 m3 ) 6.02 × 1023 mol−1 = 2.5 × 1025
pV
NA =
(
)
RT
( 8.31 J mol K )( 293 K )
According to Figure 18.2, 12% of the molecules have a speed between 700 and 800 m/s, 7% between 800 and
900 m/s, and 3% between 900 and 1000 m/s. Thus, the number of molecules in the cube with a speed between
700 m/s and 1000 m/s is (0.22)(2.51 × 1025) = 5.5 × 1024.
18.3. Solve: Nitrogen is a diatomic molecule, so r ≈ 1.0 × 10−10 m. We can use the ideal-gas law in the form
pV = NkBT and Equation 18.3 for the mean free path to obtain p:
λ=
(1.38 ×10−23 J k ) ( 293 K ) = 0.023 Pa
kBT
1
kBT
p
=
⇒
=
=
2
4 2πλ r 2 4 2π (1.0 m ) (1.0 × 10−10 m )
4 2π ( N V ) r 2 4 2π pr 2
Assess: In Example 18.1 λ = 225 nm at STP for nitrogen. λ = 1.0 m must therefore require a very small
pressure.
18.4. Solve: (a) Air is primarily comprised of diatomic molecules, so r ≈ 1.0 × 10−10 m. Using the ideal-gas
law in the form pV = NkBT, we get
N
p
=
=
V kBT
1.013 × 105 Pa
760 mm of Hg
= 3.30 × 1012 m −3
−23
(1.38 ×10 J K ) ( 293 K )
1.0 × 10−10 mm of Hg ×
(b) The mean free path is
λ=
1
1
=
= 1.71 × 106 m
2
4 2π ( N V ) ( r 2 ) 4 2π ( 3.30 × 1012 m −3 )(1.0 × 10−10 m )
Assess: The pressure p in the vacuum chamber is 1.33 × 10−8 Pa = 1.32 × 10−13 atm. A mean free path of 1.71 ×
106 m is large but not unreasonable.
18.5. Solve: (a) The mean free path of a molecule in a gas at temperature T1, volume V1, and pressure p1 is λ1 =
300 nm. We also know that
λ=
1
⇒ λ ∝V
4 2π ( N / V )r 2
Although T2 = 2T1, constant volume (V2 = V1) means that λ2 = λ1 = 300 nm.
(b) For T2 = 2 T1 and p2 = p1, the ideal gas equation gives
p1V1
pV
p1V2
⇒ V2 = 2V1
= 2 2 =
NkBT1 NkBT2 NkB ( 2T1 )
Because λ ∝ V , λ2 = 2λ1 = 2(300 nm) = 600 nm.
18.6. Solve: Neon is a monatomic gas and has a radius r ≈ 5.0 × 10−11 m. Using the ideal-gas equation,
(150 ) (1.013 × 105 Pa )
N
p
=
=
= 3.695 × 1027 m −3
V kBT (1.38 × 10−23 J/K ) ( 298 K )
Thus, the mean free path of a neon atom is
λ=
1
1
=
= 6.09 × 10−9 m
2
2
27
4 2π ( N V ) r
4 2π ( 3.695 × 10 m −3 )( 5.0 × 10−11 m )
Since the atomic diameter of neon is 2 × (5.0 × 10−11 m) = 1.0 × 10–10 m,
λ=
6.09 × 10−9 m
= 61 atomic diameters
1.0 × 10−10 m
18.7. Solve: The number density of the Ping-Pong balls inside the box is
N
2000
=
= 2000 m −3
V 1.0 m3
With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is
λ=
1
= 0.125 m = 12.5 cm
4 2π ( N V ) ( r 2 )
18.8. Solve: (a) The average speed is
vavg =
1 ⎛ n = 25 ⎞
220 m/s
= 20.0 m/s
⎜ ∑ n ⎟ m/s =
11 ⎝ n =15 ⎠
11
(b) The root-mean-square speed is
vrms =
(v2 )
1
avg
1
⎛ 1 n = 25 ⎞ 2
⎛ 4510 ⎞ 2
= ⎜ ∑ n 2 ⎟ m/s = ⎜
⎟ = 20.2 m/s
⎝ 11 ⎠
⎝ 11 n =15 ⎠
18.9. Solve: (a) In tabular form we have
Particle
vx (m/s)
vy (m/s)
1
20
30
2
70
−40
3
−80
−10
4
60
−20
5
0
−50
6
40
−20
Average
0
0
G
G
G
ˆ
The average velocity is vavg = 0 i + 0 ˆj .
vx2 (m/s)2
v y2 (m/s)2
v2 (m/s)2
v (m/s)
400
1600
6400
3600
0
1600
900
4900
100
400
2500
400
1300
6500
6500
4000
2500
2000
3800
36.06
80.62
80.62
63.25
50.00
44.72
59.20
(b) The average speed is vavg = 59 m/s .
(c) The root-mean-square speed is vrms =
(v )
2
avg
= 3800 m 2 / s 2 = 62 m/s .
18.10. Solve: (a) The most probable speed is 4.0 m/s.
(b) The average speed is
vavg =
2 × 2 m/s + 4 × 4 m/s + 3 × 6 m/s + 1 × 8 m/s
= 4.6 m/s
2 + 4 + 3 +1
(c) The root-mean-square speed is
2 × ( 2 m/s ) + 4 × ( 4 m/s ) + 3 × ( 6 m/s ) + 1 × ( 8 m/s )
= 4.9 m/s
2 + 4 + 3 +1
2
vrms =
2
2
2
18.11. Solve: (a) The atomic mass number of argon is 40. This means the mass of an argon atom is
m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg
The pressure of the gas is
2
⎛N⎞ 2
p = 13 ⎜ ⎟ mvrms
= 13 ( 2.00 × 1025 m −3 )( 6.64 × 10−26 kg ) ( 455 m/s ) = 9.16 × 104 Pa
⎝V ⎠
(b) The temperature of the gas in the container can be obtained from the ideal-gas equation in the form pV = NkBT:
T=
pV
9.16 × 104 Pa
=
= 332 K
25
NkB ( 2.00 × 10 m −3 )(1.38 × 10−23 J/K )
18.12. Model: Pressure is due to random collisions of gas molecules with the walls.
Solve:
According to Equation 18.8, the collision rate with one wall is
rate of collisions =
N coll
F
pA
= net =
Δtcoll 2mvx 2mvx
where Fnet = pA is the force exerted on area A by the gas pressure. However, this equation assumed that all
molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three
dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with
vx → (vx2)avg =
2
vrms
v
= rms
3
3
With this change,
rate of collisions =
3 pA
3(2 × 101,300 Pa)(0.10 m × 0.10 m)
=
= 6.5 × 1025 s −1
2mvrms
2(28 × 1.661× 10−27 kg)(576 m/s)
This collision rate can also be found by using the expression in Eq. 18.10, making the same change in vx, and
using the ideal-gas law to determine N/V.
18.13. Visualize: We will use Equation 18.18 to find m of the atoms.
p=
1N 2
mvrms
3V
We are given p = 2.0 atm = 202,650 Pa , N /V = 4.2 × 1025 m −3 , and vrms = 660 m/s.
Solve: Solve for m.
m=
3 p V 3(202,650 Pa)
1
=
= 3.32 × 10−26 kg = 20 u
2
vrms
N
(660 m/s) 2 4.2 ×1025 m −3
With an atomic mass of 20 the gas is likely neon.
Assess: Neon is gaseous at room temperature, so is a likely choice. As a noble gas, it also doesn’t form many
molecules.
18.14. Visualize: Equation 18.26 will give us vrms from the temperature and mass of the particles.
vrms =
3kBT
m
We are given T = 1100°C = 1373 K . The atomic mass of neon atoms is 20 u = 3.321 × 10−26 kg; the molecular
mass of oxygen molecules is 32 u = 5.31 × 10−26 kg.
Solve:
(a)
vrms =
3(1.38 × 10−23 J/K)(1373 K)
= 1310 m/s
3.321 × 10−26 kg
vrms =
3(1.38 × 10−23 J/K)(1373 K)
= 1030 m/s
5.31 × 10−26 kg
(b)
Assess: These are fast speeds, but the temperature is high.
18.15. Visualize: Use Equation 18.26.
vrms =
3kBT
m
We are given vrms = 1.5 m/s and m = 28 u = 4.650 × 10−26 kg.
Solve: Solve the equation for T .
T=
Assess:
2
vrms
m (1.5 m/s) 2 (4.650 × 10−26 kg)
=
= 2.5 mK
3kB
3(1.38 × 10−23 J/K)
2.5 mK is close to absolute zero, but that’s how cold it would have to be for vrms to be 1.5 m/s.
18.16. Solve: Because the neon and argon atoms in the mixture are in thermal equilibrium, the temperature
of each gas in the mixture must be the same. That is, using Equation 18.26,
2
2
mAr vrms
Ar = mNe vrms Ne
vrms Ar = vrms Ne
20 u
mNe
= ( 400 m/s )
= 283 m/s
40 u
mAr
18.17. Solve: The average translational kinetic energy per molecule is
3kBT
1 2
3
Pavg = mvrms
= kBT ⇒ vrms =
2
2
m
Since we want the vrms for H2 and N2 to be equal,
3kBTH2
mH2
=
mH
3kBT
⎛ 2u ⎞
⇒ TH2 = 2 TN2 = ⎜
⎟ ( 373 K ) = 27 K = −246°C
mN2
mN2
⎝ 28 u ⎠
18.18. Solve: The formula for the root-means-square speed as a function of temperature is
( vrms )T =
3kBT
m
(a) For ( vrms )T = 12 ( vrms )STP ,
3kBT 1 3kB ( 273 K )
1
⇒ T = (273 K) = 68 K= − 205°C
=2
4
m
m
(b) For ( vrms )T = 2 ( vrms )STP ,
3kB ( 273 K )
3kBT
=2
⇒ T = 4 ( 273 K ) = 1090 K = 817°C
m
m
18.19.
Solve:
Solve Equation 18.18 for vrms to see that if both p and V are doubled then vrms is also doubled.
vrms =
3 pV
Nm
So the new rms speed will be 800 m/s.
Assess: Think microscopically; for the pressure to double when the volume is doubled the particles will have to
be going a lot faster and hit the walls more often.
18.20. Solve: The formula for the root-means-square speed as a function of temperature is
( vrms )T =
3kBT
m
The ratio at 20°C and at 100°C is
( vrms )100
=
( vrms )20
373 K
= 1.13
293 K
18.21. Solve: Assuming ideal-gas behavior and ignoring relativistic effects, the root-mean-square speed of a
molecule is
vrms =
3kBT
m
The temperature where vrms is the speed of light (c) for a hydrogen molecule
( 2 u ) c 2 = 2 (1.66 ×10−27 kg )( 3.0 ×108 m/s ) = 7.22 ×1012 K
mv 2
T = rms =
3kB
3kB
3 (1.38 × 10−23 J/K )
2
18.22. Solve: (a) The average translational kinetic energy per molecule is
1 2
3
Pavg = mvrms
= kBT
2
2
This means Pavg doubles if the temperature T doubles.
(b) The root-mean-square speed vrms increases by a factor of
(c) The mean free path is
λ=
2 as the temperature doubles.
1
4 2π ( N V ) r 2
Because N/V and r do not depend on T, doubling temperature has no effect on λ.
18.23. Solve: (a) The total translational kinetic energy of a gas is K micro = 32 N A kBT = 32 nRT . For H2 gas at
STP,
K micro =
3
(1.0 mol )(8.31 J/mol K )( 273 K ) = 3400 J
2
K micro =
3
(1.0 mol )(8.31 J/mol K )( 273 K ) = 3400 J
2
(b) For He gas at STP,
(c) For O2 gas at STP, K micro = 3400 J.
Assess: The translational kinetic energy of a gas depends on the temperature and the number of molecules but
not on the molecule’s mass.
18.24. Solve: (a) The mean free path is
λ=
1
4 2π ( N V ) r 2
where r ≈ 0.5 × 10−10 m is the atomic radius for helium and N/V is the gas number density. From the ideal-gas
law,
N
p 0.10 atm × 101,300 Pa/atm
=
=
= 7.34 × 1025 m −3
V kT
(1.38 ×10−23 J/K ) (10 K )
⇒λ =
1
4 2π ( 7.34 × 10 m
25
−3
)( 0.5 × 10
−10
m)
2
= 3.1 × 10−7 m = 310 nm
(b) The root-mean-square speed is
vrms =
3 (1.38 × 10−23 J/K ) (10 K )
3kBT
=
= 250 m/s
m
4 × (1.661 × 10−27 kg )
where we used A = 4 u as the atomic mass of helium.
(c) The average energy per atom is ε avg = 32 kBT = 32 (1.38 × 10−23 J/K ) (10 K ) = 2.1 × 10−22 J .
18.25. Solve: (a) The average kinetic energy of a proton at the center of the sun is
3
Pavg = kBT ≈ 32 (1.38 × 10−23 J/K )( 2.0 × 107 K ) = 4.1× 10−16 J
2
(b) The root-mean-square speed of the proton is
vrms =
3 (1.38 × 10−23 J/K )( 2.0 × 107 K )
3kBT
≈
= 7.0 × 105 m/s
m
1.67 × 10−27 kg
18.26. Solve: (a) Since the hydrogen in the sun’s atmosphere is monatomic, the average translational kinetic
energy per atom is
3
Pavg = kBT = 32 (1.38 × 10−23 J/K ) ( 6000 K ) = 1.24 × 10−19 J
2
(b) The root-mean-square speed is
vrms =
3 (1.38 × 10−23 J/K ) ( 6000 K )
3kBT
=
= 1.22 × 104 m/s
mH
1.67 × 10−27 kg
18.27. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth
= Kmicro.
Solve: The number of atoms is
M
0.0020 kg
N=
=
= 3.01 × 1023
m 6.64 × 10−27 kg
Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is
m = 4 u = 4 (1.661 × 10−27 kg ) = 6.64 × 10−27 kg
The average kinetic energy of each atom is
2
K avg = 12 mvavg
= 12 ( 6.64 × 10−27 kg ) ( 700 m s ) = 1.63 × 10−21 J
2
Thus the thermal energy of the gas is
Eth = K micro = NK avg = ( 3.01× 1023 )(1.63 × 10−21 J ) = 490 J
18.28. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules.
Solve:
Neon atoms have an atomic mass number A = 14, so the mass of each molecule is
m = 28 u = 28(1.661 × 10−27 kg) = 4.51 × 10−26 kg
The number of molecules in the gas is
N=
M
0.010 kg
=
= 2.218 × 1023
m 4.51 × 10−26 kg
The thermal energy is
2
Eth = NK avg = N ( 12 mvavg
) ⇒ vavg =
2 Eth
=
Nm
2 (1700 J )
( 2.218 ×10 )( 4.51×10
23
−26
kg )
= 580 m/s
18.29. Solve: The volume of the air is V = 6.0 m × 8.0 m × 3.0 m = 144.0 m3, the pressure p = 1 atm = 1.013
× 105 Pa, and the temperature T = 20°C = 293 K. The number of moles of the gas is
pV
n=
= 5991 mol
RT
This means the number of molecules is
N = nN A = ( 5991 mols ) ( 6.022 × 1023 mol−1 ) = 3.61 × 1027 molecules
Since air is a diatomic gas, the room’s thermal energy is
Eth = N Pavg = N ( 52 kBT ) = 3.6 × 107 J
Assess:
The room’s thermal energy can also be obtained as follows:
Eth = nCVT = (5991 mol)(20.8 J/mol K)(293 K) = 3.6 × 107 J
18.30. Solve: The thermal energy of a solid is
Eth = 3 NkBT = 3nRT
The volume of lead V = 100 cm3 = 10−4 m3, which means the mass is
M = ρV = (11,300 kg/m3)(10−4 m3) = 1.13 kg
Because the atomic mass number of Pb is 207, the number of moles is
n=
M
1.13 kg
=
= 5.459 mol
M mol 0.207 kg/mol
⇒ Eth = 3( 5.459 mols )( 8.31 J/mol K )( 293 K ) = 3.99 × 104 J
18.31. Solve: (a) For a monatomic gas,
ΔEth = nCV ΔT = 1.0 J = (1.0 mol )(12.5 J/mol K ) ΔT ⇒ ΔT = 0.080°C or 0.080 K
(b) For a diatomic gas,
1.0 J = (1.0 mol)(20.8 J/mol K)ΔT ⇒ ΔT = 0.048°C or K
(c) For a solid,
1.0 J = (1.0 mol)(25.0 J/mol K) ΔT ⇒ ΔT = 0.040°C or K
18.32. Solve: The conservation of energy equation (ΔEth)gas + (ΔEth)solid = 0 J is
ngas ( CV )gas (Tf − Ti )gas + nsolid ( CV )solid (Tf − Ti )solid = 0 J
⇒ (1.0 mol)(12.5 J/mol K)(−50 K) + (1.0 mol)(25.0 J/mol K)(ΔT)solid = 0 ⇒ (ΔT)solid = 25°C
The temperature of the solid increases by 25°C.
Refer to Figure 18.13. At low temperatures, CV = 32 R = 12.5 J/mol K. At room
temperature and modestly hot temperatures, CV = 52 R = 20.8 J/mol K. At very hot temperatures,
CV = 72 R = 29.1 J/mol K.
Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is
18.33. Visualize:
0.20 g
= 0.10 mol
2 g/mol
The heat needed to change the temperature of the gas from 50 K to 100 K at constant volume is
Q = ΔEth = nCV ΔT = (0.10 mol)(12.5 J/mol K)(100 K – 50 K) = 62 J
(b) To raise the temperature from 250 K to 300 K,
Q = ΔEth = (0.10 mol)(20.8 J/mol K)(300 K – 250 K) = 100 J
(c) To raise the temperature from 550 K to 600 K, Q = 100 J.
(d) To raise the temperature from 2250 K to 2300 K, Q = ΔEth = nCVΔT = (0.10 mol)(29.1 J/mol K)(50 K) = 150
J.
18.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture.
We assume the starting temperature of the bowling ball to be 0°C.
Solve: The potential energy of the bowling ball is
U g = M ball gh = (11 kg)(9.8 m/s 2 )h = (107.8 kg m/s 2 )h
This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is,
(107.8 kg m/s 2 )h = ΔEth = M w Lf ⇒ h =
(0.005 kg)(3.33 × 105 J/kg)
= 15.4 m
(107.8 kg m/s 2 )
18.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and
raises the temperature of the kettle to 100°C.
Solve: The amount of heat energy from the electric stove’s output in 3 minutes is
Q = ( 2000 J s )( 3 × 60 s ) = 3.6 × 105 J
This heat energy heats the kettle and brings the water to a boil. Thus,
Q = M water cwater ΔT + M kettle ckettle ΔT
Substituting the given values into this equation,
3.6 × 105 J = M water ( 4190 J kg K )(100°C − 20°C ) + ( 0.750 kg )( 449 J kg K )(100°C − 20°C )
⇒ M water = 0.994 kg
The volume of water in the kettle is
V=
Assess:
3
3
M water
ρ water
=
0.994 kg
= 0.994 × 10−3 m3 = 994 cm3 ≈ 990 cm3
1000 kg m3
1 L = 10 cm , so V ≈ 1 L. This is a reasonable volume of water.
18.36. Solve: The equilibrium condition is
(εA)avg = (εB)avg = (εtot)avg ⇒ EAf = EBf = Etot
nA
nB nA + nB
Thus the final thermal energies are
⎛ nA ⎞
4.0 mols
⎛
⎞
EAf = ⎜
⎟ Etot = ⎜
⎟ ( 9000 J + 5000 J ) = 8000 J
n
+
n
+
4.0
mols
3.0
mols
⎝
⎠
⎝ A
B ⎠
⎛ nB ⎞
⎛ 3.0 mols ⎞
EBf = ⎜
⎟ Etot = ⎜
⎟ (14,000 J ) = 6000 J
n
+
n
⎝ 7.0 mols ⎠
⎝ A
B ⎠
Because EAi = 9000 J and EAf = 8000 J, 1000 J of heat energy is transferred from gas A to gas B.
18.37. Solve: The mean free path for a monatomic gas is
λ=
1
V
⇒ = 4π 2λ r 2
N
4π 2 ( N V ) r 2
For λ = 20r, meaning that the mean free path equals the atomic diameter,
V
V /N
⎛ 4π 3 ⎞
= 4π 2 ( 20r ) ( r 2 ) = ⎜
r ⎟ 60 2 ⇒ 4π 3 = 60 2 = 84.8
N
r
⎝ 3 ⎠
3
18.38. Visualize:
Solve:
The average energy of an oxygen molecule at 300 K is
ε avg =
Eth 5
= kBT = 52 (1.38 × 10−23 J/K ) ( 300 K ) = 1.035 × 10−20 J
N 2
The energy conservation equation Ugf + Kf = Ugi + Ki with K f = ε avg is
mgyf + ε avg = mgyi + 0 J
With yi = h and yf = 0 m, we have
mgh = 1.035 × 10−20 J ⇒ h =
1.035 × 10−20 J
= 1.99 × 104 m
( 32 ×1.66 ×10−27 kg )(9.8 m/s 2 )
18.39. Solve: (a) To identify the gas, we need to determine its atomic mass number A or, equivalently, the mass m
of each atom or molecule. The mass density ρ and the number density (N/V ) are related by ρ = m(N/V ) , so the mass
is m = ρ (V/N ) . From the ideal-gas law, the number density is
N
p
50,000 Pa
=
=
= 1.208 × 1025 m −3
V kT (1.38 × 10−23 J/K ) ( 300 K )
Thus, the mass of an atom is
m=ρ
V 8.02 × 10−2 kg/m3
=
= 6.64 × 10−27 kg
N 1.208 × 1025 m −3
Converting to atomic mass units,
A = 6.64 × 10−27 kg ×
1u
= 4.00 u
1.661 × 10−27 kg
This is the atomic mass of helium.
(b) Knowing the mass, we find vrms to be
vrms =
3 (1.38 × 10−23 J/K ) ( 300 K )
3kBT
=
= 1370 m/s
m
6.64 × 10−27 kg
(c) A typical atomic radius is r ≈ 0.5 × 10−10 m. The mean free path is thus
λ=
1
1
=
= 1.86 × 10−6 m = 1.86 μ m
2
4 2π ( N V ) r 2 4 2π (1.208 × 1025 m −3 )( 0.5 × 10−10 m )
18.40. Solve: (a) The number density is N / V = 1 cm −3 = 106 m −3 . Using the ideal-gas equation,
N
kBT ≈ (1 × 106 m −3 )(1.38 × 10−23 J/K ) ( 3 K )
V
1 atm
= 4 × 10−17 Pa ×
= 4 × 10−22 atm
1.013 × 105 Pa
p=
(b) For a monatomic gas,
vrms =
3 (1.38 × 10−23 J/K ) ( 3 K )
3kBT
= 270 m/s
=
m
1.67 × 10−27 kg
(c) The thermal energy is Eth = 32 NkBT , where N = (106 m −3 )V . Thus
Eth = 1.0 J = 32 (106 m −3 )V (1.38 × 10−23 J/K ) ( 3 K ) ⇒ V = 1.6 × 1016 m3 = L3
⇒ L = 2.5 × 105 m
18.41. Solve: Mass m of a dust particle is
3
m = ρV = ρ ( 43 π r 3 ) = ( 2500 kg/m3 ) ⎡ 43 π ( 5 × 10−6 m ) ⎤ = 1.3 × 10−12 kg
⎢⎣
⎥⎦
The root-mean-square speed of the dust particles at 20°C is
vrms =
3 (1.38 × 10−23 J/K ) ( 293 K )
3kBT
=
= 9.6 × 10−5 m/s
m
1.3 × 10−12 kg
18.42. Solve: Fluorine has atomic mass number A = 19. Thus the root-mean-square speed of 238UF6 is
vrms ( 238 UF6 ) =
3kBT
3kBT
=
m
238 u + 6 × 19 u
The ratio of the root-mean-square speed for the molecules of this isotope and the 235UF6 molecules is
vrms ( 235 UF6 )
vrms (
238
UF6 )
=
( 238 + 6 × 19 ) u =
( 235 + 6 × 19 ) u
352
= 1.0043
349
18.43. Solve: (a) If the electron can be thought of as a point particle with zero radius, then it will collide with
any gas particle that is within r of its path. Hence, the number of collisions Ncoll is equal to the number of gas
particles in a cylindrical volume of length L. The volume of a cylinder is Vcyl = AL = (π r 2 ) L . If the number
density of the gas is (N/V) particles per m3, then the number of collisions along a trajectory of length L is
N coll =
Introducing a factor of
N
N
L
1
Vcyl = (π r 2 L ) ⇒ λelectron =
=
V
V
N coll π ( N / V ) r 2
2 to account for the motion of all particles,
λelectron =
L
1
=
N coll
2π ( N / V ) r 2
(b) Assuming that most of the molecules in the accelerator are diatomic,
5.0 × 104 m =
1
2π ( N / V ) (1.0 × 10
−10
m)
2
⇒ N / V = 4.50 × 1014 m −3
From the ideal-gas equation,
p=
N
kBT = ( 4.50 × 1014 m −3 )(1.38 × 10 −23 J/K ) ( 293 K ) = 1.82 × 10 −6 Pa = 1.80 × 10 −11 atm
V
18.44. Solve: The pressure on the wall with area A = 10 cm2 = 10 × 10−4 m2 is
p=
F Δ ( mv ) N
=
A
AΔt
where N Δt is the number of N2 molecules colliding with the wall every second and Δ(mv) is the change in
momentum for one collision. The mass of the nitrogen molecule is
m = 28 u = 28 (1.66 × 10−27 kg) = 4.648 × 10−26 kg
and Δv = 400 m/s − (−400 m/s) = 800 m/s . Thus,
( 4.648 ×10
p=
−26
kg ) ( 800 m/s ) ( 5.0 × 1023 s −1 )
1.0 × 10−3 m 2
= 1.9 × 104 Pa
18.45. Solve: (a) The cylinder volume is V = πr2L = 1.571 × 10–3 m3. Thus the number density is
N
2.0 × 1022
=
= 1.273 × 1025 m −3 ≈ 1.3 × 1025 m −3
V 1.571 × 10−3 m3
(b) The mass of an argon atom is
m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg
⇒ vrms =
3 (1.38 × 10−23 J/K ) ( 323 K )
3kBT
=
= 449 m/s ≈ 450 m/s
m
6.64 × 10−26 kg
(c) vrms is the square root of the average of v2. That is,
2
vrms
= ( v2 )
avg
= ( vx2 )
avg
+ ( v y2 )
avg
+ ( vz2 )
avg
An atom is equally likely to move in the x, y, or z direction, so on average ( vx2 )
2
vrms
= 3 ( vx2 )
avg
⇒ ( vx )rms =
(v )
2
x avg
=
avg
= ( v y2 )
avg
= ( vz2 )
avg
. Hence,
vrms
= 259 m/s ≈ 260 m/s
3
(d) When we considered all the atoms to have the same velocity, we found the collision rate to be 12 ( N/V ) Avx
(see Equation 18.10). Because the atoms move with different speeds, we need to replace vx with (vx)rms. The end
of the cylinder has area A = πr2 = 7.85 × 10–3 m2. Therefore, the number of collisions per second is
1
2
( N / V ) A ( vx )rms = 12 (1.273 × 1025 m−3 )( 7.85 × 10−3 m2 ) ( 259 m/s ) = 1.3 × 1025 s−1
(e) From kinetic theory, the pressure is
2
⎛N⎞
⎛N⎞ 2
p = 13 ⎜ ⎟ m ( v 2 ) = 13 ⎜ ⎟ mvrms
= 13 (1.273 × 1025 m −3 )( 6.64 × 10−26 kg ) ( 449 m/s ) = 56,800 Pa ≈ 57,000 Pa
avg
V
V
⎝ ⎠
⎝ ⎠
(f) From the ideal-gas law, the pressure is
22
−23
NkBT ( 2.0 × 10 )(1.38 × 10 J/K ) ( 323 K )
=
= 56,700 Pa ≈ 57,000 Pa
V
1.571 × 10−3 m3
Assess: The very slight difference between parts (e) and (f) is due to rounding errors; to two significant figures
they are the same.
p=
18.46. Model: Pressure is due to random collisions of gas molecules with the walls.
Solve:
According to Equation 18.9, the collision rate with one wall is
N
1N
rate of collisions = coll =
Avx
Δtcoll 2 V
However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed
vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which
is zero, but with
vx → (vx2)avg =
2
vrms
v
= rms
3
3
rate of collisions =
1 N
Avrms
2 3V
With this change,
The molecular mass of nitrogen is A = 28 u, thus the rms speed of the molecules at 20°C is
vrms =
3kBT
3(1.38 × 10 −23 J/K)(293 K)
=
= 510 m/s
m
28(1.661 × 10 –27 kg)
With N = 0.10NA = 6.02 × 1022 molecules, the number density is
N
6.02 × 1022
=
= 6.02 × 1025 m −3
V 0.10 m × 0.10 m × 0.10 m
Thus
1
rate of collisions =
(6.02 × 1025 m −3 )(0.10 m × 0.10 m)(510 m/s)=8.9 × 1025 s −1
2 3
18.47.
Model:
Assume the gas is ideal so that Equation 18.30 will apply.
ΔEth = nCV ΔT
Visualize:
Solve:
We see from the graph that ΔT = 200 K and ΔEth = 800 J. We are also given n = 0.14 mol.
Solve for CV .
CV =
Assess:
This is about 72 R.
ΔEth
800 J
=
= 29 J/mol ⋅ K
nΔT (0.14 mol)(200 K)
18.48. Solve: (a) The number of molecules of helium is
pV ( 2.0 × 1.013 × 10 Pa )(100 × 10 m )
=
= 3.936 × 1021
kBT
(1.38 ×10−23 J/K ) ( 373 K )
−6
5
N helium =
⇒ nhelium =
3
3.936 × 1021
= 6.536 × 10−3 mol
6.022 × 1023 mol−1
The initial internal energy of helium is
Ehelium i = 32 N helium kBT = 30.4 J ≈ 30 J
The number of molecules of argon is
pV ( 4.0 × 1.013 × 10 Pa )( 200 × 10 m )
=
= 8.726 × 1021
kBT
(1.38 ×10−23 J/K ) ( 673 K )
−6
5
N argon =
⇒ nargon =
3
8.726 × 1021
= 1.449 × 10−2 mol
6.022 × 1023 mol−1
The initial thermal energy of argon is
Eargon i = 32 N argon kBT = 121.6 J ≈ 122 J
(b) The equilibrium condition for monatomic gases is
(εhelium f)avg = (εargon f)avg = (εtotal)avg
⇒
Ehelium f
nhelium
=
Eargon f
nargon
=
Etot
( 30.4 + 121.6 ) J
=
= 7228 J/mol
ntot ( 6.54 × 10−3 + 1.449 × 10−2 ) mol
⇒ Ehelium f = ( 7228 J/mol ) nhelium = ( 7228 J/mol ) ( 6.54 × 10−3 mol ) = 47.3 J ≈ 47 J
Eargon f = ( 7228 J/mol ) nargon = ( 7228 J/mol ) (1.449 × 10−2 mol ) = 104.7 J ≈ 105 J
(c) The amount of heat transferred is
Ehelium f − Ehelium i = 47.3 J − 30.4 J = 16.9 J
Eargon f − Eargon i = 104.7 J − 121.6 J = −16.9 J
The helium gains 16.9 J of heat energy and the argon loses 16.9 J. Thus approximately 17 J are transferred from
the argon to the helium.
(d) The equilibrium condition for monatomic gases is
( ε helium )avg = (ε argon )avg ⇒
Ehelium f
N helium
=
Eargon f
N argon
= 32 kBTf
Substituting the above values,
47.3 J
104.7 J
=
= 3 (1.38 × 10−23 J/K ) Tf ⇒ TF = 580 K = 307°C
3.936 × 1021 8.726 × 1021 2
(e) The final pressure of the helium and argon are
phelium f =
pargon f =
21
−23
N helium kBT ( 3.936 × 10 )(1.38 × 10 J/K ) ( 580 K )
=
= 3.15 × 105 Pa ≈ 3.1 atm
Vhelium
100 × 10−6 m 3
N argon kBT
Vargon
(8.726 ×10 )(1.38 ×10
21
=
200 × 10
−6
J/K ) ( 580 K )
−23
m
3
= 3.49 × 105 Pa ≈ 3.4 atm
18.49. Solve: (a) The number of moles of helium and oxygen are
nhelium =
2.0 g
= 0.50 mol
4.0 g/mol
noxygen =
8.0 g
= 0.25 mol
32.0 g/mol
Since helium is a monoatomic gas, the initial thermal energy is
Ehelium i = nhelium ( 32 RThelium ) = ( 0.50 mol ) ( 32 ) ( 8.31 J/mol K )( 300 K ) = 1870 J ≈ 1900 J
Since oxygen is a diatomic gas, the initial thermal energy is
Eoxygen i = noxygen ( 52 RToxygen ) = ( 0.25 mol ) ( 52 ) ( 8.31 J/mol K )( 600 K ) = 3116 J ≈ 3100 J
(b) The total initial thermal energy is
Etot = Ehelium i + Eoxygen i = 4986 J
As the gases interact, they come to equilibrium at a common temperature Tf. This means
4986 J = nhelium ( 32 RTf ) + noxygen ( 52 RTf )
⇒ Tf =
4986 J
4986 J
=1
= 436.4 K = 436 K
( R ) ( 3nhelium + 5noxygen ) 2 (8.31 J/mol K )( 3 × 0.50 mol + 5 × 0.25 mol )
1
2
The thermal energies at the final temperature Tf are
Ehelium f = nhelium ( 32 RTf ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( 436.4 K ) = 2700 J
Eoxygen f = noxygen ( 52 RTf ) = ( 52 ) ( 0.25 mol )( 8.31 J/mol K )( 436.4 K ) = 2300 J
(c) The change in the thermal energies are
Ehelium f − Ehelium i = 2720 J − 1870 J = 850 J
Eoxygen f − Eoxygen i = 2266 J − 3116 J = −850 J
The helium gains energy and the oxygen loses energy.
(d) The final temperature can also be calculated as follows:
Ehelium f = ( nhelium ) 32 RTf ⇒ 2720 J = (0.50 mol)(1.5)(8.31 J/mol K)Tf ⇒ Tf = 436.4 K ≈ 436 K
18.50. Solve: The thermal energy of any ideal gas is related to the molar specific heat at constant volume by
Eth = nCVT
Since CP = CV + R ,
20.8 J/mol K = CV + R ⇒ CV = 12.5 J/mol K
The number of moles of the gas is
n=
1.0 × 1020
= 1.66 × 10−4 mol
6.02 × 1023 mol −1
Thus
T=
(1.0 J )
= 482 K
(1.66 ×10 mols ) (12.5 J/mol K )
−4
18.51. Solve: For a system with n degrees of freedom, the molar specific heat is CV = nR/2. The specific heat
ratio is
γ=
CP CV + R
R
R
R
=
=1+
⇒
= γ − 1 = 1.29 − 1 = 0.29 ⇒ CV =
= 3.5 R = 72 R
0.29
CV
CV
CV
CV
Thus, the system has 7 degrees of freedom.
18.52. Solve: As the volume V1 of a gas increases to V2 = 2V1 at a constant pressure p1 = p2, the temperature
of the gas changes from T1 to T2 as follows:
p2V2 p1V1
V p
⇒ T2 = T1 2 2 = 2T1
=
V1 p1
T2
T1
Since the process occurs at constant pressure the heat transferred is
Q = nCP ΔT = nCP (T2 − T1 ) = nCP ( 2T1 − T1 ) = nCPT1
For a monatomic gas,
CP = CV + R = 32 R + R = 52 R
For the diatomic gas,
CP = CV + R = 52 R + R = 72 R
Thus
n ( 7 R ) T1
Qdiatomic
= 52
= 1.40
Qmonatomic n ( 2 R ) T1
18.53. Solve: Assuming that the systems are thermally isolated except from each other, the total energy for
the two thermally interacting systems must remain the same. That is,
E1i + E2i = E1f + E2f ⇒ E1f − E1i = E2i − E2f = − ( E2f − E1f ) ⇒ ΔE1 = −ΔE2
No work is done on either system, so from the first law ΔE = Q. Thus
Q1 = −Q2
That is, the heat lost by one system is gained by the other system.
18.54. Solve: (a) From Equation 18.26 vrms = 3kBT m . For an adiabatic process
γ −1
γ −1
TfVf
= TV
i i
γ −1
⎛V ⎞
⇒ Tf = Ti ⎜ i ⎟
⎝ Vf ⎠
⇒ Tf = Ti ( 8 ) 3 = 4Ti
5 −1
The root-mean-square speed increases by a factor of 2 with an increase in temperature.
(b) From Equation 18.3 λ = [4 2π ( N / V )r 2 ]−1. A decrease in volume decreases the mean free path by a factor of 1/8.
(c) For an adiabatic process,
γ −1
γ −1
Tf Vf
= TV
i i
γ −1
⎛V ⎞
⇒ Tf = Ti ⎜ i ⎟
⎝ Vf ⎠
= Ti ( 8 ) 3 = 4Ti
5 −1
Because the decrease in volume increases Tf , the thermal energy increases by a factor of 4.
(d) The molar specific heat at constant volume is CV = 32 R, a constant. It does not change.
18.55. Model: Assume the gas is monatomic.
Visualize: From the equipartition theorem there is 12 nRT of energy for each degree of freedom. For a twodimensional monatomic gas there are only two degrees of freedom.
2
Solve: (a) CV = R = R
2
(b) Equation 17.34 gives Cp = CV + R , so
Cp = R + R = 2 R
Assess:
It would be nice to measure the values and compare with these predictions.
18.56. Solve: (a) The thermal energy of a monatomic gas of N molecules is Eth = N Pavg , where Pavg = 32 kBT .
A monatomic gas molecule has 3 degrees of freedom. However, a two-dimensional monatomic gas molecule has
only 2 degrees of freedom. Thus,
⎛2
⎞
Eth = N ⎜ kBT ⎟ = NkBT = nRT
2
⎝
⎠
If the temperature changes by ΔT, then the thermal energy changes by ΔEth = nRΔT . Comparing this form with
ΔEth = nCV ΔT , we have CV = R = 8.31 J/mol K.
(b) A two-dimensional solid has 2 degrees of freedom associated with the kinetic energy and 2 degrees of
freedom associated with the potential energy (or 2 spring directions), giving a total of 4 degrees of freedom.
Thus, Eth = N Pavg and Pavg = 42 kBT . Or Eth = 2NkBT = 2nRT. For a temperature change of ΔT, ΔEth = 2nRΔT =
nCVΔT ⇒ CV = 2 R = 16.6 J/mol K.
18.57. Solve: (a) The rms speed is
vrms =
v
3kBT
32 u
⇒ rms hydrogen =
=4
vrms oxygen
2u
m
(b) The average translational energy is P = 32 kBT . Thus
Pavg hydrogen
Pavg oxygen
=
Thydrogen
Toxygen
=1
(c) The thermal energy is
Eth = 52 nRT
⇒
Eth hydrogen
Eth oxygen
=
nhydrogen
noxygen
=
mhydrogen 32.0 g/mol
= 16
2.0 g/mol moxygen
18.58. Visualize: We are given vrms = 1800 m/s. For 1.0 g of molecular hydrogen gas
⎛ 1 mol ⎞
⎛
23 molecules ⎞
23
N = 1.0 g ⎜
⎟ = 0.5 mol ⎜ 6.02 × 10
⎟ = 3.01× 10 molecules
mol ⎠
⎝
⎝ 2.0 g ⎠
−27
The mass of one molecule is 2(1.661× 10 kg) = 3.322 ×10−27 kg.
Solve:
(a)
2
⎛1⎞ 2
⎛1⎞
∈total = N ⎜ ⎟ mvrms
= ( 3.01× 1023 ) ⎜ ⎟ ( 3.322 × 10−27 kg ) (1800 m/s ) = 1.6 kJ
⎝2⎠
⎝2⎠
(b) The temperature is given by Equation 18.25.
2
2 ⎛1⎞ 2
3.322 ×10−27 kg
2
T=
∈ave =
(1800 m/s ) = 260 K
⎜ ⎟ mvrms =
−23
3kB
3kB ⎝ 2 ⎠
3 (1.38 × 10 J/K )
For diatomic gases Equation 18.37 gives the thermal energy.
5
5
Eth = nRT = ( 0.50 mol )( 8.31 J/mol ⋅ K )( 260 K ) = 2.7 kJ
2
2
(c) There is a net loss of 700 J of energy from the system, so the new Eth = 2000 J. Solve
Equation 18.37 for T.
2
2
T=
Eth =
( 2000 J ) = 192.5 K
5nR
5 ( 0.50 mol )( 8.31 J/mol ⋅ K )
Now Equation 18.26 gives the new rms speed.
3 (1.38 ×10−23 J/K ) (192.5 K )
3kBT
vrms =
=
= 1549 m/s ≈ 1500 m/s
m
3.322 ×10−27 kg
Assess: With the net loss of energy we expected vrms to decrease as the temperature
decreased.
18.59. Solve: (a) The escape speed is the speed with which a mass m can leave the earth’s surface and
escape to infinity (rf = ∞) with no left over speed (vf = 0). The conservation of energy equation Kf + Uf = Ki + Ui
is
2
0 + 0 = 12 mvesc
−
GM e m
2GM e
⇒ vesc =
Re
Re
The rms speed of a gas molecule is vrms = (3kBT/m)1/2. Equating vesc and vrms, and squaring both sides, the
temperature at which the rms speed equals the escape speed is
⎛ 2GM e ⎞
3kBT 2GM e
=
⇒ T = m⎜
⎟
m
Re
⎝ 3kB Re ⎠
For a nitrogen molecule, with m = 28 u, the temperature is
⎛ 2(6.67 × 10−11 N m 2 / kg 2 )(5.98 × 1024 kg) ⎞
T = (28 × 1.661 × 10−27 kg) ⎜
⎟ = 141,000 K
3(1.38 × 10−23 J/K)(6.37 × 106 m)
⎝
⎠
(b) For a hydrogen molecule, with m = 2 u, the temperature is less by a factor of 14, or T = 10,100 K.
(c) The average translational kinetic energy of a molecule is ε avg = 32 kBT = 6.1 × 10–21 J at a typical atmosphere
2
temperature of 20°C. The kinetic energy needed to escape is K esc = 12 mvesc
. For nitrogen molecules, Kesc =
2.9 × 10–18 J. Thus εavg/Kesc = 0.002 = 0.2%. Earth will retain nitrogen in its atmosphere because the molecules are
moving too slowly to escape. But for hydrogen molecules, with Kesc = 2.1 × 10–19 J, the ratio is εavg/Kesc = 0.03 =
3%. Thus a large enough fraction of hydrogen molecules are moving at escape speed, or faster, to allow
hydrogen to leak out of the atmosphere into space. Consequently, earth’s atmosphere does not contain hydrogen.
18.60. Solve: (a) The thermal energy of a monatomic gas of n1 moles is E1 = 32 n1RT . The thermal energy of a
diatomic gas of n2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is
Eth = 12 ( 3n1 + 5n2 ) RT ⇒ ΔEth = 12 ( 3n1 + 5n2 ) RΔT
Comparing this expression with
ΔEth = ntotal CV ΔT = ( n1 + n2 ) CV ΔT
we get
CV =
( 3n1 + 5n2 ) R
2 ( n1 + n2 )
(b) For a diatomic gas, n1 → 0, and CV = 52 R . For a monotomic gas, n2 → 0, and CV = 32 R .
18.61. Solve: (a) The thermal energy is
Eth = ( Eth ) N + ( Eth )O = 52 N N2 kBT + 52 N O2 kBT = 52 N total kBT
2
2
where Ntotal is the total number of molecules. The identity of the molecules makes no difference since both are
diatomic. The number of molecules in the room is
pV (101,300 Pa )( 2 m × 2 m × 2 m )
N total =
=
= 2.15 × 10 26
−23
kBT
1.38
×
10
J/K
273
K
)
(
)(
The thermal energy is
Eth = 52 ( 2.15 × 1026 )(1.38 × 10 −23 J/K ) ( 273 K ) = 2.03 × 106 J ≈ 2.0 × 106 J
(b) A 1 kg ball at height y = 1 m has a potential energy U = mgy = 9.8 J. The ball would need 9.8 J of initial
kinetic energy to reach this height. The fraction of thermal energy that would have to be conveyed to the ball is
9.8 J
= 4.8 × 10−6
2.03 × 106 J
(c) A temperature change ΔT corresponds to a thermal energy change ΔEth = 52 N total kBΔT . But 52 N total kB = Eth T .
Using this, we can write
E
ΔEth
−9.8 J
ΔEth = th ΔT ⇒ ΔT =
T=
273 K = −0.0013 K
T
Eth
2.03 × 106 J
The room temperature would decrease by 0.0013 K or 0.0013°C.
(d) The situation with the ball at rest on the floor and in thermal equilibrium with the air is a very probable
distribution of energy and thus a state with high entropy. Although energy would be conserved by removing
energy from the air and transferring it to the ball, this would be a very improbable distribution of energy and thus
a state of low entropy. The ball will not be spontaneously launched from the ground because this would require a
decrease in entropy, in violation of the second law of thermodynamics.
As another way of thinking about the situation, the ball and the air are initially at the same temperature.
Once even the slightest amount of energy is transferred from the air to the ball, the air’s temperature will be less
than that of the ball. Any further flow of energy from the air to the ball would be a situation in which heat energy
is flowing from a colder object to a hotter object. This cannot happen because it would violate the second law of
thermodynamics.
18.62. Solve: This is not a wise investment. Although an invention to move energy from the hot brakes to
the forward motion of the car would not violate energy conservation, it would violate the second law of
thermodynamics. The forward motion of the car is a very improbable distribution of energy. It happens only
when a force accelerates the car and then sustains the motion against the retarding forces of friction and air
resistance. The moving car is a state of low entropy. By contrast, the random motion of many atoms in the hot
brakes is a state of high probability and high entropy. To convert the random motion of the atoms in the brakes
back into the forward motion of the car would require a decrease of entropy and thus would violate the second
law of thermodynamics. In other words, the increasing temperature of the brakes as the car stops is an
irreversible process that cannot be undone.
18.63. Solve: (a) We are given that
( vrms i ) =
3kBTi
m
( vrms f ) =
3kBTf
= 2 ( vrms i )
m
This means that Tf = 4Ti. Using the ideal-gas law, it also means that pfVf = 4piVi. Since the pressure is directly
proportional to the volume during the process, we have pi / V i = pf /Vf. Combining these two equations gives pf = 2pi and
Vf = 2Vi.
(b) The change in thermal energy for any ideal gas process is related to the molar specific heat at constant
volume by ΔEth = nCV (Tf − Ti ) . The work done on the gas is
W = − ∫ pdV = − (area under the p-versus-V graph) = − 32 pV
i i
The first law of thermodynamics ΔEth = Q + W can be written
Q = ΔEth − W = nCV (Tf − Ti ) + 32 piVi = 3nCVTi + 32 pV
i i
15
3
= 3n ( 52 R ) Ti + 32 pV
i i = 2 pV
i i + 2 pV
i i = 9 pV
i i
18.64. Solve: The thermal energy of a monatomic gas of n1 moles is E1 = 32 n1RT . The thermal energy of a
diatomic gas of n2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is
Eth = E1 + E2 = 12 ( 3n1 + 5n2 ) RT ⇒ ΔEth = 12 ( 3n1 + 5n2 ) RΔT
Comparing this expression with
ΔEth = ntotal CV ΔT = ( n1 + n2 ) CV ΔT
we get
( n1 + n2 ) CV =
( 3n1 + 5n2 ) R
2
The requirement that the ratio of specific heats is 1.50 means
γ = 1.50 =
The above equation is then
( n1 + n2 )( 2 R ) =
CP CV + R
R
=
=1+
⇒ CV = 2 R
CV
CV
CV
( 3n1 + 5n2 ) R ⇒ 4n + 4n = 3n + 5n ⇒ n = n
2
1
2
1
2
1
2
Thus, monatomic and diatomic molecules need to be mixed in the ratio 1:1. Or the fraction of the molecules that
are monatomic needs to be 12 .
18.65.
Solve: (a) The thermal energy of a monatomic gas of n1 moles at an initial temperature T1i is
E1i = 32 n1RT1i . The thermal energy of a diatomic gas of n2 moles at an initial temperature T2i is E2i = 52 n2 RT2i .
Consequently, the total initial energy is
Etot = E1i + E2i =
3n1RT1i + 5n2 RT2i ( 3n1T1i + 5n2T2i ) R
=
2
2
After the gases interact, they come to equilibrium with T1f = T2f = Tf. Then their total energy is
Etot =
( 3n1 + 5n2 ) RTf
2
No energy is lost, so these two expressions for Etot must be equal. Thus,
( 3n1T1i + 5n2T2i ) R = ( 3n1 + 5n2 ) RTf ⇒ T = 3n1T1i + 5n2T2i =
2
f
2
3n1 + 5n2
2 Etot
2 ( E1i + E2i )
=
R ( 3n1 + 5n2 ) R ( 3n1 + 5n2 )
The thermal energies at the final temperature Tf are
E1f = 32 n1RTf = 32 n1R
⎛ 3n1 ⎞
2 E1i + E2i
=⎜
⎟ ( E1i + E2i )
R ( 3n1 + 5n2 ) ⎝ 3n1 + 5n2 ⎠
E2f = 52 n2 RTf = 52 n2 R
2 E1i + E2i ⎛ 5n2 ⎞
=⎜
⎟ ( E1i + E2i )
R 3n1 + 5n2 ⎝ 3n1 + 5n2 ⎠
(b) In part (a) we found that
Tf =
3n1T1i + 5n2T2i
3n1 + 5n2
(c) 2 g of He at T1i = 300 K are n1 = (2 g)/(4 g/mol) = 0.50 mol. Oxygen has an atomic mass of 16, so the
molecular mass of oxygen gas (O2) is A = 32 g/mol. 8 g of O2 at T2i = 600 K are n2 = (8 g)/(32 g/mol) = 0.25 mol.
The final temperature is
Tf =
3 ( 0.50 mol )( 300 K ) + 5 ( 0.25 mol )( 600 K )
= 436 K
3 ( 0.50 mol ) + 5 ( 0.25 mol )
The heat flows to the two gases are found from Q = nCVΔT. For helium,
Q = n1CV ΔT = ( 0.50 mol )(12.5 J/mol K )( 436 K − 300 K ) = 850 J
For O2,
Q = n2CV ΔT = ( 0.25 mol )( 20.8 J/mol K )( 436 K − 600 K ) = −850 J
So 850 J of heat is transferred from the oxygen to the helium.
Solve: (a) The engine has a thermal efficiency of η = 40% = 0.40 and a work output of 100 J per
cycle. The heat input is calculated as follows:
19.1.
η=
Wout
100 J
⇒ 0.40 =
⇒ QH = 250 J
QH
QH
(b) Because Wout = QH − QC , the heat exhausted is
QC = QH − Wout = 250 J − 100 J = 150 J
19.2.
Solve: During each cycle, the work done by the engine is Wout = 20 J and the engine exhausts
QC = 30 J of heat energy. Because Wout = QH − QC ,
QH = Wout + QC = 20 J + 30 J = 50 J
Thus, the efficiency of the engine is
η =1−
QC
30 J
=1−
= 0.40
QH
50 J
19.3. Solve: (a) During each cycle, the heat transferred into the engine is QH = 55 kJ , and the heat exhausted
is QC = 40 kJ . The thermal efficiency of the heat engine is
η = 1−
QC
40 kJ
=1−
= 0.27 = 27%
QH
55 kJ
(b) The work done by the engine per cycle is
Wout = QH − QC = 55 kJ − 40 kJ = 15 kJ
19.4. Solve: The coefficient of performance of the refrigerator is
K=
QC QH − Win 50 J − 20 J
=
=
= 1.5
Win
Win
20 J
19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows:
K=
QC
Q
⇒ 4.0 = C ⇒ QC = 200 J
Win
50 J
(b) The heat exhausted to the hot reservoir is
QH = QC + Win = 200 J + 50 J = 250 J
19.6.
Solve:
Model: Assume that the car engine follows a closed cycle.
(a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is
Wout = 500
kJ
1s
kJ
×
= 12.5
s 40 cycles
cycle
(b) The heat input per cycle is calculated as follows:
η=
Wout
12.5 kJ
⇒ QH =
= 62.5 kJ
QH
0.20
The heat exhausted per cycle is
QC = QH − Win = 62.5 kJ − 12.5 kJ = 50 kJ
19.7.
Solve: The amount of heat discharged per second is calculated as follows:
η=
⎛1 ⎞
Wout
Wout
⎛ 1
⎞
=
⇒ QC = Wout ⎜ − 1⎟ = ( 900 MW ) ⎜
− 1⎟ = 1.913 × 109 W
QH QC + Wout
0.32
η
⎝
⎠
⎝
⎠
That is, each second the electric power plant discharges 1.913 × 109 J of energy into the ocean. Since a typical
American house needs 2.0 × 104 J of energy per second for heating, the number of houses that could be heated
with the waste heat is (1.913 × 109 J ) ( 2.0 × 104 J ) = 96,000 .
19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour is QC = mwater cwater ΔT .
The mass of the water is
mwater = ρ waterVwater = (1000 kg/m3 ) (1 L ) = (100 kg/m3 )(10−3 m3 ) = 1.0 kg
⇒ QC = (1.0 kg )( 4190 J/kg K )( 20°C − 5°C ) = 6.285 × 10 4 J
The rate of heat removal from the refrigerator is
6.285 × 104 J
= 17.46 J/s
3600 s
The refrigerator does work W = 8.0. W = 8.0 J/s to remove this heat. Thus the performance coefficient of the
refrigerator is
QC =
K=
17.46 J/s
= 2.2
8.0 J/s
19.9. Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is
isobaric.
Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal
energy. Because ΔEth = Q − Ws and Ws = 0 J , Q increases for process A. Process B is isothermal, so T is constant
and hence ΔEth = 0 J. The work done Ws is positive because the gas expands. Because Q = Ws + ΔEth , Q is
positive for process B. Process C is adiabatic, so Q = 0 J. Ws is positive because of the increase in volume. Since
Q = 0 J = Ws + ΔEth , ΔEth is negative for process C. Process D is isobaric, so the decrease in volume leads to a
decrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, Ws is
negative. Because Q = Ws + ΔEth , Q also decreases for process D.
A
B
C
D
ΔEth
+
0
−
−
Ws
0
+
+
−
Q
+
+
0
−
19.10. Model: Process A is adiabatic, process B is isothermal, and process C is isochoric.
Solve: Process A is adiabatic, so Q = 0 J. Work Ws is positive as the gas expands. Since Q = Ws + ∆Eth = 0 J,
∆Eth must be negative. The temperature falls during an adiabatic expansion. Process B is isothermal so ∆T = 0
and ∆Eth = 0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, so Q is negative. Heat
energy is withdrawn during the compression to keep the temperature constant. Process C is isochoric. No work is
done (Ws = 0 J), and Q is positive as heat energy is added to raise the temperature (∆Eth positive).
∆Eth
Ws
Q
A
–
0
+
B
0
–
–
C
0
+
+
19.11.
Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area
inside the triangle is
Wout = 12 ( 3 atm − 1 atm ) ( 600 × 10−6 m3 − 200 × 10−6 m3 )
⎛
1.013 × 105 Pa ⎞
−6
3
= 12 ⎜ 2 atm ×
⎟ ( 400 × 10 m ) = 40.5 J
1
atm
⎝
⎠
19.12.
Solve:
The work done by the gas per cycle is the area enclosed within the pV curve. We have
60 J = 12 ( pmax − 100 kPa ) ( 800 cm3 − 200 cm3 ) ⇒
2 ( 60 J )
= pmax − 1.0 × 105 Pa
600 × 10−6 m3
⇒ pmax = 3.0 × 105 Pa = 300 kPa
19.13.
Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle
consists of three individual processes.
Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get
Wout = 12 ( 200 kPa ) (100 × 10−6 m3 ) = 10 J
The heat energy transferred into the engine is QH = 120 J. Because Wout = QH − QC , the heat energy exhausted is
QC = QH − Wout = 120 J − 10 J = 110 J
(b) The thermal efficiency of the engine is
η=
Assess:
Wout 10 J
=
= 0.0833
QH 120 J
Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4 .
19.14.
Solve:
Model: The heat engine follows a closed cycle, which consists of four individual processes.
(a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get
Wout = ( 400 kPa − 100 kPa ) (100 × 10−6 m3 ) = 30 J
The heat energy leaving the engine is QC = 90 J + 25 J = 115 J . The heat input is calculated as follows:
Wout = QH − QC ⇒ QH = QC + Wout = 115 J + 30 J = 145 J
(b) The thermal efficiency of the engine is
η=
Assess:
Wout 30 J
=
= 0.207
QH 145 J
Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4 .
19.15.
Solve:
Model: The heat engine follows a closed cycle.
The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
Wout = 12 ( 300 kPa − 100 kPa ) ( 600 cm3 − 300 cm3 ) = 12 ( 200 × 103 Pa )( 300 × 10−6 m3 ) = 30 J
Because Wout = QH − QC , the heat exhausted is
QC = QH − Wout = (225 J + 90 J) − 30 J = 315 J − 30 J = 285 J
19.16.
Solve:
Model: The heat engine follows a closed cycle.
(a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
Wout = 12 ( 300 kPa − 100 kPa ) ( 600 cm3 − 200 cm3 ) = 12 ( 200 × 103 Pa )( 400 × 10−6 m3 ) = 40 J
The heat exhausted is QC = 180 J + 100 J = 280 J . The thermal efficiency of the engine is
η=
Wout
Wout
40 J
=
=
= 0.125
QH QC + Wout 280 J + 40 J
(b) The heat extracted from the hot reservoir is QH = QC + Wout = 320 J.
19.17.
Model: The Brayton cycle involves two adiabatic processes and two isobaric processes.
Solve:
From Equation 19.21, the efficiency of a Brayton cycle is ηB = 1 − rp(1−γ ) γ , where rp is the pressure ratio
pmax/pmin. The specific heat ratio for a diatomic gas is
γ=
CP 72 R
=
= 1.4
CV 52 R
Solving the above equation for rp,
γ
−1.4
(1 − ηB ) = rp(1−γ ) / γ ⇒ rp = (1 − ηB )1−γ = (1 − 0.60 ) 0.4 = 25
19.18.
Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The
adiabatic processes involve compression and expansion through the turbine.
Solve: The thermal efficiency for the Brayton cycle is ηB = 1 − rp(1−γ ) / γ , where γ = CP / CV and rp is the pressure
ratio. For a diatomic gas γ = 1.4. For an adiabatic process,
p1V1γ = p2V2γ ⇒ p2 / p1 = (V1 / V2 )
γ
Because the volume is halved, V2 = 12 V1 and hence
rp = p2 / p1 = ( 2 ) = 21.4 = 2.639
γ
The efficiency is
−0.4
ηB = 1 − ( 2.639 ) 1.4 = 0.242
19.19. Model: The efficiency of a Carnot engine (ηCarnot) depends only on the temperatures of the hot and
cold reservoirs. On the other hand, the thermal efficiency (η) of a heat engine depends on the heats QH and QC.
Solve:
(a) According to the first law of thermodynamics, QH = Wout + QC . For engine (a), QH = 50 J, QC = 20 J
and Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J,
so the first law is violated. For engine (c) the first law of thermodynamics is obeyed.
(b) For the three heat engines, the maximum or Carnot efficiency is
T
TH
ηCarnot = 1 − C = 1 −
300 K
= 0.50
600 K
Engine (a) has
η = 1−
QC Wout 30 J
=
=
= 0.60
QH QH 50 J
This is larger than ηCarnot, thus violating the second law of thermodynamics. For engine (b),
η=
Wout 4 J
=
= 0.40 < ηCarnot
QH 10 J
so the second law is obeyed. Engine (c) has a thermal efficiency that is
10 J
η=
= 0.333 < ηCarnot
30 J
so the second law of thermodynamics is obeyed.
19.20. Model: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot
coefficient of performance are
K=
QC
Win
K Carnot =
TC
TH − TC
Solve: (a) For refrigerator (a) QH = QC + Win (60 J = 40 J + 20 J) , so the first law of thermodynamics is obeyed.
For refrigerator (b) 50 J = 40 J + 10 J , so the first law of thermodynamics is obeyed. For the refrigerator (c)
40 J ≠ 30 J + 20 J , so the first law of thermodynamics is violated.
(b) For the three refrigerators, the maximum coefficient of performance is
K Carnot =
TC
300 K
=
=3
TH − TC 400 K − 300 K
For refrigerator (a),
K=
QC 40 J
=
= 2 < K Carnot
Win 20 J
so the second law of thermodynamics is obeyed. For refrigerator (b),
K=
QC 40 J
=
= 4 > K Carnot
Win 10 J
so the second law of thermodynamics is violated. For refrigerator (c),
K=
so the second law is obeyed.
30 J
= 1.5 < K Carnot
20 J
19.21. Model: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and
cold reservoirs.
Solve: The efficiency of a Carnot engine is
T
TH
ηCarnot = 1 − C ⇒ 0.60 = 1 −
TC
⇒ TC = 280 K = 7°C
( 427 + 273) K
Assess: A “real” engine would need a lower temperature than 7oC to provide 60% efficiency because no real
engine can match the Carnot efficiency.
19.22. Model: Assume that the heat engine follows a closed cycle.
Solve:
(a) The engine’s efficiency is
η=
Wout
Wout
200 J
=
=
= 0.25 = 25%
QH QC + Wout 600 J + 200 J
(b) The thermal efficiency of a Carnot engine is ηCarnot = 1 − TC TH . For this to be 25%,
0.25 = 1 −
TC
( 400 + 273) K
⇒ TC = 504.8 K = 232°C
19.23. Model: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the
hot and cold reservoirs.
Solve: (a) The engine’s thermal efficiency is
η=
Wout
Wout
10 J
=
=
= 0.40 = 40%
QH QC + Wout 15 J + 10 J
(b) The efficiency of a Carnot engine is ηCarnot = 1 − TC TH . The minimum temperature in the hot reservoir is
found as follows:
0.40 = 1 −
293 K
⇒ TH = 488 K = 215°C
TH
This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than
215°C because no real engine can match the Carnot efficiency.
19.24. Solve: (a) The efficiency of the Carnot engine is
T
TH
ηCarnot = 1 − C = 1 −
300 K
= 0.40 = 40%
500 K
(b) An engine with power output of 1000 W does Wout = 1000 J of work during each Δt = 1 s. A Carnot engine
has a heat input that is
Qin =
Wout
ηCarnot
=
1000 J
= 2500 J
0.40
during each Δt = 1 s. The rate of heat input is 2500 J/s = 2500 W.
(c) Wout = Qin − Qout , so the heat output during Δt = 1 s is Qout = Qin − Wout = 1500 J . The rate of heat output is
thus 1500 J/s = 1500 W.
19.25.
Visualize:
We will use Equation 19.29 for the efficiency of a Carnot engine.
T
TH
ηCarnot = 1 − C
We are given TH = 673K and the original efficiency ηCarnot = 0.40 .
Solve:
First solve for TC .
TC = TH (1 − ηCarnot ) = (673K)(1 − 0.40) = 404K
′
= 0.60.
Solve for TC′ again with ηCarnot
′ ) = (673K)(1 − 0.60) = 269K
TC′ = TH (1 − ηCarnot
The difference of these TC ’s is 135 K, so the temperature of the cold reservoir should be decreased by 135
degrees to raise the efficiency from 40% to 60%.
Assess: We expected to have to lower TC by quite a bit to get the better efficiency.
19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine.
Solve:
The maximum efficiency is
T
TH
ηmax = ηCarnot = 1 − C = 1 −
( 273 + 20 ) K = 0.6644
( 273 + 600 ) K
Because the heat engine is running at only 30% of the maximum efficiency, η = ( 0.30 )ηmax = 0.1993 . The
amount of heat that must be extracted is
QH =
Wout
η
=
1000 J
= 5.0 kJ
0.1993
19.27. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of
the cold and hot reservoirs.
Solve: (a) The Carnot performance coefficient of a refrigerator is
K Carnot =
TC
( −20 + 273) K
=
= 6.33
TH − TC ( 20 + 273) K − ( −20 + 273) K
(b) The rate at which work is done on the refrigerator is found as follows:
K=
QC
Q
200 J/s
⇒ Win = C =
= 32 J/s = 32 W
Win
K
6.325
(c) The heat exhausted to the hot side per second is
QH = QC + Win = 200 J/s + 32 J/s = 232 J/s = 232 W
19.28. Visualize: We are given TH = 773K and TC = 273K , therefore (by Equation 19.29) the Carnot
273 K
efficiency is nCarnot = 1 − 773
K = 0.647
We are also given η = 0.60(ηCarnot ) .
Solve:
Rearrange Equation 19.7: QH = Wout (1 − η ) . Wout is the same for both engines, so it cancels.
QH
Wout (1 − η )
1 − 0.60(ηCarnot ) 1 − 0.60(0.647)
=
=
=
= 1.73
(QH )Carnot Wout (1 − ηCarnot )
1 − ηCarnot
1 − 0.647
Assess: This engine requires 1.73 times as much heat energy in during each cycle as a Carnot engine to do the
same amount of work.
19.29. Model: The minimum possible value of TC occurs with a Carnot refrigerator.
Solve:
(a) For the refrigerator, the coefficient of performance is
K=
QC
⇒ QC = KWin = ( 5.0 )(10 J ) = 50 J
Win
The heat energy exhausted per cycle is
QH = QC + Win = 50 J + 10 J = 60 J
(b) If the hot-reservoir temperature is 27°C = 300 K, the lowest possible temperature of the cold reservoir can be
obtained as follows:
K Carnot =
TC
TC
⇒ 5.0 =
⇒ TC = 250 K = −23°C
TH − TC
300 K − TC
19.30. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of
the cold and hot reservoirs.
Solve: Using the definition of the coefficient of performance for a Carnot refrigerator,
K Carnot =
TC
( −13 + 273) K = 260 K ⇒ T = 312 K
⇒ 5.0 =
H
TH − TC
TH − ( −13 + 273) K TH − 260 K
To increase the coefficient of performance to 10.0, the hot-reservoir temperature changes to TH′ . Thus,
K Carnot = 10.0 =
260 K
⇒ TH′ = 286 K
TH′ − 260 K
Since TH′ is less than TH, the hot-reservoir temperature must be decreased by 26 K or 26 ° C.
19.31.
Solve:
The work done by the engine is equal to the change in the gravitational potential energy.
Thus,
Wout = ΔUgrav = mgh = (2000 kg)(9.8 m/s2)(30 m) = 588,000 J
The efficiency of this engine is
⎛
T ⎞
TH ⎠
⎛
η = 0.40 ηCarnot = 0.40 ⎜1 − C ⎟ = 0.40 ⎜⎜1 −
⎝
⎝
( 273 + 20 ) K ⎞ = 0.3484
⎟
( 273 + 2000 ) K ⎟⎠
The amount of heat energy transferred is calculated as follows:
η=
Wout
W
588,000 J
⇒ QH = out =
= 1.7 × 106 J
QH
0.3484
η
19.32.
Solve:
The mass of the water is
100 × 10−3 L ×
10−3 m3 1000 kg
×
= 0.100 kg
1L
m3
The heat energy is removed from the water in three steps: (1) cooling from +15°C to 0°C, (2) freezing at 0°C,
and (3) cooling from 0°C to −15°C . The three heat energies are
Q1 = mcΔT = ( 0.100 kg )( 4186 J/kg K )(15 K ) = 6279 J
Q2 = mLf = ( 0.100 kg ) ( 3.33 × 105 J/kg ) = 33,300 J
Q3 = mcΔT = ( 0.100 kg )( 2090 J/kg K )(15 K ) = 3135 J
⇒ QC = Q1 + Q2 + Q3 = 42,714 J
Using the performance coefficient,
Q
42,714 J
42,714 J
K = C ⇒ 4.0 =
⇒ Win =
= 10,679 J
Win
Win
4.0
The heat exhausted into the room is thus
QH = QC + Win = 42,714 J + 10,679 J = 5.3 × 104 J
Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = –∆Eth. For any ideal-gas
process,
∆Eth = nCV∆T, so Ws = –nCV∆T. We can use the ideal-gas law to find
19.33.
T=
pV
Δ ( pV ) ( pV )f − ( pV )i pfVf − pV
i i
⇒ ΔT =
=
=
nR
nR
nR
nR
Consequently, the work is
CV
⎛ p V − pV
i i ⎞
Ws = −nCV ΔT = − nCV ⎜ f f
( pfVf − pV
i i)
⎟=−
nR
R
⎝
⎠
Because CP = CV + R, we can use the specific heat ratio γ to find
γ=
C P CV + R C V / R + 1
=
=
⇒
CV
CV
CV / R
CV
1
=
R γ −1
With this, the work done in an adiabatic process is
Ws = −
CV
1
( pfVf − pV
( pf Vf − pV
i i) = −
i i ) = ( pf Vf − pV
i i )/(1 − γ )
R
γ −1
19.34. Visualize: We
are
QH = msteam Lv = (10kg)(22.6 × 105 J/kg) = 22.6 × 106 J
given
QC = mice Lf = (55kg) (3.33 × 10 J/kg) = 18.3 × 10 J .
5
Solve:
Equation 19.6 gives Wout = QH − QC .
P=
Assess:
6
Wout QH − QC 22.6 × 106 J − 18.3 × 106 J ⎛ 1h ⎞
=
=
⎜
⎟ = 1200 W
Δt
Δt
1h
⎝ 3600s ⎠
This is a reasonable answer.
and
19.35. Solve: For any heat engine, η = 1 – QC/QH. For a Carnot heat engine, ηCarnot = 1 – TC/TH. Thus a
property of the Carnot cycle is that QC/QH = TC/TH. Consequently, the coefficient of performance of a Carnot
refrigerator is
K Carnot =
QC
QC
QC /QH
T C / TH
TC
=
=
=
=
Win QH − QC 1 − QC /QH 1 − T C / TH TH − TC
19.36. Visualize: Equation 19.29 gives ηCarnot = 1 − TT . We are given ηCarnot = 1/ 3.
C
H
Solve:
T
1
TC 2
3
= ⇒ TH = TC
TH 3 TH 3
2
Equation 19.30 gives the coefficient of performance for the Carnot refrigerator.
ηCarnot = 1 − C = ⇒
K Carnot =
Assess:
1
TC
T
=3 C
=3
=2
TH − TC 2 TC − TC 2 − 1
This result is in the ballpark for coefficients of performance.
19.37. Visualize: We are given TH = 298 K and TC = 273 K . See Figure 19.11.
Solve:
QC = mLf = (10kg)(3.33 × 105 J/kg) = 3.33 × 106 J .
(a) For a Carnot cycle ηCarnot = 1 − THC but that must also equal η = 1 − QHC , so QHC = THC .
T
QH = QC
Q
Q
T
⎛ 298K ⎞
TH
6
= (3.33 × 106 J) ⎜
⎟ = 3.63 × 10 J
TC
⎝ 273K ⎠
(b)
Win = QH − QC = 3.63 × 106 J − 3.33 × 106 J = 0.30 × 106 J = 3.0 × 105 J
Assess:
This is a reasonable amount of work to freeze 10 kg of water.
19.38. Model: We will use the Carnot engine to find the maximum possible efficiency of a floating power
plant.
Solve:
The efficiency of a Carnot engine is
T
TH
ηmax = ηCarnot = 1 − C = 1 −
( 273 + 5) K = 0.0825 ≈ 8%
( 273 + 30 ) K
19.39.
Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal
expansion at temperature TH, heat QH is transferred from the hot reservoir into the gas. During the isothermal
compression at TC, heat QC is removed from the gas. No heat is transferred during the remaining two adiabatic steps.
Solve: The thermal efficiency of the Carnot engine is
T
TH
ηCarnot = 1 − C =
Wout
323 K
Wout
⇒ Wout = 436 J
⇒ 1−
=
QH
573 K 1000 J
Using QH = QC + Wout , we obtain
Qisothermal = QC = QH − Wout = 1000 J − 436 J ≈ 560 J
19.40.
Solve:
Substituting into the formula for the efficiency of a Carnot engine,
T
TH
ηCarnot = 1 − C ⇒ 0.25 = 1 −
TC
⇒ TC = 240 K = −33°C
TC + 80 K
The hot-reservoir temperature is TH = TC + 80 K = 320 K = 47 ° C.
19.41. Model: Assume the gas is monatomic.
Visualize:
We are given TC = 273 K and QC = 15 J per cycle.
For 98 cycles Wout = mgh = (10kg)(9.8kg/m 2 )(10m) = 9800 J ; for one cycle this would be Wout = 100 J.
Solve: Do the computation for one cycle.
QH = Wout + QC = 100J + 15J = 115J
For a Carnot cycle ηCarnot = 1 − THC but that must also equal η = 1 − QHC , so QHC = THC .
T
Q
TH = TC
Assess:
Q
T
⎛ 115J ⎞
QH
= (273K) ⎜
⎟ = 2093K ≈ 2100K
QC
⎝ 15J ⎠
TH must be this high to give the 87% efficiency.
19.42.
Solve:
If QC = 23 QH, then Wout = QH – QC = 13 QC. Thus the efficiency is
η=
Wout 13 QH 1
=
=
QH
QH 3
The efficiency of a Carnot engine is η = 1 – TC/TH. Thus
T
1
TC 2
1− C =
⇒
=
TH 3
TH 3
19.43.
Solve:
(a) Q1 is given as 1000 J. Using the energy transfer equation for the heat engine,
QH = QC + Wout ⇒ Q1 = Q2 + Wout ⇒ Q2 = Q1 − Wout
The thermal efficiency of a Carnot engine is
T
TH
η = 1− C =1−
300 K
W
= 0.50 = out
600 K
Q1
⇒ Q2 = Q1 − ηQ1 = Q1 (1 − η ) = (1000 J )(1 − 0.50 ) = 500 J
To determine Q3 and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heat
engine with Win = Wout. The coefficient of performance is
K=
TC
400 K
Q
Q
Q
=
= 4.0 = C = 4 = 4
TH − TC 500 K − 400 K
Win Wout ηQ1
⇒ Q4 = KηQ1 = ( 4.0 )( 0.50 )(1000 J ) = 2000 J
Using now the energy transfer equation Win + Q4 = Q3 , we have
Q3 = Wout + Q4 = ηQ1 + Q4 = ( 0.50 )(1000 J ) + 2000 J = 2500 J
(b) From part (a) Q3 = 2500 J and Q1 = 1000 J , so Q3 > Q1 .
(c) Although Q1 = 1000 J and Q3 = 2500 J, the two devices together do not violate the second law of
thermodynamics. This is because the hot and cold reservoirs are different for the heat engine and the refrigerator.
19.44.
Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the
indoors with its exhaust heat.
Solve: (a) The coefficient of performance for this heat pump is K = 5.0 = QC Win , where QC is the amount of
heat removed from the cold reservoir. QH is the amount of heat exhausted into the hot reservoir. QH = QC + Win,
where Win is the amount of work done on the heat pump. We have
QC = 5.0Win ⇒ QH = 5.0Win + Win = 6.0Win
If the heat pump is to deliver 15 kJ of heat per second to the house, then
15 kJ
= 2.5 kJ
6.0
In other words, 2.5 kW of electric power is used by the heat pump to deliver 15 kJ/s of heat energy to the house.
(b) The monthly heating cost in the house using an electric heater is
QH = 15 kJ = 6.0Win ⇒ Win =
15 kJ
3600 s
1$
× 200 h ×
×
= $270
s
1h
40 MJ
The monthly heating cost in the house using a heat pump is
2.5 kJ
3600 s
1$
× 200 h ×
×
= $45
s
1h
40 MJ
19.45. Visualize: We are given TC = 275 K , TH = 295 K. We are also given that in one second Win = 100 J
and QC = (1 s)(100 kJ/min)(1 min/60 s) = 1667 J.
Solve: The coefficient of performance of a refrigerator is given in Equation 19.10.
K=
QC 1667 J
=
= 16.67
Win 100 J
However the coefficient of performance of a Carnot refrigerator is given in Equation 19.30.
K Carnot =
TC
275 K
=
= 13.75
TH − TC 20 K
However, informal statement #8 of the second law says that the coefficient of performance cannot exceed the
Carnot coefficient of performance, so the salesman is making false claims. You should not buy the DreamFridge.
Assess: The second law imposes real-world restrictions.
19.46. Solve: The maximum possible efficiency of the heat engine is
T
TH
ηmax = 1 − C = 1 −
300 K
= 0.40
500 K
The efficiency of the engine designed by the first student is
Wout 110 J
=
= 0.44
QH 250 J
Because η1 > ηmax, the first student has proposed an engine that would violate the second law of
thermodynamics. His or her design will not work. The efficiency of the engine designed by the second student is
η2 = 90 J 250 J = 0.36 < ηmax in agreement with the second law of thermodynamics. Applying the first law of
thermodynamics,
η1 =
QH = QC + Wout ⇒ 250 J = 170 J + 90 J
we see the first law is violated. This design will not work as claimed. The design by the third student satisfies the
first law of thermodynamics because QH = QC + Wout = 250 J . The thermal efficiency of this engine is η3 =
0.36 < ηmax , which satisfies the second law of thermodynamics. The data presented by students 1 and 2 are
faulty. Only student 3 has an acceptable design.
19.47. Model: The power plant is to be treated as a heat engine.
Solve:
(a) Every hour 300 metric tons or 3 × 105 kg of coal is burnt. The volume of coal is
3 ×105 kg
m3
×
× 24 h = 4800 m3
1h
1500 kg
The height of the room will be 48 m.
(b) The thermal efficiency of the power plant is
η=
Assess:
7.50 ×108 J/s
7.50 ×108 J
Wout
=
=
= 0.32 = 32%
5
6
1h
QH 3 ×10 kg 28 ×10 J
2.333 ×109 J
×
×
1h
kg
3600 s
An efficiency of 32% is typical of power plants.
19.48. Model: The power plant is treated as a heat engine.
Solve:
(a) The maximum possible thermal efficiency of the power plant is
T
TH
ηmax = 1 − C = 1 −
303 K
= 0.47 = 47%
573 K
(b) The plant’s actual efficiency is
η=
Wout
700 × 106 J/s
=
0.35 = 35%
QH 2000 × 106 J/s
(c) Because QH = QC + Wout,
QC = QH − Wout ⇒ QC = 2.0 × 109 J/s − 0.7 × 109 J/s = 1.3 × 109 J/s
The mass of water that flows per second through the condenser is
m = 1.2 ×108
L 1 hr 10−3 m3 1000 kg
×
×
×
= 3.333 × 104 kg
h 3600 s
1L
m3
The change in the temperature as QC = 1.3 × 109 J of heat is transferred to m = 3.333 × 104 kg of water is
calculated as follows:
QC = mcΔT ⇒ 1.3 × 109 J = ( 3.333 × 104 kg ) ( 4186 J/kg K ) ΔT ⇒ ΔT = 9°C
The exit temperature is18°C + 9°C = 27°C .
19.49. Model: The power plant is treated as a heat engine.
Solve:
The mass of water per second that flows through the plant every second is
m = 1.0 ×108
L 1 hr 10−3 m3 1000 kg
×
×
×
= 2.778 × 104 kg/s
h 3600 s
1L
m3
The amount of heat transferred per second to the cooling water is thus
QC = mcΔT = ( 2.778 × 104 kg/s ) ( 4186 J/kg K )( 27°C − 16°C ) = 1.279 × 109 J/s
The amount of heat per second into the power plant is
QH = Wout + QC = 0.750 × 109 J/s + 1.279 × 109 J/s = 2.029 × 109 J/s
Finally, the power plant’s thermal efficiency is
η=
Wout 0.750 × 109 J/s
=
= 0.37 = 37%
QH 2.029 × 109 J/s
19.50. Solve: (a) The energy supplied in one day is
J 3600 s 24 h
Wout = 1.0 ×109 ×
×
= 8.64 × 1013 J
s
1h
1d
(b) The volume of water is V = 1 km3 = 109 m3 . The amount of energy is
⎛ 1000 kg ⎞
15
QH = mcΔT = (109 m3 ) ⎜
⎟ ( 4190 J/kg K )(1.0 K ) = 4.19 × 10 J
3
⎝ m
⎠
(c) While it’s true that the ocean contains vast amounts of thermal energy, that energy can be extracted to do
useful work only if there is a cold reservoir at a lower temperature. That is, the ocean has to be the hot reservoir
of a heat engine. But there’s no readily available cold reservoir, so the ocean’s energy cannot readily be tapped.
There have been proposals for using the colder water near the bottom of the ocean as a cold reservoir, pumping it
up to the surface where the heat engine is. Although possible, the very small temperature difference between the
surface and the ocean depths implies that the maximum possible efficiency (the Carnot efficiency) is only a few
percent, and the efficiency of any real ocean-driven heat engine would likely be less than 1%—perhaps much
less. Thus the second law of thermodynamics prevents us from using the thermal energy of the ocean. Save your
money. Don’t invest.
19.51. Visualize: If we do this problem on a “per second” basis then in one second QC = (1 s)(5.0 ×105 J/min)
(1 min/60 s) = 8.33 ×103 J. QH = (1 s)(8.0 ×105 J/min)(1 min / 60 s) = 13.33 × 103 J.
Solve: (a) Again, in one second
Win = QH − QC = 13.33 × 103 J − 8.33 × 103 J = 5.0 × 103 J
Since this is per second then the power required by the compressor is P = 5.0 kW.
(b)
Assess:
K=
The result is typical for air conditioners.
QC 8.33 × 103 J
=
= 1.7
Win 5.0 × 103 J
19.52. Model: The heat engine follows a closed cycle with process 1 → 2 and process 3 → 4 being isochoric
and process 2 → 3 and process 4 → 1 being isobaric. For a monatomic gas, CV = 32 R and CP = 52 R.
Visualize: Please refer to Figure P19.52.
Solve: (a) The first law of thermodynamics is Q = ΔEth + WS . For the isochoric process 1 → 2, WS 1 → 2 = 0 J.
Thus,
Q1→ 2 = 3750 J = ΔEth = nCV ΔT
⇒ ΔT =
3750 J
3750 J
3750 J
=
=
= 301 K
nCV
(1.0 mol ) ( 32 R ) (1.0 mol ) ( 32 ) (8.31 J/mol K )
⇒ T2 − T1 = 300.8 K ⇒ T2 = 300.8 K + 300 K = 601 K
To find volume V2,
V2 = V1 =
nRT1 (1.0 mol )( 8.31 J/mol K )( 300 K )
=
= 8.31× 10−3 m3
p1
3.0 × 105 Pa
The pressure p2 can be obtained from the isochoric condition as follows:
p2 p1
T
⎛ 601 K ⎞
5
5
= ⇒ p2 = 2 p1 = ⎜
⎟ ( 3.00 × 10 Pa ) = 6.01 × 10 Pa
T2 T1
T1
300
K
⎝
⎠
With the above values of p2, V2 and T2, we can now obtain p3, V3 and T3. We have
V3 = 2V2 = 1.662 × 10−2 m3
p3 = p2 = 6.01× 105 Pa
T3 T2
V
= ⇒ T3 = 3 T2 = 2T2 = 1202 K
V3 V2
V2
For the isobaric process 2 → 3,
Q2 →3 = nCP ΔT = (1.0 mol ) ( 52 R ) (T3 − T2 ) = (1.0 mol ) ( 52 ) ( 8.31 J/mol K )( 601 K ) = 12,480 J
WS 2 →3 = p3 (V3 − V2 ) = ( 6.01× 105 Pa )( 8.31× 10−3 m3 ) = 4990 J
ΔEth = Q2 →3 − WS 2 →3 = 12,480 J − 4990 J = 7490 J
We are now able to obtain p4, V4 and T4. We have
V4 = V3 = 1.662 × 10−2 m3
p4 = p1 = 3.00 × 105 Pa
⎛ 3.00 × 105 Pa ⎞
T4 T3
p
=
⇒ T4 = 4 T3 = ⎜
⎟ (1202 K ) = 600 K
5
p4 p3
p3
⎝ 6.01 × 10 Pa ⎠
For isochoric process 3 → 4,
Q3→ 4 = nCV ΔT = (1.0 mol ) ( 32 R ) (T4 − T3 ) = (1.0 mol ) ( 32 ) ( 8.31 J/mol K )( −602 ) = −7500 J
WS 3→ 4 = 0 J ⇒ ΔEth = Q3→ 4 − WS 3→ 4 = −7500 J
For isobaric process 4 → 1,
Q4 →1 = nCP ΔT = (1.0 mol ) 52 ( 8.31 J/mol K )( 300 K − 600 K ) = −6230 J
WS 4 →1 = p4 (V1 − V4 ) = ( 3.00 × 105 Pa ) × ( 8.31 × 10−3 m3 − 1.662 × 10−2 m3 ) = −2490 J
ΔEth = Q4 →1 − WS 4 →1 = −6230 J − ( −2490 J ) = −3740 J
WS (J)
Q (J)
ΔEth (J)
1→2
0
3750
3750
2→3
4990
12,480
7490
3→4
0
–7500
–7500
4→1
–2490
–6230
–3740
Net
2500
2500
0
(b) The thermal efficiency of this heat engine is
η=
Assess:
Wout
Wout
2500 J
=
=
= 0.15 = 15%
QH Q1→ 2 + Q2 →3 3750 J + 12,480 J
For a closed cycle, as expected, (Ws)net = Qnet and (ΔEth)net = 0 J
19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, CV = 52 R and CP = 72 R.
Visualize: Please refer to Figure P19.53.
Solve: (a) Since T1 = 293 K, the number of moles of the gas is
n=
5
−6
3
p1V1 ( 0.5 × 1.013 × 10 Pa )(10 × 10 m )
=
= 2.08 × 10−4 mol
RT1
( 8.31 J/mol K )( 293 K )
At point 2, V2 = 4V1 and p2 = 3 p1 . The temperature is calculated as follows:
p1V1 p2V2
p V
=
⇒ T2 = 2 2 T1 = ( 3)( 4 )( 293 K ) = 3516 K
T1
T2
p1 V1
At point 3, V3 = V2 = 4V1 and p3 = p1 . The temperature is calculated as before:
T3 =
p3 V3
T1 = (1)( 4 )( 293 K ) = 1172 K
p1 V1
For process 1 → 2, the work done is the area under the p-versus-V curve. That is,
Ws = ( 0.5 atm ) ( 40 cm3 − 10 cm3 ) + 12 (1.5 atm − 0.5 atm ) ( 40 cm3 − 10 cm3 )
⎛ 1.013 × 105 Pa ⎞
= ( 30 × 10−6 m3 ) (1 atm ) ⎜
⎟ = 3.04 J
1 atm
⎝
⎠
The change in the thermal energy is
ΔEth = nCV ΔT = ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )( 3516 K − 293 K ) = 13.93 J
The heat is Q = Ws + ΔEth = 16.97 J . For process 2 → 3, the work done is Ws = 0 J and
Q = ΔEth = nCV ΔT = n ( 52 R ) (T3 − T2 )
= ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )(1172 K − 3516 K ) = −10.13 J
For process 3 → 1,
Ws = ( 0.5 atm ) (10 cm3 − 40 cm3 ) = ( 0.5 × 1.013 × 105 Pa )( −30 × 10−6 m3 ) = −1.52 J
ΔEth = nCV ΔT = ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )( 293 K − 1172 K ) = −3.80 J
The heat is Q = ΔEth + Ws = −5.32 J .
1→2
2→3
3→1
Net
Ws (J)
Q (J)
ΔEth
3.04
0
−1.52
1.52
16.97
−10.13
−5.32
1.52
13.93
−10.13
−3.80
0
(b) The efficiency of the engine is
η=
Wnet 1.52 J
=
= 0.090 = 9.0%
QH 16.97 J
(c) The power output of the engine is
500
Assess:
revolutions 1 min
500
Wnet
×
×
=
× 1.52 J/s = 13 W
min
60 s revolution 60
For a closed cycle, as expected, (Ws)net = Qnet and (ΔEth)net = 0 J.
19.54. Model: For the closed cycle, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and process 3 →
1 is isochoric.
Solve: (a) We first need to find the conditions at points 1, 2, and 3. We can then use that information to find WS and
Q for each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas
is
n=
5
−6
3
p1V1 (1.013 × 10 Pa )( 600 × 10 m )
=
= 0.0244 mol
RT1
(8.31 J/mol K )( 300 K )
We are given that γ = 1.25, which means this is not a monatomic or a diatomic gas. The specific heats are
CV =
R
= 4R
1−γ
C P = CV + R = 5 R
At point 2, process 1 → 2 is isothermal, so we can find the pressure p2 as follows:
p1V1 = p2V2 ⇒ p2 =
V1
6.00 × 10−4 m3
p1 =
p1 = 3 p1 = 3 atm = 3.039 × 105 Pa
V2
2.00 × 10−4 m3
At point 3, process 2 → 3 is isobaric, so we can find the temperature T3 as follows:
V2 V3
V
6.00 × 10−4 m 3
= ⇒ T3 = 3 T2 =
T2 = 3T2 = 900 K
T2 T3
V2
2.00 × 10−4 m3
Point
V (m3 )
P (Pa)
1.0 atm = 1.013 × 10
3.0 atm = 3.039 × 105
3.0 atm = 3.039 × 105
1
2
3
T (K)
−4
5
6.00 × 10
2.00 × 10−4
6.00 × 10−4
300
300
900
Process 1 → 2 is isothermal:
(WS )12 = p1V1 ln (V2 V1 ) = −66.8 J
Q12 = (WS )12 = −66.8 J
Process 2 → 3 is isobaric:
(WS )23 = p2ΔV = p2 (V3 − V2 ) = 121.6 J
Q23 = nCP ΔT = nCP (T3 − T2 ) = 608.3 J
Process 3 → 1 is isochoric:
(WS )31 = 0 J
Q31 = nCV ΔT = nCV (T1 − T3 ) = −486.7 J
We find that
(WS )cycle = −66.8 J + 121.6 J + 0 J = 54.8 J
Qcycle = −66.8 J + 608.3 J − 486.7 J = 54.8 J
These are equal, as they should be. Knowing that the work done is Wout = (WS)cycle = 54.8 J/cycle, an engine
operating at 20 cycles/s has power output
Pout =
54.8 J 20 cycle
J
×
= 1096 = 1096 W ≈ 1.10 kW
cycle
s
s
(b) Only Q23 is positive, so Qin = Q23 = 608 J. Thus, the thermal efficiency is
η=
Wout 54.8 J
=
= 0.090 = 9.0%
Qin 608.3 J
19.55. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is
adiabatic, and process 3 → 1 is isobaric. CV = 32 R and CP = 52 R for a monatomic gas.
Visualize: Please refer to Figure P19.55.
Solve: (a) Using the ideal-gas equation, the number of moles is
n=
5
−6
3
p1V1 (1.0 × 10 Pa ) (100 × 10 m )
=
= 0.00401 mol
RT1
(8.31 J/mol K)(300 K)
We can use the adiabat 2 → 3 to calculate p2 as follows:
γ
γ
⎛V ⎞
⎛V ⎞
⎛ 600 cm3 ⎞
p3V3 = p2V2 ⇒ p2 = p3 ⎜ 3 ⎟ = p1 ⎜ 3 ⎟ = (1.0 × 105 Pa ) ⎜
3 ⎟
⎝ 100 cm ⎠
⎝ V2 ⎠
⎝ V2 ⎠
γ
5/3
γ
= 1.981 × 106 Pa
T2 can be determined from the ideal-gas equation and is
T2 =
6
−6
3
p2V2 (1.981 × 10 Pa )(100 × 10 m )
=
= 5943 K
nR
(8.31 J/mol K)(0.00401 mol)
At point 3, p3 = p1 and
T3 T1
V
600 × 10−6 m3
= ⇒ T3 = 3 T1 =
( 300 K ) = 1800 K
V3 V1
V1
100 × 10−6 m3
Now we can calculate Ws, Q, and ΔEth for the three processes involved in the cycle. For process 1 → 2, Ws 1→2 =
0 J and
ΔEth = Q1→ 2 = nCV (T2 − T1 ) = n ( 32 R ) (T2 − T1 ) = 282.2 J
For process 2 → 3, Q2→3 = 0 J and
ΔEth = nCV (T3 − T2 ) = n ( 32 R ) (T3 − T2 ) = –207.2 J
Because, ΔEth = W + Q, W = −Ws, so Ws 2→3 = −ΔEth = +207.2 J for process 2 → 3. For process 3 → 1,
ΔEth = nCV (T1 − T3 ) = n ( 32 R ) (T1 − T3 ) = −75.0 J
Q3→1 = nCP (T1 − T3 ) = n ( 52 R ) (T1 − T3 ) = −125.0 J
The work done Ws 3→1 is the area under the p-versus-V graph. We have
Ws 3→1 = (100 × 103 Pa )(100 × 10−6 m3 − 600 × 10−6 m3 ) = −50.0 J
1→2
2→3
3→1
Net
Ws (J)
Q (J)
ΔEth (J)
0
207.2
−50.0
157.2
282.2
0
−125.0
157.2
282.2
−207.2
−75.0
0
(b) The thermal efficiency of the engine is
Wout 157.2 J
=
= 0.52 = 52%
QH 282.2 J
As expected for a closed cycle, (Ws)net = Qnet and (ΔEth)net = 0 J.
η=
Assess:
19.56. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is
adiabatic, and process 3 → 1 is isothermal. For a diatomic gas CV = 52 R and γ = 75 .
Solve:
(a) From the graph V2 = 1000 cm3 .
The pressure p2 lies on the adiabat from 2 → 3. We can find the pressure as follows:
γ
⎛V ⎞
⎛ 4000 cm3 ⎞
p2V2γ = p3V3γ ⇒ p2 = p3 ⎜ 3 ⎟ = (1.0 ×105 Pa ) ⎜
3 ⎟
⎝ 1000 cm ⎠
⎝ V2 ⎠
7/5
= 6.964 ×105 Pa ≈ 700 kPa
The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2:
⎛ 6.964 × 105 Pa ⎞
p1V1 p2V2
p V
=
⇒ T2 = T1 2 2 = ( 300 K ) ⎜
⎟ (1) = 522.3 K ≈ 522 K
5
T1
T2
p1 V1
⎝ 4.0 × 10 Pa ⎠
(b) The number of moles of the gas is
R=
5
−3
3
p1V1 ( 4.0 × 10 Pa )(1.0 × 10 m )
=
= 0.1604 mol
RT1
(8.31 J/mol K)(300 K)
For isochoric process 1 → 2 , Ws = 0 J and
Q = ΔEth = nCV ΔT = n ( 52 R ) ΔT = 741.1 J
For adiabatic process 2 → 3, Q = 0 J and
ΔEth = nCV ΔT = n ( 52 R ) (T3 − T2 ) = −741.1 J
Using the first law of thermodynamics, ΔEth = ΔWs + Q, which means Ws = −ΔEth = +741.1 J. Ws can also be
determined from
Ws =
p3V3 − p2V2 nR (T3 − T2 ) ( 34 J/K ) ( 300 K − 522.3 K )
=
=
= 741.1 J
1−γ
1−γ
( − 52 )
For isothermal process 3 → 1, ΔEth = 0 J and
V
Ws = nRT1 ln 1 = −554.5 J
V3
Using the first law of thermodynamics, ΔEth = −Ws + Q , Q = Ws = −554.5 J .
1→2
2→3
3→1
Net
ΔEth (J)
Ws (J)
Q (J)
741.1
−741.1
0
0
0
741.1
−554.5
186.6
741.1
0
−554.5
186.6
(c) The work per cycle is 186.6 J and the thermal efficiency is
η=
Ws 186.6 J
=
= 0.25 = 25%
QH 741.1 J
19.57. Model: For the closed cycle of the heat engine, process 1 → 2 is adiabatic, process 2 → 3 is
isothermal, and process 3 → 1 is isochoric. For a diatomic gas CV = 52 R and γ = 75 .
Solve: (a) From the graph V1 = 4000 cm3 .
The number of moles of gas is
p2V2 ( 4.0 × 10 Pa )(1000 × 10 m )
=
= 0.1203 mol
RT2
(8.31 J/mol K)(400 K)
5
n=
−6
3
Using p1V1γ = p2V2γ , and reading V1 from the graph,
γ
⎛V ⎞
7/5
p1 = p2 ⎜ 2 ⎟ = ( 4.0 × 105 Pa ) ( 14 ) = 5.743 × 104 Pa ≈ 5.7 kPa
⎝ V1 ⎠
With p1, V1, and nR having been determined, we can find T1 using the ideal-gas equation:
T1 =
4
−6
3
p1V1 ( 5.743 × 10 )( 4000 × 10 m )
=
= 229.7 K ≈ 230 K
nR
(8.31 J/mol K)(0.1203 mol)
(b) For adiabatic process 1 → 2, Q = 0 J and
Ws =
5
−3
3
4
−3
3
p2V2 − p1V1 ( 4.0 × 10 Pa )(1.00 × 10 m ) − ( 5.743 × 10 Pa )( 4.00 × 10 m )
=
= −425.7 K
1−γ
1 − 1.4
Because ΔEth = −Ws + Q , ΔEth = −Ws = 425.7 K . For isothermal process 2 → 3, ΔEth = 0 J . From Equation 17.15,
Ws = nRT2 ln
V3
= 554.5 J
V2
From the first law of thermodynamics, Q = Ws = 554.5 J for process 2 → 3. For isochoric process 3 → 1, Ws = 0 J and
Q = ΔEth = nCV ΔT = n ( 52 R ) (T1 − T3 ) = −425.7 J
1→2
2→3
3→1
Net
ΔEth (J)
Ws (J)
Q (J)
425.7
0
−425.7
0
−425.7
554.5
0
128.8
0
554.5
−425.7
128.8
(c) The work per cycle is 128.8 J and the thermal efficiency of the engine is
η=
Assess:
Ws 128.8 J
=
= 0.23 = 23%
QH 554.5 J
As expected, for a closed cycle (Ws )net = Qnet and ( ΔEth )net = 0 J.
19.58. Model: In the Brayton cycle, process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and
process 4 → 2 are isobaric. We will assume the gas to be diatomic, with CP = 72 R and γ = 1.40.
Visualize: Please refer to Figure P19.58.
Solve: The number of moles of gas is
p1V1 (1.013 × 10 Pa )( 2.0 m )
=
= 81.27 mol
RT1
(8.31 J/mol K)(300 K)
5
n=
3
We can find the volume at 2 from the adiabatic equation p1V1γ = p2V2γ ,
1
1
⎛ p ⎞γ
⎛ 1 ⎞1.4
V2 = V1 ⎜ 1 ⎟ = ( 2.0 m3 ) ⎜ ⎟ = 0.3861 m3
⎝ 10 ⎠
⎝ p2 ⎠
The temperature T2 is determined from the ideal-gas equation as follows:
2
p1V1 p2V2
p V
⎛ 10 atm ⎞ ⎛ 0.3861 m ⎞
=
⇒ T2 = T1 2 2 = ( 300 K ) ⎜
⎟ = 579.2 K
⎟⎜
3
T1
T2
p1 V1
⎝ 1 atm ⎠ ⎝ 2.0 m ⎠
To find T3 we use the heat input:
QH = 2.0 × 106 J = nCP ΔT = 72 n RΔT = 72 ( 675.33 J/K ) ΔT
⇒ ΔT = 846.1 K ⇒ T3 = T2 + ΔT = 579.2 K + 846.1 K = 1425.3 K
We are now able to find V3 using the ideal-gas law. We have
V3 V2
T
⎛ 1425.3 K ⎞
3
= ⇒ V3 = V2 3 = ( 0.3861 m3 ) ⎜
⎟ = 0.9501 m
T3 T2
T2
⎝ 579.2 K ⎠
The volume at 4 is found by again using the adiabatic equation p3V3γ = p4V4γ ,
1
1
⎛ p ⎞γ
V4 = V3 ⎜ 3 ⎟ = ( 0.9501 m 3 ) (10 )1.4 = 4.921 m3
⎝ p4 ⎠
T4 is now obtained from
T4 =
5
3
p4V4 (1.013 × 10 Pa )( 4.921 m )
=
= 738.2 K
nR
(81.27 mol)(8.31 J/mol K)
To find the engine’s efficiency, we must first find the total work done per cycle which is
Wnet = W1→ 2 + W2 →3 + W3→ 4 + W4 →1
The four contributions are
W1→ 2 =
p2V2 − p1V1 nR (T2 − T1 ) ( 81.27 mol )( 8.31 J/mol K )( 579.2 K − 300 K )
=
=
= −4.714 × 105 J
1−γ
1−γ
1 − 1.4
W2 →3 = p2 ΔV = (1.013 × 106 Pa )( 0.9501 m3 − 0.3861 m3 ) = 5.713 × 105 J
W3→ 4 =
nR (T4 − T3 ) ( 81.27 mol )( 8.31 J/mol K )( 738.2 K − 1425.3 K )
=
= +1.160 × 106 J
1−γ
1 − 1.4
W4 →1 = p4 ΔV = (1.013 × 105 Pa )( 2.0 m3 − 4.921 m3 ) = −2.959 × 105 J
Thus, Wnet = 9.64 × 105 J . The thermal efficiency is
η=
Wnet
9.64 × 105 J
⇒η =
= 0.482 = 48.2%
QH
2.0 × 106 J
As a comparison, we can calculate η from Equation 19.21 which is
η =1−
1
(r )
p
where rp = pmax pmin = 10 .
( )
γ −1
γ
=1−
1
(10 )
0.4
1.4
= 48.2%
19.59. Model: Process 1 → 2 of the cycle is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is
isobaric. For a monatomic gas, CV = 32 R and CP = 52 R.
Visualize: Please refer to Figure P19.59.
Solve: (a)
At
point
1:
The
pressure
p1 = 1.0 atm = 1.013 × 105 Pa
and
the
volume
V1 = 1000 ×10−6 m 3 = 1.0 × 10−3 m3 . The number of moles is
n=
0.120 g
= 0.030 mol
4 g/mol
Using the ideal-gas law,
T1 =
5
−3
3
p1V1 (1.013 × 10 Pa )(1.0 × 10 m )
=
= 406 K
nR
( 0.030 mol )(8.31 J/mol K )
At point 2: The pressure p2 = 5.0 atm = 5.06 × 105 Pa and V2 = 1.0 × 10−3 m3 . The temperature is
T2 =
5
−3
3
p2V2 ( 5.06 × 10 Pa )(1.0 × 10 m )
=
= 2030 K
nR
( 0.030 mol )(8.31 J/mol K )
At point 3: The pressure is p3 = 1.0 atm = 1.013 × 105 Pa and the temperature is T3 = T2 = 2030 K . The volume is
V3 = V2
p2
⎛ 5 atm ⎞
−3
3
= (1.0 × 10−3 m3 ) ⎜
⎟ = 5.0 × 10 m
p3
1
atm
⎝
⎠
(b) For isochoric process 1 → 2, W1→2 = 0 J and
Q1→ 2 = nCV ΔT = ( 0.030 mol ) ( 32 R ) ( 2030 K − 406 K ) = 607 J
For isothermal process 2 → 3, ΔEth 2 →3 = 0 J and
Q2 →3 = W2 →3 = nRT2 ln
⎛ 5.0 × 10−3 m3 ⎞
V3
= ( 0.030 mol )( 8.31 J/mol K )( 2030 K ) ln ⎜
= 815 J
−3
3 ⎟
V2
⎝ 1.0 × 10 m ⎠
For isobaric process 3 → 1,
W3→1 = p3ΔV = (1.013 × 105 Pa )(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) = −405 J
Q3→1 = nCP ΔT = ( 0.030 mol ) ( 52 ) ( 8.31 J/mol K )( 406 K − 2030 K ) = −1012 J
The total work done is Wnet = W1→ 2 + W2 →3 + W3→1 = 410 J. The total heat input is QH = Q1→ 2 + Q2 →3 = 1422 J . The
efficiency of the engine is
η=
Wnet 410 J
=
= 20%
QH 1422 J
(c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is
ηmax = 1 −
Assess:
406 K
Tmin
=1−
= 80%
Tmax
2030 K
The actual efficiency of an engine is less than the maximum possible efficiency.
19.60. Model: The process 2 → 3 of the heat engine cycle is isochoric and the process 3 → 1 is isobaric. For
a monatomic gas CV = 32 R and CP = 52 R .
Solve: (a) The three temperatures are
T1 =
5
3
p1V1 ( 4.0 × 10 Pa )( 0.025 m )
=
= 601.7 K ≈ 602 K
nR ( 2.0 mol )( 8.31 J/mol K )
T2 =
5
3
p2V2 ( 6.0 × 10 Pa )( 0.050 m )
=
= 1805.1 K ≈ 1805 K
nR
( 2.0 mol )( 8.31 J/mol K )
T3 =
p3V3 ( 4.0 × 10 Pa )( 0.050 m )
=
= 1203.4 K ≈ 1203 K
nR
( 2.0 mol )( 8.31 J/mol K )
5
3
(b) For process 1 → 2, the work done is the area under the p-versus-V graph. The work and the change in internal
energy are
Ws = 12 ( 6.0 × 105 Pa − 4.0 × 105 Pa )( 0.050 m3 − 0.025 m3 ) + ( 4.0 × 105 Pa )( 0.050 m3 − 0.025 m3 )
= 1.25 × 104 J
ΔEth = nCV ΔT = ( 2.0 mol ) ( 32 R ) (T2 − T1 )
= ( 2.0 mol ) ( 32 ) ( 8.31 J/mol K )(1805.1 K − 601.7 K ) = 3.00 × 104 J
The heat input is Q = Ws + ΔEth = 4.25 × 104 J . For isochoric process 2 → 3, Ws = 0 J and
Q = ΔEth = nCV ΔT = ( 2.0 mol ) 32 ( 8.31 J/mol K )(1203.4 K − 1805.1 K ) = −1.50 × 104 J
For isobaric process 3 → 1, the work done is the area under the p-versus-V curve. Hence,
Ws = ( 4.0 × 105 Pa )( 0.025 m3 − 0.050 m3 ) = −1.0 × 104 J
ΔEth = nCV ΔT = n ( 32 R ) (T1 − T3 ) = ( 2.0 mol ) 32 ( 8.31 J/mol K )( 601.7 K − 1203.4 K ) = −1.5 × 104 J
The heat input is Q = WS + ΔEth = −2.50 × 104 J .
ΔEth (J)
1→2
2→3
3→1
Net
Ws (J)
4
3.0 × 10
−1.5 × 104
−1.5 × 104
0
1.25 × 10
0
−1.0 ×104
2.5 × 103
(c) The thermal efficiency is
η=
Q (J)
4
Wnet
2.5 × 103 J
=
= 5.9%
QH 4.25 × 104 J
4.25 × 104
−1.50 × 104
−2.50 × 104
2.5 × 103
19.61. Model: The closed cycle in this heat engine includes adiabatic process 1 → 2, isobaric process 2 → 3,
and isochoric process 3 → 1. For a diatomic gas, CV = 52 R, CP = 72 R, and γ = 75 =1.4.
Visualize: Please refer to Figure P19.61.
Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows:
T2 =
5
−3
3
p2V2 ( 4.0 × 10 Pa )(1.0 × 10 m )
=
= 2407 K
nR
( 0.020 mol )( 8.31 J/mol K )
We can use the equation p2V2γ = p1V1γ to find V1,
1/ γ
⎛p ⎞
V1 = V2 ⎜ 2 ⎟
⎝ p1 ⎠
1/1.4
⎛ 4.0 × 105 Pa ⎞
= (1.0 × 10−3 m3 ) ⎜
⎟
5
⎝ 1.0 × 10 Pa ⎠
= 2.692 × 10−3 m3
The ideal-gas equation can now be used to find T1,
T1 =
5
−3
3
p1V1 (1.0 × 10 Pa )( 2.692 × 10 m )
=
= 1620 K
nR
( 0.020 mol )(8.31 J/mol K )
T3 =
5
−3
3
p3V3 ( 4 × 10 Pa )( 2.692 × 10 m )
=
= 6479 K
nR
( 0.020 mol )( 8.31 J/mol K )
At point 3, V3 = V1 so we have
(b) For adiabatic process 1 → 2, Q = 0 J, ΔEth = −Ws , and
Ws =
p2V2 − p1V1 nR (T2 − T1 ) ( 0.020 mol )( 8.31 J/mol K )( 2407 K − 1620 K )
=
=
= −327.0 J
1−γ
1−γ
(1 − 1.4 )
For isobaric process 2 → 3,
Q = nCP ΔT = n ( 72 R ) ( ΔT ) = ( 0.020 mol ) 72 ( 8.31 J/mol K )( 6479 K − 2407 K ) = 2369 J
ΔEth = nCV ΔT = n ( 52 R ) ΔT = 1692 J
The work done is the area under the p-versus-V graph. Hence,
Ws = ( 4.0 × 105 Pa )( 2.692 × 10−3 m3 − 1.0 × 10−3 m3 ) = 677 J
For isochoric process 3 → 1, Ws = 0 J and
ΔEth = Q = nCV ΔT = ( 0.020 mol ) ( 52 ) ( 8.31 J/mol K )(1620 K − 6479 K ) = −2019 J
1→2
2→3
3→1
Net
ΔEth (J)
Ws (J)
Q (J)
327
1692
−2019
0
−327
677
0
350
0
2369
−2019
350
η=
350 J
= 0.15 = 15%
2369 J
(c) The engine’s thermal efficiency is
19.62. Model: The closed cycle of the heat engine involves the following processes: one isothermal
compression, one isobaric expansion, and one isochoric cooling.
Visualize:
Solve:
The number of moles of helium is
n=
2.0 g
= 0.50 mol
4.0 g/mol
The total work done by the engine per cycle is Wnet = W1→ 2 + W2 →3 + W3→1 . For the isothermal process,
W1→ 2 = nRT1 ln
V2
= ( 0.50 mol )( 8.31 J/mol K )( 273 K ) ln ( 0.5) = −786.2 J
V1
For the isobaric process,
⎛V ⎞
W2 →3 = p2 ( ΔV ) = ( 2 p1 ) ⎜ 1 ⎟ = p1V1 = nRT1 = ( 0.50 mol )( 8.31 J/mol K )( 273 K ) = 1134.3 J
⎝2⎠
For the isochoric process, W3→1 = 0 J. Thus, Wnet = 348.1 J ≈ 350 J .
For calculating heat transfers, note that T2 = T1 = 273 K and, since process 3 → 1 is isochoric,
T3 = (2 atm/1 atm) T1 = 546 K. For the isothermal process, Q1→ 2 = W1→ 2 = −786.2 J because ΔEth 1→ 2 = 0 J. For
the isobaric process,
Q2 →3 = nCP ΔT = n ( 52 R ) (T3 − T2 ) = 52 ( 0.50 mol )( 8.31 J/mol K )( 273 K ) = 2835.8 J
For the isochoric process,
Q3→1 = nCv ΔT = n ( 32 R ) (T1 − T3 ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( –273 K ) = −1701.5 J
From Q1→2, Q2→3, and Q3→1, we see that QH = 2835.8 J. Thus, the thermal efficiency is
η=
348.1 J
= 0.12 = 12%
2835.8 J
19.63. Model: The closed cycle of the heat engine involves the following four processes: isothermal
expansion, isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas CV = 32 R .
Visualize:
Solve:
Using the ideal-gas law,
p1 =
nRT1 ( 0.20 mol )( 8.31 J/mol K )( 600 K )
=
= 4.986 × 105 Pa
V1
2.0 × 10−3 m3
At point 2, because of the isothermal conditions, T2 = T1 = 600 K and
⎛ 2.0 × 10−3 m3 ⎞
V
p2 = p1 1 = ( 4.986 × 105 Pa ) ⎜
= 2.493 × 105 Pa
−3
3 ⎟
V2
⎝ 4.0 × 10 m ⎠
At point 3, because it is an isochoric process, V3 = V2 = 4000 cm3 and
p3 = p2
T3
⎛ 300 K ⎞
5
= ( 2.493 × 105 Pa ) ⎜
⎟ = 1.247 × 10 Pa
T2
⎝ 600 K ⎠
Likewise at point 4, T4 = T3 = 300 K and
p4 = p3
⎛ 4.0 × 10−3 m3 ⎞
V3
= (1.247 × 105 Pa ) ⎜
= 2.493 × 105 Pa
−3
3 ⎟
V4
×
2.0
10
m
⎝
⎠
Let us now calculate Wnet = W1→2 + W2→3 + W3→4 + W4→1. For the isothermal processes,
W1→ 2 = nRT1 ln
W3→ 4 = nRT3 ln
V2
= ( 0.20 mol )( 8.31 J/mol K )( 600 K ) ln ( 2 ) = 691.2 J
V1
V4
= ( 0.20 mol )( 8.31 J/mol K )( 300 K ) ln ( 12 ) = −345.6 J
V3
For the isochoric processes, W2→3 = W4→1 = 0 J. Thus, Wnet = 345.6 J ≈ 350 J .
Because Q = Ws + ΔEth,
Q1→ 2 = W1→ 2 + ( ΔEth )1→ 2 = 691.2 J + 0 J = 691.2 J
For the first isochoric process,
Q2 →3 = nCV ΔT = ( 0.20 mol ) ( 32 R ) (T3 − T2 )
= ( 0.20 mol ) 32 ( 8.31 J/mol K )( 300 K − 600 K ) = −747.9 K
For the second isothermal process
Q3→ 4 = W3→ 4 + ( ΔEth )3→ 4 = −345.6 J + 0 J = −345.6 J
For the second isochoric process,
Q4 →1 = nCV ΔT = n ( 32 R ) (T1 − T4 )
= ( 0.20 mol ) ( 32 ) ( 8.31 J/mol K )( 600 K − 300 K ) = 747.9 K
Thus, QH = Q1→ 2 + Q4 →1 = 1439.1 J . The thermal efficiency of the engine is
η=
Wnet 345.6 J
=
= 0.24 = 24%
QH 1439.1 J
19.64. Model: Processes 2 → 1 and 4 → 3 are isobaric. Processes 3 → 2 and 1 → 4 are isochoric.
Visualize:
Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus
heat is transferred in during processes 4 → 3 and 3 → 2. This is the reverse of the heat engine in Example 19.2.
(b) Heat flows from hot to cold. Since heat energy is transferred into the gas during processes 4 → 3 and 3 → 2,
which end with the gas at temperature 2700 K, the reservoir temperature must be T > 2700 K. This is the hot
reservoir, so the heat transferred is QH. Similarly, heat energy is transferred out of the gas during processes 2 → 1
and 1 → 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energy
transferred during these two processes is QC.
(c) The heat energies were calculated in Example 19.2, but now they have the opposite signs.
QH = Q43 + Q32 = 7.09 × 105 J + 15.19 × 105 J = 22.28 × 105 J
QC = Q21 + Q14 = 21.27 × 105 J + 5.06 × 105 J = 26.33 × 105 J
(d) For a counterclockwise cycle in the pV diagram, the work is Win. Its value is the area inside the curve, which is
Win = (∆p)(∆V) = (2 × 101,300 Pa)(2 m3) = 4.05 × 105 J. Note that Win = QC – QH, as expected from energy
conservation.
(e) Since QC > QH, more heat is exhausted to the cold reservoir than is extracted from the hot reservoir. In this
device, work is used to transfer energy “downhill,” from hot to cold. The exhaust energy is QC = QH + Win > QH.
This is the energy-transfer diagram of Figure 19.19.
(f) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. This
device uses work input to transfer heat energy from the hot reservoir to the cold reservoir.
19.65. Solve: (a) If you wish to build a Carnot engine that is 80% efficient and exhausts heat into a cold
reservoir at 0°C, what temperature (in °C) must the hot reservoir be?
(b)
0.80 = 1 −
( 0°C + 273) ⇒ 273 = 0.20 ⇒ T = 1092°C
H
(TH + 273) TH + 273
19.66. Solve: (a) A refrigerator with a coefficient of performance of 4.0 exhausts 100 J of heat in each cycle.
What work is required each cycle and how much heat is removed each cycle from the cold reservoir?
(b) We have 4.0 = QC Win ⇒ QC = 4Win . This means
QH = QC + Win = 4Win + Win = 5Win ⇒ Win = QH = 100 J = 20 J
5
5
Hence, QC = QH − Win = 100 J − 20 J = 80 J.
19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What is
the net heat extracted from the hot reservoir and the net heat exhausted in each cycle?
(b) We have 0.20 = 1 − QC QH . Using the first law of thermodynamics,
Wout = QH − QC = 20 J ⇒ QC = QH − 20 J
Substituting into the definition of efficiency,
0.20 = 1 −
QH − 20 J
20 J 20 J
20 J
= 1−1+
=
⇒ QH =
= 100 J
0.20
QH
QH
QH
The heat exhausted is QC = QH − 20 J = 100 J − 20 J = 80 J .
19.68.
Solve: (a)
In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure?
(b)
1
( pmax − 1.0 ×105 Pa )( 2.0 m3 ) = 4.0 × 105 J ⇒ pmax = 5.0 ×105 Pa = 500 kPa
2
19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state.
Visualize: The figure indicates the following seven steps. First, the pin is inserted when the heat engine has the
initial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 atm to 3 atm.
Third, the pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from
50 cm3 to 100 cm3. Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on
ice and the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still on ice, the pin is
removed. The gas continues to cool at constant pressure to a volume of 50 cm3. Seventh, with no ice or flame, the
pin is inserted back in and the weights returned bringing the engine back to the initial conditions and ready to start
over.
Solve:
(a)
(b) The work done per cycle is the area inside the curve:
Wout = (∆p)(∆V) = (2 × 101,300 Pa)(50 × 10–6 m3) = 10.13 J
(c) Heat energy is input during processes 1 → 2 and 2 → 3, so QH = Q12 + Q23. This is a diatomic gas, with CV =
3
R and CP = 52 R. The number of moles of gas is
2
n=
p1V1 (101,300 Pa)(50 × 10−6 m3 )
=
= 0.00208 mol
RT1
(8.31 J/mol K) (293 K)
Process 1 → 2 is isochoric, so T2 = (p2/p1)T1 = 3T1 = 879 K. Process 2 → 3 is isobaric, so T3 = (V3/V2)T2 = 2T2 =
1758 K. Thus
Q12 = nCV ΔT = 52 nR(T2 − T1 ) = 52 (0.00208 mol)(8.31 J/mol K)(586 K) = 25.32 J
Similarly,
Q23 = nCP ΔT = 72 nR (T3 − T2 ) = 72 (0.00208 mol)(8.31 J/mol K)(879 K) = 53.18 J
Thus QH = 25.32 J + 53.18 J = 78.50 J and the engine’s efficiency is
η=
Wout 10.13 J
=
= 0.13 = 13%
QH 78.50 J
19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process.
Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat input
from (or loss to) the outside world, we have Q1 + Q2 = 0 J. Heat Q1, which is negative, will change the
temperature of system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston.
But these consequences of heat flow don’t change the fact that Q1 + Q2 = 0 J. System 1 undergoes constant
volume cooling from T1i = 600 K to Tf. System 2, whose pressure is controlled by the weight of the piston,
undergoes constant pressure heating from T2i = 300 K to Tf. Thus,
Q1 + Q2 = 0 J = n1CV (Tf − T1i ) + n2CP (Tf − T2i ) = n1 ( 32 R ) (Tf − T1i ) + n2 ( 52 R ) (Tf − T2i )
Solving this equation for Tf gives
Tf =
3n1T1i + 5n2T2i 3 ( 0.060 mol )( 600 K ) + 5 ( 0.030 mol )( 300 K )
=
= 464 K
3n1 + 5n2
3 ( 0.060 mol ) + 5 ( 0.030 mol )
(b) Knowing Tf , we can compute the heat transferred from system 1 to system 2:
Q2 = n2CP (Tf − T2i ) = n2 ( 52 R ) (Tf − T2i ) = 102 J
(c) The change of thermal energy in system 2 is
ΔEth = n2CV ΔT = n2 ( 32 R ) (Tf − T2i ) = 53 Q2 = 61.2 J
According to the first law of thermodynamics, Q2 = Ws + ΔEth. Thus, the work done by system 2 is Ws = Q − ΔEth
=
102.0 J − 61.2 J = 40.8 J. The work is done to lift the weight of the cylinder and the air above it by a height Δy.
The weight of the air is wair = pA = pπ r 2 = (101.3 × 10 N/m 2 )π (0.050 m) 2 = 795.6 N. So,
Ws = ( wcyl + wair )Δy ⇒ Δy =
Ws
40.8 J
=
= 0.050 m
( wcyl + wair ) ( 2.0 kg ) ( 9.8 m/s 2 ) + 795.6 N
(d) The fraction of heat converted to work is
Ws 40.8 J
=
= 0.40 = 40%
Q2 102.0 J
19.71. Model: Process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 1 are
isochoric.
Visualize: Please refer to Figure CP19.71.
Solve: (a) For adiabatic process 1 → 2, Q12 = 0 J and
W12 =
p2V2 − p1V1 nR (T2 − T1 )
=
1−γ
1−γ
For isochoric process 2 → 3, W23 = 0 J and Q23 = nCV (T3 − T2 ) . For adiabatic process 3 → 4, Q34 = 0 J and
W34 =
p4V4 − p3V3 nR (T4 − T3 )
=
1−γ
1−γ
For isochoric process 4 → 1, W41 = 0 J and Q41 = nCV (T1 − T4 ) . The work done per cycle is
Wnet = W12 + W23 + W34 + W41 =
nR (T2 − T1 )
nR (T4 − T3 )
nR
+0 J+
+0 J =
(T2 − T1 + T4 − T3 )
1− γ
1−γ
1−γ
(b) The thermal efficiency of the heat engine is
η=
Q
nC (T − T )
Wout
Q
= 1 − C = 1 − 41 = 1 − V 4 1
QH
QH
Q23
nCV (T3 − T2 )
The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as
follows:
γ −1
⎛V ⎞
γ −1
= nRT2V2γ −1 ⇒ T2 = T1 ⎜ 1 ⎟
pV γ = pVV γ −1 = nRTV γ −1 ⇒ nRTV
1 1
⎝ V2 ⎠
Similarly, T3 = T4 r γ −1 . The equation for thermal efficiency now becomes
η =1−
(c)
1
T4 − T1
= 1 − γ −1
γ −1
r
−
T4 r
T1r
γ −1
= T1r γ −1
19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaric
expansion, process 3 → 4 is an adiabatic expansion, and process 4 → 1 is isochoric.
Visualize: Please refer to CP19.72.
Solve: (a) It will be useful to do some calculations using the compression ratio, which is
r=
Vmax V1 1050 cm3
= =
= 21
Vmin V2
50 cm3
The number of moles of gas is
n=
5
−6
3
p1V1 (1.013 × 10 Pa )(1050 × 10 m )
=
=0.0430 mol
RT1
( 8.31 J/mol K ) ⎡⎣( 25+273) K ⎤⎦
For an adiabatic process,
γ
⎛V ⎞
p1V1 = p2V2 ⇒ p2 = ⎜ 1 ⎟ p1 = r γ p1 = 211.40 × 1 atm = 71.0 atm = 7.19 × 106 Pa
⎝ V2 ⎠
γ
γ
γ −1
⎛V ⎞
γ −1
= T2V2γ −1 ⇒ T2 = ⎜ 1 ⎟
TV
1 1
⎝ V2 ⎠
T1 = r γ −1T1 = 210.41 × 298 K = 1007 K
Process 2 → 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys
Q = nCP ΔT ⇒ ΔT =
Q
nCP
The gas has a specific heat ratio γ = 1.40 = 7 / 5, thus CV = 52 R and CP = 72 R. Knowing CP, we can calculate first
ΔT =
800 K and then T3 = T2 + ΔT = 1807 K. Finally, for an isobaric process we have
V2 V3
T
1807 K
= ⇒ V3 = 3 V2 =
( 50 ×10−6 m3 ) = 89.7 × 10−6 m3
T2 T3
T2
1007 K
Process 3 → 4 is an adiabatic expansion to V4 = V1. Thus,
γ
1.4
⎛V ⎞
⎛ 89.7 ×10−6 m 3 ⎞
p3V3γ = p4V4γ ⇒ p4 = ⎜ 3 ⎟ p3 = ⎜
3 ⎟
−6
⎝ 1050 × 10 m ⎠
⎝ V4 ⎠
γ −1
γ −1
T3V3
= T4V4
Point
1
2
3
4
γ −1
⎛V ⎞
⇒ T4 = ⎜ 3 ⎟
⎝ V4 ⎠
p
( 7.19 ×10 Pa ) = 2.30 ×10 Pa = 2.27 atm
6
⎛ 89.7 ×10−6 m3 ⎞
T3 = ⎜
−6
3 ⎟
⎝ 1050 × 10 m ⎠
5
0.4
V
5
1.00 atm = 1.013 × 10 Pa
6
71.0 atm = 7.19 × 10 Pa
6
71.0 atm = 7.19 × 10 Pa
5
2.27 atm = 2.30 × 10 Pa
T
−6
3
298 K = 25°C
−6
3
1007 K = 734°C
−6
3
1807 K = 1534°C
−6
3
675 K = 402°C
1050 × 10 m
50.0 × 10 m
89.7 × 10 m
1050 × 10 m
(b) For adiabatic process 1 → 2,
(Ws )12 =
p2V2 − p1V1
= −633 J
1−γ
For isobaric process 2 → 3,
(Ws )23 = p2ΔV = p2 (V3 − V2 ) = 285 J
For adiabatic process 3 → 4,
(1807 K ) = 675 K
(Ws )34 =
p4V4 − p3V3
= 1009 J
1−γ
For isochoric process 4 → 1, (Ws )41 = 0 J. Thus,
(Ws )cycle = Wout = (Ws )12 + (Ws )23 + (Ws )34 + (Ws )41 = 661 J
(c) The efficiency is
η=
Wout 661 J
=
= 66.1% ≈ 66%
QH 1000 J
(d) The power output of one cylinder is
661 J 2400 cycle 1 min
J
×
×
= 26,440 = 26.4 kW
cycle
min
60 sec
s
For an 8-cylinder engine the power will be 211 kW or 283 horsepower.
20.1.
Model: The wave is a traveling wave on a stretched string.
Solve: The wave speed on a stretched string with linear density μ is vstring = TS / μ . The wave speed if the
tension is doubled will be
′ =
vstring
2TS
μ
= 2vstring = 2 ( 200 m/s ) = 283 m/s
20.2.
Solve:
Model: The wave is a traveling wave on a stretched string.
The wave speed on a stretched string with linear density μ is
T
75 N
⇒ μ = 3.333 × 10−3 kg/m
vstring = S ⇒ 150 m/s =
μ
μ
For a wave speed of 180 m/s, the required tension will be
2
2
TS = μ vstring
= ( 3.333 × 10−3 kg/m ) (180 m/s ) = 110 N
20.3.
Model: The wave pulse is a traveling wave on a stretched string.
The wave speed on a stretched string with linear density μ is
( 2.0 m )( 20 N ) ⇒ m = 0.025 kg = 25 g
T
TS
LTS
2.0 m
=
⇒
=
vstring = S =
μ
m/L
m
50 × 10−3 s
m
Solve:
20.4. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.
Visualize: The snapshot graph shows the wave at all points on the x-axis at t = 0 s. The wave is just reaching x
= 5.0. The first part of the wave causes an upward displacement of the medium. The rising portion of the wave is
2 m wide, so it will take 2 s to pass the x = 5.0 m point. The constant part of the wave, whose width is 2 m, will
take 2 seconds to pass x = 5.0 m and during this time the displacement of the medium will be a constant (Δy = 1
cm). The trailing edge of the pulse arrives at t = 4 s at x = 5.0 m. The displacement now becomes zero and stays
zero for all later times.
20.5.
Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second.
Visualize: This snapshot graph shows the wave at all points on the x-axis at t = 2 s. You can see that the
leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first 2 seconds, the displacement is zero
at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2
s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 m wide and takes 3 s
to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 s the displacement of
the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for all later times.
20.6. Model: This is a wave traveling at constant speed to the right at 1 m/s.
Visualize: This is the history graph of a wave at x = 0 m. The graph shows that the x = 0 m point of the
medium first sees the negative portion of the pulse wave at t = 1.0 s. Thus, the snapshot graph of this wave at t =
1.0 s must have the leading negative portion of the wave at x = 0 m.
20.7. Model: This is a wave traveling at constant speed to the left at 1 m/s.
Visualize: This is the history graph of a wave at x = 2 m. Because the wave is moving to the left at 1 m/s, the
wave passes the x = 2 m position a distance of 1 m in 1 s. Because the flat part of the history graph takes 2 s to
pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearly increasing part of the history graph
is 2 m. The center of the flat part of the history graph corresponds to both t = 0 s and x = 2 m.
20.8.
Visualize:
Figure EX20.8 shows a snapshot graph at t = 0 s of a longitudinal wave. This diagram shows a row of particles
with an inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude
of 0.5 cm between x = 3 cm and x = 8 cm, the particles at x = 3, 4, 5, 6, 7 and 8 cm are displaced to the right by
0.5 cm.
20.9.
Visualize:
We first draw the particles of the medium in the equilibrium positions, with an inter-particle spacing of 1.0 cm. Just
underneath, the positions of the particles as a longitudinal wave is passing through are shown at time t = 0 s. It is
clear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively on
the right side. For example, the particles at x = 0 cm and x = 1 cm are at equilibrium, the particle at x = 2 cm is
displaced left by 0.5 cm, the particle at x = 3 cm is displaced left by 1.0 cm, the particle at x = 4 cm is displaced left
by 0.5 cm, and the particle at x = 5 cm is undisplaced. The behavior of particles for x > 5 cm is opposite of that for x
< 5 cm.
20.10. Solve: (a) The wave number is
k=
2π
λ
=
2π
= 3.1 rad/m
2.0 m
(b) The wave speed is
⎛ω ⎞
⎛ 30 rad/s ⎞
v = λ f = λ⎜
⎟ = (2.0 m) ⎜
⎟ = 9.5 m/s
⎝ 2π ⎠
⎝ 2π ⎠
20.11. Solve: (a) The wavelength is
λ=
2π
2π
=
= 4.2 m
k 1.5 rad/m
(b) The frequency is
f =
v
λ
=
200 m/s
= 48 Hz
4.19 m
20.12. Model: The wave is a traveling wave.
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 3.5 cm, k = 2.7 rad/m, ω =
124 rad/s, and φ0 = 0 rad. The frequency is
f =
ω
2π
=
124 rad/s
= 19.7 Hz ≈ 20 Hz
2π
(b) The wavelength is
λ=
(c) The wave speed v = λ f = 46 m/s .
2π
2π
=
= 2.33 m ≈ 2.3 m
k
2.7 rad/m
20.13. Model: The wave is a traveling wave.
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 5.2 cm, k = 5.5 rad/m, ω = 72
rad/s, and φ0 = 0 rad. The frequency is
f =
ω
2π
=
72 rad/s
= 11.5 Hz ≈ 11 Hz
2π
(b) The wavelength is
λ=
(c) The wave speed v = λ f = 13 m/s.
2π
2π
=
= 1.14 m ≈ 1.1 m
k
5.5 rad/m
20.14. Solve: The amplitude of the wave is the maximum displacement, which is 6.0 cm. The period of the
wave is 0.60 s, so the frequency f = 1 T = 1 0.60 s = 1.67 Hz . The wavelength is
λ=
v
2 m/s
=
= 1.2 m
f 1.667 Hz
20.15.
Solve:
According to Equation 20.28, the phase difference between two points on a wave is
Δφ = φ2 − φ1 = 2π
Δr
λ
=
2π
λ
( r2 − r1 ) ⇒ ( 3π rad − 0 rad ) =
2π
λ
( 80 cm − 20 cm ) ⇒ λ = 40 cm
20.16.
Solve:
According to Equation 20.28, the phase difference between two points on a wave is
Δφ = φ2 − φ1 =
2π
λ
( r2 − r1 )
If φ1 = π rad at r1 = 4.0 m, we can determine φ2 at any r value at the same instant using this equation. At r2 = 3.5
m,
φ2 = φ1 +
At r2 = 4.5 m, φ = 32 π rad.
2π
λ
( r2 − r1 ) = π rad +
2π
π
( 3.5 m − 4.0 m ) = rad
2.0 m
2
20.17.
Visualize:
Solve:
For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28:
Δφ = φ2 − φ1 =
2π
λ
( r2 − r1 ) =
2π
λ
( 40 m − 30 m ) ⇒ λ =
2π
(10 m )
Δφ
Δφ = 2π for two points on adjacent wavefronts and Δφ = 4π for two points separated by 2λ. Thus, λ = 10 m
when Δφ = 2π , and λ = 5 m when Δφ = 4π . The crests corresponding to these two wavelengths are shown in
the figure. One can see that a crest of the wave passes the 40 m–listener and the 30 m–listener simultaneously.
The lowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5
m. Thus, the lowest frequency is
f1 =
The next highest frequency is f 2 = 68 Hz.
v
λ
=
340 m/s
= 34 Hz
10 m
20.18. Visualize:
Solve:
(a) Because the same wavefront simultaneously reaches listeners at x = −7.0 m and x = + 3.0 m,
Δφ = 0 rad =
2π
λ
( r2 − r1 ) ⇒ r2 = r1
Thus, the source is at x = −2.0 m, so that it is equidistant from the two listeners.
(b) The third person is also 5.0 m away from the source. Her y-coordinate is thus y = (5 m) 2 − (2 m) 2 = 4.6 m .
20.19.
Solve: Two pulses of sound are detected because one pulse travels through the metal to the
microphone while the other travels through the air to the microphone. The time interval for the sound pulse
traveling through the air is
Δtair =
Δx
4.0 m
=
= 0.01166 s = 11.66 ms
vair 343 m/s
Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone
before the pulse traveling through the air. Because the pulses are separated in time by 11.0 ms, the pulse
traveling through the metal takes Δtmetal = 0.66 ms to travel the 4.0 m to the microphone. Thus, the speed of
sound in the metal is
vmetal =
Δx
4.0 m
=
= 6060 m/s ≈ 6100 m/s
Δtmetal 0.00066 s
20.20.
Solve:
(a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to
λ=
v
6420 m/s
=
= 3.21 × 10−3 m = 3.21 mm ≈ 3.2 mm
f 2.0 × 106 Hz
(b) The speed of an electromagnetic wave is c. The frequency would be
f =
c
λ
=
3.0 × 108 m/s
= 9.3 × 1010 Hz
3.21 × 10−3 m
20.21.
Solve:
(a) The frequency is
f =
vair
λ
=
343 m/s
= 1715 Hz ≈ 1700 Hz
0.20 m
(b) The frequency is
f =
c
λ
=
3.0 × 108 m/s
= 1.5 × 109 Hz = 1.5 GHz
0.20 m
(c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be
v
1480 m/s
λ = water =
= 9.87 × 10−7 m ≈ 990 nm
f
1.50 × 109 Hz
20.22.
Model: Light is an electromagnetic wave that travels with a speed of 3 × 108 m/s.
Solve:
(a) The frequency of the blue light is
f blue =
c
λ
=
3.0 × 108 m/s
= 6.67 × 1014 Hz
450 × 10−9 m
(b) The frequency of the red light is
3.0 × 108 m/s
= 4.62 × 1014 Hz
650 × 10−9 m
(c) Using Equation 20.30 to calculate the index of refraction,
f red =
λmaterial =
λvacuum
n
⇒n=
λvacuum 650 nm
=
= 1.44
λmaterial 450 nm
20.23.
Solve:
Model: Microwaves are electromagnetic waves that travel with a speed of 3 × 108 m/s.
(a) The frequency of the microwave is
f microwaves =
c
λ
=
3.0 × 108 m/s
= 1.0 × 1010 Hz = 10 GHz
3.0 × 10−2 m
(b) The refractive index of air is 1.0003, so the speed of microwaves in air is vair = c /1.00 ≈ c. The time for the
microwave signal to travel is
t=
50 km
50 × 103 m
=
= 0.167 ms ≈ 0.17 ms
vair
( 3.0 ×108 m 1.00 )
Assess: A small time of 0.17 ms for the microwaves to cover a distance of 50 km shows that the
electromagnetic waves travel very fast.
20.24.
Solve:
Model: Radio waves are electromagnetic waves that travel with speed c.
(a) The wavelength is
λ=
c 3.0 × 108 m/s
=
= 2.96 m
f
101.3 MHz
(b) The speed of sound in air at 20°C is 343 m/s. The frequency is
f =
vsound
λ
=
343 m/s
= 116 Hz
2.96 m
20.25.
Solve:
Model: Light is an electromagnetic wave.
(a) The time light takes is
3.0 mm 3.0 × 10−3 m
3.0 × 10−3 m
=
=
= 1.5 × 10−11 s
8
vglass
cn
3.0
×
10
m/s
1.50
(
)
t=
(b) The thickness of water is
d = vwatert =
c
nwater
t=
3.0 × 108 m/s
(1.5 ×10−11 s ) = 3.4 mm
1.33
20.26.
Solve:
(a) The speed of light in a material is given by Equation 20.29:
n=
c
c
⇒ vmat =
vmat
n
The refractive index is
n=
λvac
λ
420 nm
⇒ vsolid = c solid = ( 3.0 × 108 m/s )
= 1.88 × 108 m/s
670 nm
λmat
λvac
(b) The frequency is
f =
vsolid
λsolid
=
1.88 × 108 m/s
= 4.48 × 1014 Hz
420 nm
20.27. Model: Assume
that
the
glass
has
index
of
refraction
n = 1.5.
This
means
that
vglass = c/n = 2 × 108 m/s.
Visualize: We apply v = λ f twice, once in air and then in the glass. The frequency will be the same in both
cases.
Solve: (a) In the air
f air =
vair
λair
=
3.0 ×108 m/s
= 8.57 × 108 Hz ≈ 8.6 × 108 Hz
0.35 m
The frequency is the same in both media, so f glass = 8.6 × 108 Hz.
(b) Now that we know f glass and vglass , we can find λglass .
λglass =
Assess:
vglass
f glass
=
2.0 × 108 m/s
= 23 cm
8.57 ×108 Hz
We get the same answer from λglass = λair /nglass = 35 cm/1.5 = 23 cm .
20.28.
Solve: The energy delivered to the eardrum in time t is E = Pt, where P is the power of the wave. The
intensity of the wave is I = P / a where a is the area of the ear drum. Putting the above information together, we
have
E = Pt = ( Ia ) t = Iπ r 2t = ( 2.0 × 10−3 W/m 2 )π ( 3.0 × 10−3 m ) ( 60 s ) = 3.4 × 10−6 J
2
20.29.
Solve: The energy delivered to an area a in time t is E = Pt, where the power P is related to the
intensity I as I = P / a. Thus, the energy received by your back is
E = Pt = Iat = (0.80)(1400 W/m2)(0.30 × 0.50 m2)(3600 s) = 6.0 × 105 J
20.30.
Solve: If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities at
distances r1 and r2 is
2
I1 r22
I
⎛ 2m ⎞
−3
= 2 ⇒ 50 m = ⎜
⎟ = 1.6 × 10
I 2 m ⎝ 50 m ⎠
I 2 r1
⇒ I 50 m = I 2 m (1.6 × 10−3 ) = ( 2.0 W/m 2 )(1.6 × 10−3 ) = 3.2 × 10−3 W/m 2
Assess: The power generated by the sound source is P = I2m [4π(2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is a
significant amount of power.
20.31.
Solve: (a) The intensity of a uniform spherical source of power Psource a distance r away is
I = Psource / 4π r 2 . Thus, the intensity at the position of the microphone is
I 50 m =
35 W
4π ( 50 m )
2
= 1.1 × 10−3 W/m 2
(b) The sound energy impinging on the microphone per second is
P = Ia = (1.1× 10−3 W/m 2 )(1.0 × 10−4 m 2 ) = 1.1× 10−7 W = 1.1× 10−7 J/s
⇒ Energy impinging on the microphone in 1 second = 1.1×10−7 J
20.32. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun’s rays
when they impinge upon the earth is
I=
Psun
P
4 × 1026 W
⇒ I earth = sun2 =
= 1400 W/m 2
2
4π r
4π rearth 4π (1.496 × 1011 m )2
With rsun-Venus = 1.082 × 1011 m and rsun-Mars = 2.279 × 1011 m, the intensities of electromagnetic waves at these
planets are I venus = 2700 W/m 2 and I Mars = 610 W/m 2 .
20.33. Visualize: Equation 20.35 gives the sound intensity level as
⎛ I ⎞
⎟
⎝ I0 ⎠
β = (10 dB)log10 ⎜
where I 0 = 1.0 ×10−12 W/m 2 .
Solve:
(a)
⎛I ⎞
⎛ 5.0 ×10−8 W/m 2 ⎞
= 47 dB
⎟ = (10 dB)log10 ⎜
−12
2 ⎟
⎝ 1.0 ×10 W/m ⎠
⎝ I0 ⎠
β = (10 dB)log10 ⎜
(b)
⎛I ⎞
⎛ 5.0 ×10−2 W/m 2 ⎞
= 107 dB
⎟ = (10 dB)log10 ⎜
−12
2 ⎟
⎝ 1.0 ×10 W/m ⎠
⎝ I0 ⎠
β = (10 dB)log10 ⎜
Assess: As mentioned in the chapter, each factor of 10 in intensity changes the sound intensity level by 10 dB;
between the first and second parts of this problem the intensity changed by a factor of 106 , so we expect the
sound intensity level to change by 60 dB.
20.34. Visualize: We can solve Equation 20.35 for the sound intensity, finding I = I 0 ×10 β /10dB .
Solve:
(a)
I = I 0 ×10 β /10dB = (1.0 ×10−12 W/m 2 ) × 103.6 = 4.0 × 10−9 W/m 2
(b)
I = I 0 ×10 β /10dB = (1.0 ×10−12 W/m 2 ) × 109.6 = 4.0 ×10−3 W/m 2
Assess:
Since the sound intensity levels in the two parts of this problem differ by 60dB we expect the sound
intensities to differ by a factor of 106 .
20.35. Model: Assume the pole is tall enough that we don’t have to worry about the ground absorbing or
reflecting sound.
Visualize: The area of a sphere of radius R is A = 4π R 2 . Also recall that I = P/A ; we are given P = 5.0 W .
We seek R for β = 90dB .
Solve:
A=
P
P
5.0 W
=
=
= 5000 m 2
I I 0 ×10 β/10dB (1.0 ×10−12 W/m 2 ) ×1090dB/10dB
R=
A
5000 m 2
=
= 20 m
4π
4π
Assess: This is a reasonable distance from the loudspeaker for a moderately loud sound.
20.36. Model: The frequency of the opera singer’s note is altered by the Doppler effect.
Solve:
(a) Using 90 km/h = 25 m/s, the frequency as her convertible approaches the stationary person is
f+ =
f0
600 Hz
=
= 650 Hz
1 − vS v 1 − 25 m/s
343 m/s
(b) The frequency as her convertible recedes from the stationary person is
f0
600 Hz
f− =
=
= 560 Hz
1 + vS v 1 + 25 m/s
343 m/s
20.37. Model: Your friend’s frequency is altered by the Doppler effect. The frequency of your friend’s note
increases as he races towards you (moving source and a stationary observer). The frequency of your note for your
approaching friend is also higher (stationary source and a moving observer).
Solve: (a) The frequency of your friend’s note as heard by you is
f+ =
f0
400 Hz
=
= 432 Hz
vS
25.0 m/s
1−
1−
v
340 m/s
(b) The frequency heard by your friend of your note is
⎛ v ⎞
⎛ 25.0 m/s ⎞
f + = f 0 ⎜1 + 0 ⎟ = ( 400 Hz ) ⎜1 +
⎟ = 429 Hz
v⎠
340 m/s ⎠
⎝
⎝
20.38. Model: Sound frequency is altered by the Doppler effect. The frequency increases for an observer
approaching the source and decreases for an observer receding from a source.
Solve: You need to ride your bicycle away from your friend to lower the frequency of the whistle. The
minimum speed you need to travel is calculated as follows:
v0 ⎞
⎛ v ⎞
⎛
f − = ⎜1 − 0 ⎟ f 0 ⇒ 20 kHz = ⎜1 −
⎟ ( 21 kHz ) ⇒ v0 = 16.3 m/s ≈ 16 m/s
v⎠
⎝
⎝ 343 m/s ⎠
Assess: A speed of 16.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a
bicycle.
20.39.
Solve:
Model: The mother hawk’s frequency is altered by the Doppler effect.
The frequency is f + as the hawk approaches you is
f+ =
Assess:
f0
800 Hz
⇒ 900 Hz =
⇒ vS = 38.1 m/s
vS
1 − vS v
1−
343 m/s
The mother hawk’s speed of 38.1 m/s ≈ 80 mph is reasonable.
20.40. Visualize:
changing shape.
Solve:
The function D(x, t) represents a pulse that travels in the positive x-direction without
(a)
(b) The leading edge of the pulse moves forward 3 m each second. Thus, the wave speed is 3 m/s.
(c) x − 3t is a function of the form D(x – vt), so the pulse moves to the right at v = 3 m/s.
Solve: (a) We see from the history graph that the period T = 0.20 s and the wave speed v = 4.0 m/s.
Thus, the wavelength is
20.41.
λ=
v
= vT = ( 4.0 m/s )( 0.20 s ) = 0.80 m
f
(b) The phase constant φ0 is obtained as follows:
D ( 0 m, 0 s ) = A sin φ0 ⇒ −2 mm = ( 2 mm ) sin φ0 ⇒ sin φ0 = −1 ⇒ φ0 = − 12 π rad
(c) The displacement equation for the wave is
2π t
π⎞
⎛ 2π x
⎞
⎛ 2π x
D ( x, t ) = A sin ⎜
− 2π ft + φ0 ⎟ = ( 2.0 mm ) sin ⎜
−
− ⎟ = ( 2.0 mm ) sin ( 2.5π x − 10π t − 12 π )
0.80
m
0.20
s
2⎠
λ
⎝
⎠
⎝
where x and t are in m and s, respectively.
20.42. Solve: (a) We can see from the graph that the wavelength is λ = 2.0 m. We are given that the wave’s
frequency is f = 5.0 Hz. Thus, the wave speed is v = λf = 10 m/s.
(b) The snapshot graph was made at t = 0 s. Reading the graph at x = 0 m, we see that the displacement is
D ( x = 0 m, t = 0 s ) = D ( 0 m, 0 s ) = 0.5 mm = 12 A
Thus
D ( 0 m,0 s ) = 12 A = A sin φ0 ⇒ φ0 = sin −1 ( 12 ) =
π
6
rad or
5π
rad
6
Note that the value of D(0 m, 0 s) alone gives two possible values of the phase constant. One of the values will
cause the displacement to start at 0.5 mm and increase with distance—as the graph shows—while the other will
cause the displacement to start at 0.5 mm but decrease with distance. Which is which? The wave equation for t =
0 s is
⎛ 2π x
⎞
D ( x, t = 0 ) = A sin ⎜
+ φ0 ⎟
λ
⎝
⎠
If x is a point just to the right of the origin and is very small, the angle (2π x / λ + φ0 ) is just slightly bigger than
the angle φ0 . Now sin 31° > sin 30°, but sin151° < sin150°, so the value φ0 = 16 π rad is the phase constant for
which the displacement increases as x increases.
(c) The equation for a sinusoidal traveling wave can be written as
⎡ ⎛x
⎤
⎛ 2π x
⎞
⎞
− 2π ft + φ0 ⎟ = A sin ⎢ 2π ⎜ − ft ⎟ + φ0 ⎥
D ( x, t ) = A sin ⎜
⎝ λ
⎠
⎠
⎣ ⎝λ
⎦
Substituting in the values found above,
⎡ ⎛ x
⎞ π⎤
− ( 5.0 s −1 ) t ⎟ + ⎥
D ( x, t ) = (1.0 mm ) sin ⎢ 2π ⎜
2.0
m
⎠ 6⎦
⎣ ⎝
20.43.
Solve:
The time for the sound wave to travel down the tube and back is t = 440 μ s since 1 division
is equal to 100 μs. So, the speed of the sound wave in the liquid is
v=
2 × 25 cm
= 1140 m/s ≈ 1100 m/s
440 μ s
20.44.
Model: The wave is a traveling wave on a stretched string.
Solve:
The wave speed on a string whose radius is R, length is L, and mass density is ρ is vstring = TS / μ with
μ=
m ρV ρπ R 2 L
=
=
= ρπ R 2
L
L
L
If the string radius doubles, then
′ =
vstring
TS
ρπ ( 2R )
2
=
vstring
2
=
280 m/s
= 140 m/s
2
20.45.
Solve:
Model: The wave pulse is a traveling wave on a stretched string.
While the tension TS is the same in both the strings, the wave speeds in the two strings are not. We have
v1 =
TS
μ1
and
v2 =
TS
μ2
⇒ v12 μ1 = v22 μ 2 = TS
Because v1 = L1 / t1 and v2 = L2 / t2 , and because the pulses are to reach the ends of the string simultaneously, the
above equation can be simplified to
L12 μ1 L22 μ 2
L
μ2
4.0 g/m
= 2 ⇒ 1=
=
= 2 ⇒ L1 = 2 L2
t2
t
L2
2.0 g/m
μ1
Since L1 + L2 = 4 m,
2 L2 + L2 = 4 m ⇒ L2 = 1.66 m ≈ 1.7 m
and
L1 = 2 (1.66 m ) = 2.34 m ≈ 2.3 m
Solve: Δt is the time the sound wave takes to travel down to the bottom of the ocean and then up to
the ocean surface. The depth of the ocean is
20.46.
2d = ( vsound in water ) Δt ⇒ d = ( 750 m/s ) Δt
Using this relation and the data from Figure P20.46, we can generate the following table for the ocean depth (d )
at various positions (x) of the ship.
x (km)
d (km)
Δt (s)
0
20
40
45
50
60
6
4
4
8
4
2
4.5
3.0
3.0
6.0
3.0
1.5
20.47. Visualize:
Solve: The explosive’s sound travels down the lake and into the granite, and then it is reflected by the oil
surface. The echo time is thus equal to
techo = t water down + tgranite down + tgranite up + twater up
0.94 s =
d granite
d granite
500 m
500 m
+
+
+
⇒ d granite = 790 m
1480 m/s 6000 m/s 6000 m/s 1480 m/s
20.48. Model: Assume a room temperature of 20°C.
Visualize:
Solve:
The distance between the source and the left ear (EL) is
d L = x 2 + ( y + 0.1 m ) = ⎡⎣( 5.0 m ) cos45° ⎤⎦ + ⎡⎣( 5.0 m ) sin45° + 0.1 m ⎤⎦ = 5.0712 m
2
2
2
Similarly d R = 4.9298 m. Thus,
d L − d R = Δd = 0.1414 m
For the sound wave with a speed of 343 m/s, the difference in arrival times at your left and right ears is
Δt =
Δd
0.1414 m
=
= 410 μ s
343 m/s 343 m/s
20.49.
Solve:
Model: The laser beam is an electromagnetic wave that travels with the speed of light.
The speed of light in the liquid is
vliquid =
30 × 10−2 m
= 2.174 × 108 m/s
1.38 × 10−9 s
The liquid’s index of refraction is
n=
c
vliquid
=
3.0 × 108
= 1.38
2.174 × 108
Thus the wavelength of the laser beam in the liquid is
λliquid =
λvac
n
=
633 nm
= 459 nm
1.38
20.50.
Solve:
Model: The temperature is 20°C for both air and water.
For the sound speed of vair = 343 m/s, the wavelength of the sound wave in air is
λair =
343 m/s
= 1.340 m
256 Hz
On entering water the frequency does not change, so fwater = fair and fwater /fair = 1. The wave speed in water is vwater
=
1480 m/s, so
vwater 1480 m/s
=
= 4.31
vair
343 m/s
Finally, the wavelength in water is
v
1480 m/s
λ
5.781 m
λwater = water =
= 5.781 m ⇒ water =
= 4.31
f water
256 Hz
λair 1.340 m
Assess: This last result is expected because v = f λ and the frequency remains unchanged as the wave enters
from air into water.
20.51.
Solve:
The difference in the arrival times for the P and S waves is
Δt = tS − tP =
Assess:
d d
1
1
⎛
⎞
6
⇒ 120 s = d ⎜
−
−
⎟ ⇒ d = 1.23 × 10 m=1230 km
vS vP
⎝ 4500 m/s 8000 m/s ⎠
d is approximately one-fifth of the radius of the earth and is reasonable.
20.52.
Solve:
Model: This is a sinusoidal wave.
(a) The equation is of the form D ( y, t ) = A sin(ky + ω t + φ0 ), so the wave is traveling along the y-axis.
Because it is +ω t rather than −ω t the wave is traveling in the negative y-direction.
(b) Sound is a longitudinal wave, meaning that the medium is displaced parallel to the direction of travel. So the
air molecules are oscillating back and forth along the y-axis.
(c) The wave number is k = 8.96 m -1 , so the wavelength is
λ=
2π
2π
=
= 0.701 m
k
8.96 m −1
The angular frequency is ω = 3140 s −1 , so the wave’s frequency is
f =
ω 3140 s −1
=
= 500 Hz
2π
2π
Thus, the wave speed v = λ f = (0.70 m)(500 Hz) = 350 m/s. The period T = 1/f = 0.00200 s = 2.00 ms.
(d) The interval t = 0 s to t = 4 ms is exactly 2 cycles of the wave. The initial value at y = 1 m is
D ( y = 1 m, t = 0 s ) = ( 0.02 mm ) sin ( 8.96 + 14 π ) = −0.0063 mm
Assess: The wave is a sound wave with speed v = 350 m/s. This is greater than the room-temperature speed of
343 m/s, so the air temperature must be greater than 20°.
20.53.
Model: This is a sinusoidal wave.
Solve: (a) The displacement of a wave traveling in the positive x-direction with wave speed v must be of the form
D(x, t) = D(x− vt). Since the variables x and t in the given wave equation appear together as x + vt, the wave is
traveling toward the left, that is, in the −x direction.
(b) The speed of the wave is
v=
ω
k
=
2π 0.20 s
= 12 m/s
2π rad 2.4 m
The frequency is
f =
ω 2π rad 0.20 s
=
= 5.0 Hz
2π
2π
The wave number is
k=
2π rad
= 2.6 rad/m
2.4 m
(c) The displacement is
⎡ ⎛ 0.20 m 0.50 s ⎞ ⎤
D ( 0.20 m, 0.50 s ) = ( 3.0 cm ) sin ⎢ 2π ⎜
+
+ 1⎟ ⎥ = −1.5 cm
⎣ ⎝ 2.4 m 0.20 s ⎠ ⎦
20.54. Model: This is a sinusoidal wave traveling on a stretched string in the +x direction.
Solve: (a) From the displacement equation of the wave, A = 2.0 cm, k = 12.57 rad/m, and ω = 638 rad/s. Using
the equation for the wave speed in a stretched string,
vstring =
2
2
⎛ω ⎞
⎛ 638 rad/s ⎞
2
⇒ TS = μ vstring
= μ ⎜ ⎟ = ( 5.00 × 10−3 kg/m3 ) ⎜
⎟ = 12.6 N
μ
⎝k⎠
⎝ 12.57 rad/m ⎠
TS
(b) The maximum displacement is the amplitude Dmax ( x, t ) = 2.00 cm .
(c) From Equation 20.17,
v y max = ω A = ( 638 rad/s ) ( 2.0 × 10−2 m ) = 12.8 m/s
20.55.
Solve:
The wave number and frequency are calculated as follows:
2π rad
= 4π rad/m ⇒ ω = vk = ( 4.0 m/s )( 4π rad/m ) = 16π rad/s
0.50 m
Thus, the displacement equation for the wave is
k=
2π
λ
=
D ( y, t ) = ( 5.0 cm ) sin ⎣⎡( 4π rad/m ) y + (16π rad/s ) t ⎦⎤
Assess:
The positive sign in the sine function’s argument indicates motion along the −y direction.
20.56.
Solve:
The angular frequency and wave number are calculated as follows:
ω = 2π f = 2π ( 200 Hz ) = 400π rad/s ⇒ k =
ω
v
=
400π rad/s
= π rad/m
400 m/s
The displacement equation for the wave is
D ( x, t ) = ( 0.010 mm ) sin ⎣⎡(π rad/m ) x − ( 400π rad/s ) t + 12 π rad ⎦⎤
Assess: Note the negative sign with ω t in the sine function’s argument. This indicates motion along the +x
direction.
20.57.
Solve:
A sinusoidal traveling wave is represented as D ( x, t ) = A sin ( kx − ω t ) . Replacing t with t + T
and using the relationship ω = 2π/T between the angular frequency and period,
D ( x, t + T ) = A sin ( kx − ω ( t + T ) ) = A sin ( kx − ω t − ωT ) = A sin ( kx − ω t − 2π )
= A ⎡⎣sin ( kx − ω t ) cos ( 2π ) − cos ( kx − ω t ) sin ( 2π ) ⎤⎦ = A sin ( kx − ω t ) = D ( x, t )
20.58. Solve: According to Equation 20.28, the phase difference between two points on a wave is Δφ =
2π ( r2 − r1 ) λ . For the first point and second point,
r1 =
(1.00 cm − 0 cm ) + ( 3.00 cm − 0 cm ) + ( 2.00 cm − 0 cm ) = 3.742 cm
r2 =
( −1.00 cm − 0 cm ) + (1.50 cm − 0 cm ) + ( 2.50 cm − 0 cm ) = 3.082 cm
2
2
2
2
2
2
The wavelength is
λ=
⇒ Δφ =
v
346 m/s
=
= 0.02641 m = 2.641 cm
f 13,100 Hz
2π ( 3.742 cm − 3.082 cm ) π
π
180°
= rad = rad ×
= 90°
2.641 cm
2
2
π rad
20.59.
Solve:
Model: We have a sinusoidal traveling wave on a stretched string.
(a) The wave speed on a string and the wavelength are calculated as follows:
v=
TS
μ
=
v 100.0 m/s
20 N
= 100 m/s ⇒ λ = =
= 1.0 m
f
0.002 kg/m
100 Hz
(b) The amplitude is determined by the oscillator at the end of the string and is A = 1.0 mm. The phase constant
can be obtained from Equation 20.15 as follows:
D ( 0 m, 0 s ) = A sin φ0 ⇒ −1.0 mm = (1.0 mm ) sin φ0 ⇒ φ0 = −
π
2
rad
(c) The wave (as distinct from the oscillator) is described by D ( x, t ) = A sin ( kx − ω t + φ0 ) . In this equation the
wave number and angular frequency are
k=
2π
λ
=
2π
= 2π rad/m
1.0 m
ω = vk = (100.0 m/s )( 2π rad/m ) = 200π rad/s
Thus, the wave’s displacement equation is
D ( x, t ) = (1.0 mm ) sin ⎡⎣( 2π rad/m ) x − ( 200π rad/s ) t − 12 π rad ⎤⎦
(d) The displacement is
D ( 0.50 m, 0.015 s ) = (1.0 mm ) sin ⎡⎣( 2π rad/s )( 0.50 m ) − ( 200π rad/s )( 0.015 s ) − 12 π ⎤⎦ = −1.0 mm
20.60. Model: We have a wave traveling to the right on a string.
Visualize:
Solve: The snapshot of the wave as it travels to the right for an infinitesimally small time Δt shows that the
velocity at point 1 is downward, at point 3 is upward, and at point 2 is zero. Furthermore, the speed at points 1
and 3 is the maximum speed given by Equation 20.17: v1 = v3 = ω A . The frequency of the wave is
ω = 2π f = 2π
v
λ
=
2π ( 45 m/s )
= 300π rad/s ⇒ ω A = ( 300π rad/s ) ( 2.0 × 10−2 m ) = 19 m/s
0.30 m
Thus, v1 = −19 m/s, v2 = 0 m/s, and v3 = +19 m/s.
20.61.
Model: This is a wave traveling to the left at a constant speed of 50 cm/s.
Solve: The particles at positions between x = 2 cm and x = 7 cm have a speed of 10 cm/s, and the particles
between x = 7 cm and x = 9 cm have a speed of −25 cm/s. That is, at the time the snapshot of the velocity is shown,
the particles of the medium have upward motion for 2 cm ≤ x ≤ 7 cm, but downward motion for 7 cm ≤ x ≤ 9 cm.
The width of the front section of the wave pulse is 7 cm – 2 cm = 5 cm and the width of the rear section is 9 cm – 7
cm = 2 cm. With a wave speed of 50 cm/s, the time taken by the front section to pass through a particular point is
5 cm 50 cm/s = 0.1 s and the time taken by the rear section of the wave to pass through a point is
2 cm 50 cm/s = 0.04 s . Thus the wave causes the upward moving particles to go through a displacement of
A = (10 cm/s )( 0.1 s ) = 1.0 cm . The downward moving particles have a maximum displacement of
( −25 cm/s )( 0.04 s ) = −1.0 cm .
20.62.
Solve:
Model: The wave is traveling on a stretched string.
The wave speed on the string is
v=
TS
μ
=
50 N
= 100 m/s
0.005 kg/m
The speed of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculated
as follows:
v y = −ω A cos ( kx − ω t + φ0 ) ⇒ v y max = ω A = 2π fA = 2π
⎛ 100 m/s ⎞
A = 2π ⎜
⎟ ( 0.030 m ) = 9.4 m/s
λ
⎝ 2.0 m ⎠
v
20.63. Model: A sinusoidal wave is traveling along a stretched string.
Solve: From Equation 20.17 and Equation 20.20, vy max = ωA and ay max = ω2A. These two equations can be
combined to give
ω=
a y max
v y max
=
v
200 m/s 2
ω
2.0 m/s
= 100 rad/s ⇒ f =
= 15.9 Hz ≈ 16 Hz ⇒ A = y max =
= 2.0 cm
100 rad/s
ω
2.0 m/s
2π
20.64.
Solve: (a) At a distance r from the bulb, the 5 watts of visible light have spread out to cover the
surface of a sphere of radius r. The surface area of a sphere is a = 4πr2. Thus, the intensity at a distance of 2 m is
I=
P
P
5.0 W
=
=
= 0.095 W/m 2
a 4π r 2 4π ( 2.0 m )2
Note that the presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but
the light wave has the same intensity at all points 2 m from the bulb whether it is striking a surface or moving
through empty space.
(b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread of
the beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full 5 watts of light is
2
concentrated in this dot of area a = π r 2 = π ( 0.001 m ) = 3.14 × 10−6 m 2 . The intensity is
P
5W
=
= 1.6 MW/m 2
a 3.14 × 10−6 m 2
Although the power of the light source is the same in both cases, the laser produces light on the wall whose
intensity is over 16 million times that of the light bulb.
I=
20.65. Model: The radio wave is an electromagnetic wave.
Solve: At a distance r, the 25 kW power station spreads out waves to cover the surface of a sphere of radius r.
The surface area of a sphere is 4πr2. Thus, the intensity of the radio waves is
I=
Psource
25 × 103 W
=
= 2.0 × 10−5 W/m 2
2
2
4π r
4π (10 × 103 m )
20.66.
Solve:
(a) The peak power of the light pulse is
Ppeak =
ΔE 500 mJ
0.50 J
=
=
= 5.0 × 107 W
Δt
10 ns
1.0 × 10−8 s
(b) The average power is
Pavg =
Etotal 10 × 500 mJ 5.0 J
=
=
= 5.0 W
1.0 s
1.0 s
1.0 s
The laser delivers pulses of very high power. But the laser spends most of its time “off,” so the average power is
very much less than the peak power.
(c) The intensity is
I laser =
P 5.0 × 107 W
5.0 × 107 W
=
=
= 6.4 × 1017 W/m 2
2
a π ( 5.0 μ m )
7.85 × 10−11 m 2
(d) The ratio is
I laser 6.4 × 1017 W/m 2
=
= 5.8 × 1014
I sun
1.1 × 103 W/m 2
20.67.
Solve:
Model: We have a traveling wave radiated by the tornado siren.
(a) The power of the source is calculated as follows:
I 50 m = 0.10 W/m 2 =
Psource
Psource
2
=
⇒ Psource = ( 0.10 W/m 2 ) 4π ( 50 m ) = (1000π ) W
4π r 2 4π ( 50 m )2
The intensity at 1000 m is
I1000 m =
Psource
4π (1000 m )
2
=
(1000π ) W = 250 μ W m 2
2
4π (1000 m )
(b) The maximum distance is calculated as follows:
I=
Psource
(1000π ) W ⇒ r = 16 km
⇒ 1.0 × 10−6 W/m 2 =
4π r 2
4π r 2
20.68. Model: Assume the saw is far enough off the ground that we don’t have to worry about reflected
sound.
Visualize:
First note that β1 − β 2 = 20dB ⇒ I1/I 2 = 10 ⋅ 10 = 100 (a change of 10 dB corresponds to a change in
intensity by a factor of 10). Then use I1 A1 = P and then P = I 2 A2 ⇒ A2 = P/I 2 , and finally solve for
R2 = A2 / 4π .
Solve:
Put all of the above together.
P
R2 =
Assess:
I1 A1
A2
I2
I2
=
=
=
4π
4π
4π
I1
I2
(4π R12 )
4π
= R1
The scaling laws help and the answer is reasonable.
I1
= R1 100 = (5.0 m)(10) = 50 m
I2
20.69. Model: Assume the two loudspeakers broadcast the same power and that the platforms are high
enough off the ground that we don’t have to worry about reflected sound.
Visualize: Call the distance between the loudspeakers d . Call the intensity halfway between the speakers (at
d / 2 ) I1 and the sound intensity level there β1 (= 75 dB ); call them I 2 and β 2 at 1/4 the distance from one pole
and 3/4 the distance from the other pole on the line between them. We seek β 2 .
We first apply a general approach for different sound intensity levels:
⎡
⎛I ⎞
⎛ I ⎞⎤
⎛ I /I ⎞
⎛I ⎞
Δβ = β 2 − β1 = (10dB) ⎢ log10 ⎜ 2 ⎟ − log10 ⎜ 1 ⎟ ⎥ = (10 dB)log10 ⎜ 2 0 ⎟ = (10 dB)log10 ⎜ 2 ⎟
I
I
I
I
/
⎢⎣
⎝ I1 ⎠
⎝ 0⎠
⎝ 0 ⎠ ⎥⎦
⎝ 1 0⎠
Solve: Recall that for the general case of spherical symmetry I = P/A , where P is the power emitted by the
source and A = 4π R 2 is the area of the sphere. Now we find the ratio of the intensities I 2 /I1 and then plug it in
the formula above and add it to 75 dB.
I1 =
I2 =
P
P
2P
+
=
4π ( d / 2) 2 4π (d / 2) 2 π d 2
P
P
4P
4P
(36 + 4) P 40 P 20
+
=
+
=
=
=
I1
4π ( d / 4) 2 4π (3d / 4) 2 π d 2 9π d 2
9π d 2
9π d 2 9
⎛I ⎞
⎛ 20 ⎞
Δβ = (10 dB)log10 ⎜ 2 ⎟ = (10 dB)log10 ⎜ ⎟ = 3.48 dB
I
⎝ 9 ⎠
⎝ 1⎠
β 2 = β1 + Δβ = 75 dB + 3.48 dB = 78 dB
Assess:
An increase of about 3dB corresponds to a doubling of the intensity. 20/9 is close to double.
20.70.
Model:
As suggested, model the bald head as a hemisphere with radius R = 0.080 m. This means the
surface area of the bald head (hemisphere) is A = 2π R 2 = 0.0402m 2 .
Visualize: We are given β = 93 dB and ΔE = 0.10 J . We also know that I = I 0 × 10 β /10 dB and P = IA . Also
recall P = ΔE Δt .
Solve: Put all of the above together to find Δt .
Δt =
ΔE ΔE
ΔE
0.10 J
=
=
=
= 1250 s ≈ 21 min
P
IA ( I 0 × 10 β /10 dB )(2π R 2 ) (10−12 W/m 2 × 109.3 )(0.0402 m 2 )
Assess: 21 min seems like quite a while to deliver 0.10 J of energy, but sounds waves don’t carry a lot of
energy unless the intensity is high.
20.71. Model: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as
the bat approaches and it decreases as the bat recedes away.
Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. From
Equation 20.38,
f− =
25,000 Hz
f0
⇒ vS = 85.8 m/s ≈ 86 m/s
⇒ 20,000 Hz =
1 + vS / v
⎛ vS ⎞
1+ ⎜
⎟
⎝ 343 m/s ⎠
Assess: This is a rather large speed: 85.8 m/s ≈ 180 mph. This is not possible for a bat.
20.72. Model: The sound generator’s frequency is altered by the Doppler effect. The frequency increases as
the generator approaches the student, and it decreases as the generator recedes from the student.
Solve: The generator’s speed is
⎛ 100
⎞
vS = rω = r ( 2π f ) = (1.0 m ) 2π ⎜
rev/s ⎟ = 10.47 m/s
⎝ 60
⎠
The frequency of the approaching generator is
f0
600 Hz
f+ =
=
= 619 Hz ≈ 620 Hz
1 − vS v 1 − 10.47 m/s
343 m/s
Doppler effect for the receding generator, on the other hand, is
f0
600 Hz
=
= 582 Hz ≈ 580 Hz
1 + vS v 1 + 10.47 m/s
343 m/s
Thus, the highest and the lowest frequencies heard by the student are 620 Hz and 580 Hz.
f− =
20.73. Solve: We will closely follow the details of section 20.7 in the textbook. Figure 20.29 shows that the
wave crests are stretched out behind the source. The wavelength detected by Pablo is λ− = 13 d , where d is the
distance the wave has moved plus the distance the source has moved at time t = 3T. These distances are
Δxwave = vt = 3vT and Δxsource = vSt = 3vST . The wavelength of the wave emitted by a receding source is thus
d Δxwave + Δxsource 3vT + 3vST
=
=
= ( v + vS ) T
3
3
3
The frequency detected in Pablo’s direction is thus
v
v
f0
f− =
=
=
λ− ( v + vs ) T 1 + vS v
λ− =
20.74. Model: We are looking at the Doppler effect for the light of an approaching source.
Solve:
(a) The time is
54 × 106 km
= 180 s = 3.0 min
3 × 105 km/s
(b) Using Equation 20.40, the observed wavelength is
t=
λ=
1 − vs c
1 − 0.1c c
λ0 =
( 540 nm ) = (0.9045)(540 nm) = 488 nm ≈ 490 nm
1 + vs c
1 + 0.1c c
Assess: 490 nm is slightly blue shifted from green.
20.75. Model: We are looking at the Doppler effect for the light of a receding source.
Visualize:
Note that the daredevil’s tail lights are receding away from your rocket’s light detector with a relative speed of 0.2c.
Solve: Using Equation 20.40, the observed wavelength is
λ=
1 + vS c
1 + 0.2c c
λ0 =
( 650 nm ) = 796 nm ≈ 800 nm
1 − vS c
1 − 0.2c c
This wavelength is in the infrared region.
20.76. Model: The Doppler effect for light of a receding source yields an increased wavelength.
Solve: Because the measured wavelengths are 5% longer, that is, λ = 1.05λ0, the distant galaxy is receding
away from the earth. Using Equation 20.40,
λ = 1.05λ0 =
1 + vs c
1 + vs c
2
⇒ vs = 0.049 c = 1.47 × 107 m/s
λ0 ⇒ (1.05) =
1 − vs c
1 − vs c
20.77. Model: The Doppler effect for light of an approaching source leads to a decreased wavelength.
Solve: The red wavelength (λ0 = 650 nm) is Doppler shifted to green (λ = 540 nm) due to the approaching light
source. In relativity theory, the distinction between the motion of the source and the motion of the observer
disappears. What matters is the relative approaching or receding motion between the source and the observer.
Thus, we can use Equation 20.40 as follows:
λ = λ0
1 − vs c
1 − vs c
⇒ 540 nm = ( 650 nm )
1 + vs c
1 + vs c
⇒ vs = 5.5 × 104 km/s = 2.0 × 108 km/h
The fine will be
1$ ⎞
( 2.0 ×10 km/hr − 50 km/hr ) ⎛⎜⎝ 1 km/hr
⎟ = $200 million
⎠
8
Assess:
The police officer knew his physics.
20.78. Model: The wave pulse is a traveling wave on a stretched string. The two masses hanging from the
steel wire are in static equilibrium.
Visualize:
Solve:
The wave speed along the wire is
vwire =
4.0 m
= 166.7 m/s
0.024 s
T1
T1
Using Equation 20.2,
vwire = 166.7 m/s=
=
μ
( 0.060 kg 8.0 m )
G
G
Because point 1 is in static equilibrium, with Fnet = 0,
( Fnet ) x = T1 − T2 cos 40° ⇒ T2 =
⇒ T1 = 208.4 N
T1
= 2721 N
cos 40°
( Fnet ) y = T2 sin 40° − w = 0 N ⇒ w = mg = T2 sin 40° ⇒ m =
( 272.1 N ) sin 40° = 17.8 kg
9.8 m/s 2
20.79. Solve: The time for the wave to travel from California to the South Pacific is
t=
d 8.00 × 106 m
=
= 5405.4 s
v
1480 m/s
A time decrease to 5404.4 s implies the speed has changed to v =
d
= 1480.28 m/s.
t
Since the 4.0 m/s increase in velocity is due to an increase of 1°C, an increase of 0.28 m/s occurs due to a
temperature increase of
⎛ 1°C ⎞
⎜
⎟ ( 0.28 m/s ) = 0.07°C
⎝ 4.0 m/s ⎠
Thus, a temperature increase of approximately 0.07°C can be detected by the researchers.
20.80. Solve: The wave speeds along the two metal wires are
v1 =
T
μ1
=
2250 N
= 500 m/s
0.009 kg/m
T
v2 =
=
μ2
2250 N
= 300 m/s
0.025 kg/m
The wavelengths along the two wires are
v
f
λ1 = 1 =
500 m/s 1
= m
1500 Hz 3
v
f
λ2 = 2 =
300 m/s 1
= m
1500 Hz 5
Thus, the number of wavelengths over two sections of the wire are
1.0 m
λ1
=
1.0 m
( 13 m )
=3
1.0 m
λ2
=
1.0 m
( 15 m )
The number of complete cycles of the wave in the 2.00-m-long wire is 8.
=5
20.81. Model: The wave pulse is a traveling wave on a stretched wire.
Visualize:
Solve: (a) At a distance y above the lower end of the rope, the point P is in static equilibrium. The upward
tension in the rope must balance the weight of the rope that hangs below this point. Thus, at this point
T = w = Mg = ( μ y ) g
where μ = m/L is the linear density of the entire rope. Using Equation 20.2, we get
v=
T
μ
=
μ yg
= gy
μ
(b) The time to travel a distance dy at y, where the wave speed is v = gy , is
dt =
dy
dy
=
v
gy
Finding the time for a pulse to travel the length of the rope requires integrating from one end of the rope to the other:
T
L
L
dy
1
2
L
Δt = ∫ dt = ∫
=
L ⇒ Δt = 2
2 y =
0
g
gy
g
g
0
0
(
)
20.82. Visualize:
Solve:
as
(a) Using the graph, the refractive index n as a function of distance x can be mathematically expressed
n = n1 +
n2 − n1
x
L
At position x, the light speed is v = c / n. The time for the light to travel a distance dx at x is
dt =
dx n
1⎛
n −n ⎞
= dx = ⎜ n1 + 2 1 x ⎟ dx
v c
c⎝
L
⎠
To find the total time for the light to cover a thickness L of a glass we integrate as follows:
T
T = ∫ dt =
0
2
n
n
1 ⎛
n2 − n1 ⎞
(n − n )
⎛n −n ⎞L ⎛n +n ⎞
x ⎟ dx = 1 ∫ dx + 2 1 ∫ x dx = 1 L + ⎜ 2 1 ⎟ = ⎜ 1 2 ⎟ L
⎜ n1 +
∫
c 0
cL 0
c0⎝
L
c
⎝ cL ⎠ 2 ⎝ 2c ⎠
⎠
L
L
(b) Substituting the given values into this equation,
(1.50 + 1.60 ) × 0.010 m = 5.17 × 10−11 s
T=
2 ( 3.0 × 108 m/s )
L
21.1. Model: The principle of superposition comes into play whenever the waves overlap.
Visualize:
The graph at t = 1.0 s differs from the graph at t = 0.0 s in that the left wave has moved to the right by 1.0 m and
the right wave has moved to the left by 1.0 m. This is because the distance covered by the wave pulse in 1.0 s is
1.0 m. The snapshot graphs at t = 2.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right moving waves.
The overlapping parts of the two waves are shown by the dotted lines.
21.2. Model: The principle of superposition comes into play whenever the waves overlap.
Visualize:
The snapshot graph at t = 1.0 s differs from the graph t = 0.0 s in that the left wave has moved to the right by 1.0
m and the right wave has moved to the left by 1.0 m. This is because the distance covered by each wave in 1.0 s
is 1.0 m. The snapshot graphs at t = 2.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right moving
waves. The overlapping parts of the two waves are shown by the dotted lines.
21.3. Model: The principle of superposition comes into play whenever the waves overlap.
Visualize:
At t = 4.0 s the shorter pulses overlap and cancel. At t = 6.0 s the longer pulses overlap and cancel.
21.4. Model: The principle of superposition comes into play whenever the waves overlap.
Solve:
4 s.
(b)
(a) As graphically illustrated in the figure below, the snapshot graph of Figure EX21.5b was taken at t =
21.5. Model: A wave pulse reflected from the string-wall boundary is inverted and its amplitude is
unchanged.
Visualize:
The graph at t = 2 s differs from the graph at t = 0 s in that both waves have moved to the right by 2 m. This is
because the distance covered by the wave pulse in 2 s is 2 m. The shorter pulse wave encounters the boundary
wall at 2.0 s and is inverted on reflection. This reflected pulse wave overlaps with the broader pulse wave, as
shown in the snapshot graph at t = 4 s. At t = 6 s, only half of the broad pulse is reflected and hence inverted; the
shorter pulse wave continues to move to the left with a speed of 1 m/s. Finally, at t = 8 s both the reflected pulse
waves are inverted and they are both moving to the left.
21.6. Model: Reflections at both ends of the string cause the formation of a standing wave.
Solve: Figure EX21.6 indicates 5/2 wavelengths on the 2.0-m-long string. Thus, the wavelength of the standing
wave is λ = 52 ( 2.0 m ) = 0.80 m . The frequency of the standing wave is
f =
v
λ
=
40 m/s
= 50 Hz
0.80 m
21.7. Model: Reflections at the string boundaries cause a standing wave on the string.
Solve: Figure EX21.7 indicates two full wavelengths on the string. Hence λ = 12 (60 cm) = 30 cm = 0.30 m.
Thus
v = λ f = ( 0.30 m )(100 Hz ) = 30 m/s
21.8. Model: Reflections at the string boundaries cause a standing wave on the string.
Solve: (a) When the frequency is doubled ( f ′ = 2 f 0 ) , the wavelength is halved ( λ ′ = 12 λ0 ) . This halving of the
wavelength will increase the number of antinodes to six.
(b) Increasing the tension by a factor of 4 means
v=
T
μ
⇒ v′ =
T′
μ
=
4T
μ
= 2v
For the string to continue to oscillate as a standing wave with three antinodes means λ ′ = λ0 . Hence,
v′ = 2v ⇒ f ′λ ′ = 2 f 0λ0 ⇒ f ′λ0 = 2 f 0λ0 ⇒ f ′ = 2 f 0
That is, the new frequency is twice the original frequency.
21.9. Model: A string fixed at both ends supports standing waves.
Solve: (a) We have fa = 24 Hz = mf1 where f1 is the fundamental frequency that corresponds to m = 1. The next
successive frequency is fb = 36 Hz = (m + 1) f1. Thus,
24 Hz
f b ( m + 1) f1 m + 1 36 Hz
= 12 Hz
=
=
=
= 1.5 ⇒ m + 1 = 1.5m ⇒ m = 2 ⇒ f1 =
2
fa
mf1
m
24 Hz
The wave speed is
v = λ1 f1 =
2L
f1 = ( 2.0 m )(12 Hz ) = 24 m/s
1
(b) The frequency of the third harmonic is 36 Hz. For m = 3, the wavelength is
λm =
2 L 2 (1 m ) 2
=
= m
3
3
m
21.10. Model: A string fixed at both ends supports standing waves.
Solve:
(a) A standing wave can exist on the string only if its wavelength is
λm =
2L
m
m = 1, 2, 3, …
The three longest wavelengths for standing waves will therefore correspond to m = 1, 2, and 3. Thus,
λ1 =
2 ( 2.40 m )
= 4.80 m
1
λ2 =
2 ( 2.40 m )
= 2.40 m
2
λ3 =
2 ( 2.40 m )
= 1.60 m
3
(b) Because the wave speed on the string is unchanged from one m value to the other,
f 2λ2 = f3λ3 ⇒ f 3 =
f 2λ2
λ3
=
( 50 Hz )( 2.40 m ) = 75 Hz
1.60 m
21.11. Model: A string fixed at both ends forms standing waves.
Solve:
(a) The wavelength of the third harmonic is calculated as follows:
λm =
2L
2 L 2.42 m
⇒ λ3 =
=
= 0.807 m ≈ 0.81 m
m
3
3
(b) The speed of waves on the string is v = λ3 f3 = (0.807 m)(180 Hz) = 145.3 m/s. The speed is also given by
v = TS / μ , so the tension is
TS = μ v 2 =
m 2 0.004 kg
2
v =
(145.3 m/s ) = 69.7 N ≈ 70 m
1.21 m
L
21.12. Model: For the stretched wire vibrating at its fundamental frequency, the wavelength of the standing
wave is λ1 = 2 L.
Visualize:
Solve:
The wave speed on the steel wire is
vwire = f λ = f ( 2 L ) = ( 80 Hz )( 2 × 0.90 m ) = 144 m/s
and is also equal to
TS μ , where
μ=
mass 5.0 × 10−3 kg
=
= 5.555 × 10−3 kg/m
length
0.90 m
The tension TS in the wire equals the weight of the sculpture or Mg. Thus,
2
2
5.555 × 10−3 kg/m ) (144 m/s )
(
Mg
μ vwire
vwire =
⇒M =
=
= 12 kg
μ
g
9.8 m/s 2
21.13. Model: The laser light forms a standing wave inside the cavity.
Solve:
The wavelength of the laser beam is
λm =
2 ( 0.5300 m )
2L
⇒ λ100,000 =
= 10.60 μ m
m
100,000
The frequency is
f100,000 =
c
λ100,000
=
3.000 × 108 m/s
= 2.830 × 1013 Hz
10.60 × 10−6 m
21.14. Solve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The
possible wavelengths for this case are
λm =
2L
m
λ2 =
2 (1.21 m )
= 1.21 m
2
λm =
4L
m
λ2 =
4 (1.21 m )
= 1.61 m
3
m = 1, 2, 3, …
The three longest wavelengths are
λ1 =
2 (1.21 m )
= 2.42 m
1
λ3 =
2 (1.21 m )
= 0.807 m
3
λ3 =
4 (1.21 m )
= 0.968 m
5
(b) In the case of an open-closed tube,
m = 1, 3, 5, …
The three longest wavelengths are
λ1 =
4 (1.21 m )
= 4.84 m
1
21.15. Model: We have an open-open tube that forms standing sound waves.
Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends.
There is another antinode in the middle of the tube. Thus, this is the m = 2 mode and the wavelength of the
standing wave is equal to the length of the tube, that is, λ = 0.80 m. Since the frequency f = 500 Hz, the speed of
=
sound
in
this
case
is
v
=
fλ
(500 Hz)(0.80 m) = 400 m/s.
Assess: The experiment yields a reasonable value for the speed of sound.
21.16. Solve: For the open-open tube, the fundamental frequency of the standing wave is f1 = 1500 Hz when
the tube is filled with helium gas at 0°C. Using λm = 2 L m ,
f1 helium =
vhelium
λ1
=
970 m/s
2L
Similarly, when the tube is filled with air,
f1 air =
Assess:
vair
λ1
=
331 m/s
f
331 m/s
⎛ 331 m/s ⎞
⇒ 1 air =
⇒ f1 air = ⎜
⎟ (1500 Hz ) = 512 Hz
2L
f1 helium 970 m/s
⎝ 970 m/s ⎠
Note that the length of the tube is one-half the wavelength whether the tube is filled with helium or air.
21.17. Model: An organ pipe has a “sounding” hole where compressed air is blown across the edge of the
pipe. This is one end of an open-open tube with the other end at the true “end” of the pipe.
Solve: For an open-open tube, the fundamental frequency is f1 = 16.4 Hz. We have
λ1 =
⎞ 1 ⎛ 343 m/s ⎞
2L
λ 1⎛v
⇒ L = 1 = ⎜ sound ⎟ = ⎜
⎟ = 10.5 m
1
2 2 ⎝ f1 ⎠ 2 ⎝ 16.4 Hz ⎠
Assess: The length of the organ pipe is ≈ 34.5 feet. That is actually somewhat of an overestimate since the
antinodes of real tubes are slightly outside the tube. The actual length in a real organ is about 32 feet, and this is
the tallest pipe in the so called “32 foot rank” of pipes.
21.18. Model: Reflections at the string boundaries cause a standing wave on a stretched string.
Solve: Because the vibrating section of the string is 1.9 m long, the two ends of this vibrating wire are fixed,
and the string is vibrating in the fundamental harmonic. The wavelength is
λm =
2L
⇒ λ1 = 2 L = 2 (1.90 m ) = 3.80 m
m
The wave speed along the string is v = f1λ1 = (27.5 Hz)(3.80 m) = 104.5 m/s. The tension in the wire can be
found as follows:
⎛ mass ⎞ 2 ⎛ 0.400 kg ⎞
T
2
v = S ⇒ TS = μ v 2 = ⎜
⎟v = ⎜
⎟ (104.5 m/s ) = 2180 N
length
2.00
m
μ
⎝
⎠
⎝
⎠
21.19. Model: A string fixed at both ends forms standing waves.
Solve: A simple string sounds the fundamental frequency f1 = v/2L. Initially, when the string is of length LA =
30 cm, the note has the frequency f1A = v/2LA. For a different length, f1B = v/2LB. Taking the ratio of each side of
these two equations gives
f1A v / 2 LA LB
f
=
=
⇒ LB = 1A LA
f1B v / 2 LB LA
f1B
We know that the second frequency is desired to be f1B = 523 Hz. The string length must be
LB =
440 Hz
( 30 cm ) = 25.2 cm
523 Hz
The question is not how long the string must be, but where must the violinist place his finger. The full string is
30 cm long, so the violinist must place his finger 4.8 cm from the end.
Assess: A fingering distance of 4.8 cm from the end is reasonable.
21.20. Model: Interference occurs according to the difference between the phases ( Δφ ) of the two waves.
Solve: (a) A separation of 20 cm between the speakers leads to maximum intensity on the x-axis, but a
separation of 60 cm leads to zero intensity. That is, the waves are in phase when (Δ x)1 = 20 cm but out of phase
when (Δ x) 2 = 60 cm. Thus,
( Δx)2 − ( Δ x)1 =
λ
2
⇒ λ = 2 ( 60 cm − 20 cm ) = 80 cm
(b) If the distance between the speakers continues to increase, the intensity will again be a maximum when the
separation between the speakers that produced a maximum has increased by one wavelength. That is, when the
separation between the speakers is 20 cm + 80 cm = 100 cm.
21.21. Model: The interference of two waves depends on the difference between the phases (Δφ ) of the two
waves.
Solve: (a) Because the speakers are in phase, Δφ0 = 0 rad. Let d represent the path-length difference. Using m
= 0 for the smallest d and the condition for destructive interference, we get
Δφ = 2π
⇒ 2π
Δx
λ
Δx
λ
+ Δφ0 = 2 ( m + 12 )π rad
+ Δφ0 = π rad ⇒ 2π
d
λ
+ 0 rad = π rad ⇒ d =
m = 0, 1, 2, 3 …
λ
2
1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞
= ⎜ ⎟= ⎜
⎟ = 0.25 m
2 ⎝ f ⎠ 2 ⎝ 686 Hz ⎠
(b) When the speakers are out of phase, Δφ0 = π . Using m = 1 for the smallest d and the condition for
constructive interference, we get
Δx
m = 0, 1, 2, 3, …
Δφ = 2π
+ Δφ0 = 2mπ
λ
λ 1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞
d
⇒ 2π + π = 2π ⇒ d = = ⎜ ⎟ = ⎜
⎟ = 0.25 m
2 2 ⎝ f ⎠ 2 ⎝ 686 Hz ⎠
λ
21.22. Model: We assume that the speakers are identical and that they are emitting in phase.
Solve: Since you don’t hear anything, the separation between the two speakers corresponds to the condition of
destructive interference. With Δφ0 = 0 rad, Equation 21.23 becomes
2π
d
λ
= 2 ( m + 12 )π rad ⇒ d = ( m + 12 ) λ ⇒ d =
Since the wavelength is
λ=
v 340 m/s
=
= 2.0 m
f 170 Hz
three possible values for d are 1.0 m, 3.0 m, and 5.0 m.
λ 3λ 5λ
2
,
2
,
2
21.23. Model: Reflection is maximized if the two reflected waves interfere constructively.
Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32:
λC =
−9
λ m ( 600 × 10 m ) (1)
2nd
⇒d = C =
= 216 nm
m
2n
( 2 )(1.39 )
where we have used m = 1 to calculate the thinnest film.
Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable
because
nair < nfilm < nglass.
21.24. Model: Reflection is maximized if the two reflected waves interfere constructively.
Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32:
λC =
−9
λ m ( 500 × 10 m ) (1)
2nd
⇒d = C =
= 200 nm
m
2n
( 2 )(1.25)
where we have used m = 1 to calculate the thinnest film.
Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable
because
nair < noil < nwater.
21.25. Solve: (a) The circular wave fronts emitted by the two sources show that the two sources are in phase.
This is because the wave fronts of each source have moved the same distance from their sources.
(b) Let us label the top source as 1 and the bottom source as 2. Since the sources are in phase, Δφ0 = 0 rad. For
the point P, r1 = 3λ and r2 = 4λ. Thus, Δr = r2 − r1 = 4λ − 3λ = λ. The phase difference is
Δφ =
2πΔr
λ
2π ( λ )
=
λ
= 2π
This corresponds to constructive interference.
For the point Q, r1 = 72 λ and r2 = 2λ. The phase difference is
Δφ =
2πΔr
λ
=
2π ( 32 λ )
λ
= 3π
This corresponds to destructive interference.
For the point R, r1 = 52 λ and r2 = 72 λ . The phase difference is
Δφ =
2π ( λ )
λ
= 2π
This corresponds to constructive interference.
P
Q
R
r1
3λ
λ
5
λ
2
7
2
Δr
r2
4λ
2λ
3
2
λ
λ
C/D
C
D
λ
λ
C
7
2
21.26. Solve: (a) The circular wave fronts emitted by the two sources indicate the sources are out of phase.
This is because the wave fronts of each source have not moved the same distance from their sources.
(b) Let us label the top source as 1 and the bottom source as 2. The phase difference between the sources is
Δφ0 = π . For the point P, r1 = 2λ and r2 = 3λ. The phase difference is
Δφ =
2πΔr
λ
+ Δφ0 =
2π ( 3λ − 2λ )
λ
+ π = 3π
This corresponds to destructive interference.
For the point Q, r1 = 3λ and r2 = 32 λ. The phase difference is
Δφ =
2π ( 32 λ )
λ
+ π = 4π
This corresponds to constructive interference.
For the point R, r1 = 52 λ and r2 = 3λ. The phase difference is
Δφ =
2π ( 12 λ )
λ
+ π = 2π
This corresponds to constructive interference.
Assess:
P
Q
r1
2λ
3λ
r2
3λ
R
5
2
λ
3λ
3
2
λ
Δr
λ
λ
1
λ
2
3
2
Note that it is not r1 or r2 that matter, but the difference Δr between them.
C/D
D
C
C
21.27. Model: The two speakers are identical, and so they are emitting circular waves in phase. The overlap
of these waves causes interference.
Visualize:
Solve:
From the geometry of the figure,
r2 = r12 + ( 2.0 m ) =
2
( 4.0 m ) + ( 2.0 m ) = 4.472 m
2
2
So, Δr = r2 − r1 = 4.472 m − 4.0 m = 0.472 m. The phase difference between the sources is Δφ0 = 0 rad and the
wavelength of the sound waves is
v 340 m/s
λ= =
= 0.1889 m
f 1800 Hz
Thus, the phase difference of the waves at the point 4.0 m in front of one source is
2π ( 0.472 m )
+ 0 rad = 5π rad = 2.5(2π rad)
λ
0.1889 m
This is a half-integer multiple of 2π rad, so the interference is perfect destructive.
Δφ = 2π
Δr
+ Δφ0 =
21.28. Model: The two radio antennas are emitting out-of-phase, circular waves. The overlap of these waves
caus