Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
This print-out should have 24 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
Chapters 6 and 7 problems.
001 (part 1 of 1) 0 points
A cheerleader lifts his 70.9 kg partner straight
up off the ground a distance of 0.594 m before
releasing her.
The acceleration of gravity is 9.8 m/s2 .
If he does this 27 times, how much work has
he done?
Correct answer: 11143.5 J.
Explanation:
The work done in lifting the cheerleader
once is
W1 = m g h
= (70.9 kg)(9.8 m/s2 )(0.594 m)
= 412.723 J .
The work required to lift her n = 27 times is
W = n W1 = (27)(412.723 J) = 11143.5 J.
002 (part 1 of 5) 0 points
In the parallel spring system, the springs
are positioned so that the 44 N weight
stretches each spring equally. The spring constant for the left-hand spring is 3.5 N/cm and
the spring constant for the righ-hand spring
is 5.5 N/cm .
k2 = 5.5 N/cm ,
W = 44 N ,
1
and
The springs stretch the same amount x because of the way they were positioned.
Then F1 = k1 x and F2 = k2 x, so the force
equation for the suspended mass is
X
X
Fup =
Fdown
k1 x + k 2 x = W
x=
W
k1 + k 2
44 N
3.5 N/cm + 5.5 N/cm
= 4.88889 cm .
=
Dimensional analysis for x:
N
cm
=N·
= cm
N/cm
N
003 (part 2 of 5) 0 points
In this same parallel spring system, what is
the effective combined spring constant kparallel
of the two springs?
Correct answer: 9 N/cm.
Explanation:
If the springs were one spring, that spring
would react with a force F = −k x where
F = −W due the law of action and reaction,
we have
−W
−x
W
=
x
44 N
=
4.88889 cm
= 9 N/cm ,
5.5 N/cm
3.5 N/cm
kparallel =
44 N
How far down will the 44 N weight stretch
the springs?
Correct answer: 4.88889 cm.
Explanation:
Let :
k1 = 3.5 N/cm ,
which happens to be the sum of the individual
constants
kparallel = k1 + k2
= 3.5 N/cm + 5.5 N/cm
= 9 N/cm .
Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
5.5 N/cm 3.5 N/cm
004 (part 3 of 5) 0 points
Now consider the same two springs in series.
Explanation:
xtotal = x1 +x2 , so W = kseries xtotal , which
is equivalent to
W
x1 + x 2
44 N
=
12.5714 cm + 8 cm
= 2.13889 N/cm , or
1
= x +x
1
2
W
1
= x1
x2
+
W
W
1
=
1
1
+
k1 k2
1
=
1
1
+
3.5 N/cm 5.5 N/cm
= 2.13889 N/cm .
kseries =
kseries
44 N
What distance will the spring of constant
3.5 N/cm stretch?
Correct answer: 12.5714 cm.
Explanation:
In the series system, the springs stretch
a different amount, but each carries the full
weight W = 44 N. W = k1 x1 and
x1 =
2
W
k1
44 N
3.5 N/cm
= 12.5714 cm .
=
005 (part 4 of 5) 0 points
In this same series spring system, what distance will the spring of constant 5.5 N/cm
stretch?
Correct answer: 8 cm.
Explanation:
W = k2 x2 and
x2 =
007 (part 1 of 2) 5 points
The pulley system is in equilibrium, the spring
constant k1 = 7 N/cm and the suspended
mass m = 16 kg.
The acceleration of gravity is 9.8 m/s2 .
W
k2
44 N
5.5 N/cm
= 8 cm .
k1
=
006 (part 5 of 5) 0 points
In this same series spring system, what is the
effective combined spring constant kseries of
the two springs?
Correct answer: 2.13889 N/cm.
m
How much will the spring stretch?
Correct answer: 7.46667 cm.
Explanation:
The existence of a spring in a string defines
the tension in the string because the force
exerted by a spring is F = kx.
Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
3
k2
T1
T1
T1
T3
T3
T3
T2
T3
T4
k1
W1
W2
T5
m
T2 = 2 T1 . Thus at the suspended mass,
T2 + T1 = mg
For the pulley suspended from the spring,
k2 x2 = 2 T3 so that
T3 =
3 T1 = mg
k 2 x2
2
Working up from the suspended weight W2 ,
T3 = W2 + T5 so that
3 k1 x1 = mg
T5 = T 3 − W 2 =
mg
3 k1
¡
¢
(16 kg) 9.8 m/s2
=
3(7 N/cm)
= 7.46667 cm
x1 =
k 2 x2
− W2
2
For the lowest pulley,
T4 = 2 T 3 = k 2 x 2
Thus at the weight W1 ,
T3 + T 4 = W 1 + T 5
008 (part 2 of 2) 5 points
The pulley system is in equilibrium, the spring
constant k2 = 18 N/cm, the suspended weight
W1 = 39 N, and the suspended weight W2 =
11 N.
k 2 x2 = W 1 − W 2
W1 − W 2
k2
39 N − 11 N
=
18 N/cm
= 1.55556 cm
x2 =
k2
W1
k 2 x2
k 2 x2
+ k 2 x2 = W 1 +
− W2
2
2
W2
How much will the spring stretch?
Correct answer: 1.55556 cm.
Explanation:
009 (part 1 of 2) 5 points
A crate is pulled by a force (parallel to the
incline) up a rough incline. The crate has
an initial speed shown in the figure below.
The crate is pulled a distance of 6.3 m on the
incline by a 150 N force.
The acceleration of gravity is 9.8 m/s2 .
Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
The work due to the applied force is
/s
6m
0N
15
1.5
13
µ
4
Wappl = F d
= (150 N) (6.3 m)
= 945 J ,
kg
28
0.3
29◦
=
and the work due to gravity is
a) What is the change in kinetic energy of
the crate?
Correct answer: 325.63 J.
Explanation:
Let : F = 150 N ,
d = 6.3 m ,
θ = 29◦ ,
m = 13 kg ,
g = 9.8 m/s2 ,
µ = 0.328 , and
v = 1.56 m/s .
v
N
Wgrav = −m g d sin θ
= −(13 kg) (9.8 m/s2 )
× (6.3 m) sin 29◦
= −389.118 J ,
so that
∆K = Wf ric + Wappl + Wgrav
= (−230.252 J) + (945 J)
+ (−389.118 J)
= 325.63 J .
F
µN
θ
010 (part 2 of 2) 5 points
b) What is the speed of the crate after it is
pulled the 6.3 m?
Correct answer: 7.2478 m/s.
Explanation:
Since
1
m (vf2 − vi2 ) = ∆K
2
2 ∆K
vf2 − vi2 =
m
mg
The work-energy theorem with nonconservative forces reads
Wf ric + Wappl + Wgravity = ∆K
To find the work done by friction we need the
normal force on the block from Newton’s law
X
Fy = N − m g cos θ = 0
⇒ N = m g cos θ .
Thus
vf =
=
r
2 ∆K
+ vi2
m
s
2(325.63 J)
+ (1.56 m/s)2
13 kg
= 7.2478 m/s .
Wf ric = −µ m g d cos θ
= −(0.328) (13 kg) (9.8 m/s2 )
× (6.3 m) cos 29◦
= −230.252 J .
011 (part 1 of 4) 0 points
You drag a suitcase of mass 8.1 kg with a
force of F at an angle 31.6 ◦ with respect to
Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
the horizontal along a surface with kinetic
coefficient of friction 0.36.
The acceleration of gravity is 9.8 m/s2 .
If the suitcase is moving with constant velocity 2.99 m/s, what is F ?
Correct answer: 27.4681 N.
Explanation:
F
5
014 (part 4 of 4) 0 points
If you are accelerating the suitcase with acceleration 0.947 m/s2 what is F ?
Correct answer: 34.8412 N.
Explanation:
If the suitcase is accelerating with an acceleration a, the force F is
F cos θ − µ (m g − F sin θ) = m a
mg
N
If the suitcase is moving at a constant velocity,
F cos θ = µ N = µ (m g − F sin θ)
µmg
cos θ + µ sin θ
0.36 (8.1 kg) (9.8 m/s2 )
=
cos 31.6◦ + 0.36 sin 31.6◦
= 27.4681 N .
ma + µmg
cos θ + µ sin θ
(8.1 kg) (0.947 m/s2 )
=
cos 31.6◦ + 0.36 sin 31.6◦
0.36 (8.1 kg) (9.8 m/s2 )
+
cos 31.6◦ + 0.36 sin 31.6◦
= 34.8412 N .
F =
F =
012 (part 2 of 4) 0 points
What is the normal force on the suitcase?
Correct answer: 64.9871 N.
Explanation:
You are helping support the weight, so the
normal force is
N = m g − F sin θ
= (8.1 kg) (9.8 m/s2 )
− (27.4681 N) sin 31.6◦
= 64.9871 N .
013 (part 3 of 4) 0 points
If you pull the suitcase 98.9 m, what work
have you done?
Correct answer: 2313.8 J.
Explanation:
w = F d cos θ
= (27.4681 N) (98.9 m) cos 31.6◦
= 2313.8 J .
015 (part 1 of 4) 3 points
A 52.7 kg box initially at rest is pushed 2.36 m
along a rough, horizontal floor with a constant
applied horizontal force of 261.303 N.
The acceleration of gravity is 9.8 m/s2 .
If the coefficient of friction between box
and floor is 0.468, find the work done by the
applied force.
Correct answer: 616.676 J.
Explanation:
The work done by the applied force is given
by
WF = F ·s = (261.303 N)(2.36 m) = 616.676 J
016 (part 2 of 4) 3 points
Find the work done by the friction.
Correct answer: −570.42 J.
Explanation:
The work done by the friction is given by
Wf = f · s = −µ m g s
Wf = −0.468(52.7 kg)(9.8 m/s2 )(2.36 m) = −570.42
017 (part 3 of 4) 2 points
Find the change in kinetic energy of the box.
Correct answer: 46.256 J.
Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
Explanation:
The net work done is given by
Then, using Kepler’s third law, these four
quantities are related by
Tc2
a3c
=
,
Te2
a3e
Wnet = WF +Wf = 616.676 J+(−570.42 J) = 46.256 J
018 (part 4 of 4) 2 points
Find the the final speed of the box.
Correct answer: 1.32493 m/s.
Explanation:
The change in kinetic energy is equal to the
net work done, so
∆K = Wnet =
m vf2
2
−
since the factor of proportionality cancels out.
Hence, solving for ac , one gets
ac = a e
µ
Tc2
Te2
¶1/3
µ
75.6 years
= (1 AU )
1 years
= 17.8792 AU .
m vi2
2
The initial velocity is zero, so
r
2 Wnet
vf =
p m
2 (46.256 J)
=
52.7 kg
= 1.32493 m/s
6
¶2/3
020 (part 2 of 3) 3 points
Denote the length of the semi-major axis of
the orbit by a, and the distance of the closest
approach to the sun by d. The maximum
distance between the Halley’s comet and the
sun is given by
1. 2 a − d correct
019 (part 1 of 3) 4 points
The distance of closest approach of Halley’s
comet to the sun is 8.55 × 107 km ( 0.57 AU).
The period of the comet is 75.6 years. The
radius of the earth’s orbit around the sun is
1.5 × 108 km (1 AU) (assume the earth’s orbit
is circular). Find the length of the semi-major
axis of the comet’s orbit.
Correct answer: 17.8792 AU.
Explanation:
Basic Concepts: Kepler’s Third Law.
Solution: to the semi-major axis, a, by Kepler’s third law,
¶
µ
4 π2
2
a3 .
T =
G Msun
Let
2.
1
(a − d)
2
3. a − d
4.
1
(a + d)
2
5. a + d
6. 2 a + d
Explanation:
s
d
x
2a
Te
ae
Tc
and ac
= Period of Earth’s orbit,
= Semi-major axis of Earth’s orbit,
= Period of comet’s orbit,
= Semi-major axis of comet’s orbit.
(Refer to the above figure.)
021 (part 3 of 3) 3 points
What is the maximum distance between the
Answer, Key – Homework 7 – David McIntyre – 45123 – Mar 25, 2004
Halley’s comet and the sun?
Correct answer: 35.1884 AU.
Explanation:
x = 2(17.8792 AU ) − 0.57 AU = 35.1884 AU .
022 (part 1 of 2) 0 points
Given: G = 6.672 × 10−11 N m2 /kg2 .
The planet Mars requires 2.4 years to orbit
the sun, which has a mass of 2 × 1030 kg, in
an almost circular trajectory.
Calculate the radius of the orbit of Mars as
it circles the sun.
Correct answer: 2.68527 × 1011 m.
Explanation:
Fr = G
Mm
= m ar = m ω 2 r ,
r2
2π
T
r=
=
·
µ
GM T2
4 π2
¶ 31
6.67 × 10−11 N m2 /kg2 × M × (35640 s)2
=
4 × (3.1415926536 )2
= 1.59759 × 108 m .
¶ 13
Now, the altitude h of the satellite (measured
from the surface of Jupiter) is
¶ 13
2 × 1030
4 π2 3
r
GM
µ
Thus,
GM
ω2
T2 =
r=
2π
(7.56864 × 107 s)
= 8.3016 × 10−8 rad/s .
=
µ
A “synchronous” satellite, which always remains above the same point on a planet’s
equator, is put in orbit about Jupiter to study
the Great Red Spot. Jupiter rotates once
every 9.9 h, has a mass of 1.9 × 1027 kg
and a radius of 6.99 × 107 m. Given that
G = 6.67 × 10−11 N m2 /kg2 , calculate how
far above Jupiter’s surface the satellite must
be.
Correct answer: 8.98589 × 107 m.
Explanation:
Basic Concepts: Solution:
According to Kepler’s third law:
where r is the radius of the satellite’s orbit.
Thus, solving for r:
and
ω=
7
kg
rad/s)2
(8.3016 × 10−8
¸ 31
h = r − R = 8.98589 × 107 m
1
× (6.672 × 10−11 N m2 /kg2 ) 3
= 2.68527 × 1011 m .
023 (part 2 of 2) 0 points
Calculate the orbital speed of Mars as it circles the sun.
Correct answer: 22292 m/s.
Explanation:
v = ωr
= (8.3016 × 10−8 rad/s) (2.68527 × 1011 m)
= 22292 m/s .
024 (part 1 of 1) 0 points