HOMEWORK 8
Chapter 8
10. Picture the Problem: The climber stands at the top of Mt. Everest.
Strategy: Find the gravitational potential energy by using equation 8-3.
Solution: Calculate U
mgy :
U
88 kg 9.81 m/s2 8848 m
mgy
7.6 106 J
Insight: You are free to declare that the climber’s potential energy is zero at the top of Mt.
Everest and −7.2 MJ at sea level!
13. Picture the Problem: A mass is suspended from a vertical spring. As the spring is stretched
it stores potential energy.
Strategy: Doubling the mass doubles the force exerted on the spring, and therefore doubles the stretch
distance due to Hooke’s Law F
kx. Use a ratio to find the increase in spring potential energy
U 12 kx 2 when the stretch distance is doubled.
Solution: 1. (a) If the mass attached to the spring is doubled the stretch distance will double.
Doubling the extension of the spring will increase the potential energy by a factor of 4.
2. (b) Doubling the mass
doubles the force
and therefore doubles the
stretch distance:
x1
mg k
x2
2mg k
3. Calculate the ratio U 2 U1 :
U2
U1
1
2
1
2
U2
4U1
kx22
kx12
2 x1
2 x1
2
x12
4
4 0.962 J
3.85 J
Insight: Note that the change in gravitational potential energy also quadruples as the new
mass is hung on the spring. It doubles because there is twice as much mass and it doubles
again because the spring stretches twice as far. If you calculate the stretch distances you’ll
find the spring stretched x1 2U1 m1 g 5.6 cm when the 3.5-kg mass was suspended and it
stretched 11.2 cm when the 7.0-kg mass was suspended from it.
15. Picture the Problem: The spring is stretched by the applied force and stores potential
energy.
Strategy: Use equation 6-4 F
kx to find the spring constant, and then equation 8-5
2
1
U 2 kx to find the stretch distance.
Solution: 1. (a) Solve equation 6-4 for k:
k
F
x
2. Solve equation 8-5 for x:
x
2U
k
4.1 N
0.014 m
290 N/m
2 0.020 J
290 N/m
1.2 cm
7.6 MJ
3. (b) Repeat step 2 with the new U:
2 0.080 J
2U
k
x
290 N/m
2.3 cm
Insight: Notice that in part (b) the stretch distance doubled but the stored potential energy
quadrupled because U is proportional to x2.
30. Picture the Problem: A block slides on a frictionless, horizontal surface and encounters a
horizontal spring. It compresses the spring and briefly comes to rest.
Strategy: Set the mechanical energy when sliding freely equal to the mechanical energy
when the spring is fully compressed and the block is at rest. Solve the resulting equation for
the spring constant k, then repeat the procedure to find the initial speed required to compress
the spring only 1.2 cm before coming to rest.
Solution: 1. (a) Set Ei Ef where the
initial
state is when it is sliding freely and the
final
state is when it is at rest, having
compressed
the spring.
Ki U i
1
2
2. (b) Solve the equation from step 1 for
vi :
2
i
mv
Kf
0
0
k
mvi2
2
xmax
2
kxmax
m
vi
Uf
1
2
2
kxmax
2.9 kg 1.6 m/s
0.048 m
2
3200 N/m
2
3200 N/m 0.012 m
2.9 kg
3.2 kN/m
2
0.40 m/s
Insight: The kinetic energy of the sliding block is stored as potential energy in the spring.
Moments later the spring will have released all its potential energy, the block would have
gained its kinetic energy again, and would then be sliding at the same speed but in the
opposite direction.
31. Picture the Problem: The trajectory of the rock is depicted at
right.
Strategy: The rock starts at height h, rises to ymax , comes briefly to
rest, then falls down to the base of the cliff at y 0. Set the
mechanical energy at the point of release equal to the mechanical
energy at the base of the cliff and at the maximum height ymax in order
to find vi and ymax .
Solution: 1. (a) Set
Ei Ef and solve for
vi :
1
2
Ki U i
mv mgh
2
i
vi
2. (b) Now set
Eymax Ef
and solve for ymax :
K ymax U ymax
0 mgymax
ymax
h
Kf U f
1
mvf2 0
2
vf 2
2 gh
29 m/s
vi
vi
2
2 9.81 m/s 2 32 m
15 m/s
Kf
1
2
Uf
mvf2
0
ymax
vf2
2g
29 m/s
2
2 9.81 m/s 2
43 m
Insight: In part (a) the initial energy is a combination of potential and kinetic, but becomes all
kinetic just before impact with the ground. In part (b) the rock at the peak of its flight has zero
kinetic energy; all of its energy is potential energy.
32. Picture the Problem: The block slides on a frictionless, horizontal surface, encounters a
spring, compresses it, and briefly comes to rest when the spring compression is 4.15 cm.
Strategy: As the block compresses the spring its kinetic energy is converted into spring potential
energy. The sum of the spring potential and kinetic energies equals the mechanical energy, which
remains constant throughout. Use equations 7-6 and 8-5 to find the kinetic and spring potential
energies, respectively.
Solution: 1. (a) Find K a when va
0.950 m/s: Ka
1
2
mva2
kxa2
2. The spring is not compressed so
xa 0 cm:
Ua
1
2
3. The total energy is the sum of K and U:
Ea
Ka U a
4. (b) Find U b when xb
Ub
1
2
Kb
E Ub
1.00 cm:
5. The total energy remains 0.632 J always
so find K b :
6. (c), (d), (e) Repeat steps 4 and 5:
kxb2
1
2
1.40 kg 0.950 m/s
2
0.632 J
0
E
1
2
0.632 J 0
0.632 J
734 N/m 0.0100 m
0.632 0.0367 J
2
0.0367 J
0.595 J
y (cm)
0.00
1.00
2.00
3.00
4.00
U (J)
0.00
0.037
0.147
0.330
0.587
K (J)
0.632
0.595
0.485
0.302
0.045
E (J)
0.632
0.632
0.632
0.632
0.632
Insight: The initial kinetic energy of the sliding block is stored as potential energy in the
spring when it comes to rest. Moments later the spring will have released all its potential
energy, the block would have gained its kinetic energy again, and would then be sliding at the
same speed but in the opposite direction.
35. Picture the Problem: The pendulum bob swings from point B to
point A and gains altitude and thus gravitational potential energy.
See the figure at right.
Strategy: Use equation 7-6 to find the kinetic energy of the bob at
point B. Use the geometry of the problem to find the maximum
change in altitude ymax of the pendulum bob, and then use equation
8-3 to find its maximum change in gravitational potential energy.
Apply conservation of energy between points B and the endpoint of
its travel to find the maximum angle max the string makes with the
vertical.
Solution: 1. (a) Use equation K
B
7-6 to find K B :
1
2
mvB2
1
2
0.33 kg 2.4 m/s
2
0.95 J
2. (b) Since there is no friction, mechanical energy is conserved. If
we take the potential energy at point B to be zero, we can say that all
of the bob’s kinetic energy will become potential energy when the
bob reaches its maximum height and comes momentarily to rest.
Therefore the change in potential energy between point B and the
point where the bob comes to rest is 0.95 J.
3. (c) Find the height
ymax
change ymax of the pendulum
bob:
4. Use equation 8-3 and the
result of part (b) to solve for
max :
U
max
L L cos
mg ymax
cos
1
1
max
L 1 cos
mgL 1 cos
U
mgL
cos
1
max
max
1
0.95 J
0.33 kg 9.81 m/s2 1.2 m
Insight: The pendulum bob cannot swing any farther than 41° because there is not enough
energy available to raise the mass to a higher elevation.
41
36. Picture the Problem: The motions of the masses in
the Atwood’s machine are depicted in the figure at
right:
Strategy: Mechanical energy is conserved because there is
no friction. Set Ei Ef and solve for vf . The speeds of
each mass must always be the same because they are
connected by a rope.
Solution:
1. (a) Set
Ei Ef and
solve for vf :
Ki U i
0 0
0
m2 m1 gh
vf
vf
2. (b) Use the
expression
from part (a) to find vf
:
Kf U f
1
m v 2 12 m2 vf2 m1 gy1 m2 gy2
2 1 f
1
m1 m2 vf2 m1 gh m2 g h
2
1
m1 m2 vf 2
2
2 gh
m2 m1
m1 m2
2 9.81 m/s 2 1.2 m
4.1 kg 3.7 kg
3.7 kg 4.1 kg
1.1 m/s
Insight: The mass m2 loses more gravitational potential energy than m1 gains, so there is
extra energy available to give the system kinetic energy. We bent the rules for significant
figures a bit in step 2 because by the rules of subtraction,
4.1 − 3.7 kg = 0.4 kg, only one significant figure.