Chapter 8 Solutions
3.
Picture the Problem: The three paths of the sliding box are depicted at right.
Strategy: The work done by friction is W  k mgd , where d is the distance the box is pushed
irregardless of direction, because the friction force always acts in a direction opposite the motion.
Sum the work done by friction for each segment of each path.
Solution: 1. Calculate
the work for path 1:
W1   k mg  d1  d 2  d3  d 4  d5 
  k mg  4.0  4.0  1.0  1.0  1.0 m 
W1   0.26  3.7 kg   9.81 m/s 2  11.0 m   100 J
14.
2. Calculate W for path 2:
W2   k mg  d 6  d 7  d8 
3. Calculate the work for path 3:
W3   k mg  d9  d10  d11 
  0.26  3.7 kg   9.81 m/s 2   2.0 m    2.0 m   1.0 m    47 J
  0.21 3.2 kg   9.81 m/s 2  1.0 m    3.0 m    3.0 m     66 J
Picture the Problem: The spring in the soap dispenser is compressed by the applied force.
Strategy: Use equation 8-5 to find the spring constant using the given energy and compression distance data. Solve the same
equation for x in order to answer part (b).
24.
Solution: 1. (a) Solve equation 8-5 for k:
k
2. (b) Solve equation 8-5 for x:
x
2U 2  0.0025 J 

 200 N/m  0.20 kN/m
x 2  0.0050 m 2
2U

k
2  0.0084 J 
200 N/m
 0.92 cm
Picture the Problem: A swimmer descends through a vertical height of 2.31 m as she slides without friction.
Strategy: As the swimmer descends the slide her gravitational potential energy is converted into kinetic energy. Set the loss in
gravitational potential energy equal to the gain in kinetic energy by setting her change in mechanical energy equal to zero, so
that E  Ef  Ei  0 or Ef  Ei . Let y  0 at the bottom of the slide, v  0 at the top.
Ebottom  Etop
Solution: Set Ebottom  Etop and solve for vbottom :
K bottom  U bottom  K top  U top
1
2
2
mvbottom
 0  0  mgytop
vbottom  2 gytop  2  9.81 m/s 2   2.31 m   6.73 m/s
25.
Picture the Problem: A swimmer descends through a vertical height of 2.31 m as she slides without friction.
Strategy: As the swimmer descends the slide her gravitational potential energy is converted into kinetic energy. Set the loss in
gravitational potential energy equal to the gain in kinetic energy by setting her change in mechanical energy equal to zero, so
that E  Ef  Ei  0 or Ef  Ei . Let y  0 at the bottom of the slide, v  0.840 m/s at the top.
Ebottom  Etop
Solution: Set Ebottom  Etop and solve for vbottom :
K bottom  U bottom  K top  U top
1
2
2
2
mvbottom
 0  12 mvtop
 mgytop
2
vbottom  vtop
 2 gytop

 0.840 m/s 
2
 2  9.81 m/s 2   2.31 m 
 6.78 m/s
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  12 mva2  12 1.40 kg  0.950 m/s   0.632 J
2. The spring is not compressed so xa  0 cm:
U a  12 kxa2  0
3. The total energy is the sum of K and U:
Ea  Ka  Ua  E  0.632 J  0  0.632 J
4. (b) Find U b when xb  1.00 cm:
U b  12 kxb2  12  734 N/m 0.0100 m   0.0367 J
5. The total energy remains 0.632 J always so find K b :
Kb  E  Ub  0.632  0.0367 J  0.595 J
6. (c), (d), (e) Repeat steps 4 and 5:
2
2
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
47.
Picture the Problem: The seal slides down the ramp, changing altitude and
speed, and loses mechanical energy along the way due to friction.
Strategy: The nonconservative work done by friction changes the mechanical
energy of the seal. Use equation 8-9 and the given information to find the
nonconservative work. Then use Newton’s Second Law to find the normal
force on the seal, and use the normal force to find the force of kinetic friction.
Solve the resulting expression for k . Let y = 0 at the water’s surface.
Solution: 1. (a) Use equations 8-9, 7-6,
and 8-3 to find Wnc :
Wnc  E  Ef  Ei
  12 mvf2  mgyf    12 mvi2  mgyi 
 12 m  vf2  vi2   mg  yf  yi 
  42.0 kg 
  4.40 m/s   0   9.81 m/s   0 1.75 m 
2
1
2
2
2
Wnc  314 J
2. Use Newton’s Second Law to find the
normal force:
F
3. Use the right triangle formed by the ramp
to find the distance the seal slides:
sin  
4. (b) Set Wnc equal to the work done by
friction and solve for k :
49.
y
 N  mg cos  0  N  mg cos
h
d
 d
h
sin 
Wnc   f k d    k Nd    k  mg cos   h sin  
Wnc sin 
 314 J  tan 35.0
k  

 0.305
 mg cos   h  42.0 kg   9.81 m/s 2  1.75 m 
Picture the Problem: The car drives up the hill, changing its kinetic and gravitational potential energies, while both the engine
force and friction do nonconservative work on the car.
Strategy: The total nonconservative work done on the car changes its mechanical energy according to equation 8-9. This
nonconservative work includes the positive work Wnc1 done by the engine and the negative work Wnc2 done by the friction. Use
this relationship and the known change in potential energy to find  K .
Solution: Set the nonconservative
work equal to the change in mechanical energy and solve for K :
Wnc  Wnc1  Wnc2  E  Ef  Ei
Wnc1  Wnc2   K f  U f    K i  U i   K  mg  yf  yi 
K  Wnc1  Wnc2  mg  yf  yi 
  6.44  105 J    3.11105 J   1250 kg   9.81 m/s 2  16.2 m 
K  1.34  105 J  134 kJ
53.
Picture the Problem: The block slides horizontally on a rough surface, encounters a spring, and compresses it a distance of 11.0
cm before coming to rest.
Strategy: The nonconservative work done by friction changes the mechanical energy of the system. Use equation 8-9 to find
E and set it equal to the work done by friction. Solve the resulting expression for the spring constant k.
Solution: 1. Write equation 8-9 to obtain
an expression for Wnc :
Wnc  Ef  Ei   K f  U f    K i  U i 
2. The nonconservative work is done by
friction as the block travels a distance x:
Wnc   f k d  k mg x
3. Substitute the expression from step 2 into
step 1 and solve for k:
 k mg x  12 kx 2  12 mvi2
  0  12 kx 2    12 mvi2  0 
2
mvi2  2k mg x m vi  2k g x 

x2
x2
2
1.80 kg   2.00 m/s   2  0.560   9.81 m/s 2   0.110 m 

2
 0.110 m 
k
k  415 N/m
77. Picture the Problem: The child slides from rest at point A and
lands at point B as indicated in the figure at right.
Strategy: Use the conservation of mechanical energy to find
the horizontal speed of the child at the bottom of the slide in
terms of h. Then use equation 4-9, the landing site of a
projectile launched horizontally, to find the speed the child
should have in order to land 2.50 m down range. Set the speeds
equal to each other and solve for h.
Solution: 1. Use equation 8-3 and let EA  Ebottom to find
vbottom :
K A  U A  K bottom  U bottom
1
2
2
mv  mgyA  12 mvbottom
 mgybottom
2
A
2
0  mg  h  1.50 m   12 mvbottom
 mg 1.50 m 
2 gh  vbottom
2. Use equation 4-9 to find the appropriate vbottom for the
child to land 2.50 m down range:
3. Set the two velocities equal to each other and solve for h:
x  vbottom 2 ybottom g  vbottom  x g 2 ybottom
2 gh  x g 2 ybottom
2 gh  x 2 g 2 ybottom
 2.50 m 
x2
h

 1.04 m
4 ybottom 4 1.50 m 
2
86.
Picture the Problem: The block slides from rest at point A and is launched
horizontally at point B as indicated in the figure at right.
Strategy: Use the conservation of mechanical energy to find the horizontal
speed of the block at the bottom of the ramp. Then use equation 4-9 to find the
landing site of the block because it is launched horizontally.
Solution: 1. Set EA  EB and use
equation 8-3 to find vB :
KA  U A  KB  U B
0  mgyA  12 mvB2  mgyB
2 g  yA  yB   vB
2. Use equation 4-9 to find d:
d  vB 2 yB g   2 g  yA  yB    2 yB g   4 yB  yA  yB 
 4  0.25 m 1.5  0.25 m 
d  1.1 m