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CHAPTER 8: Conservation of Energy
Solutions to Assigned Problems
8.
The force is found from the relations on page 189.
U
U
Fx  
  6x  2 y 
Fy  
   2 x  8 yz 
x
y

F  ˆi  6 x  2 y   ˆj  2 x  8 yz   k 4 y 2
Fz  
U
z
 4 y 2

19. Use conservation of energy. The level of the ball on the uncompressed spring is taken as the zero
location for both gravitational potential energy  y  0  and elastic potential energy  x  0 . It is
diagram 2 in the figure. Take “up” to be positive for both x and y.
(a) Subscript 1 represents the ball at the launch point, and subscript 2
represents the ball at the location where it just leaves the spring, at the uncompressed length.
We have v1  0, x1  y1  0.160 m, and x2  y2  0. Solve for v2 .
E1  E2 
1
2
mv12  mgy1  12 kx12  12 mv22  mgy2  12 kx22 
0  mgy1  12 kx12  12 mv22  0  0  v2 
v2 
kx12  2mgy1
m
875 N m  0.160 m 2  2  0.380 kg   9.80 m
 0.380 kg 
s2
  0.160 m  
7.47 m s
(b) Subscript 3 represents the ball at its highest point. We have v1  0, x1  y1  0.160 m, v3  0,
and x3  0. Solve for y3 .
E1  E3 
1
2
mv12  mgy1  12 kx12  12 mv32  mgy3  12 kx32 
0  mgy1  kx  0  mgy2  0  y2  y1 
1
2
2
1
kx12
2mg
875 N m  0.160 m 2


2  0.380 kg   9.80 m s 2 
3.01m
20. Since there are no dissipative forces present, the mechanical energy of the roller coaster will be
conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero
location for gravitational potential energy. We have v1  0 and y1  32 m.
Point 2:
1
2
mv12  mgy1  12 mv22  mgy2 ; y2  0  mgy1  12 mv22 

v2  2 gy1  2 9.80 m s2
Point 3:
1
2
  32 m   25m s
mv12  mgy1  12 mv32  mgy3 ; y3  26 m  mgy1  12 mv32  mgy3 

v3  2 g  y1  y3   2 9.80 m s2
Point 4:
1
2
  6 m  11m s
mv12  mgy1  12 mv42  mgy4 ; y4  14 m  mgy1  12 mv42  mgy1 


v4  2 g  y1  y4   2 9.80 m s2 18 m   19 m s
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218
Physics for Scientists & Engineers, 4th Edition
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27. The maximum acceleration of 5.0 g occurs where the force is at a maximum. The
maximum force occurs at the bottom of the motion, where the spring is at its
maximum compression. Write Newton’s second law for the elevator at the bottom of
the motion, with up as the positive direction.
Fnet  Fspring  Mg  Ma  5.0Mg  Fspring  6.0Mg
Now consider the diagram for the elevator at various
points in its motion. If there are no non-conservative
forces, then mechanical energy is conserved. Subscript 1
represents the elevator at the start of its fall, and subscript
2 represents the elevator at the bottom of its fall. The
bottom of the fall is the zero location for gravitational
potential energy  y  0 . There is also a point at the top
of the spring that is the zero location for elastic potential
energy (x = 0). We have v1  0, y1  x  h, x1  0,
Mg
Fspring
Start of fall
h
Contact with spring,
0 for elastic PE
x
Bottom of fall, 0 for
gravitational PE
v2  0, y2  0, and x2  x. Apply conservation of
energy.
E1  E2 
1
2
Mv12  Mgy1  12 kx12  12 Mv22  Mgy2  12 kx22 
0  Mg  x  h   0  0  0  12 kx 2  Mg  x  h   12 kx 2
Fspring  6.0 Mg  kx  x 
 6 Mg  h   1 k  6 Mg 
 Mg 
 2 

 k

 k 
6.0 Mg
k
2

k
12 Mg
h
48. Note that the difference in the two distances from the center of the Earth, r2  r1 , is the same as the
height change in the two positions, y2  y1. Also, if the two distances are both near the surface of
the Earth, then r1r2  rE2 .
 GM E m   GM E m  GM E m GM E m
 1 1  GM E m



 GM E m    
 r2  r1 


r2  
r1 
r1
r2
r1r2

 r1 r2 
U   

GM E m
2
E
r
 y2  y1   m
GM E
rE2
 y2  y1  
mg  y2  y1 
49. The escape velocity for an object located a distance r from a mass M is given by Eq. 8-19,
vesc 
2 MG
r
(a) vesc at 
Sun's
surface
(b) vesc at 
Earth
orbit
. The orbit speed for an object located a distance r from a mass M is vorb 
2 M SunG
rSun
2 M SunG
rEarth orbit




2 2.0  1030 kg 6.67  1011 N m 2 kg 2
7.0  10 m
8


2 2.0  1030 kg 6.67  1011 N m 2 kg 2
11
1.50  10 m


MG
r
.
6.2  105 m s
4.2  104 m s
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
219
Chapter 8
Conservation of Energy
vesc at
Earth
orbit
vEarth

2 M SunG
rEarth orbit

2 
M SunG
rEarth orbit
orbit
vesc at 
2vEarth
Earth
orbit
orbit
Since vesc at  1.4vEarth , the orbiting object will not escape the orbit.
Earth
orbit
orbit
52. (a) With the condition that U  0 at r  , the potential energy is given by U  
GM E m
r
. The
kinetic energy is found from the fact that for a circular orbit, the gravitational force is a
centripetal force.
2
GM E m mvorbit
GM E m
GM E m
2
2

 mvorbit

 K  12 mvorbit
 12
2
r
r
r
r
E  K U 
1
2
GM E m

GM E m
  12
GM E m
r
r
r
(b) As the value of E decreases, since E is negative, the radius r must get smaller. But as the radius
1
gets smaller, the kinetic energy increases, since K  . If the total energy decreases by 1
r
Joule, the potential energy decreases by 2 Joules and the kinetic energy increases by 1 Joule.
70. The force to lift the water is equal to its weight, and so the work to lift the water is equal to the
weight times the vertical displacement. The power is the work done per unit time.
2
W mgh  21.0 kg  9.80 m s  3.50 m 
P


 12.0 W
t
t
60 sec


85. (a) The tension in the cord is perpendicular to the path at all times, and so the tension in the cord
does not do any work on the ball. Thus only gravity does work on the ball, and so the
mechanical energy of the ball is conserved. Subscript 1 represents the ball when it is horizontal,
and subscript 2 represents the ball at the lowest point on its path. The lowest point on the path
is the zero location for potential energy  y  0 . We have v1  0 , y1  l , and y2  0. Solve
for v2 .
E1  E2 
1
2
mv12  mgy1  12 mv22  mgy2  mg l  12 mv22  v2 
2g l
(b) Use conservation of energy, to relate points 2 and 3. Point 2 is as described above. Subscript 3
represents the ball at the top of its circular path around the peg. The lowest point on the path is
the zero location for potential energy  y  0 . We have v2  2gl , y2  0, and
y3  2  l  h   2  l  0.80l
E 2  E3 
1
2
  0.40l
. Solve for v3 .
mv  mgy2  12 mv32  mgy3 
2
2
1
2
m  2 gl
  12 mv32  mg  0.40l 

v3  1.2 gl
86. The ball is moving in a circle of radius  l  h  . If the ball is to complete the circle with the string
just going slack at the top of the circle, the force of gravity must supply the centripetal force at the
top of the circle. This tells the critical (slowest) speed for the ball to have at the top of the circle.
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220
Physics for Scientists & Engineers, 4th Edition
Giancoli
mg 
2
mvcrit
2
 vcrit
 gr  g  l  h 
r
To find another expression for the speed, we use energy conservation. Subscript 1 refers to the ball
at the launch point, and subscript 2 refers to the ball at the top of the circular path about the peg. The
zero for gravitational potential energy is taken to be the lowest point of the ball’s path. Let the speed
at point 2 be the critical speed found above.
E1  E2  12 mv12  mgy1  12 mv22  mgy2  mg l  12 mg  l  h   2mg  l  h  
h  0.6l
If h is any smaller than this, then the ball would be moving slower than the critical speed when it
reaches the top of the circular path, and would not stay in centripetal motion.
88. The spring constant for the scale can be found from the 0.5 mm compression due to the 760 N force.
F
760 N
k 
 1.52  106 N m. Use conservation of energy for the jump. Subscript 1
x 5.0  104 m
represents the initial location, and subscript 2 represents the location at maximum compression of the
scale spring. Assume that the location of the uncompressed scale spring is the 0 location for
gravitational potential energy. We have v1  v2  0 and y1  1.0 m. Solve for y 2 , which must be
negative.
E1  E2  12 mv12  mgy1  12 mv22  mgy2  12 ky22 
mgy1  mgy2  12 ky22  y22  2
2
mg
k
y2  2
mg
k
y1  y22  1.00  103 y2  1.00  103  0
2
y2  3.21  10 m, 3.11  10 m



Fscale  k x  1.52  106 N m 3.21  102 m  4.9  104 N
90. (a) Draw a free-body diagram for the block at the top of the curve. Since the
block is moving in a circle, the net force is centripetal. Write Newton’s
second law for the block, with down as positive. If the block is to be on
the verge of falling off the track, then FN  0.
F
R
2
 FN  mg  m v 2 r  mg  m vtop
r  vtop 
FN
mg
gr
Now use conservation of energy for the block. Since the track is frictionless, there are no nonconservative forces, and mechanical energy will be conserved. Subscript 1 represents the block
at the release point, and subscript 2 represents the block at the top of the loop. The ground is
the zero location for potential energy  y  0 . We have v1  0, y1  h, v2  gr , and y2  2r.
Solve for h.
E1  E2  12 mv12  mgy1  12 mv22  mgy2  0  mgh  12 mgr  2mgr 
h  2.5 r
(b) See the free-body diagram for the block at the bottom of the loop. The net
force is again centripetal, and must be upwards.
2
r
 FR  FN  mg  m v2 r  FN  mg  m vbottom
The speed at the bottom of the loop can be found from energy conservation,
similar to what was done in part (a) above, by equating the energy at the
FN
mg
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221
Chapter 8
Conservation of Energy
release point (subscript 1) and the bottom of the loop (subscript 2). We now have v1  0,
y1  2h  5r, and y2  0. Solve for v2 .
E1  E2 
1
2
2
mv12  mgy1  12 mv22  mgy2  0  5mgr  12 mvbottom
0 
2
2
vbottom
 10 gr  FN  mg  m vbottom
r  mg  10mg  11mg
(c) Again we use the free body diagram for the top of the loop, but now the normal force does not
vanish. We again use energy conservation, with v1  0, y1  3r , and y2  0. Solve for v2 .
F
R
2
 FN  mg  m v 2 r  FN  m vtop
r  mg
E1  E2 
1
2
2
mv12  mgy1  12 mv22  mgy2  0  3mgr  12 mvtop
0 
2
2
vtop
 6 gr  FN  m vtop
r  mg  6mg  mg  5mg
(d) On the flat section, there is no centripetal force, and FN  mg .
91. (a) Use conservation of energy for the swinging motion. Subscript 1
represents the student initially grabbing the rope, and subscript 2
represents the student at the top of the swing. The location where the
student initially grabs the rope is the zero location for potential
energy  y  0 . We have v1  5.0 m s , y1  0, and v2  0. Solve

l
l-h
for y 2 .
E1  E2 
y2 = h
1
2
mv  mgy1  mv  mgy2 
2
1
2
2
v12
h
2g
Calculate the angle from the relationship in the diagram.
l h
h
v2
cos  
 1  1 1

l
l
2gl
1
2
mv12  mgy2  y2 
1
2


v2 

 5.0 m s 2
  cos 1  1  1   cos 1  1 
 29
2
 2  9.80 m s  10.0 m  
 2gl 


(b) At the release point, the speed is 0, and so there is no radial acceleration,
since a R  v 2 r . Thus the centripetal force must be 0. Use the free-body
diagram to write Newton’s second law for the radial direction.
 FR  FT  mg cos   0 


FT
mg 

FT  mg cos    56 kg  9.80 m s 2 cos 29 o  480 N
(c) Write Newton’s second law for the radial direction for any angle, and solve for the tension.
 FR  FT  mg cos  mv2 r  FT  mg cos  mv2 r
As the angle decreases, the tension increases, and as the speed increases, the tension increases.
Both effects are greatest at the bottom of the swing, and so that is where the tension will be at
its maximum.
56 kg  5.0 m s 
  10.0
m

FT  mg cos 0  m v12 r   56 kg  9.80 m s 2 
max
2
 690 N
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222
Physics for Scientists & Engineers, 4th Edition
Giancoli
98. It is shown in problem 52 that the total mechanical energy for a satellite orbiting in a circular orbit of
GmM E
radius r is E   12
. That energy must be equal to the energy of the satellite at the surface of
r
the Earth plus the energy required by fuel.
(a) If launched from the equator, the satellite has both kinetic and potential energy initially. The
kinetic energy is from the speed of the equator of the Earth relative to the center of the Earth. In
problem 53 that speed is calculated to be 464 m/s.
GmM E
GmM E
Esurface  Efuel  Eorbit  12 mv02 
 Efuel   12

RE
r
 1
Efuel  GmM E 
 RE

1 
2
11
2
2
24
1
  2 mv0   6.67  10 N m kg  1465 kg   5.98  10 kg 
2r 

  1
1
1
2


   2 1465 kg   464 m s 
6
6
6
 6.38  10 m 2  6.38  10 m  1.375  10 m  
 5.38  1010 J
(b) If launched from the North Pole, the satellite has only potential energy initially. There is no
initial velocity from the rotation of the Earth.
GmM E
GmM E
Esurface  Efuel  Eorbit  
 Efuel   12

RE
r
 1
Efuel  GmM E 
 RE

1 
   6.67  10
11
2r 


N m 2 kg 2 1465 kg  5.98  10 24 kg


 
1
1


 
6
6
6
 6.38  10 m 2  6.38  10 m  1.375  10 m  
 5.39  1010 J
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
223