Download SolutionstoAssignedProblemsChapter7

CHAPTER 7: Work and Energy
Solutions to Problems
11. The piano is moving with a constant velocity down the plane. FP is the
FN
force of the man pushing on the piano.
F
x
y
FP
(a) Write Newton’s second law on each direction for the piano, with an
acceleration of 0.
 Fy  FN  mg cos   0  FN  mg cos 
 mg

x
 mg sin   FP  0 
FP  mg sin   mg sin 

  380 kg  9.80 m s 2
 sin 27  1691N  1700 N
(b) The work done by the man is the work done by FP . The angle between FP and the direction of
motion is 180. Use Eq. 7-1.
WP  FP d cos180   1691N  3.9 m   6595J  6600 J .
(c) The angle between the force of gravity and the direction of motion is 63. Calculate the work
done by gravity.
WG  FG d cos 63  mgd cos 63   380 kg   9.80 m s 2   3.9 m  cos 63
 6594 N  6600 J
(d) Since the piano is not accelerating, the net force on the piano is 0, and so the net work done on
the piano is also 0. This can also be seen by adding the two work amounts calculated.
Wnet  WP  WG  6.6  103 J  6.6  103 J  0 J
18. Use Eq. 7.4 and Eq. 7.2 to calculate the dot product, and then solve for the angle.
A B  Ax Bx  Ay B y  Az Bz   6.8 8.2    3.4  2.3   6.2  7.0   91.34
A
 6.8    3.4    6.2 
2
2
A B  AB cos     cos 1
2
 9.81
AB
AB
B
 cos 1
8.2    2.3   7.0
2
2
91.34
 9.8111.0 
2
 11.0
 32


19. We utilize the fact that if B  Bx ˆi  By ˆj  Bz kˆ , then  B    Bx  ˆi   B y ˆj    Bz  kˆ .
A  B   Ax   Bx   Ay   B y   Az   Bz 
   Ax  Bx     Ay  B y     Az  Bz    A B
21. If A is perpendicular to B , then A B  0. Use this to find B.
A B  Ax Bx  Ay By   3.0  Bx  1.5 By  0  By  2.0Bx
Any vector B that satisfies By  2.0 Bx will be perpendicular to A . For example, B  1.5ˆi  3.0ˆj .
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
194
Physics for Scientists & Engineers, 4th Edition
Giancoli
26. We are given that the magnitudes of the two vectors are the same, so Ax2  Ay2  Az2  Bx2  By2  Bz2 .
If the sum and difference vectors are perpendicular, their dot product must be zero.
A  B   Ax  Bx  ˆi   Ay  B y  ˆj   Az  Bz  kˆ
A  B   Ax  Bx  ˆi   Ay  By  ˆj   Az  Bz  kˆ
A  B A  B   A
x
 Bx  Ax  Bx    Ay  B y  Ay  B y    Az  Bz  Az  Bz 

 

 Ax2  Bx2  Ay2  B y2  Az2  Bz2  Ax2  Ay2  Az2  Bx2  B y2  Bz2  0
35. The force exerted to stretch a spring is given by Fstretch  kx
(the opposite of the force exerted by the spring, which is
given by F   kx. A graph of Fstretch vs. x will be a
straight line of slope k through the origin. The stretch from
x1 to x2, as shown on the graph, outlines a trapezoidal area.
This area represents the work.
W  12  kx1  kx2  x2  x1   12 k  x1  x2  x2  x1 

1
2
 65 N
F = kx
kx2
Force
kx1
x1
Stretch distance
m  0.095 m  0.035 m   0.11J
x2
40. The work done will be the area under the Fx vs. x graph.
(a) From x  0.0 to x  10.0 m, the shape under the graph is trapezoidal. The area is
Wa   400 N  12 10 m  4 m   2800 J .
(b) From x  10.0 m to x  15.0 m, the force is in the opposite direction from the direction of
motion, and so the work will be negative. Again, since the shape is trapezoidal, we find
Wa   200 N  12  5m  2 m   700 J.
Thus the total work from x  0.0 to x  15.0 m is 2800 J  700 J  2100 J .
53. The work done on the car is equal to the change in its kinetic energy.
2

 1m s  
W  K  mv  mv  0  1300 kg   95 km h  
 4.5  105 J


 3.6 km h  

1
2
2
2
1
2
2
1
1
2
Note that the work is negative since the car is slowing down.
62. (a) From the free-body diagram for the load being lifted, write Newton’s second law for
the vertical direction, with up being positive.
 F  FT  mg  ma  0.150mg 

FT

FT  1.150mg  1.150  265 kg  9.80 m s 2  2.99 103 N
mg
(b) The net work done on the load is found from the net force.

Wnet  Fnet d cos 0o   0.150mg  d  0.150  265 kg  9.80 m s 2
 8.96  103 J
(c) The work done by the cable on the load is as follows.

Wcable  FT d cos 0o  1.150mg  d  1.15  265 kg  9.80 m s 2
  23.0 m 
  23.0 m   6.87 10 J
4
(d) The work done by gravity on the load is as follows.
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
195
Chapter 7
Work and Energy

WG  mgd cos180o  mgd    265 kg  9.80 m s 2
  23.0 m  
5.97 104 J
(e) Use the work-energy theorem to find the final speed, with an initial speed of 0.
Wnet  K 2  K1  mv  mv
1
2
2
2
1
2
 v2 
2
1
2Wnet
m
v 
2
1

2 8.96  103 J
265kg
  0  8.22 m s
63. (a) The angle between the pushing force and the displacement is 32.
WP  FP d cos   150 N  5.0 m  cos 32  636.0 J  640 J
(b) The angle between the force of gravity and the displacement is 122.

WG  FG d cos   mgd cos   18 kg  9.80 m s 2
 5.0 m  cos122  467.4 J 
470 J
(c) Because the normal force is perpendicular to the displacement, the work done by the normal
force is 0 .
(d) The net work done is the change in kinetic energy.
W  WP  Wg  WN  K  12 mv 2f  12 mvi2 
vf 
2W
m

2  636.0 J  467.4 J 
 4.3m s
18 kg 
75. Since the forces are constant, we may use Eq. 7-3 to calculate the work done.
W   F  F  d  1.50ˆi  0.80ˆj  0.70kˆ  N   0.70ˆi  1.20ˆj N    8.0ˆi  6.0ˆj  5.0kˆ  m 
net
1

2






  0.80ˆi  0.40ˆj  0.70kˆ N   8.0ˆi  6.0ˆj  5.0kˆ m    6.4  2.4  3.5 J  12.3J
84. The spring must be compressed a distance such that the work done by the spring is equal to the
change in kinetic energy of the car. The distance of compression can then be used to find the spring
constant. Note that the work done by the spring will be negative, since the force exerted by the
spring is in the opposite direction to the displacement of the spring.
Wspring  K 

 12 kx 2  0  12 mvi2  x  vi
F  ma   kx  m  5.0 g    kvi
m
k
m
k

 9.80 m s 
 5.0 g 
k  m
 1300 kg  25
 5.0  103 N m

2
v

 i 
 1m s  
90 km h  3.6 km h  



2
2
2
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
196