Homework #3 Answers
9.(c) Since the net force is now zero, Newton’s first law requires that the object will move in a straight
line at constant speed. A net force would be needed to bring the object to rest.
10. (d) By Newton’s third law, the force you exert on the box must be equal in magnitude to the
force the box exerts on you. The box accelerates forward because the force you exert on the
box is greater than other forces (such as friction) that are also exerted on the box.
11. (b) The maximum static friction force is 25 N. Since the applied force is less than this
maximum, the crate will not accelerate, Newton’s second law can be used to show that the
resulting friction force will be equal in magnitude but opposite in direction to the applied
force.
12. (b, d) The normal force between the skier and the snow is a contact force preventing the skier from
passing through the surface of the snow. The normal force requires contact with the surface and
an external net force toward the snow. The normal force does not depend upon the speed of the
skier. Any slope less than 90° will have a component of gravity that must be overcome by the
normal force.
13. (a) If the two forces pulled in the same direction, then the net force would be the maximum and
equal to the sum of the two individual forces, or 950 N. Since the forces are not parallel, the
net force will be less than this maximum.
Problems pg. 101
5.Find the average acceleration from Eq. 2–11c, and then find the force needed from Newton’s second
law. We assume the train is moving in the positive direction.
 2  02
 1 m/s 
0  (120 km/h) 
aavg 
  33.33 m/s
2( x  x0 )
 3.6 km/h 
 0  (33.33 m/s) 2 
 2  02
6
6
 maavg  m
 (3.6  105 kg) 
  1.333  10 N  1.3  10 N
2( x  x0 )
 2 (150 m) 
 0
Favg
The negative sign indicates the direction of the force, in the opposite direction to the initial
velocity.
We compare the magnitude of this force to the weight of the train.
Favg
mg

1.333 106 N
(3.6 105 kg)(9.80 m/s 2 )
 0.3886
Thus the force is 39% of the weight of the train.
By Newton’s third law, the train exerts the same magnitude of force on Superman that Superman
exerts on the train, but in the opposite direction. So the train exerts a force of 1.3  106 N in the
forward direction on Superman.
7. Find the average acceleration from Eq. 2–4. The average force on the car is found from Newton’s
second law.
  0 0  26.39 m/s
 1 m/s 

 3.299 m/s2
  26.39 m/s aavg 
t
8.0 s
 3.6 km/h 
  0 0  (95 km/h) 
Favg  maavg  (950 kg)(3.299 m/s2 )  3134 N  3100 N
The negative sign indicates the direction of the force, in the opposite direction to the initial
velocity.
9. The problem asks for the average force on the glove, which in a direct calculation would require
knowledge about the mass of the glove and the acceleration of the glove. But no information
about the glove is given. By Newton’s third law, the force exerted by the ball on the glove is
equal and opposite to the force exerted by the glove on the ball. So we calculate the average force
on the ball, and then take the opposite of that result to find the average force on the glove. The
average force on the ball is its mass times its average acceleration. Use Eq. 2–11c to find the
acceleration of the ball, with   0, 0  35.0 m/s, and x  x0  0.110 m. The initial velocity of the ball
is the positive direction.
aavg 
 2  02
2( x  x0 )

0  (35.0 m/s)2
 5568m/s2
2(0.110 m)
Favg  maavg  (0.140 kg)(5568 m/s2 )  7.80 102 N