A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N) as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J, respectively, how much work is done on the block by the force P between A and B?
A)2.7 J
B)1.0 J
C)2.2 J
D)1.6 J
E)3.2 J

Question

A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N) as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J, respectively, how much work is done on the block by the force P between A and B? 
A)2.7 J 
B)1.0 J 
C)2.2 J 
D)1.6 J 
E)3.2 J

Quizplus · Accepted Answer

The work done on the block by the force P is given by:
work done = force x distance moved in the direction of force
= P x AB cos(theta)
where theta is the angle between the force P and the displacement AB. Since the force P is horizontal and AB is also horizontal, we have theta = 0. Therefore, cos(theta) = 1.
So, work done = 5.4 N x 0.5 m x 1 = 2.7 J
But we need to subtract the work done by friction because it is opposite to the direction of motion. The work done by friction is given by:
work done by friction = force of friction x distance moved in the direction of force of friction
= f x AB
= 1.2 N x 0.5 m
= 0.6 J
Therefore, the net work done on the block by the force P between A and B is:
net work done = work done - work done by friction
= 2.7 J - 0.6 J
= 2.1 J, which is closest to choice C (2.2 J).

A Block Is Pushed Across a Rough Horizontal Surface from Point