Problem: As you can see from the diagram, the 3 kg block falls through a 1.5 m distance, causing the 4 kg block to move across the rough surface (kinetic coefficient = 0.38). The pulley is very light and frictionless. Find:

a) the work done by gravity
b) the work done by the tension on the 3 kg block
c) the work done by the tension on the 4 kg block
d) the work done by friction

Solution:

a) The work done by gravity is only on the 3 kg block, since the angle is 90^o for the 4 kg block:

b) Gotta find the tension first...

Adding the two equations that contain the tension, T is eliminated and I get a = 2.07 m/s². Substituting this back into either equation, T = 23.2 N. Now for the work done by the tension...

c) The work done by the tension on the 4 kg block is 23.2 N times 1.5 m = 34.8 J.

d) The work done by friction is 14.9 N times 1.5 times -1 (cos 180^o) = -22.4 J

The applied force is the gravitational force on the 3 kg block; its work is 44.1 J. Subtracting the work done by friction, 21.7 J remains to increase the speed (kinetic energy) of the system. Let's look at the net work:

This is another way to look at the energy acquired by the system. To summarize, the net work equals the energy acquired. We can either take the product of the sum of the forces on the system and the distance to get this, or compute it individually for each part of the system. Or we can find it by finding the work done by applied force and subtracting the work done by the frictional force.