Work Done by Gravity and Tension in a Pulley System

[solarcat]

Homework Statement

Two blocks are attached to a rope across a pulley. Block A is on the ground and has a mass of 4 kg. Block B is 2 meters above the ground and has a mass of 12 kg. When the blocks are released, block B hits the floor and block A rises to 2 meters; each block has a final speed of 4.43 m/s. Find the work done by gravity and tension for each block.

Homework Equations

Total Work = Change in kinetic energy = Work done by gravity + work done by tension
Work done by gravity = mg(Hf - H0)

The Attempt at a Solution

Block A
KE 0 = 0
KEf = (1/2) (4 kg) ( 4.43 m/s)² = 39.2 J
Total work = KEf - KE0 = 39.2 J
Work done by gravity = 4 kg * 9.8 m/s² * 2m = 78.4 J
Total work = work done by gravity + work done by tension
39.2 = 78.4 + Wt
Wt = -39.2 J
Similarly, Wg for block B is -235.2 J and Wt = 352.9 J
But shouldn't the work done by tension be 0 because the work done by tension on B is equal to the work done by tension on A?

 

[Doc Al]

Total work = work done by gravity + work done by tension
39.2 = 78.4 + Wt
Wt = -39.2 J

Careful with signs. The work done by gravity on block A is negative, since gravity acts down while the block rises. Similarly, the work done by tension on that block must be positive.

But shouldn't the work done by tension be 0 because the work done by tension on B is equal to the work done by tension on A?

The net work done by tension on both masses will be zero, yes. But if you analyze each block separately, the tension does non-zero work on each.

 

[haruspex]

Work done by gravity = 4 kg * 9.8 m/s2 * 2m = 78.4 J

Be careful with signs.
First, decide whether positive is up or down, then use signs for acceleration and displacement accordingly.

 

[neilparker62]

You could solve:

$$M_bg-T=M_ba...1$$ and
$$ T-M_ag=M_aa...2$$

After which work done by tension on B is just -TΔx and on A +TΔx summing to zero as you expect.