Calculating Work Done on Cart During Fall?

[TG3]

Homework Statement

A cart of mass 5 kg is attached to a block of mass 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and we will assume that friction can be ignored. The block falls a distance of 1 m.
What is the work done on the cart by the string during this fall?

Homework Equations

W= FD
W= Change in Kinetic Energy
K = mgh

The Attempt at a Solution

M=3
G=9.81
H=1
So 3 x 9.81 x 1= 29.43 J
But this answer is not correct. What am I doing wrong?

 

[tiny-tim]

A cart of mass 5 kg is attached to a block of mass 3 kg by a string …
What is the work done on the cart by the string during this fall?
…
M=3
G=9.81
H=1
So 3 x 9.81 x 1= 29.43 J

HI TG3!

You're saying that the work done on the cart is the change in KE of the cart, which is correct.

But you've used the mass of the block.

 

[TG3]

Erm... as opposed to what? I've tried the mass of the block, the mass of the cart, and the mass of both of them both put together x 9.81, and nothing works...

--Edit---
Never mind, I figured it out. g x ((m1xm2)/(m1+m2)).