How to Calculate Work, Kinetic Energy, and Tension in a Block-Cart-Pulley System

[Awwnutz]

http://img339.imageshack.us/img339/5208/blockcartpulleyyb1.gif

A cart of mass M1 = 3 kg is attached to a block of mass M2 = 4 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 0.8 m:

a) What is the work Wg done by gravity on the system?

b) What is the increase in kinetic energy DK of the cart-plus-block system?

c) What is the speed |v| of the cart-plus-weight system?

d) What is the work Ws done on the cart (not the block!) by the string?

e) What is the tension T in the string?


Relevant equations
Work Energy Theorem, Kinetic Energy: (1/2)mv^2, Potential Energy

The thing that's tripping me up is how to incorporate M1 with the work done by gravity on M2. This question confuses me because of the cart. I know that the work done by gravity for this conservative system is the negative change in potential energy, but what is the potential energy of M1? I'm just having trouble figuring out all these parts. Sorry for being so vague.

 

[skatj]

I don't think gravity does any work on M1, (mgh)final = (mgh)initial. No change in potential energy, no work done.

 

[Awwnutz]

Thats right, since the force is working in a perpendicular direction to M1 then it does no work on it.

I'm having a problem with c though, I found the work done by gravity on the system is 31.392J. So to find the speed i put it in an equation like this:

(1/2)mv(final)^2 - (1/2)mv(initial)^2 = Work done total

(1/2)(3kg)(v(final))^2 - 0 = 31.392

v(final) = 4.58m/s, but this is not the right answer. What did i do wrong?

 

[skatj]

Mass is of the entire system, which is 7kg, not 3kg.

 

[Awwnutz]

That's exactly it. This problem just combined too many ideas i guess i just made it harder than it really was. Thanks for your help!