# Homework Help: Block, Cart, and Pulley Question

1. Oct 7, 2008

### r34racer01

A cart of mass M1 = 2 kg is attached to a block of mass M2 = 3 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 = 0.7 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?

I'm really Lost. I can't even figure out what I did wrong when solving for Wg. I did
Wg = mgh = (5kg)(9.81)(0.7) = 34.335 and that wasn't right.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 7, 2008
2. Oct 7, 2008

### tiny-tim

Welcome to PF!

Hi r34racer01! Welcome to PF!

Hint: work done by gravity = change in PE …

so how much mass is having its PE changed?

3. Oct 7, 2008

### r34racer01

Re: Welcome to PF!

Is it m2 or both masses?

4. Oct 7, 2008

### tiny-tim

Gravitational PE (potential energy) is caused by change in height.

So how much mass is changing height?

5. Oct 7, 2008

### r34racer01

Oh so it is m2, I forgot also that forces perpendicular to the object perform 0 work, so gravity does no work on m1 right?

6. Oct 7, 2008

### tiny-tim

Bingo!

7. Oct 12, 2008

### r34racer01

Ok so I got Wg =20.601, But now I'm a little confused with finding the change in KE.

8. Oct 12, 2008

### tiny-tim

work done by gravity = change in PE = change in KE

9. Oct 12, 2008

### r34racer01

Wow I didn't even realize that. So Wg = 20.601. And with that I was able to figure out that |v|=2.87.

So now I'm trying to figure out the work done by the string. We know that the speed is 2.87, so the KE of the cart is 8.32J. I thought I could do 20.601-8.32 to get Ws, but apparantly that's wrong. So what should I be doing?

10. Oct 13, 2008

### tiny-tim

HI r34racer01!

You didn't realize that? But it's so important

work and energy are the same ("work-energy theorem") …

KE and PE are the same ("conservation of energy") …

see how neatly it all fits together?

It means that if you want to calculate the KE, say, it's sometimes easier to calculate either the PE or the work done, and use that instead!
i] it's 8.24, not 8.32

ii] why do you want the work done … the question doesn't ask for it?

but it's equal to the KE anyway, which you now have.

Then you can find the tension by "reverse-engineering" from the work done.

11. Oct 13, 2008

### r34racer01

Wow thx, but how is 8.24 and not 8.32?

12. Oct 13, 2008

### aimslin22

part e.

combine the two masses to find the acceleration using F=ma because acceleration is uniform through the system.

Then use one of the masses and F=ma to determine tension.

This equation would be set up like this:

Winner force - loser force = ma

13. Oct 14, 2008

### tiny-tim

2.872 = 8.24

how did you get 8.32?