# GPE and KE of two unequal masses over pulley

1. Sep 28, 2011

### ethex

1. The problem statement, all variables and given/known data
Two unequal masses are connected by a light cord passing over a smooth pulley. State the gravitational potential energy $\Delta$U(> or < or = 0) and the kinetic energy $\Delta$K(> or < or = 0) of the system after the masses are released from rest?

2. Relevant equations

3. The attempt at a solution
Both masses were originally at the same level. After released, both went in an opposite direction away from the equilibrium level. Thus, $\Delta$U=0?

Both gained speed, thus $\Delta$K>0?

2. Sep 28, 2011

### grzz

Note that potential evergy is a scalar and hence has not direction.

3. Sep 28, 2011

### ethex

Hi grzz, So there is a difference in height, thus there is a change in GPE, How do i determine if change in GPE is >0 or <0?

4. Sep 28, 2011

### grzz

To use grav p e = mgh one needs a zero reference level. To make things simple you can choose the lowest position reached as this level. A diagram always helps.

5. Sep 28, 2011

### ethex

I am confused. So which is the initial height and final since there are two masses?

6. Sep 28, 2011

### grzz

You have to assume the initial (equal) height, say h. Continue from there.

7. Sep 28, 2011

### ethex

For heavier mass, $\Delta$U = mg(0-h) = -mgh
For lighter mass, $\Delta$U = mg (2h- h) = mgh

Thus for the whole system = 0?

8. Sep 28, 2011

### grzz

What height are you taking as the reference for GPE = 0?

9. Sep 28, 2011

### ethex

Lowest point reached as you have suggested.

10. Sep 28, 2011

### grzz

Your only mistake is that the masses are not equal. Hence use m and M.

11. Sep 28, 2011

### grzz

Any way. The bigger one goes down,hence U decreased and the other goes ...

12. Sep 28, 2011

### ethex

That's right! I missed out the difference in masses.

For heavier mass, change in U = -Mgh
For lighter mass, change in U = mgh

Since M>m, total change in U < 0.

As for KE, as long as there is a movement, does it mean KE is affected?

So the ans to this question is change in u< 0 , change in K>0, right?

And is there any situation change in KE<0?

I am quite confused over all these.

Thank you so much grzz.

13. Sep 28, 2011

### grzz

initial KE = 0 since they were at rest.
Hence KE increased for both.

14. Sep 28, 2011

### ethex

I got it, thank you.

Would you mind helping me out another question relating to work energy?

A 3 kg ball swings rapidly in a complete vertical circle of radius 2 m by a light string. The ball moves so fast that the string is always taut. As the ball swings from its lowest point to its highest point,

the work done on it by gravity is ___ and the work done on it by the tension in the string is ___.

The first blank i found out to be -188J. How do it work on the second part?

15. Sep 28, 2011

### grzz

Can you show the working for -188J?

16. Sep 28, 2011

### ethex

Ok.

Change in GPE = mgh = 3 x 4 x 9.8 = 188J
Since work done is by the gravity on the system, therefore -188J.

17. Sep 28, 2011

### grzz

-118j

18. Sep 28, 2011

### grzz

Now can you state the definition of work?

19. Sep 28, 2011

### grzz

The CORRECT defn of work will give you the answer for the work by the tension.

20. Sep 29, 2011

### ethex

Work done is the product of force that acts on the object with the displacement along the action of force.

The top tension is pointing down while tension at bottom is pointing up with larger magnitude. So how do i cal the work done by the tension base on definition?