# One more: three block, a table, and a pulley

1. Dec 19, 2006

### dkgojackets

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

There is a table with a 6.9 kg box and a 5 kg box on it, connected by a string. The string goes around a pulley on the corner of the table, between the 5 kg box and a 35 kg box which it holds up. The coefficient of friction is .55 b/w the table and 6.9 kg box and .33 b/w table and 5 kg box. After releasing the system from rest, the 35 kg box descends .2 m. I need to find the speed of the 35 kg mass and the time it takes to drop .2 m.

2. Relevant equations

more work-energy

3. The attempt at a solution

I solved the first part of the question, which asked for the work done against friction. The answer is 10.6722 J. In my attempt at speed, I took the potential energy of the mass (35 x 9.8 x .2), subtracted the work done against friction, and set the answer equal to the kinetic energy at the end, but the resulting velocity was wrong. I'm sure I can just use kinematics to find time once I get the final velocity?

2. Dec 19, 2006

### dkgojackets

I tried the problem using forces to get the acceleration and then kinematics to find velocity and got the same wrong answer. What am I missing?

3. Dec 19, 2006

### Saketh

You have this expression:

$$\Delta KE + \Delta PE = W^{\mathrm{NC}}$$

We can rewrite this as (as you have, intuitively):

$$KE_{\mathrm{hanging block}} + KE_{\mathrm{sliding blocks}} = PE_{\mathrm{hanging block}} - W_{\mathrm{friction}}$$

I suspect that in your calculations you have forgotten to take into account the kinetic energy of the blocks still on the table. However, this would be more straightforward to solve with forces.

Last edited: Dec 19, 2006
4. Dec 19, 2006

### dkgojackets

That would be it. Thank you. I assume standard kinematics can be used to find the time?

5. Dec 19, 2006

### Saketh

By "standard," if you mean constant acceleration, then yes.