# Homework Help: Blocks on a Pulley System

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1. Oct 20, 2014

### minimario

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

2. Relevant equations
$W_{nc} = \Delta KE + \Delta PE$
$PE = mgh$
$KE = \frac{1}{2} mv^2$
3. The attempt at a solution
For Block B, $\Delta KE + \Delta PE = \frac{1}{2} 100 v^2 + (-20)(g)(100)$

For Block A, $\Delta KE + \Delta PE = \frac{1}{2} 50 v^2 + (20 \sin 37^{\circ})(g)(100)$

Therefore, we have the total change in energy is $75 v^2 - 13702$. This is the total work done by nonconservative forces.

The only nonconservative force is friction on the A block. The normal force on the A block is $50g \cos 37^{\circ}$, so the friction force is $(0.25)(50g \cos 37^{\circ})$ The work done by friction is then $- 20 \cdot (0.25)(50g \cos 37^{\circ}) = 1956.66$, so $75v^2-13702 = 1956.66 \Rightarrow v^2 = 208.77$, so the Kinetic Energy change is $\frac{1}{2} (50)(208.77) = 5219.25$

This is incorrect, can anyone find what's wrong?

2. Oct 20, 2014

### Staff: Mentor

Well, for one thing, in the PE of A, the mass is 50 kg, not 100 kg. Is this a typo, or did you really use 100?

Chet

3. Oct 20, 2014

### NTW

I suggest a simpler approach: determine the net force; from that and from the masses involved, find the acceleration, and you have solved a half of the problem...

4. Oct 20, 2014

### minimario

That was a typo.

5. Oct 20, 2014

### Staff: Mentor

There should be a sin 37 in the potential energy term of block B also.

Chet

6. Oct 20, 2014

### minimario

Then the P.E.s of A and B cancel out?

That doesn't give the right ans either... (do you get v^2 = 52.1776)

7. Oct 20, 2014

### Staff: Mentor

No, they don't cancel out. Don't forget your typo on the masses.

Chet

8. Oct 20, 2014

### minimario

So now v^2 = 104.73, is that right?

9. Oct 20, 2014

### Staff: Mentor

Shouldn't the friction decrease the kinetic energy?

Chet

10. Oct 20, 2014

### minimario

Yes, but the 100 kg block provides the energy and accelerates it.

11. Oct 20, 2014

### Staff: Mentor

If there were no friction, the velocity of the blocks would be higher.

Chet

12. Oct 20, 2014

### minimario

What do you mean? The 100 kg provides a force to counteract the friction...

13. Oct 20, 2014

### Staff: Mentor

What I mean is that you have the wrong sign on the friction term.

One way to be sure is to solve the problem using force balances rather than the energy balance. At the very least, you should check to see that they both give the same answer.

Chet