# Energy and WET (Incline pulley system)

1. Oct 17, 2008

### closer

A m1 = 45.0 kg block and a m2 = 105.0 kg block are connected by a string as in Figure P7.44. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 45.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 45.0 kg block as it moves from A to B, a distance of 20.0 m.

$$\Delta$$KE = Wnet
$$\Delta$$KE = Wg - Wfk
$$\Delta$$KE = mgh - (uk)(N)(x)
$$\Delta$$KE = mgh - (uk)(mgcos$$\theta$$)(x)

2. Oct 17, 2008

### krausr79

Maybe switch the trig and not trig around?

Gain in energy = potential gain: 20*sin(37)*9.8*45
loss in energy = friction work: (M1 Normal)*uk*20

3. Oct 17, 2008

### Staff: Mentor

If you are going to apply the WET to m1 separately, you must include the work done by all forces, including string tension.

Also: Express h in terms of x. Be careful of the sign of the work contributions.

4. Oct 17, 2008

### closer

(delta KE) = (m2)(g)(h) - (m1)(g)(h)(sin(theta)) - (m)(g)(cos(theta))(uk)(x)

I added in the potential energy for the second block. Still incorrect. I don't understand why I would need tension, nor would I know how to incorporate it into the formula.

Edit: Isn't (m2)(g)(h) the string tension?

5. Oct 17, 2008

### Staff: Mentor

If you are attempting to apply the WET to mass 1, then you need to include the work done by all forces on mass 1, including the string tension. Of course, this is the hard way to do this problem.

No.

I recommend that you use energy conservation on the system as a whole:
Ei + Wf = Ef

Where E is the total mechanical energy of both masses.

6. Oct 17, 2008

### krausr79

Oops! Gain in energy = potential gain + kinetic gain; you would need to know initial and final kinetic for mine to work, I think.