# Work w/ Pulleys

1. May 30, 2014

### Coop

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

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin60 = 0.87; cos60 = 0.5. Ignore friction and the weights of the pulleys.)

A) 50 J
B) 100 J
C) 174 J
D) 200 J

2. Relevant equations

$$W=Fdcos(\theta)$$

3. The attempt at a solution

I get that it takes 200 J of energy to raise a 4-kg weight 5 m. But I don't see how the force F does 200 J of work when referring to the relevant equation. Since theta is the angle between the force F and velocity of object, why would you not say $$W=(4 kg)(10\frac{m}{s^2})(5 m)cos(150)$$? This would give 174 J. I guess what I am confused about is why is the theta said to be 0?

Thanks

2. May 30, 2014

### Nathanael

Well first, θ would be the angle between the F and d, which in your picture would be 30°

But you don't know what "F" is so how would you use that equation?

Think of it in terms of gravitational energy

(It said ignore friction, I'm also going to make the assumption that once it's at a height of 5 meters it has no velocity. Maybe someone can correct me but I think you have to make this assumption?)

EDIT:
Your question is basically just that you're confused because this seems to imply theta=0 (right?)

Well the answer is that it doesn't imply that because the Force is unknwon, therefore the equation doesn't apply (there's 2 unknowns in it)

3. May 30, 2014

### CAF123

Yes, if the force applied is parallel to the displacement of the mass. This is not the case in your problem.
If F and d are parallel, the angle between them is zero. You want to find the [STRIKE]force[/STRIKE] work required to raise the mass 5m from the ground. What component of the force is responsible for vertical displacement of the mass?

Last edited: May 30, 2014
4. May 30, 2014

### Coop

@Nathanael

Ah, I think I was just making a stupid mistake. I was letting F = the force of gravity on the weight. Thanks for pointing that out.

5. May 30, 2014

### Nathanael

The problem just said find the work not the force

6. May 30, 2014

### Coop

So only the parallel component is responsible for doing the work, correct?

7. May 30, 2014

### CAF123

Typo corrected.

Work required to raise the mass 5m from the ground means you are interested in the vertical component of the force. Can you write an explicit equation for this?