Solve Work w/ Pulleys Homework: Answer & Explanation

[Coop]

Homework Statement

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

http://www.bestsamplequestions.com/mcat-sample-questions/mcat-sample-questions-physical-sciences/images/mcat-sample-questions-physical-science-ques-57.gif

Homework Equations

W=Fdcos(\theta)

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

 

[Nathanael]

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?

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:
Rereading it I see that you said you understand the answer:

I get that it takes 200 J of energy to raise a 4-kg weight 5 m

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)

 

[CAF123]

I get that it takes 200 J of energy to raise a 4-kg weight 5 m.

Yes, if the force applied is parallel to the displacement of the mass. This is not the case in your problem.

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?

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?

 

[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.

 

[Nathanael]

You want to find the force required to raise the mass 5m from the ground.

The problem just said find the work not the force

 

[Coop]

If F and d are parallel, the angle between them is zero. You want to find the force required to raise the mass 5m from the ground. What component of the force is responsible for vertical displacement of the mass?

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

 

[CAF123]

The problem just said find the work not the force

Typo corrected.

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

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?