Kinetic Energy and Work of a crate

[Kruz87]

Homework Statement

Can someone just give me some kind of idea of where to start, I've really been stressing over this alot... A 303 kg crate hangs from the end of a rope of length L = 12.1 m. You push horizontally on the crate with a varying force to move it distance d = 3.56 m to the side (a) What is the magnitude of when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force does on the crate.




So I tried to out start by establishing some relationships...
(1) W= (Delta)KE=0 (b/c motionless before and after and therefore no work done)= W(gravity) +W(Tension in rope) + W(applied).

So I know that no work occurs in the tension of the rope b/c there's no motion along the axis of the rope.

Therefore, W(applied)= -W(gravity), RIGHT?


I also know that W(gravity)= -mg(dy), but there's no displacement given in the y- direction!

Finally, W(applied)= F(applied)*(dx)

And we know from the given variables that the displacement in the x-direction is 3.56m.

Our proffesor did a similar problem in which a crate was pulled up a hill, and used trigonometric identies to get angles from the right triangle, but that doesn't work in my case because a crate swinging doesn't form a right triangle. The initial height will be lower than final height...

I'm so lost right now, absolutely anything would help at this point...

 

[Doc Al]

So I tried to out start by establishing some relationships...
(1) W= (Delta)KE=0 (b/c motionless before and after and therefore no work done)= W(gravity) +W(Tension in rope) + W(applied).

So I know that no work occurs in the tension of the rope b/c there's no motion along the axis of the rope.

Therefore, W(applied)= -W(gravity), RIGHT?

So far, so good.

I also know that W(gravity)= -mg(dy), but there's no displacement given in the y- direction!

You'll need to figure out the crate's y displacement. It's attached to a rope--use a little trig.

 

[Kruz87]

Thanks, I got the crate's y displacement to be .686 meters and from there I was able to get the correct work of the force and gravity. However, I'm still having trouble finding the force magnitude of the F(applied) when the crate is in the last position.

I assume the last position,
I know F= m(a)= sum of the forces.
I want to say there are three forces acting on the crate when it is in the final position, F(applied), F(tension), and F(gravity). I believe gravity and tension cancel each other out, but that still leaves two unknowns because we don't know the acceleration of the body in the final position, or does it equal zero b/c its motionless. I'm not quite sure...
I can't use the relationship W=Fd b/c the force is a varying one

 

[Doc Al]

I want to say there are three forces acting on the crate when it is in the final position, F(applied), F(tension), and F(gravity).

Good.

I believe gravity and tension cancel each other out,

Why would you think that? In what directions do those forces act?

but that still leaves two unknowns because we don't know the acceleration of the body in the final position, or does it equal zero b/c its motionless.

The acceleration is zero. You may assume that at all points the applied force is just enough to balance the net force on the crate.