Kinetic Energy and Work of a crate

In summary, the problem involves a crate hanging from a rope with a varying force applied to move it a distance of 3.56m to the side. The work done on the crate is equal to the work done by gravity and the tension in the rope, and the work done by the applied force is equal to the negative of the work done by gravity. The y displacement of the crate must be calculated using trigonometry, and the acceleration in the final position is zero. The magnitude of the applied force when the crate is in the final position is still unknown and cannot be calculated using the work formula.
  • #1
Kruz87
17
0

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...
 
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  • #2
Kruz87 said:
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.
 
  • #3
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
 
  • #4
Kruz87 said:
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.
 

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is calculated by multiplying the mass of the object by the square of its velocity, and is measured in joules (J).

2. How is work related to kinetic energy?

Work is the transfer of energy from one object to another. In the context of kinetic energy, work is done on an object when a force is applied to it, causing it to move and thus increasing its kinetic energy.

3. How is the kinetic energy of a crate calculated?

The kinetic energy of a crate is calculated by multiplying its mass (in kilograms) by the square of its velocity (in meters per second) and dividing by 2. The formula is KE = 1/2 * m * v^2.

4. Can the kinetic energy of a crate be negative?

No, kinetic energy cannot be negative. It is always a positive value, as it represents the amount of energy an object has due to its motion.

5. How does the height of a crate affect its kinetic energy?

The height of a crate does not directly affect its kinetic energy. However, if the crate is lifted to a higher position, it gains potential energy, which can then be converted into kinetic energy when the crate is released and falls to a lower position.

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