# Homework Help: Help in Homework - Kinetic Energy

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1. May 2, 2017

### Yuval M

1. The problem statement, all variables and given/known data
Problem: http://imgur.com/a/Sw2zA

Hi all,
I was given this problem as homework and I have almost no clue on how to solve it. I have tried for some time with no luck.

2. Relevant equations
We learned about kinetic and potential energy as well as work.
W = F*d
W = delta KE
PE = m*g*h

3. The attempt at a solution
I tried using the second work formula I have written, yet in the problem the ring stars with Kinetic energy of 0 and ends with Kinetic energy of 0, yet the ring still traveled some distance. This is another part I don't understand. If the ring traveled a distance, how come the work on the ring is 0?

Last edited by a moderator: May 2, 2017
2. May 2, 2017

### haruspex

What will/might happen at first? Did you draw a diagram to represent that? What forces act on the ring and in what directions?

W is the net work done on the ring. If you are considering the interval from starting to stopping there will be net positive work done on the ring initially to get it moving, then net negative work to bring it to rest.

3. May 3, 2017

### Yuval M

First, the ring will fall down a bit (a distance I need to find), because the Force of Friction and the Force of the spring x cos@ (where @ is the angle between the pole and the spring) will be bigger than mg.
The ring should later go up, to above its starting point (I think)

I drew the Diagram, here's my best representation of it:
N acts Right.
mg acts Down
Kinetic Fiction acts Up
Spring Force acts Left+Up

4. May 3, 2017

### haruspex

Not necessarily. What forces act on it initially?
Again, not necessarily.
That really would be surprising. What would that say about the net change in energy?

5. May 3, 2017

### Yuval M

The forces initially acting on the ring are:
mg DOWN
friction UP
normal RIGHT
spring LEFT

If the ring does not move, that means that Fd = mg.
I already found that the normal force acting on the ring is equal to F * sinϴ, where ϴ is the angle between the pole and the spring.
So, Fd = u*N = u * F * sinϴ.

I'm having a hard time understand why the ring does not necessarily fall down the pole.

That would say the net change would be negative, meaning it's not possible.

6. May 3, 2017

### haruspex

Where Fd is frictional force and F is tension, I assume. What is the tension in terms of k, l and θ?
No, it would mean there is a net gain in energy, which is not possible.