Help in Homework - Kinetic Energy

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SUMMARY

The discussion revolves around a homework problem involving kinetic energy, work, and forces acting on a ring. The key equations referenced include W = F*d, W = delta KE, and PE = m*g*h. Participants clarify that while the ring starts and ends with zero kinetic energy, it can still travel a distance due to the net work done on it, which includes both positive and negative contributions. The forces acting on the ring are identified as gravitational force (mg), friction, normal force, and spring force, leading to a deeper understanding of energy changes in the system.

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Yuval M
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Homework Statement


Problem: http://imgur.com/a/Sw2zA
rkmVMNj.jpg

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.

Homework Equations


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

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

Yuval M said:
If the ring traveled a distance, how come the work on the ring is 0?
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.
 
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
 
Yuval M said:
First, the ring will fall down a bit
Not necessarily. What forces act on it initially?
Yuval M said:
The ring should later go up
Again, not necessarily.
Yuval M said:
to above its starting point
That really would be surprising. What would that say about the net change in energy?
 
haruspex said:
Not necessarily. What forces act on it initially?
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.
haruspex said:
That really would be surprising. What would that say about the net change in energy?
That would say the net change would be negative, meaning it's not possible.
 
Yuval M said:
Fd = u*N = u * F * sinϴ.
Where Fd is frictional force and F is tension, I assume. What is the tension in terms of k, l and θ?
Yuval M said:
That would say the net change would be negative, meaning it's not possible.
No, it would mean there is a net gain in energy, which is not possible.
 

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