Energy Conservation of a Vertical Spring

[Physicsboi123]

Homework Statement

A 9-kg stone is at rest on top of a spring, which causes it to compress 15cm.
What is the spring constant of the spring?

Relevant Equations

Ei=Ef, F=kx, W=mg, Ek=0.5kx^2

Using conservation of energy,

0.5kx^2=mgh=mgx

0.5kx=mg

0.5kx=mg, x=0.15, m=9, g= 9.8

So isn't it k= 1176N/m?

For this problem, I understand that you can't use conservation of energy, but why? There is gravitational potential energy at the top and spring elastic energy at the bottom, and no kinetic energy at both points since it is at rest on top of the spring?

Any help would be much appreciated!

Thanks!

 

[haruspex]

Homework Statement:: A 9-kg stone is at rest on top of a spring, which causes it to compress 15cm.
What is the spring constant of the spring?
Homework Equations:: Ei=Ef, F=kx, W=mg, Ek=0.5kx^2

you can't use conservation of energy, but why?

If you were to place the stone carefully on top of the relaxed spring and let go, what would happen?

 

[Physicsboi123]

If you were to place the stone carefully on top of the relaxed spring and let go, what would happen?

I suppose it would slowly descend to resting place and stop? I know you have to use Fnet=0 to calculate k, but wouldn't that just be where acceleration is 0, not velocity?

 

[haruspex]

I suppose it would slowly descend to resting place and stop?

That's not what springs do usually.

wouldn't that just be where acceleration is 0, not velocity?

Exactly! All the while the stone is above the equilibrium position the net force is downward, so the speed is increasing. When it reaches the equilibrium position it is at maximum speed, so a lot of KE.
In the question you were given, the stone is at rest in the equilibrium position, so all that energy which would have been KE at that position has somehow been dissipated.

 

[Physicsboi123]

That's not what springs do usually.

Exactly! All the while the stone is above the equilibrium position the net force is downward, so the speed is increasing. When it reaches the equilibrium position it is at maximum speed, so a lot of KE.
In the question you were given, the stone is at rest in the equilibrium position, so all that energy which would have been KE at that position has somehow been dissipated.

Oh, I understand, so the spring oscillates for a while before coming to rest.

I understand that KE is greatest at equilibrium and that energy, but the question didn't mention that the stone is at rest in equilibrium position, which is mg=kx. How would you deduce that v=0 when x=mg/k, an energy isn't conserved?

 

[haruspex]

the question didn't mention that the stone is at rest in equilibrium position

It says "A 9-kg stone is at rest on top of a spring". How can it be at rest if not at the equilibrium position?