Spring stretch - Do I use Force or Potential Energy

[Steve1971]

I have read posts about this but still don't have a good handle on it. I am confused about something that I know is simple. If a mass is attached to a spring, the spring will stretch according to Hooke,s law, correct? So won't the weight, (mg) balance out the spring force of -kx? So in other words, won't x=mg/k?

My confusion is when I look at the same problem using a potential energy balance. the balance of the change in spring potential energy of 1/2kx^2 with the change in potential energy of the hanging mass mgx will result in x=2mg/k.

So the energy equation shows twice as much stretch as the force equation. Can someone please clear this up for me?

 

[BvU]

Hi Steve,

Well observed !
The mass, when gently moving from position with spring unstretched to equilibrium, can do useful work (in theory e.g. run a clock or something).
Conversely, when lifting the weight from equilibrium to spring unstretched position, you have to add some energy (you have to do work).

 

[Steve1971]

maybe it's equilibrium vs. total stretch that is getting me. So will the equilibrium position be mg/k?

but prior to that equilibrium, the spring will stretch as far as 2mg/k as it is oscillating?

 

[nasu]

Yes, if don't have any external force to slowly lower the weight, it will oscillate. At equilibrium position you have kinetic energy as well.
What you found from energy conservation is the maximum displacement and not the equilibrium one.

 

[mfig]

I think you may have left out a term. If I am not mistaken, an energy balance should include the work done in stretching the spring.

$W = -\frac{1}{2}k{x_d}^2$

So the balance looks like:

$-\frac{1}{2}k{x_d}^2 = \frac{1}{2}k{x_d}^2 - mgx_d$

This can be solved to yield the same answer you got from a force balance.