Exploring Energy of a Vertical Spring System

[nothing123]

Hi,

If we have a vertical spring with a mass attached to it, how does the energy of the system work. Let's say we set the zero point as the point where the spring is unstretched (basically with no mass attached). Then at the top we have zero kinetic, zero spring potential as well as zero gravitational potential. As it passes through its new equilibrium point, it will have negative gravitational and positive elastic potential and kinetic energy so 1/2kx^2 + 1/2mv^2 - mgh which rearranged is mgh - 1/2kx^2 = 1/2mv^2. Now let's say we take the zero point to be where the new equilibrium point is. At the top, we have gravitational potential as well as spring potential. As it passes through the equilibrium point, it would only have kinetic. By conservation of energy, mgh + 1/2kx^2 = 1/2mv^2. Why are the two equations different if we change the reference point?

Thanks.

 

[Kurdt]

Does this applet help?

http://webphysics.davidson.edu/physlet_resources/gustavus_physlets/VerticalSpring.html

 

[tiny-tim]

zero point

Hi nothing123!

… Now let's say we take the zero point to be where the new equilibrium point is …

Your equations are completely confusing me (especially since h is related to x).

But if by "zero point", you mean for calculating the elastic restoring force, it has to be the "old" equilibrium point … that's how springs work!

(unless, I suppose, you take g out of the equation completely )