- #1
nothing123
- 97
- 0
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.
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.