Height of falling object vs. compression of spring

AI Thread Summary
When a mass is dropped onto a spring from a height H1, it compresses the spring a distance D. If the mass is dropped from a greater height H2, the gravitational potential energy increases, suggesting greater compression of the spring. While the force due to gravity remains constant, the energy perspective indicates that the spring must compress more to match the increased potential energy. Therefore, the spring will compress a greater distance when the mass is dropped from a higher elevation. The maximum compression point is not the same as the equilibrium point of the spring.
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1. Question
Given an idea spring with constant K, a mass of M is dropped onto the spring from a height of H1. The spring compresses D distance.

We take the mass off, and drop the mass again at a height of H2, where H2 is significantly greater than H1. This time, will the spring compress a greater distance or the same distance?

Homework Equations


F = - k x
and Conservation of energy:
U = mgh, Es = ½kx2, where Es is elastic potential of spring
U = Es

3. Attempt at solution

So if we think about it only in terms of F = -kx, then no matter how high we drop the mass, it should compress the same distance D, because F (gravity) remains the same.

But if we consider it from an energy perspective, then U (gravitational protential) would be greater if the height is greater. And thus the compression must also be greater to match this greater potential energy.

So which one is correct?
 
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The point of maximum compression is not the equilibrium point!
 
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