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Also would a spring under tension greater than the force of gravity pulling the spring downward shrink in both directions until it has depleted it's stored energy while in free fall?

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Could the physical situation you are thinking of be described by hanging a spring which stretches under the action of gravity, then dropping it from this position? It seems like that is what you are asking about, but it's not obvious from the wording (esp. "under tension greater than the force of gravity"). – tmac May 18 '12 at 18:08
Sorry, this is more like a 241 question. The initial question situation is for a slinky hanging from a fixed anchor point which stretches under the action of gravity. The second question is for a spring which is anchored to a fixed point, stretched to but not beyond the elastic limit, under tension greater than the force of gravity. Once in that state anchors are release and the spring is in free fall towards the surface of the earth. – Kyle May 19 '12 at 13:08
up vote 4 down vote accepted

As long as the total weight does not exceed the elastic limit, then yes.

Why? For any given length element, the weight that it is supporting doubles, so the change in effective length of that segment also doubles as per Hooke's law.

The limit is when the most strained element passes it's elastic limit. That will be the topmost bit which is holding the weight of the whole thing.

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This is true unless the spring has a finite equilibrium length (which just gets added in). Also note that this seems to be an answer to the title question. In the body, a "spring under tension greater than the force of gravity" is clearly not in equilibrium, but the question wording is unclear to me. – tmac May 18 '12 at 17:19
Yeah. @Tmac's right about. The extension doubles. – dmckee May 18 '12 at 17:21

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