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A spring hangs vertically in its relaxed state. A block of mass m is attached to the spring, but the block is held in place so that the spring at first does not stretch. Now the hand holding the block is slowly lowered, so that the block descends at a constant speed until it reaches the point at which it hangs at equilibrium with the hand removed.

On calculating the work done by the spring, the hand, and gravity separately, it is found that the work done by the hand and the spring is half that of the work done by gravity/ΔU each. The sum of all the work done is zero.

Now imagining a situation where the hand was not present I would assume that all the change in potential would be stored in the spring's new state (no friction, air resistance, etc).

What fundamentally changed due to the presence of the hand and why was only half the change in gravitational potential energy stores in the spring in this case?

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In order to lower the block slowly and so prevent it gaining a significant amount of kinetic energy, energy must be dissipated into the environment. If the block is lowered by hand, this energy is dissipated as heat in the muscles attached to the hand. If the block is slowed by a friction brake then energy is dissipated as heat in the friction brake etc.

It is this this energy "lost" to the environment that accounts for the difference in the energy of the block that is lowered slowly and the block that is allowed to fall freely, retarded only by the tension in the spring.

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A general answer was pointed out in the comments for questions regarding these restoring forces and why the factor of 2 comes into play in these cases here

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What changed is: the block wasn't slowly lowered, meaning it acquired a velocity while freely falling. What will happen now is it will oscillate up and down. In the first case elongation is $\frac{mg}k$ and in second case, the point of maximum elongation is $\frac{2mg}k$.

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