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One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If S be gain in spring energy and G be loss in gravitational potential energy in the process, then

This was the problem I was trying to solve. I first did it without the law of energy conservation. For equilibrium position we get, $mg = kx$ (gravitational force balances the spring force at equilibrium)

Putting the value of x in $1/2kx^2$... (where x is the deformation in the spring at equilibrium) I end up with the relation G = 2S (the right answer).

How do I go about applying energy conservation in this situation?

$mgx = 1/2kx^2$ (K.E. not considered as the block is lowered slowly)

This gives S=G as the answer. I know I'm making some error while applying energy conservation.

Is there some non-conservative force acting on the block? In that case we can't apply energy conservation, right? But I can only see the spring force and the gravitational force (and also the force applied by the external agent to slowly lower it). Should I also be considering the force applied by the external agent? If so, how do I go about this?

This problem may be silly but it has been troubling me for quite a while. Any help will be appreciated.

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OK I wrote the wrong answer previously, here is a corrected version.

The main observation is that conservation of energy doesn't apply. At least not in the way you are writing it.

Why? Because if you cut the support from the initial condition without any loss of energy, the mass will oscillate about the new equilibrium position. For conservation of energy to apply, you need to take kinetic energy into account, or else consider the bottom point of spring travel, when the mass velocity is zero (twice the $x$ you are considering).

The only way to get the mass into the equilibrium position you are considering is to somehow dissipate half the energy.

Sorry for my previous error and wrong comments. You were correct about the factor of two in the energy accounting.

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  • $\begingroup$ At the point of equilibrium the spring force must balance the gravitational force. That is what I did. From the relation I get the value of x for which there's equilibrium. So the first answer should be right. I'm sorry if this isn't what you're trying to say. $\endgroup$
    – entropy
    Commented Sep 3 at 13:15
  • $\begingroup$ At least that is what my book says. I didn't quite get why my first answer is incorrect as per you. $\endgroup$
    – entropy
    Commented Sep 3 at 13:21
  • $\begingroup$ Ok I understand your confusion now. Need to temporarily delete my answer and re-write. $\endgroup$
    – Mariano G
    Commented Sep 3 at 14:47
  • $\begingroup$ Thank you for the answer. It helped a lot. $\endgroup$
    – entropy
    Commented Sep 3 at 16:26

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