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I had the question-

A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be?
a. $\frac{Mg}{k}$
b. $\frac{2Mg}{k}$
c. $\frac{Mg}{2k}$

I had assumed that initially there would be no force, and when released from rest, equilibrium would be reached when $Mg = kx$ where $x$ is the extension of the srping, hence $x$ = $\frac{Mg}{k}$

However, the answer is $x = \frac{2Mg}{k}$ and Conservation of Energy ($Mgx = \frac{1}{2}kx^2$) is used instead.

I would be grateful if someone could explain why my process is wrong.

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    $\begingroup$ In your analysis, you assusme that the block is at rest when it reaches the new equilibrium position. But it's not. It's still moving down. It has kinetic energy that it does not have if the block is lowered gently to the new equilibrium question. $\endgroup$
    – garyp
    Commented Aug 27, 2021 at 1:12
  • $\begingroup$ Thank you very much, I understand. If this was an answer I could have marked it as the right one, however since it is just a comment I cannot do so :/ $\endgroup$
    – m-Xylene
    Commented Aug 27, 2021 at 1:18
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    $\begingroup$ @Metaxylene We don't normally answer HW&E questions on this site. $\endgroup$
    – Gert
    Commented Aug 27, 2021 at 1:36
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    $\begingroup$ Also related: physics.stackexchange.com/q/278462/179151 $\endgroup$ Commented Aug 27, 2021 at 4:50

1 Answer 1

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The block oscillates around its equilibrium position. When the block is at its equilibrium position, it's either moving up or moving down, so it still has some kinetic energy, which is yet to be transformed to either elastic or potential energy once it reaches the max/min displacement.

Since the block starts from rest, you know the amplitude of its motion is $A=\dfrac{mg}k$, so the maximum displacement must be twice the amplitude, $\dfrac{2mg}k$.

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