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A block of mass 'm' is attached to one end of a string and a floor-fixed spring with spring constant 'k' is attached to the other. The string goes over a frictionless pulley. Initially, the spring is unstretched when the system is released from rest. Find the maximum elongation of the spring.

I tried to solve the problem with two different approaches:

Conservation of energy: Let's assume, initially, the block has potential energy which is equal to $mgx$ and the spring has no potential energy as it is not elongated, then the system is released and the block reaches 'x' unit distance down so that the spring stretches to 'x' unit. As a result, spring gains P.E. = $\frac{1}{2} kx^2$ According to conservation of energy this follows $\frac{1}{2} kx^2$ = $mgx$ therefore, elongation is $x = \frac{2mg}{k}$ so far so good.

Newton's Second Law: First the block accelerates and then stops when it has reached $x$ unit down we conclude that the gravity on the box equals the tension produced by the spring in the string so that $kx = mg$ which gives us $x = \frac{mg}{k}$

Huh? Why is that? Two distinct approaches provide us with two different answers!

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  • $\begingroup$ You assume that mg=kx at maximum elongation. This is not so. Wrong assumption produces wrong result. $\endgroup$
    – nasu
    Commented May 23, 2023 at 10:37
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    $\begingroup$ @Amit We are told the spring is initially unstretched. I think this means that the spring starts stretching as soon as the block is released, so the displacement of the block and the extension of the spring are indeed equal. $\endgroup$
    – gandalf61
    Commented May 23, 2023 at 10:54
  • $\begingroup$ Thanks @gandalf61 -- I've deleted my comment a few seconds before you replied because I realized I didn't imagine this correctly, this is the kind of cases where a diagram would have helped :) $\endgroup$
    – Amit
    Commented May 23, 2023 at 10:56
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    $\begingroup$ List of dozens of discussions of the same factor-of-two issue, which also arises when charging a capacitor with a constant voltage, pushing down a buoyant object with a constant force, and dropping a mass on a moving surface, for example. $\endgroup$
    – Chemomechanics
    Commented May 23, 2023 at 14:49
  • $\begingroup$ Nice reference @Chemomechanics, this one is almost identical $\endgroup$
    – Amit
    Commented May 23, 2023 at 14:55

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Your first method gives the correct result, although it is probably simpler to argue that the block loses PE $mgx$ during the motion, the spring gains PE $\frac 1 2 kx^2$, and since the KE of the system is zero both initially and at maximum displacement, we have

$mgx = \frac 1 2 kx^2$

Your second method is incorrect because the point at which the weight of the block equals the force from the spring is the point at which the block stops accelerating and starts to decelerate. As you say, the displacement of the block is $\frac {mg} k$ at this point but the block is still moving downwards, so this is not the maximum displacement.

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