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The question is:

The lower end of a spring of spring constant $k$ is fixed on the ground and its other end passes through a light pulley and is connected to a block of mass $m$ after passing through the pulley. The system is released from rest. Find the maximum elongation of the spring. Initially the spring is unstretched.

I solved it using two different methods and I got different answers.

  1. Let the maximum elongation of the spring be $x$. Then, the block also descends by a distance x. So, increase in PE of the spring = decrease in PE of the block. So, $1/2kx^2=mgx$. This gives $x=2mg/k$.

  2. At the maximum elongation of the spring, the system is in equilibrium. Let the module of the tension in the string be $T$. Then, since the block is at rest, so $T=mg$ when elongation is maximum. Also, the forces on the spring at equilibrium. are $kx$( due to elongation $x$) and $T$. So, $T=kx$, at equilibrium. So, $mg=kx$ or $x=mg/k$.

So, which of these two methods is wrong and why? Please answer.

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closed as off-topic by Kyle Kanos, Jon Custer, Emilio Pisanty, Sebastian Riese, John Rennie May 28 '18 at 14:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

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Your mistake is saying

At the maximum elongation of the spring, the system is in equilibrium.

At the maximum elongation, the velocity of the mass is zero, but its acceleration is not zero. If you hold the mass at that position and then release it, it will move upwards. The tension in the string at the maximum elongation is not $T = mg$.

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  • $\begingroup$ I still don't understand why the block won't achieve equilibrium. When the block is released from rest, the only force on it is mg. But as it descends, the spring exerts a forces -kx on the string, and in turn, the string exerts the same force on the block. Then, why won't the force on the block become zero when the elongation is x, such that mg=kx? $\endgroup$ – Ribosome Dec 13 '16 at 5:53
  • $\begingroup$ @Ribosome : When the net force on the block becomes zero ($mg=kx$) the block continues moving down because of inertia - no force is required to keep it moving. The net force on it is now upwards, it slows down and stops instantaneously before it moves back up again, and continues oscillating. $\endgroup$ – sammy gerbil Dec 13 '16 at 6:07
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    $\begingroup$ Note that you can use > to get a block quote, rather than using a code block. $\endgroup$ – Kyle Kanos Dec 13 '16 at 11:09

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