A mass $m$ is attached to a vertical spring of elastic constant $K$ and length $L$. The spring is supposed to be of negligible mass. Due to the attached mass $m$ the spring reaches a new equilibrium configuration at a length $L+x$.

According to Hooke's law the spring elongation is given by $x= \frac{mg}{K}$, since the weight has to be equilibrated by the elastic force.

On the other hand, by applying the principle of energy conservation, the initial gravitational potential energy is transformed in elastic potential energy say $mgh$ is transformed in $\frac{1}{2}Kx^2+mg(h-x)$ which leads to $mgx= \frac{1}{2}Kx^2$. The spring elongation is then $x= \frac{2mg}{K}$. It seems a contradiction.

What is wrong in these arguments?

  • $\begingroup$ or this one physics.stackexchange.com/q/551295/179151 $\endgroup$ May 3, 2021 at 17:54
  • $\begingroup$ or this one physics.stackexchange.com/questions/336702/… $\endgroup$ May 3, 2021 at 18:17
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    $\begingroup$ The error lies in the statement "the initial gravitational potential energy is transformed in elastic potential energy". It is transformed into elastic potential energy and kinetic energy as the mass overshoots and then oscillates around the equilibrium position. $\endgroup$ May 3, 2021 at 18:21
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    $\begingroup$ @Chemomechanics Or if the mass is brought to rest at equilibrium, work is being done to bring it to rest there. $\endgroup$ May 3, 2021 at 19:15
  • $\begingroup$ Thank you very much everybody for these explations and to the references to similar questions. Therefore this is a common problem. Actually, the matter, if I correctly understand, is that in the static derivation of Hooke' Law some extra work is done to produce equilibrium between the applied weight and the elastic force. Otherwise we get an energy contradiction. However this crucial point I believe is missing in many textbooks. Thank you again $\endgroup$ May 5, 2021 at 8:53