An ideal spring with spring-constant $K$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is
IIT 2002, Medium
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(b) Let $x$ be the maximum extension of the spring. From energy conservation
Loss in gravitational potential energy = Gain in potential energy of spring
$Mgx = \frac{1}{2}K{x^2}$
$ \Rightarrow x = \frac{{2Mg}}{K}$
Loss in gravitational potential energy = Gain in potential energy of spring
$Mgx = \frac{1}{2}K{x^2}$
$ \Rightarrow x = \frac{{2Mg}}{K}$
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