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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 rnass is released with the spring initially unstretched. Then the maximum extension in the spring is

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$\frac{4 Mg}{K}$

$\frac{2 Mg}{K}$

$\frac{Mg}{K}$

$\frac{Mg}{2 K}$

answer is B.

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Detailed Solution

Let x be the maximum extension of the spring. From energy conservation

Loss in gravitational potential energy = Gain in potential energy of spring

$Mgx = \frac{1}{2} {Kx}^{2}$

$\Rightarrow x = \frac{2 Mg}{K}$

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