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An elastic spring of spring constant k hangs vertically with one end fixed to a support and the other end fixed to a block of mass m. The block is held such that the spring is at its natural length. If the block is released, the maximum extension of the spring is

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$\frac{2 m g}{k}$

$\frac{m g}{2 k}$

m g k

$\frac{m g}{k}$

answer is B.

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Detailed Solution

Let x be the maximum extension of the spring. The block loses potential energy and the spring gains potential energy, i.e.
$m g x = \frac{1}{2} k x^{2} \Rightarrow x = \frac{2 m g}{k}$

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