A mass $M$ is allowed to fall on a pedestal fixed on the top of a vertical spring. If the height of the mass was $H$ from the pedestal and the compression of the spring is $d$ then the spring's force factor is given by :

M g \frac{(H + d)}{d^{2}}

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2 M g \frac{(H - d)}{d^{2}}

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\frac{M g}{2} \frac{H}{d^{2}}

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2 M g \frac{(H + d)}{d^{2}}

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Solution

The correct option is D $2 M g \frac{(H + d)}{d^{2}}$
A loss in PE of the mass is the gain in PE of the spring.

$M g (H + d) = \frac{1}{2} k d^{2}$

$\Rightarrow k = 2 \frac{M g (H + d)}{d^{2}}$
Option D.

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