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(B) Case $1$: When the block is lowered very gradually,the acceleration is zero at all times. The spring force balances the weight of the block at equ

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$2d$

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(B) Case $1$: When the block is lowered very gradually,the acceleration is zero at all times. The spring force balances the weight of the block at equilibrium.$mg = kd$$	herefore d = \frac{mg}{k}$Case $2$: When the block is released suddenly from the natural length of the spring,the block starts from rest and reaches its maximum expansion $x_{max}$ where its velocity again becomes zero.Applying the Work-Energy Theorem between the initial position (natural length) and the final position (maximum extension):$W_{gravity} + W_{spring} = \Delta K$$mgx_{max} - \frac{1}{2}kx_{max}^2 = 0 - 0$$mgx_{max} = \frac{1}{2}kx_{max}^2$$x_{max} = \frac{2mg}{k}$Since $d = \frac{mg}{k}$,we have $x_{max} = 2d$.

$A$ block is attached to a spring as shown and very gradually lowered so that finally the spring expands by $d$. If the same block is attached to the spring and released suddenly,then the maximum expansion in the spring will be:

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