A block is attached to a spring as shown | vedclass

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Question

In first case acceleration of block is zero always.

therefore $\mathrm{mg}=\mathrm{kx}$

$	herefore \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}=\m

Vedclass · Accepted Answer

$2\,d$

Vedclass · Answer

In first case acceleration of block is zero always.

therefore $\mathrm{mg}=\mathrm{kx}$

$	herefore \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}=\mathrm{d}(\mathrm{given})$

In second case block variably accelerate from rest
to rest

$0 =-\operatorname{mg} \mathrm{x}+\frac{1}{2} \mathrm{Kx}^{2}$

$	herefore \quad \mathrm{x} =\frac{2 \mathrm{mg}}{\mathrm{K}}=2 \mathrm{d}$

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

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