To the free end of a spring hanging from a ri | vedclass

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(B) Case $1$: When the block is lowered slowly,it reaches equilibrium where the spring force balances the gravitational force. $k d = m g$,so $d = \fr

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$2 d$

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(B) Case $1$: When the block is lowered slowly,it reaches equilibrium where the spring force balances the gravitational force. $k d = m g$,so $d = \frac{m g}{k}$.Case $2$: When the block is allowed to fall suddenly from the unstretched position,the block undergoes simple harmonic motion. Let the maximum extension be $x$. By the principle of conservation of energy,the loss in gravitational potential energy equals the gain in elastic potential energy of the spring.$m g x = \frac{1}{2} k x^2$.Solving for $x$,we get $x = \frac{2 m g}{k}$.Since $d = \frac{m g}{k}$,we have $x = 2 d$.

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