When a body is suspended from a spring and re | vedclass

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(B) When the body is released slowly,the spring reaches equilibrium where the spring force equals the gravitational force: $kx = mg$,which gives $x =

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$2x$

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(B) When the body is released slowly,the spring reaches equilibrium where the spring force equals the gravitational force: $kx = mg$,which gives $x = \frac{mg}{k}$.When the body is released suddenly,the potential energy lost by the body is stored as elastic potential energy in the spring at the maximum extension $y$. By the law of conservation of energy: $mgy = \frac{1}{2}ky^2$.Solving for $y$: $y = \frac{2mg}{k}$.Since $x = \frac{mg}{k}$,we substitute this into the equation for $y$ to get $y = 2x$.

When a body is suspended from a spring and released slowly,the extension in the spring is $x$. If the body is released suddenly,the extension in the spring is:

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