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Question

(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}

Vedclass · Accepted Answer

$\frac{1}{2} \frac{M^2 g^2 L}{A Y}$

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(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} 	imes 	ext{volume}$.We know that $	ext{stress} = \frac{F}{A} = \frac{Mg}{A}$ and $	ext{strain} = \frac{	ext{stress}}{Y} = \frac{Mg}{AY}$.The volume of the wire is $V = A 	imes L$.Substituting these values into the energy formula:$U = \frac{1}{2} 	imes \left( \frac{Mg}{A} ight) 	imes \left( \frac{Mg}{AY} ight) 	imes (AL)$.Simplifying the expression:$U = \frac{1}{2} 	imes \frac{M^2 g^2}{A^2 Y} 	imes AL = \frac{1}{2} \frac{M^2 g^2 L}{AY}$.

$A$ block of mass $M$ is suspended from a wire of length $L$,area of cross-section $A$ and Young's modulus $Y$. The elastic potential energy stored in the wire is

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