A wire 2 ,m in length suspended vertically st | vedclass

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

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$0.5$

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(A) The elastic potential energy $(U)$ stored in a stretched wire is given by the formula:$U = \frac{1}{2} 	imes F 	imes \Delta L$Where $F$ is the applied force and $\Delta L$ is the extension.Given:Mass $m = 10 \,kg$Acceleration due to gravity $g = 10 \,m/s^2$Force $F = m 	imes g = 10 	imes 10 = 100 \,N$Extension $\Delta L = 10 \,mm = 10 	imes 10^{-3} \,m = 0.01 \,m$Substituting these values into the formula:$U = \frac{1}{2} 	imes 100 	imes 0.01$$U = 50 	imes 0.01$$U = 0.5 \,J$Therefore,the elastic potential energy gained by the wire is $0.5 \,J$.

$A$ wire $2 \,m$ in length suspended vertically stretches by $10 \,mm$ when a mass of $10 \,kg$ is attached to the lower end. The elastic potential energy gained by the wire is ...... $J$ (take $g=10 \,m/s^2$)

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