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

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$0.1$

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(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula:$U = \frac{1}{2} 	imes F 	imes x$Where $F$ is the applied force and $x$ is the extension produced.Given:Force $F = 200 \ N$Extension $x = 1 \ mm = 10^{-3} \ m$Substituting these values into the formula:$U = \frac{1}{2} 	imes 200 	imes 10^{-3}$$U = 100 	imes 10^{-3}$$U = 0.1 \ J$Therefore,the elastic energy stored in the wire is $0.1 \ J$.

$A$ wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \ N$ to the lower end. The weight stretches the wire by $1 \ mm$. The elastic energy stored in the wire is ......... $J$.

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