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 $1\ mm$. Then the elastic energy stored in the wire ......... $J$
Medium
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Elastic energy $=\frac{1}{2} \times F \times x$
$F=200 N, x=1 m m=10^{-3} m$
$\therefore E=\frac{1}{2} \times 200 \times 1 \times 10^{-3}=0.1 J$
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