When a $4\, kg$ mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by $2\, cms$. The work required to be done by an external agent in stretching this spring by $5\, cms$ will be ......... $joule$ $(g = 9.8\,metres/se{c^2})$
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(b) $K = \frac{F}{x}$$ = \frac{{40}}{{2 \times {{10}^{ - 2}}}} = 0.2\;N/m$
Work done$ = \frac{1}{2}K{x^2} = \frac{1}{2} \times (0.2) \times {(0.05)^2} = 2.5\;J$
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