The work done in stretching a spring of natural length 25 $\mathrm{cm}$ and spring constant $50 \mathrm{Nm}^{-1}$ from $50 \mathrm{~cm}$ to $60 \mathrm{~cm}$ is

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Aicharya · Accepted Answer

$\Delta \mathrm{x}_{\mathrm{i}}=(50-25) \mathrm{cm}=25 \mathrm{~cm}$ $\Delta \mathrm{x}_{\mathrm{f}}=(60-25) \mathrm{cm}=35 \mathrm{~cm}$ So, work done $=\Delta U$ $\begin{aligned} & =\frac{1}{2} \mathrm{~K}\left[\Delta \mathrm{x}_{\mathrm{f}}^2-\Delta \mathrm{x}_{\mathrm{i}}^2ight] \ & =\frac{1}{2} 	imes 50\left[35^2-25^2ight] 	imes 10^{-4} \ & =1.5 \mathrm{~J}\end{aligned}$

The work done in stretching a spring of natural length 25 $\mathrm{cm}$ and spring constant $50…

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