Physics - W.E.P. Question with Solution | TestHub

The potential energy of a stretched spring is $$U=\frac{1}{2}k{x}^2$$ Here, $k=$ spring constant, $x=$ elongation in spring. But given that, the elongation is $2\mathrm{cm}$. So, $U=\frac{1}{2}k(2{)}^2$ $$\Rightarrow U=\frac{1}{2}k\times \mathrm{4...}(i)$$ If elongation is $10\mathrm{cm}$ then potential energy $$\begin{array}{rl}{U}^{\prime }& =\frac{1}{2}k(10{)}^2\\ U^{\prime }& =\frac{1}{2}k\times \mathrm{100...}(ii)\end{array}$$ On dividing Eq. (ii) by Eq. (i), we have $$\frac{U^{\prime }}{U}=\frac{\frac{1}{2}k\times 100}{\frac{1}{2}k\times 4}$$ or $\frac{U^{\prime }}{U}=25\Rightarrow U^{\prime }=25U$