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The potential energy of a long spring when stretched by $2\, cm$ is $U.$ If the spring is stretched by $8\, cm$ the potential energy stored in it is
$\frac{U}{4}$
$4U$
$8U$
$16U$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,Potential\,energy\,of\,a\,\,{\rm{spring}}\\
{\rm{ = }}\frac{1}{2} \times \,force{\rm{ constant}}\, \times {\left( {extension} \right)^2}\\
\therefore \,Potential\,energy \propto \,{\left( {extension} \right)^2}.\\
or,\,\frac{{{U_1}}}{{{U_2}}} = {\left( {\frac{{{x_1}}}{{{x_2}}}} \right)^2}\,\,\,\,or,\,\,\frac{{{U_1}}}{{{U_2}}} = {\left( {\frac{2}{8}} \right)^2}\\
or,\,\frac{{{U_1}}}{{{U_2}}} = \frac{1}{{16}}\,\,or,\,{U_2} = 16{U_1} = 16U.\,\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{U_1} = U} \right)
\end{array}$
