The potential energy of a long spring when stretched by $2 \mathrm{~cm}$ is $U$. If the spring is stretched…

The potential energy of a long spring when stretched by $2 \mathrm{~cm}$ is $U$. If the spring is stretched by $8 \mathrm{~cm}$ the potential energy stored in it is:
  1. $U / 4$
  2. $4 U$
  3. $8 U$
  4. $16 U$

Solution

Potential energy of a spring
$\begin{aligned}
& \begin{aligned}
&= \frac{1}{2} \times \text { force constant } \\
& \times(\text { extension })^2 \\
& \Rightarrow \text { Potential energy } \propto(\text { extension })^2
\end{aligned} \\
& \text {Hence } \frac{U_1}{V_2}=\left(\frac{x_1}{x_2}\right)^2 \\
& \Rightarrow \frac{V_1}{V_2}=\left(\frac{2}{8}\right)^2 \\
& \Rightarrow \frac{U_1}{U_2}- \frac{1}{16}\left(\because U_1=V\right)
\end{aligned}$

Asked in: NEET 2006

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