Q. A long spring, when stretched by $x$ cm has a potential energy $U$. On increasing the length of spring by stretching to $nx$ cm, the potential energy stored in the spring will be
A) $\frac{U}{n}$
B) $nU$
C) $n^{2}U$
D) $\frac{U}{n^{2}}$
Explanation
(c) Initially potential energy $= \frac{1}{2}kx^{2}$
$\Rightarrow \ \ U = \frac{1}{2}kx^{2}$
or $2U = kx^{2} \Rightarrow k = \frac{2U}{x^{2}}$
When it is stretched to $nx$ cm, then
$PE\ = \ \frac{1}{2}kx_{1}^{2} = \frac{1}{2} \times \frac{2U}{x^{2}} \times n^{2}x^{2} = n^{2}U$
$\therefore$ Potential energy stored in the spring $= n^{2}U$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.