Q 11-05-008NEETNEET 2023Top questionEasy
The potential energy of a long spring when stretched by $2$ cm is U. If the spring is stretched by $8$ cm, potential energy stored in it will be :
Answer: (D) $16$U
$U = \dfrac{1}{2}kx^2$, so $U \propto x^2$.
$$U' = U\left(\frac{8}{2}\right)^2 = 16U$$
Solution by Sreeraj P, M.Sc Physics