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 & explanation
Correct answer: option 1
$ U = \frac{1}{2} Kx^2$
$\text{ At x = 2 cm, } U = \frac{1}{2} K\times 2^2 = 2K$
$ \text{ At x = 8cm, } U'= \frac{1}{2} K 8^2 = 32 K = 16U$