A long elastic spring is stretched by `2 cm` and its potential energy is `U`. If the spring is stretched by `10 cm`, the `PE` will be

Options

A. U/5
B. U/25
C. 5 U
D. 25 U

Answer

d

Explanation

The potential energy of a stretched spring is
`U= (1)/(2) kx^(2)`
For x = 2, `U = (1)/(2) k xx 4`
For x = 10 cm, potential energy ,
`U' = (1)/(2) k xx 100`
On dividing Eq. (ii) by Eq. (i), we have
`(U')/(U) = ((1)/(2) k xx 100)/((1)/(2) k xx 4)`
or `(U')/(U) = 25 implies U' = 25 U`