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)`
Here, k = spring constant,
x = elongataion is spring. But given that, the elongation is 2 cm.
So, ` U = (1)/(2)k(2)^(2)`
`rArr U = (1)/(2)k xx4 " "...(i)`
If elongation is 10 cm , then potential energy
`U' = (1)/(2)k(10)^(2)`
`U' = (1)/(2)k xx 100 " "...(ii)`
On dividing Eq. (ii) by Eq. (i), we have
`(U')/(U) = ((1)/(2)kxx100)/((1)/(2)kxx4)`
or `(U')/(U) = 25 rArr U' = 25U`
SO, the correct option is (d).