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=elongation in spring.
But given that, the elongation is 2 cm.
So, `U=(1)/(2)k(2)^(2)`
`impliesU=(1)/(2)kxx4" "......(i)`
If elongation is 10 cm then potential energy
`U'=(1)/(2)k(10)^(2)`
`U'=(1)/(2)kxx100" "......(ii)`
On dividing Eq. (ii) by Eq. (i), we have
`(U')/(U)=((1)/(2)kxx100)/((1)/(2)kxx4)`
or `(U')/(U)=25impliesU'=25U`