The elastic potential enegry of a stretched spring is given by `E = 50 x^(2)`. Where `x` is the displacement in meter and and `E` is in joule, then the force constant of the spring is

Options

A. 50 Nm
B. `100 N m^(-1)`
C. `100 N//m^(2)`
D. 100 Nm

Answer

B

Explanation

`U = (1)/(2) Kx^(2) "___"(1)`, `U = 50 x^(2) "_____"(2)`, compare equation (1) and (2) to find `K`.