A particle, which is constrained to move along the x-axis, is subjected to a force from the origin as `F(x) = -kx + ax^(3)`. Here `k` and a are origin as `F(x) = -kx + ax^(3)`. Here `k` and a are positive constants. For `x=0`, the functional form of the potential energy `U(x)` of particle is.

Options

A.
B.
C.
D.

Answer

D

Explanation

`F(x)=-kx +ax^(3) ,F_((x)) =0,` for `x = 0 x = sqrt((k)/(a)),`
So slope is zero at `x = 0 x = sqrt((k)/(a))`.