A particle, which is constrained to move along x-axis, is subjected to a force in the some direction which varies with the distance x of the particle from the origin an `F (x) =-kx + ax^(3)`. Here, k and a are positive constants. For `x(ge0,` the functional form of the potential energy (u) U of the U (x) the particle is.
(a) , (b) , (c) , (d) .

Explanation

`F=-(dU)/(dx)`
:. `dU=-F.dx` or `U(x)=-int_(0)^x(-kx + ax^(3))dx`
`U(x)=-(kx_(2))/(2)-(ax^(4))/4`
`U(x)=0` at `x=0` and `x=sqrt((2k)/a)`
`U (x) =0` at `x=0` and `x=sqrt((2k)/a)`
Further, `F=0` at `x=0`. Therefore slope of `U-x` graph should be zero at `x=0`.
Hence, the correct answer is (d).