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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 positive constants. For `x=0`, the functional form of the potential energy `U(x)` of particle is.

A

(a)

B

(b)

C

(c)

D

(d)

Text Solution

Verified by Experts

The correct Answer is:
D
`dU_((x))=-Fdx`
`:. U_x=underset0oversetx intFdx=(kx^2)/(2)-(ax^4)/(4)`
`U=0` at `x=0` and at `x=sqrt((2k)/(a))`
`U(x)` is negative for `xgtsqrt((2k)/(a))`
`F=0` at `x=0`. It means slope of `U-x` graph is zero at `x=0`.
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