The potential energy of a 1kg particle free to move along the x-axis is given by `V(x)=(x^(4)/4-x^(2)/2)J` The total mechanical energy of the particle is 2J then the maximum speed `(in m//s)` is

Options

A. `sqrt2`
B. `1/sqrt2`
C. `2`
D. `3/sqrt2`

Answer

`(4)`

Explanation

`F = - dU/dx = 0
x(x^(2) - 1) = 0 Rightarrow x = pm1`
`d^(2)U/dx^(2) = 3x^(2) - 1 gt 0`
`therefore U is minimum and U_(min) = 1/4-1/2 = -1/4`
`TE = KE _(max) + U_(min)`
`KE_(max = TE - U_(min) `
`1/2 mv_(max)^(2) = TE - U_(min)`