The PE of a 2 kg particle, free to move along x-axis is given by `V(x)=((x^(3))/(3)-(x^(2))/(2))J.` The total mechanical energy of the particle is 4 J. Maximum speed (in `ms^(-1)`) is

Options

A. `1/sqrt2`
B. `sqrt2`
C. `3/sqrt2`
D. `5/sqrt6`

Answer

`(4)`

Explanation

`F = -(dv)/(dx) = 0`
`x^(2) - x = 0 Rightarrow x(x-1) = 0 Rightarrow x = 1`
At` x = 1, ((d^(2)v)/(dx^(2)))` = + ve i.e., PE is min.
`V_(min) = 1/3-1/2 = -1/6`
`TE = KE _(max) + V_(min) `
`1/2mV_(max)^(2) = TE -V_(min)`
`V_(max) = 5/sqrt6`