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

Explanation

For maximu valu of U , `(dU)/(dx) = 0`
`therefore (4x^(3))/(4) - (2x)/(2) = 0 or x = 0, x = pm1`
At `x =0 , (d^(2)U)/(dx^(2)) = -1`
and At `x = pm 1, (d^(2)U)/(dx^(2)) = 2`
Hence U is minimum at `x = pm 1` with value
`U_(min) = 1/4 - 1/2 = - 1/4 J`
`K_(max) + U_(min) = E or K_(max) - 1/4 = 2 or K_(max) = 9/4`
`Rightarrow 1/2 mv^(2) = 9/4 Rightarrow v_(max) = 3/sqrt2 ms^(-1)`