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

Answer

`1.5`

Explanation

Total energy `E_(T) = 2J` It is fixed. For maximum speed, kinetic energy is maximum The potential energy should, therefore, be minimum.
`therefore V(x) = x^(4)/4 -x^(2)/2`
or `(dV)/(dx) = (4x^(3))/4 - (2x)/2 = x(x^(2)-1)`
For V to be minimum, `(dV)/(dx) =0`
`therefore x(x^(2)-1) = 0`, or `x =0, +-1`
at x=0, `V(x) = 0`
At `x = +- 1, V(x) = -1/4 J`
`therefore ("Kinetic energy")_("max") = E_(T) -V_("min")`
or `1/2 mv_(m)^(2) = 9/4`
`v_(m)^(2) = (9 xx 2)/(m xx 4)`
or `v_(m)^(2) = (9 xx 2)/(m xx 4) = (9 xx 2)/(2 xx 4)`
`therefore v_(m) = 1.5 ms^(-1)`