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

Options

A. `sqrt(2)`
B. `1//sqrt(2)`
C. `2`
D. `3//sqrt(2)`

Answer

C

Explanation

Max. `K.E.=(1)/(2)m upsilon_(max)^(2)0`total mechanical energy
`:. (1)/(2)xx1xxupsilon_(max)^(2)=2`
`upsilon_(max)=sqrt(2xx2)=2m//s`