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. `(3)/(sqrt2)m//s`
B. `sqrt2m//s`
C. `(1)/(sqrt2)m//s`
D. `2 m//s`

Answer

A

Explanation

At max. speed , `KE` is max. and `PE` is min. `PE`
is min at `x=+-1` so `PE_(min)=-(1)/(4)`
`KE_(max)=TE_(min)-PE_(min)rArr(1)/(2)mV_(max)^(2)=(9)/(4)rArrV_(max)=(3)/(sqrt2)m//s`