The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) =
Channel: Doubtnut
Date: 2020-07-12

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