The potential energy of 2kg particle, free to move along x axis is given by $ U (X) = \left( { x^4 \over 4 } - { x^2 \over 2} \right) J $ .. If its mechanical energy is 2 J, its maximum speed is……….. m/s
K.E. is maximum than P.E. minimum. $ so { du \over dx } = 0 \Rightarrow x = 0 OR $ $ For x = \pm 4 U(x) = - { 1 \over 4} = Umin$