\(U(x)=\frac{x^3}{3}-\frac{x^2}{2} \ldots (1)\)

\(F=\frac{-d U}{d x}=\frac{3 x^2}{3}-\frac{2 x}{2}=0\)

\(x^2-x=0\)

\(\Rightarrow x=1,0\)

Potential energy is minimum at \(x=1 m\) and the value of this minimum \(P.E.\) will be \(U=\frac{-1}{6} J\) (Putting \(x=1\) in \((1)\))

Now, \(E=U+K\)

Kinetic energy will be maximum, when potential energy will be minimum

\(4=\frac{-1}{6}+K\)

\(K=\frac{25}{6}\)

\(\frac{1}{2} m v_m^2=\frac{25}{6}\)

\(v_m=\frac{5}{\sqrt{6}}\)