The potential energy of a particle of mass $1 \mathrm{kg}$ free to move along the $x$-axis is given by $\cup \left(\mathrm{x}\right)=\left(\frac{{\mathrm{x}}^2}{2}-\mathrm{x}\right) \mathrm{joules}$. If total mechanical energy of the particle is $2 \mathrm{J}$, then find its maximum speed.

Potential energy $\mathrm{U}=\left(\frac{{\mathrm{x}}^2}{2}-\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{2}-\mathrm{x}$

For minimum $\mathrm{U},\frac{\mathrm{dU}}{\mathrm{dx}}=\frac{2\mathrm{x}}{2}-1=0 \& \frac{{\mathrm{d}}^2\mathrm{U}}{{\mathrm{dx}}^2}=1 =$positive

so at $\mathrm{x} = 1, \mathrm{U}$ is minimum. Hence ${\mathrm{U}}_{\min }=-\frac{1}{2}\mathrm{J}$

Total mechanical energy = $\mathrm{Max} \mathrm{KE} + \mathrm{Min} \mathrm{PE}$

$$\begin{gathered} \Rightarrow \mathrm{Max}.\mathrm{KE}=\frac{1}{2}{{\mathrm{mv}}^2}_{\max }=2-\left(-\frac{1}{2}\right)=\frac{5}{2} \\ \Rightarrow {\mathrm{v}}_{\max }=\sqrt{\frac{5}{2}\times \frac{2}{\mathrm{m}}}=\sqrt{5} {\mathrm{ms}}^{-1} \end{gathered}$$