Question

The potential energy of a 1 kg particle free to move along the x-axis is given by
$V\left(x\right)=\frac{x^2}{4}-\frac{x^2}{2}$ (in joule)
The total mechnical energy of the particle is 2J. Then, the maximum speed (in m$s^{-1}$) is

Answer 1

$\begin{array}{c}V=\frac{x^2}{4}-\frac{x^2}{2}m=1kg\\ TE=PE+KE\\ KE\to maxPE\to 0\\ \Rightarrow 2=KE\\ \Rightarrow \frac{1}{2}\times 1\times V^2=2\\ \Rightarrow V^2=4\\ \Rightarrow V=2m/s\end{array}$

$\therefore$ Max speed $=2m/s$.