Question

The potential energy of a particle of mass $1kg$ moving along x-axis given by $U\left(x\right)=[\frac{x^2}{2}-x]J$. If total mechanical energy of particle is $2J$, then find its maximum speed.

Answer 1

$U\left(x\right)=[\frac{x^2}{2}-x]\frac{dU}{dx}=\frac{2x}{2}-1=x-1$

For maximum speed, U must be minimum

$=\frac{dU}{dx}=0\Rightarrow x=1$

$U=\frac{1}{2}-1=-\frac{1}{2}J\therefore KE=2-\frac{1}{2}=\frac{3}{2}J\Rightarrow \frac{1}{2}\times 1\times v^2=\frac{3}{2}\Rightarrow v^2=3\Rightarrow v=\sqrt{3}m/s$