Question

The potential energy of a particle of mass $1\text{kg}$ free to move along $x-$axis is given by $U\left(x\right)=(\frac{x^2}{2}-x)$ joule. If total mechanical energy of the particle is $2\text{J}$, then the maximum speed of the particle is $v$.The value of $v^2$ (in Sl units) is
(Assuming only conservative force acts on particle)

Answer 1

The total mechanical energy of the particle at any instant is sum of kinetic and potential energy hence we use
$KE+PE=2\text{J}$
$\Rightarrow \frac{1}{2}m{v}^2+\frac{x^2}{2}-x=2$
$\Rightarrow v^2=2x-x^2+4$
For maximum velocity,
we use $\frac{d\left(v^2\right)}{dx}=0$
$\Rightarrow 2-2x=0$
$\Rightarrow x=1$
Thus maximum speed is at $x=1$, which is given as
$v_{max}^2=2-1+4$
$v_{max}=\sqrt{5}{\text{ms}}^{-1}$