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 find the maximum speed of the particle. (Assuming that only conservative force is acting on the particle)

• A. $\sqrt{5}{\text{ms}}^{-1}$
• B. $\sqrt{4}{\text{ms}}^{-1}$
• C. $\sqrt{3}{\text{ms}}^{-1}$
• D. $\sqrt{2.5}{\text{ms}}^{-1}$

Answer 1

The correct option is A $\sqrt{5}{\text{ms}}^{-1}$

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 max v; 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
$v_{max}^2=2-1+4$
$v_{max}=\sqrt{5}{\text{ms}}^{-1}$