Question

The Potential energy of a particle of mass $1\text{kg}$ free to mone 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

(Assume only conservative forces acts on the particle)

• A. $\sqrt{2}{\text{ms}}^{-1}$
• B. $\sqrt{5}{\text{ms}}^{-1}$
• C. $5{\text{ms}}^{-1}$
• D. $2\sqrt{5}{\text{ms}}^{-1}$

Answer 1

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

Since, only conservative forces are acting, mechanical energy is conserved.
Total mechnical energy at any instant is given by, $T.E=K.E+P.E$
Also, maximum speed of the particle implies maximum kinetic energy and minimun potential energy i.e.
$T.E=\frac{1}{2}m{v}_{max}^2+(P.E{)}_{min}...(i)$
Now, for minimum potential energy,
$\frac{dU}{dx}=0$
$\Rightarrow \frac{d(\frac{x^2}{2}-x)}{dx}=x-1=0$
$\Rightarrow x=1$
also, $\frac{d^{2}U}{d{x}^2}=1>0$
$\Rightarrow (P.E{)}_{min}=-\frac{1}{2}\text{J}$
Putting this value in eq $\left(i\right)$, we get
$2=\frac{1}{2}{v}_{max}^2-\frac{1}{2}\Rightarrow v_{max}=\sqrt{5}\text{m/s}$