Question

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

• A. $\sqrt{5}$
• B. $\sqrt{7}$
• C. $\sqrt{3}$
• D. None

Answer 1

The correct option is A $\sqrt{5}$

Mechanical energy $=K.E+P.E$
Kinetic energy to be maximum when potential energy is minimum,
Given,
$U\left(x\right)=[\frac{x^2}{2}-x]\text{J}$
For $U$ to be minimum
$\frac{dU}{dx}=0\&\frac{d^{2}U}{d{x}^2}>0\Rightarrow \frac{dU}{dx}=x-1=0\Rightarrow x=1$
$U_{min}=(\frac{1}{2}-1)=-\frac{1}{2}$

So,
$2=\frac{1}{2}m{v}_{max}^2-\frac{1}{2}\Rightarrow v_{max}=\sqrt{5}\text{m/s}$
Hence option $\left(1\right)$