Question

Potential energy of a body of mass $1kg$ free to move along X-axis is given by $U(x)=\left(\frac{x^2}{2}-x\right)J.$ If the total mechanical energy of the body is $2J$. then the maximum speed of the body is (Assume only conservative force acts on the body)

• A. $\sqrt{5}m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $3.5m{s}^{-1}$
• D. $\sqrt{8}m{s}^{-1}$

Answer 1

Answer: (A)

Total mechanical energy of a system is the addition of potential and kinetic energy
$E=U+KE$
Here, mass of the body, $m=1kg$,
$E_{\text{mech }}=2J$
So, for $U_{\min }=\frac{dU(x)}{dx}=\frac{d}{dx}\left[\frac{x^2}{2}-x\right]=0$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Hence, $U_{\min }=\frac{(1{)}^2}{2}-1=-\frac{1}{2}$
Kinetic energy. $KF=\frac{1}{2}m{v}^2=\frac{v^2}{2}(\because m=1kg)$
Now, putting the values in above expression,
$\Rightarrow 2=-\frac{1}{2}+\frac{v^2}{2}$
$\Rightarrow v^2=5$
$\Rightarrow v=\sqrt{5}m/s$