Question

Potential energy of a body of mass $1\, kg$ 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 $2\, J$. then the maximum speed of the body is (Assume only conservative force acts on the body)

• A. $\sqrt{5}\, ms^{-1} $
• B. $ 5 \; ms^{-1} $
• C. $ 3.5 \; ms^{-1} $
• D. $ \sqrt{8} \; ms^{-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, $m=1 \,kg$,
$E_{\text {mech }}=2 \,J$
So, for $U_{\min }=\frac{d U(x)}{d x}=\frac{d}{d x}\left[\frac{x^{2}}{2}-x\right]=0$
$\Rightarrow x - 1 = 0$
$\Rightarrow x = 1$
Hence, $U_{\min }=\frac{(1)^{2}}{2}-1=-\frac{1}{2}$
Kinetic energy. $KF =\frac{1}{2} m v^{2}=\frac{v^{2}}{2} \,\,(\because m=1 \,kg )$
Now, putting the values in above expression,
$\Rightarrow 2=-\frac{1}{2}+\frac{v^{2}}{2}$
$\Rightarrow v^{2}=5 $
$\Rightarrow v=\sqrt{5}\, m / s$