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Q. A particle is moving in a circular path of radius a under the action of an attractive potential $U = - \frac{k}{2r^2}$. Its total energy is -

JEE MainJEE Main 2018Work, Energy and Power

Solution:

$F=\frac{-d U}{d r} \,\,\,\,\,\left[U=-\frac{k}{2 r^{2}}\right]$
$\frac{m v^{2}}{r}=\frac{k}{r^{3}} \,\,\,\,\,$ [This force provides necessary centripetal force]
$\Rightarrow m v^{2}=\frac{k}{r^{2}}$
$\Rightarrow K . E=\frac{k}{2 r^{2}}$
$\Rightarrow P . E=-\frac{k}{2 r^{2}}$
Total energy $=$ Zero