# A particle is moving in a circular path of radius a under the action of an attractive potential U =?K/2r^2 . what is total energy

P Pankaj Tamang

$F=- \frac{\mathrm{d} U}{\mathrm{d} r}= -k r$

This force provides the necessary centripetal force. So,

$k r =\frac{m v^{2}}{r}$

Kinetic energy:

$K=\frac{1}{2} m v^{2}=\frac{1}{2}k r^{2}$

Total energy = kinetic energy + Potential energy

$E=K+U=\frac{1}{2} k r^{2}+\frac{1}{2} k r^{2}= k r^{2}$

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