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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A particle moves in a circle of radius rr under the action of a centripetal force equal to βˆ’kr2r^-\frac{k}{r^2} \hat{r}, where kk is a constant. The total energy of the particle is
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Correct option: C
βˆ’k2r-\frac{k}{2 r}
Option C (-k/2r) is correct because the total energy is the sum of kinetic and potential energies, where kinetic energy is k/2r and potential energy is -k/r.

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