Question

# A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to −Kr2 , where K is a constant. The total energy of the particle is

Solution

## The correct option is B Here mv2r=Kr2∴K.E=12mv2=K2r U=−r∫∞F.dr=−∫r∞(−Kr2)dr=−kr Total energy E=K.E+P.E=K2r−Kr=−K2r

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