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Question

A particle is moving in a circular path of radius a with constant speed under the action of an attractive conservative force. Potential energy of the particle is given by the relation U=K2r2, where r is the radial distance of the particle from the centre of the circular path. Its total energy will be:

A
K2a2
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B
Zero
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C
−32Ka2
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D
−K4a2
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Solution

The correct option is B Zero
Relation between potential energy and conservative force is given by
F=−dUdr
where r is the radial distance of the particle describing circular path.

Here U=K2r2
⇒F=−dUdr=−K2×(−2r−3)=Kr3
Since it is performing circular motion, force F must provide centripetal force.
⇒F=mac=mv2r
⇒mv2r=Kr3
∴mv2=Kr2

Kinetic energy of particle is given by,
K.E=12mv2=K2r2
Total energy of particle is given by:
T.E=P.E.+K.E.
∴T.E=K2r2+K2r2=Kr2

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