Laws of Motion 5 Question 2

2. A particle is moving in a circular path of radius a under the action of an attractive potential energy U=k2r2. Its total energy is

(2018 Main)

(a) 32ka2

(b) k4a2

(c) k2a2

(d) zero

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Answer:

Correct Answer: 2. (d)

Solution:

  1. Given, U=k2r2Fr=dUdr=kr3

Since, the particle moves in a circular path of radius ’ a ‘, the required centripetal force is provided by the above force.

Hence, mv2a=ka3mv2=ka2

Kinetic energy, K=12mv2=k2a2

Total energy =K+U=k2a2+k2a2=0



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