JEE Main2018PhysicsWork, Power and EnergyActual
A particle is moving in a circular path of radius a under the action of an attractive potential U = - k 2 r 2 . Its total energy is:
Options
- A- 3 2 k a 2
- B– k 4 a 2
- Ck 2 a 2
- DZero
Correct answer
D. Zero
Step-by-step solution
U = - K 2 r 2 ⇒ F = - d U d r = 2 K 2 r 3 = K r 3 ⇒ m v 2 r = K r 3 ⇒ v 2 = K m r 2 ⇒ 1 2 m v 2 = K 2 r 2 ⇒ E = K 2 r 2 - K 2 r 2 = 0 .