JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A particle is moving in a circle of radius \(r\) under the action of a force \(F\, = \alpha r^2\) which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy \(= 0\) for \(r\, = 0\))
- A \(\frac{1}{2}\alpha {r^3}\)
- B \(\frac{5}{6}\alpha {r^3}\)
- C \(\frac{4}{3}\alpha {r^3}\)
- D \(\alpha {r^3}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{6}\alpha {r^3}\)
Step-by-step Solution
Detailed explanation
As we Known, \(dU = F.dr\)…
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