Work Energy And Power Question 294

Question: A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration $ a _{c} $ is varying with time t as $ a _{c}=k^{2}rt^{2} $ where k is a constant. The power delivered to the particles by the force acting on it is

Options:

A) $ 2\pi mk^{2}r^{2}t $

B) $ mk^{2}r^{2}t $

C) $ \frac{( mk^{4}r^{2}t^{5} )}{3} $

D) zero

Show Answer

Answer:

Correct Answer: B

Solution:

[b] The centripetal acceleration $ a _{c}=k^{2}rt^{2}or\frac{v^{2}}{r}=k^{2}rt^{2}\therefore v=krt $

So, tangential acceleration, $ {a _{t}} = \frac{dv}{dt} = kr $

Work is done by tangential force. Power $ = F _{i}.v.cos 0{}^\circ = ( ma _{t} )( krt ) = ( mkr )( krt ) $

$ =mk^{2}r^{2}t $



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