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 $
BETA
