Work Energy Power

JEE Main level, single correct, medium

156 of 183

Question 156 of 183

A particle of mass mm is moving in a circular path of constant radius rr such that its centripetal acceleration ac{a}_{c} is varying with time tt as ac=k2rt2{a}_{c}={k}^{2}r{t}^{2} , where kk is a constant. The power delivered to the particle by the force acting on it is -

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Answer: (C) mk2r2tm{k}^{2}{r}^{2}t

ac=k2rt2{a}_{c}={k}^{2}r{t}^{2}
Or v2r=k2rt2\dfrac{{v}^{2}}{r}={k}^{2}r{t}^{2} Or v=krtv=krt
Therefore, tangential acceleration, at=dvdt=kr{a}_{t}=\dfrac{dv}{dt}=kr
Therefore, tangential force, Ft=mat=mkr{F}_{t}=m{a}_{t}=mkr

Only tangential force does work.

Power=Ftv=(mkr)(krt)\text{Power}={F}_{t}v=\left(mkr\right)\left(krt\right) ⇒Power=mk2r2t\text{⇒Power}=m{k}^{2}{r}^{2}t

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