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Question

Physics Question on work, energy and power

A particle of mass mm is moving in a circular path of constant radius rr such that centripetal acceleration aca_{c} varying with time is ac=k2rt2a_{c}=k^{2}\, r \,t^{2} where kk is a constant. What is the power delivered to the particle by the force acting on it?

A

2πmk2r22 \pi \, mk^2 \, r^2

B

mk2r2tmk^2 \, r^2 t

C

(mk4r2t5)3\frac{(mk^4 \, r^2 t^5)}{3}

D

zero

Answer

mk2r2tmk^2 \, r^2 t

Explanation

Solution

ac=v2r=k2rt2a_{c}=\frac{v^{2}}{r}=k^{2}\, r\, t^{2}
v2=k2r2t2\Rightarrow v^{2}=k^{2} \,r^{2}\, t^{2}
KE=12mv2=12mk2r2t2K E=\frac{1}{2} m v^{2}=\frac{1}{2} m k^{2} \,r^{2}\, t^{2}
According to work-energy theorem,
change in kinetic energy is equal to work done.
W=12mk2r2t2\therefore W=\frac{1}{2} m k^{2}\, r^{2}\, t^{2}
Thus, power delivered to the particle,
P=dWdt=mk2r2tP=\frac{d W}{d t}=m k^{2} \,r^{2} \,t