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, \(a_c = k^2rt^2\). The power delivered to the particle by the forces acting on it is (\(K\) = constant)

• A. \(m^2k^2r^2t^2\)
• B. \(mk^2r^2t\)
• C. \(2πmk^2r^2t\)
• D. \(\frac{mk^4r^2t^5}{3}\)

Hint:

Find $v$ from $a_c=v^2/r$, then use $P=F_tv$ with $F_t=m\,dv/dt$.

Answer 1

The Correct Option is B

Step 1: Use the work-energy theorem
Kinetic energy $=\frac12mv^2=\frac12mk^2r^2t^2$ since $v=krt$.
Power equals the rate of change of kinetic energy: $P=\frac{d}{dt}\left(\frac12mk^2r^2t^2\right)=mk^2r^2t$. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]