Question

A particle of mass $m$ moves along a circular path of radius $r$ with a centripetal acceleration $a_n$ changing with time $t$ as $a_n=k{t}^2$, where $k$ is a positive constant. The average power developed by all the forces acting on the particle during the first $t_0$ seconds is

• A. $mkr{t}_0$
• B. $\frac{mkr{t}_0^2}{2}$
• C. $\frac{mkr{t}_0}{2}$
• D. $\frac{mkr{t}_0}{4}$

Answer 1

Answer: (B)

Given $a_n=k{t}^2$
or $\frac{v^2}{r}=k{t}^2$ or $v^2=kr{t}^2$
Therefore, average power delivered $=$ Total work done/Total time elapsed $=$ Increase in KE/Total time elapsed
or $⟨P⟩\frac{\frac{1}{2}m\left(v^2-0^2\right)}{t_0}=\frac{m}{2}\frac{kr{t}_0^2}{t_0}=\frac{mkr{t}_0^2}{2}$ Alternative: $v=\sqrt{krt}\Rightarrow a_1=\sqrt{kr}$
$P=F_{t}v=m{a}_{t}v=mkrt$
$\begin{array}{rl}P=F_{t}v& =m{a}_{t}v=mkrt\\ & ⟨P⟩=\frac{{\int }_0^{t_0}Pdt}{t_0}=\frac{1}{2}mkr{t}_0^2\end{array}$