Work Energy And Power Question 139

Question: A particle of mass m moves along a circular path of radius r with centripetal acceleration $ a _{n} $ changing with time t as $ a _{n}=kt^{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

Options:

A) $ mkrt_0 $

B) $ \frac{mkrt_0^{2}}{2} $

C) $ \frac{mkrt_0^{{}}}{2} $

D) $ \frac{mkrt_0^{{}}}{4} $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Given $ a _{n}=kt^{2} $

Or $ \frac{v^{2}}{r}=kt^{2} $

or $ v^{2}=krt^{2} $ Therefore, average power delivered = $ \frac{Totalworkdone}{Totaltimeelaped} $

$ =\frac{increaseinKE}{Totaltimeelapsed} $

Or $ =\frac{\frac{1}{2}m(v^{2}-0^{2})}{t_0}=\frac{m}{2}\frac{krt_0^{2}}{t_0}=\frac{mkrt_0^{2}}{2} $

Alternative: $ v=\sqrt{krt}\Rightarrow a _{t}=\sqrt{kr} $

$ P=F _{t}v=ma _{t}v=mkrt $

$ =\frac{\int\limits_0^{t_0}{Pdt}}{t_0}=\frac{1}{2}mkrt_0^{2} $



NCERT Chapter Video Solution

Dual Pane