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} $
BETA
