Work Power And Energy Ques 6
- 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 $a_c=k^2 r t^2$, where $k$ is a constant. The power delivered to the particle by the force acting on it is
(1994, 1M)
(a) $2 \pi m k^2 r^2$
(b) $m k^2 r^2 t$
(c) $\frac{\left(m k^4 r^2 t^5\right)}{3}$
(d) zero
Show Answer
Answer:
Correct Answer: 6.( b )
Solution:
Formula:
$ \begin{aligned} a_c & =k^2 r t^2 \\ \text { or } \quad \frac{v^2}{r} & =k^2 r t^2 \end{aligned} $
$ v=k r t $
Therefore, tangential acceleration, $a_t=\frac{d v}{d t}=k r$
or $\quad$ Tangential force, $F_t=m a_t=m k r$
Only tangential force does work.
Power $=F_t \quad v=(m k r)(k r t)$
or
Power $=m k^2 r^2 t$
BETA
