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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A particle of mass m is moving in a circular path of constant radius 'r'. Such that the centripetal acceleration $a_c$ is varying with time as $a_c=k^2rt$ where R is a constant. What is the power on the particle

\[(a)\;2 \pi mk^2r^2t\quad (b)\;mk^2r^2t \quad (c)\; \frac{mk^4r^2t}{3} \quad (d)\;zero\]

1 Answer

Since $a_c=k^2rt^2$
$\large\frac{v^2}{r}$$=k^2rt^2$
$v^2=k^2r^2t^2$
$v=krt$
$F=m \large\frac{dv}{dt}$
$\quad=m kr$
Power $=F.v$
$\qquad=(mkr) \times (krt)$
$\qquad=mk^2r^2t$
Hence b is the correct answer.
answered Aug 8, 2013 by meena.p
edited Jun 13, 2014 by lmohan717
 

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