JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 2)
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as
zero
mk2r2t2
mk2r2t
mk2rt
설명
$${a_r} = {k^2}r{t^2} = {{{v^2}} \over r}$$
$$ \Rightarrow {v^2} = {k^2}{r^2}{t^2}$$ or $$v = krt$$
and $${{d|v|} \over {dt}} = kr$$
$$ \Rightarrow {a_t} = kr$$
$$ \Rightarrow |\overline F \,.\,\overline v | = (mkr)(krt)$$
$$ = m{k^2}{r^2}t = $$ power delivered