Question

A particle slides down a smooth inclined plane of elevation a fixed in an elevator going up with an acceleration $a_0$ as shown in figure. The base of the incline has a length $l$. Prove that the time taken by the particle of reach the bottom is given by $t=\sqrt{2l/(g+a_0)\sin \alpha \cos \alpha }$.

Answer 1

Pseudo force: Pseudo force is the friction force arises in non-inertial frame.

$m{a}_0$= Pseudo force acting on the block.

eq of motion,

$m{a}_0\sin \alpha +mgsin\alpha =ma$

$\Rightarrow a=(a_0+g)\sin \alpha$

Again,

$S=ut+\frac{1}{2}a{t}^2$

here,

$S=AB=\frac{l}{\cos }\alpha$

$u=o=initialvel.$

$\therefore t=\sqrt{\frac{25}{a}}=\sqrt{\frac{2l}{(a_0+g)\sin \alpha \cos \alpha }}$