Question

A particle slides down a smooth inclined palns of elevation $\theta$, fixed in an elevator going up with on acceleration ${\alpha }_0$ (figure 5 W12). The base of the incline has a length L. Find the time taken by the particle to reach the bottom.

Answer 1

In lift frame $(mg+m{a}_0)sin\theta =ma$
$\Rightarrow a=(g+a)sin\theta$
Distance moved = $\frac{L}{cos\theta }$
$\therefore \frac{L}{cos\theta }=\frac{1}{2}(g+a_0)sin\theta .t^2$
$\therefore t=\sqrt{\frac{2L}{(g+a_0)sin\theta cos\theta }}$