A block slides down from top of a smooth inclined plane of elevation ● fixed in an elevator going up with an acceleration `a_(0)` The base of incline hs length L Find the time taken by the block to reach the bottom
.

Explanation

Let us solve the problem in the elevator frame. The free body force diagram is shown. The forces are

(i) N normal reaction to the plane,
(ii) mg acting vertically down,
(iii) `ma_(0)` (pseudo force). acting vertically down
If a is the acceleration of the body with respect to incline, taking components of forces
parallel to the incline.
`mg sin theta + ma_(0)sin theta = ma " " therefore " " a = (g+a_(0))sin theta`
This is the acceleration with respect to the elevator
The distance travelled is `(L)/(cos theta)`. If t is the time for reaching the bottom of icline.
`(L)/(cos theta)=0+(1)/(2)(g+a_(0))sin theta.t^(2)`
`t=[(2L)/((g+a_(0))sin theta cos theta)]^((1)/(2))=[(4L)/((g+a_(0))sin 2 theta)]^((1)/(2))`