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
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Explanation

Let us solve the problem in the frame The free body diagram is shown The forces are
(i) N normal reacion to the plane
(ii) mg acting vertically downwards
(iii) `ma_(0)` (pseudo force) acting vertically down if a is acceleration of the body with respect to inclined pane, taking components of forces parallel to the inclined plane
`mg sin theta + ma_(0) sin theta = ma :. a = (g +a_(0)) sin theta`
This is the acceleration with respect to the elevator THe distance travelled is `(L)/(costheta)` if 't' is the time for reaching the bottom of inclined plane
`(L)/(costheta) = 0 +(1)/(2) (g +a_(0))sin thetat^(2)`
`t =[(2L)/((g +a_(0))sin theta cos theta)]^(1/2)= [(4L)/((g +a_(0))sin2theta)]^(1/2)`
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