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

A block slides down from top of a smooth inclined plane of elevation t

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