A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a0. The angle between the inclined plane and ground is and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is . [figure]
- 2Lg - a0
- 4Lg 2 - a0(1 + 2 )
- 2Lg 2 - a0(1 + 2 )
- 4Lg 2 - a0
Answer
(2) 4Lg 2 - a0(1 + 2 )
Solution
In the frame of the incline (acceleration a0 towards left), a pseudo force ma0 acts on the block towards right.
The block moves down the incline towards the left, so along the incline (downwards):
arel = g - a0
Length of the incline: s = L
Starting from rest relative to the incline: s = 12arelt2
t = 2L (g - a0 ) = 2Lg - a0 2
Using 2 = 2 and 2 2 = 1 + 2 :
t = 4Lg 2 - a0(1 + 2 )
Answer: (2)