Question

A lift l is moving upwards with constant acceleration a=g

Answer 1

your question is -> A lift L is moving upward with a constant acceleration a=g. a small block A of mass 'm' is kept on a wedge B of same mass 'm'. the height of the vertical face of the wedge is 'h'. A is released from the top most point of wedge. find the time taken by A to reach the bottom of B. All surfaces are smooth and B is also free to move.

solution : wedge is located in non-inertial frame of reference. so, net acceleration acting on block placed on wedge = (a + g) [ downward ]

now, net acceleration of block along plane = (a + g)sinθ,

where θ is inclination angle of wedge.

see figure, sinθ = h/AB

so, net acceleration along plane = (a + g)h/AB

as it is given, a = g

so, net acceleration along plane = 2gh/AB

using formula, s = ut + 1/2 at²

here, a = 2gh/AB , u = 0, s = AB

so, AB = 1/2 (2gh/AB)t²

⇒2AB² = (2gh)t²

⇒AB² = (gh)t²

⇒t = AB/√(gh)

Answer 2

$\huge\bold\purple{Answer:-}$

your question is -> A lift L is moving upward with a constant acceleration a=g. a small block A of mass 'm' is kept on a wedge B of same mass 'm'. the height of the vertical face of the wedge is 'h'. A is released from the top most point of wedge.