Question

A smooth inclined plane of length l having inclination theta with the horizontal is inside a lift ,which is movung inside an retardation a .the time taken by a body to slide down the inclined plane from rest will be?

Answer 1

Let the mass of the block be m.  Length of the incline = L.

  Let the lift move with a uniform velocity u upwards and then start to decelerate at a rate "- a ".  A Pseudo force equal to m a acts on the block vertically upwards.  There is a force mg acts on the block vertically downwards.  The angle between the vertical and the incline is (90°-θ).
==== LIFT MOVING UP

  In the non-inertial frame of reference of the lift, the acceleration of the block m down the incline :
   a' = g sinθ - a sinθ = (g - a) sinθ

Equation of motion :
   L = u t +1/2 a t²
   L = 0 + 1/2 * (g - a) sin
   Time to slide down the incline plane from rest = t
$t=\sqrt{\frac{2L}{(g-a)\ Sin\theta}}$

==========LIFT  MOVING DOWN ==

The lift is decelerating at a rate "a".  In this case the Pseudo force on the block m will act downwards. Then the net acceleration of the block along the incline :

  a' = (g + a) sinθ

Consequently the time t taken to slide down the incline: t
$t=\sqrt{\frac{2L}{(g+a)\ Sin\theta}}$

So in this case the block slides down faster.

Answer 2

Answer:

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