A frictionless inclined plane of length l having inclination θ is placed inside a lift which is accelerating downward with an acceleration a(

Question

A frictionless inclined plane of length l having inclination θ is placed inside a lift which is accelerating downward with an acceleration a(< g). If a block is allowed to move, down the inclined plane, from rest, then the time taken by the block to slide from top of the inclined plane to the bottom of the inclined plane is (A) √(2 l/g) (B) √(2 l/g-a) (C) √(2 l/g+a) (D) √(2 l/(g-a) sin θ)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5cb6020f005e6595c5ead3c330dbc719-.png" /> Along x -axis: F=m g sin θ-m a sin θ (F/m)=(g-a) sin θ Acceleration along the inclined plane =(g-a) sin θ Using S=u t+(1/2) a t2 l=0+(1/2)(g-a) sin θ t2 or t=√(2 l/(g-a) sin θ)

$\frac{2 l}{g}$

$\frac{2 l}{g - a}$

$\frac{2 l}{g + a}$

$\frac{2 l}{( g - a ) s i n θ}$

Along $x$ -axis: $F = m g sin θ - ma sin θ$
$\frac{F}{m} = (g - a) sin θ$
Acceleration along the inclined plane $= (g - a) sin θ$
Using $S = u t + \frac{1}{2} a t^{2}$
$l = 0 + \frac{1}{2} (g - a) sin θ t^{2}$ or $t = \frac{2 l}{( g - a ) s i n θ}$

g2l​​

g−a2l​​

g+a2l​​

(g−a)sinθ2l​​

Along x -axis: F=mgsinθ−masinθ mF​=(g−a)sinθ Acceleration along the inclined plane =(g−a)sinθ Using S=ut+21​at2 l=0+21​(g−a)sinθt2 or t=(g−a)sinθ2l​​

$\frac{2 l}{g}$

$\frac{2 l}{g - a}$

$\frac{2 l}{g + a}$

$\frac{2 l}{( g - a ) s i n θ}$

Along $x$ -axis: $F = m g sin θ - ma sin θ$
$\frac{F}{m} = (g - a) sin θ$
Acceleration along the inclined plane $= (g - a) sin θ$
Using $S = u t + \frac{1}{2} a t^{2}$
$l = 0 + \frac{1}{2} (g - a) sin θ t^{2}$ or $t = \frac{2 l}{( g - a ) s i n θ}$