Question

A smooth inclined plane of length $L$ having inclination $\theta$ with the horizontal is inside a lift which is moving down with a retardation $a$. The time taken by a body to slide down the inclined plane from rest will be :

• A. $\sqrt{\frac{2L}{(g+a)\sin \theta }}$
• B. $\sqrt{\frac{2L}{(g-a)\sin \theta }}$
• C. $\sqrt{\frac{2L}{a\sin \theta }}$
• D. $\sqrt{\frac{2L}{g\sin \theta }}$

Answer 1

Answer: (B)

Net acceleration in downward motion decreases.
The free body diagram of the set up is shown.
Net acceleration is $(g-a)\sin \theta$

Using equation of motion
$s=ut+\frac{1}{2}{a}^{\prime }{t}^2$
where, $s$ is displacement, $u$ is initial velocity, $a$ is acceleration and $t$ is time.
We have, $u=0,a^{\prime }=(g-a)\sin \theta ,s=L$
$\therefore L=0+\frac{1}{2}(g-a)\sin \theta \cdot t^2$
$\Rightarrow t=\sqrt{\frac{2L}{(g-a)\sin \theta }}$