Question

A block slides down from the top of a smooth inclined place of elevation $\theta$ fixed in an elevator going up with an acceleration $a_0$. The base of incline has a length L. The time taken by the block to reach the bottom is?

• A. $2\sqrt{\frac{L}{(g+a_0)\sin \theta }}$
• B. $\frac{1}{2}\sqrt{\frac{L}{(g+a_0)\sin 2\theta }}$
• C. $2\sqrt{\frac{L}{(g-a_0)\sin 2\theta }}$
• D. $2\sqrt{\frac{L}{(g+a_0)\sin 2\theta }}$

Answer 1

Answer: (D)

$2\sqrt{\frac{L}{(g+a_0)\sin 2\theta }}$

From the diagram,
Let a' be the net acceleration of the block
$\therefore \sum F_{real}+F_{pseudo}=m{a}^{\prime }$
$\Rightarrow mg\sin \theta +m{a}_0\sin \theta =m{a}^{\prime }$
$\Rightarrow a^{\prime }=(g+a_0)\sin \theta$
Here, $u=0$
$\because S=ut+\frac{1}{2}a{t}^2$
So, $\frac{L}{\cos \theta }=\frac{1}{2}(g+a_0)\cdot \sin \theta t^2$
$\Rightarrow t^2=\frac{2L}{(g+a_0)\sin \theta \cos \theta }$
$\therefore t^2=\frac{4L}{(g+a_0)2\sin \theta \cos \theta }$
$\therefore t=2\sqrt{\frac{L}{(g+a_0)\sin 2\theta }}$.