Question

A block of mass $m$ starting from rest slides down a smooth inclined plane of gradient $\theta$, fixed in a lift moving upwards with an acceleration $a_0$ as shown in figure. If the base of the inclined plane has length $L$, the time taken by the block to slide from top to bottom of the inclined plane, will be

• A. $\frac{L}{\left(g+a_0\right)\sin 2\theta }$
• B. ${\left[\frac{4L}{\left(g+a_0\right)\sin 2\theta }\right]}^{1/2}$
• C. $\frac{\left(g+a_0\right)\sin 2\theta }{2L}$
• D. ${\left[\frac{2L}{\left(g+a_0\right)\sin \theta }\right]}^2$

Answer 1

Answer: (B)

Acceleration along the inclined plane,
$a=\left(g+a_0\right)\sin \theta$
Slope length $=\frac{L}{\cos \theta }$
Using $s=ut+\frac{1}{2}a{t}^2,$ weget $\frac{L}{\cos \theta }=0\times t+\frac{1}{2}\left(g+a_0\right)\sin \theta t^2$
or $t=\sqrt{\frac{2L}{\left(g+a_0\right)\sin \theta \cos \theta }}$ or $t=\sqrt{\frac{4L}{\left(g+a_0\right)\sin 2\theta }}$