Question

A particle slides down on a smooth incline of inclination ${30}^o$, fixed in an elevator going up with an acceleration $2m/s^2$. The box of incline has a length $4m$. The time taken by the particle to reach the bottom will be:

• A. $\frac{8}{9}\sqrt{3}s$
• B. $\frac{9}{8}\sqrt{3}s$
• C. $\frac{4}{3}\sqrt{\frac{\sqrt{3}}{2}}s$
• D. $\frac{3}{4}\sqrt{\frac{\sqrt{3}}{2}}s$

Answer 1

The correct option is C $\frac{4}{3}\sqrt{\frac{\sqrt{3}}{2}}s$

In the frame of elevator,
$a=$ acceleration of the particle with respect to the elevator
$\therefore m\sin 30(g+2)=ma$
$a=(g+2)\sin {30}^o$
$(10+2)\cdot \frac{1}{2}$
$=6m/s^2$
$\therefore$ The distance travelled by the particle from the top to the bottom.
$d=\frac{4}{\cos {30}^o}=\frac{4}{\sqrt{3}/2}=\frac{8\sqrt{3}}{3}m$
Using $s=ut+\frac{1}{2}{at}^2$
$\frac{8\sqrt{3}}{3}=0\times t+\frac{1}{2}6t^2$
$\Rightarrow t^2=\frac{16\sqrt{3}}{3\times 6}$
$\therefore t=\frac{4}{3}\sqrt{\frac{\sqrt{3}}{2}}s$