A person climbs up a conveyor belt with a constant acceleration. The speed of the belt is $\sqrt{\frac{g…
A person climbs up a conveyor belt with a constant acceleration. The speed of the belt is $\sqrt{\frac{g h}{6}}$ and coefficient of friction is $\frac{5}{3 \sqrt{3}}$. The time taken by the person to reach from A to B with maximum possible acceleration is
$\sqrt{\frac{h g}{6}}$
$\sqrt{6 g h}$
$\sqrt{\frac{2 h}{g}}$
$\sqrt{\frac{6 h}{g}}$
Solution
Maximum possible acceleration of the person on the conveyor belt is
$a_{\max }=\frac{\mu m g \cos \theta-m g \sin \theta}{m}=g(\mu \cos \theta-\sin \theta)$
$\begin{aligned}
& =g\left(\frac{5}{3 \sqrt{3}} \cos 30^{\circ}-\sin 30^{\circ}\right)=g\left(\frac{5}{3 \sqrt{3}} \times \frac{\sqrt{3}}{2}-\frac{1}{2}\right) \\
& =\frac{g}{3}
\end{aligned}$
$\therefore$ Time taken to climb the conveyer belt,
$\begin{aligned}
& S=u t+\frac{1}{2} a t^2 \\
& \Rightarrow 2 h=\left(\frac{\sqrt{g h}}{6}\right) t+\frac{1}{2} \cdot \frac{g}{3} t^2
\end{aligned}$ By solving, we get
$t=\sqrt{\frac{6 h}{g}} s$
