Question

In the figure shown, an inclined plane makes an angle ${30}^{\circ }$ with the horizontal. 𝐴 groove $OA=5m$ cut in the plane makes an angle ${30}^{\circ }$ with $OX$. A short smooth cylinder is free to slide down in the groove due to the influence of gravity. The time (in $s$ taken by the cylinder to reach from $A$ to $O$, if released from rest from $A$, is $g=10m/s^2$

Answer 1

Given,
Inclination angle $\theta ={30}^{\circ }$
Length of groove $s=5m$
Resolve the component of gravity,
Component of gravity along the incline

$=g\sin {30}^{\circ }$
$=10\times 0.5$
$=5m/s^2$

Component of gravity along lone
$OA=\left(g\sin {30}^{\circ }\right)\sin {30}^{\circ }$
$=10\times 0.5\times 0.5$
$=2.5m/s^2$

Thus, acceleration of the cylinder is $2.5m/s^2$ along line $OA$.

Using,
$s=ut+\frac{1}{2}a{t}^2$

$5=\frac{1}{2}\times 2.5\times t^2$

Solving,

$t=2sec$

Final anwer: 2