First slide

Friction on inclined plane of angle more than angle of repose

Question

An inclined plane of height h and length l have the angle of inclination $θ$ . The time taken by a body to come from the top to the bottom of this inclined plane will be

Moderate

A

$\sin θ \sqrt{\frac{2 h}{g}}$
B

$\frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$
C

$\sqrt{\frac{2 h}{g}}$
D

$\sqrt{\frac{2 l}{g}}$

Solution

Force down the plane $= m g \sin θ$
$∴$ Acceleration down the plane $= g \sin θ$
Using the relation, $h = u t + \frac{1}{2} g t^{2}$
$l = 0 + \frac{1}{2} (g \sin θ) t^{2}$
$\Rightarrow t^{2} = \frac{2 l}{g \sin θ} = \frac{2 h}{g \sin^{2} θ}$ $(∵ l = \frac{h}{\sin θ})$
$t = \frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App