Question

A block slides down on an inclined plane of slope angle $\theta$ with constant velocity. It is then projected up on the same plane with initial velocity $V_0$. How far up the incline will it move before coming to rest

• A. $\frac{V_0^2}{gsin\theta }$
• B. $\frac{V_0^2}{2gsin\theta }$
• C. $\frac{V_0^2}{4gsin\theta }$
• D. $\frac{V_0^2}{8gsin\theta }$

Answer 1

Answer: (C)

$\frac{V_0^2}{4gsin\theta }$

As the block is sliding down with constant velocity. Therefore friction force along the slope must be equal to the component of gravity along the slope.
$\therefore \mu N=mgsin\theta \Rightarrow \mu mgcos\theta =mgsin\theta \Rightarrow \mu =tan\theta$
When thrown up:
Initial Velocity is $V_o$
Acceleration (both friction and component of gravity acting downwards the slope as relative motion is upwards) is $\mu N+mgsin\theta =tan\theta mgcos\theta +mgsin\theta =2mgsin\theta$
Final Velocity is zero.
Therefore distance traveled up the incline is:
$v^2-u^2=2as\Rightarrow 0-{V_o}^2=2(-2mgsin\theta )s\Rightarrow s=\frac{{V_o}^2}{4mgsin\theta }$