Question

A block slides down an inclined plane of slope $\theta$ with constant velocity. It is then projected up the plane with an initial speed $v_0$. How far up the incline will it move before coming to rest?

• A. $\frac{v_0^2}{4g\sin \theta }$
• B. $\frac{v_0^2}{g\sin \theta }$
• C. $\frac{v_0^2}{2g\sin \theta }$
• D. $\frac{v_0^2}{2g}$

Answer 1

The correct option is A $\frac{v_0^2}{4g\sin \theta }$

Since block is moving down the plane with constant speed

So here friction force on the block is counter balanced by weight along the incline plane while it is sliding down

now when block is pushed up the plane with initial speed "v"

then it has two forces along the incline plane opposite to the velocity

1. friction force

2. component of weight along the incline

so net force along the incline opposite to velocity

deceleration of the block is given by

now we have

$0-V^2=2\times (-2gsin\theta )\times d$

$V^2=4g\left(sin\theta \right)\times d$

$d=\frac{V^2}{4g\sin \theta }$