Question

A block slides down an inclined plane of slope of angle $\theta$ with a constant velocity $v$. It is then projected up the plane with an initial velocity $u$. What is the distance upto which it will rise before coming to rest?

Answer 1

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

$mg\sin \theta =F_f$

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

$F_{net}=-mg\sin \theta -Ff=-2mg\sin \theta$

deceleration of the block is given by

$a=-2g\sin \theta$

we have

$v_f^2-v_i^2=2ad$

$0-v^2=2\times (-2g\sin \theta )\times d$

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