Question

A block slides down a rough inclined plane of slope angle θ with a constant velocity. It is then projected up the same plane with an initial velocity v. The distance travelled by the block up the plane before coming to rest is
A. $\dfrac{{{v}^{2}}}{4g\sin \theta }$
B. $\dfrac{{{v}^{2}}}{2g\sin \theta }$
C. $\dfrac{{{v}^{2}}}{g\sin \theta }$
D. $\dfrac{4g{{v}^{2}}}{\sin \theta }$

Answer 1

Inclined plane is a surface which is elevated at some angle with respect to the horizontal. Given that the inclined is not smooth but rough. Thus, frictional force will be present when the block slides this inclined. The body moves with the constant velocity as it comes down. We need to find out the distance travelled by the block.

Complete step by step answer:
Since, the block is moving down with a constant velocity it means the acceleration of the body is zero because acceleration is given by the differential of velocity with time and if velocity is a constant, then the result will be zero. So, the net force acting on the body is zero.Thus, the force acting downwards is equal to the frictional force.

We know the force which moves the body down is $mg\sin \theta$ and the frictional force always opposes the motion and in this case it can be given as the product of coefficient of friction and the normal friction, $\mu mg\cos \theta$.
$mg\sin \theta =\mu mg\cos \theta \\\ \Rightarrow \sin \theta =\mu \cos \theta \\\ \therefore \mu =\tan \theta \\\$
When the block is pushed up the ramp the net force on the block is given by:
$F=mg\sin \theta +\mu mg\cos \theta \\\ \Rightarrow ma=mg\sin \theta +\mu mg\cos \theta \\\ \Rightarrow ma=mg\sin \theta +\tan \theta mg\cos \theta [\because \tan \theta ] \\\ \Rightarrow ma=2mg\sin \theta \\\ \therefore a=2g\sin \theta \\\$
Let the distance covered be x, Using the third equation of motion,
$v{{'}^{2}}={{v}^{2}}-2as \\\ \Rightarrow {{v}^{2}}=2(2g\sin \theta )x \\\ \therefore x=\dfrac{{{v}^{2}}}{4g\sin \theta }$

Hence, option A is the correct answer.

Note: : When the body moves up, the final velocity will be zero at the highest point when it has covered distance, x. We should keep in mind that the value of static friction is always bigger than the kinetic friction. And since, the force on the block due to gravity is equal to the friction force, therefore the gravitational force on the block is not enough to move the block from the rest.