Question

A given object takes $n$ times more time to slide down a ${45^0}$ rough inclined plane as it takes to slide down a perfectly smooth ${45^0}$ incline. The coefficient of kinetic friction between the object and the inclined is:
A. $\sqrt {1 - \dfrac{1}{{{n^2}}}} $
B. $1 - \dfrac{1}{{{n^2}}}$
C. $\dfrac{1}{{2 - {n^2}}}$
D. $\sqrt {\dfrac{1}{{1 - {n^2}}}} $

Answer 1

Hint: The coefficient of friction (μ) is the ratio defining the force that resists the motion of one body in relation to another body in contact with it. This ratio is dependent on material properties and most materials have a value between 0 and 1.

Formulas used:
$mg\sin \theta - \mu mg\cos \theta = ma$ is used in case of a rough surface.
$mg\sin \theta = ma'$ is used in case of a smooth surface.
Where $m$ is the mass, $g$ is the acceleration due to gravity, $\theta $ is the angle of inclination, $\mu $ is the coefficient of friction, $a$ is the acceleration of the block on the rough inclined plane and $a'$is the acceleration of the block on the smooth inclined plane.

Complete step by step answer:
It is given in the question that the angle of inclination $\theta $ is equal to ${45^0}$.
From the formula $mg\sin \theta - \mu mg\cos \theta = ma$, we get
$ \Rightarrow \dfrac{g}{{\sqrt 2 }} - \dfrac{{\mu g}}{{\sqrt 2 }} = a$
(since, $\theta $ is ${45^0}$)
From the formula $mg\sin \theta = ma'$, we get
$ \Rightarrow \dfrac{g}{{\sqrt 2 }} = a'$
As in both cases, the object started from the same point from rest and travels the same distance, that is
 \[
{S_1} = {S_2} \\
\Rightarrow \dfrac{1}{2}a{t^2} = \dfrac{1}{2}a{\left( {nt} \right)^2} \\
\Rightarrow \dfrac{g}{{\sqrt 2 }}{t^2} = \dfrac{g}{{\sqrt 2 }}(1 - \mu ){n^2}{t^2} \\
\Rightarrow {n^2}\left[ {1 - \mu } \right] = 1 \\
\Rightarrow 1 - \mu = \dfrac{1}{{{n^2}}} \\
\therefore \mu = (1 - \dfrac{1}{{{n^2}}}) \\
\]
Therefore, the correct option is (B).

Note: The frictional force is supposed to be proportional to the coefficient of friction. Although, the amount of force required to move an object starting from rest is usually greater than the force required to keep it moving at constant velocity once it is started.