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Question: A given object takes \(n\) times more time to slide down a \({45^0}\) rough inclined plane as it tak...

A given object takes nn times more time to slide down a 450{45^0} rough inclined plane as it takes to slide down a perfectly smooth 450{45^0} incline. The coefficient of kinetic friction between the object and the inclined is:
A. 11n2\sqrt {1 - \dfrac{1}{{{n^2}}}}
B. 11n21 - \dfrac{1}{{{n^2}}}
C. 12n2\dfrac{1}{{2 - {n^2}}}
D. 11n2\sqrt {\dfrac{1}{{1 - {n^2}}}}

Explanation

Solution

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:
mgsinθμmgcosθ=mamg\sin \theta - \mu mg\cos \theta = ma is used in case of a rough surface.
mgsinθ=mamg\sin \theta = ma' is used in case of a smooth surface.
Where mm is the mass, gg is the acceleration due to gravity, θ\theta is the angle of inclination, μ\mu is the coefficient of friction, aa is the acceleration of the block on the rough inclined plane and aa'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 θ\theta is equal to 450{45^0}.
From the formula mgsinθμmgcosθ=mamg\sin \theta - \mu mg\cos \theta = ma, we get
g2μg2=a\Rightarrow \dfrac{g}{{\sqrt 2 }} - \dfrac{{\mu g}}{{\sqrt 2 }} = a
(since, θ\theta is 450{45^0})
From the formula mgsinθ=mamg\sin \theta = ma', we get
g2=a\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

S1=S2 12at2=12a(nt)2 g2t2=g2(1μ)n2t2 n2[1μ]=1 1μ=1n2 μ=(11n2) {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.