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Question: A body takes \(n\) times the time to slide down a rough inclined plane as it takes to slide down the...

A body takes nn times the time to slide down a rough inclined plane as it takes to slide down the same inclined plane when it is perfectly frictionless. The coefficient of kinetic friction between the body and the plane for an angle of inclination of 45o{45^o}is given by μ\mu . Then find μ=?\mu = ?
A) 1−1n1 - \dfrac{1}{n}
B) 1n\dfrac{1}{n}
C) (1−1n2)\left( {1 - \dfrac{1}{{{n^2}}}} \right)
D) (1n2−1)\left( {\dfrac{1}{{{n^2}}} - 1} \right)

Explanation

Solution

When two particles are moving relative to each other and rub together. The friction force between two surfaces after sliding begins is the product of the coefficient of kinetic friction and the normal reaction and solving by the second equation of motion.

Complete step by step solution:
Given that,
A body take n time to slide down, angle of inclination is 45o{45^o}
Take a case, for smooth surface when timett,
Say, ma=mgsin⁡θma = mg\sin \theta
In terms of gravity,
a=gsin⁡θa = g\sin \theta ……………….(1)
Using Second Equation of Motion,
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2} …………………...(2)
At time t=0t = 0, then equation (2) becomes
s=0+12at2s = 0 + \dfrac{1}{2}a{t^2}
Where, s=d=12at2s = d = \dfrac{1}{2}a{t^2} ……………..(3)
Now substitute the value of acceleration from equation (1) into the equation (3)
We get here,
d=12×gsin⁡θ×t2d = \dfrac{1}{2} \times g\sin \theta \times {t^2} ……………...(4)
Now solving for rough surface,
At time t=ntt = nt
ma=mgsin⁡θ−μmgcos⁡θma = mg\sin \theta - \mu mg\cos \theta
Taking common mgmg
ma=mg(sin⁡θ−μcos⁡θ)ma = mg(\sin \theta - \mu \cos \theta )
Acceleration becomes, a=g(sin⁡θ−μcos⁡θ)a = g(\sin \theta - \mu \cos \theta ) ……………..(5)
Using Second Equation of Motion,
d=12at2d = \dfrac{1}{2}a{t^2} ……………….(6)
Putting the value of equation (5) in equation (6)
d=12×g(sin⁡θ−μcos⁡θ)×t2×n2d = \dfrac{1}{2} \times g(\sin \theta - \mu \cos \theta ) \times {t^2} \times {n^2}
Solving equation (4)and (6)
12gsin⁡θ×t2=12×g(sin⁡θ−μcos⁡θ)n2×t2\dfrac{1}{2}g\sin \theta \times {t^2} = \dfrac{1}{2} \times g(\sin \theta - \mu \cos \theta ){n^2} \times {t^2}
sin⁡θ=n2sin⁡θ−μ×n2×cos⁡θ\sin \theta = {n^2}\sin \theta - \mu \times {n^2} \times \cos \theta (equation 7)
Now, we have an angle which is θ=45o\theta = {45^o}
sin⁡45o=n2sin⁡45o−μn2×cos⁡45o\sin {45^o} = {n^2}\sin {45^o} - \mu {n^2} \times \cos {45^o}
12=12n2−12μn2\dfrac{1}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}{n^2} - \dfrac{1}{{\sqrt 2 }}\mu {n^2}
1=n2(1−μ)1 = {n^2}(1 - \mu )
μ=1−1n2\mu = 1 - \dfrac{1}{{{n^2}}}

Hence, the correct option of this question is C

Note: The ratio of the force of friction between two bodies and the force them together and the coefficient of friction depends on the materials used. There is no frictionless surface in the real world.