Question

A body takes $n$ 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 ${45^o}$is given by $\mu $. Then find $\mu = ?$
A) $1 - \dfrac{1}{n}$
B) $\dfrac{1}{n}$
C) $\left( {1 - \dfrac{1}{{{n^2}}}} \right)$
D) \[\left( {\dfrac{1}{{{n^2}}} - 1} \right)\]

Answer 1

Hint: 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 ${45^o}$
Take a case, for smooth surface when time$t$,
Say, $ma = mg\sin \theta $
In terms of gravity,
$a = g\sin \theta $ ……………….(1)
Using Second Equation of Motion,
$s = ut + \dfrac{1}{2}a{t^2}$ …………………...(2)
At time $t = 0$, then equation (2) becomes
$s = 0 + \dfrac{1}{2}a{t^2}$
Where, $s = d = \dfrac{1}{2}a{t^2}$ ……………..(3)
Now substitute the value of acceleration from equation (1) into the equation (3)
We get here,
$d = \dfrac{1}{2} \times g\sin \theta \times {t^2}$ ……………...(4)
Now solving for rough surface,
At time $t = nt$
$ma = mg\sin \theta - \mu mg\cos \theta $
Taking common $mg$
$ma = mg(\sin \theta - \mu \cos \theta )$
Acceleration becomes, $a = g(\sin \theta - \mu \cos \theta )$ ……………..(5)
Using Second Equation of Motion,
\[d = \dfrac{1}{2}a{t^2}\] ……………….(6)
Putting the value of equation (5) in equation (6)
$d = \dfrac{1}{2} \times g(\sin \theta - \mu \cos \theta ) \times {t^2} \times {n^2}$
Solving equation (4)and (6)
$\dfrac{1}{2}g\sin \theta \times {t^2} = \dfrac{1}{2} \times g(\sin \theta - \mu \cos \theta ){n^2} \times {t^2}$
\[\sin \theta = {n^2}\sin \theta - \mu \times {n^2} \times \cos \theta \] (equation 7)
Now, we have an angle which is $\theta = {45^o}$
$\sin {45^o} = {n^2}\sin {45^o} - \mu {n^2} \times \cos {45^o}$
$\dfrac{1}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}{n^2} - \dfrac{1}{{\sqrt 2 }}\mu {n^2}$
$1 = {n^2}(1 - \mu )$
$\mu = 1 - \dfrac{1}{{{n^2}}}$

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.