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Physics Question on Newtons Laws of Motion

A given object takes nn times the time to slide down 45∘45^\circ rough inclined plane as it takes the time to slide down an identical perfectly smooth 45∘45^\circ inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is:

A

1−1n21 - \frac{1}{n^2}

B

1−n21 - n^2

C

1−1n2\sqrt{1 - \frac{1}{n^2}}

D

1−n2\sqrt{1 - n^2}

Answer

1−1n21 - \frac{1}{n^2}

Explanation

Solution

For the smooth inclined plane, the acceleration is:
asmooth=gsin⁡45∘=g2.a_{\text{smooth}} = g \sin 45^\circ = \frac{g}{\sqrt{2}}.
For the rough inclined plane, the acceleration is:
arough=g(sin⁡45∘−μkcos⁡45∘)=g2(1−μk).a_{\text{rough}} = g (\sin 45^\circ - \mu_k \cos 45^\circ) = \frac{g}{\sqrt{2}} (1 - \mu_k).
The time taken is inversely proportional to the square root of acceleration:
trough=n⋅tsmooth  ⟹  asmootharough=n.t_{\text{rough}} = n \cdot t_{\text{smooth}} \implies \sqrt{\frac{a_{\text{smooth}}}{a_{\text{rough}}}} = n.
Substituting:
g2g2(1−μk)=n.\sqrt{\frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}} (1 - \mu_k)}} = n.
Simplify:
11−μk=n  ⟹  1−μk=1n2.\sqrt{\frac{1}{1 - \mu_k}} = n \implies 1 - \mu_k = \frac{1}{n^2}.
Solving for μk\mu_k:
μk=1−1n2.\mu_k = 1 - \frac{1}{n^2}.
Thus, the coefficient of kinetic friction is:
μk=1−1n2.\mu_k = 1 - \frac{1}{n^2}.