Laws of Motion

JEE Main level, single correct, medium

147 of 210

Question 147 of 210

The time taken by an object to slide down 45∘45^{\circ} rough inclined plane is nn times as it takes to slide down a perfectly smooth 45∘45^{\circ} incline plane. The coefficient of kinetic friction between the object and the incline plane is:

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Answer: (D) 1−1n21-\dfrac{1}{{n}^{2}}

Let angle of inclination is θ=45∘\theta =45^{\circ}.
Then, acceleration along the incline in case of smooth incline will be, a1=gsin⁡θ=g2{a}_{1}=g\sin \theta =\dfrac{g}{\sqrt{2}}.
In case of rough incline, kinetic friction will act against the direction of motion. Normal force acting on the object will be mgcos⁡θmg\cos \theta. Therefore, acceleration along the incline in case of rough incline will be,
a2=gsin⁡θ−μgcos⁡θ=g2−μg2{a}_{2}=g\sin \theta -\mu g\cos \theta =\dfrac{g}{\sqrt{2}}-\dfrac{\mu g}{\sqrt{2}}.
Now as per question,
t2=nt1{t}_{2}=n{t}_{1}
and as the displacement covered is the same in both case, 0+12a1t12=0+12a2t220+\dfrac{1}{2}{a}_{1}{t}_{1}^{2}=0+\dfrac{1}{2}{a}_{2}{t}_{2}^{2}
⇒g2t12=(g2−μg2)n2t12\Rightarrow \dfrac{g}{\sqrt{2}}{t}_{1}^{2}=\left(\dfrac{g}{\sqrt{2}}-\dfrac{\mu g}{\sqrt{2}}\right){n}^{2}{t}_{1}^{2}
⇒μ=1−1n2\Rightarrow \mu =1-\dfrac{1}{{n}^{2}}

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