The ratio of time taken by a body to cover an equal distance while sliding down a smooth to rough identical inclined plane of angle θ is ‘n′. Find the coefficient of friction.
A
tanθ(1−n2)
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B
cotθ(1−n2)
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C
sinθ(1−n2)
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D
tanθ(1−n2)
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Solution
The correct option is Dtanθ(1−n2) For a smooth inclined plane friction is zero and, F=ma⇒mgsinθ=ma a1=gsinθ
Using the second equation of motion, s1=12g(sinθ)t21
For rough inclined plane: F=ma2 mgsinθ−f=ma2
Since f=μN=μ(mgcosθ) mgsinθ−μ(mgcosθ)=ma2 ⇒a2=g(sinθ−μcosθ)
Hence, s2=12g(sinθ−μcosθ)t22; Given: s1=s2 and t1t2=n⇒(1−n2)sinθ=μcosθ⇒μ=tanθ(1−n2)