A body of mass m is launched up a rough inclined plane of angle 45∘ with the horizontal. If the time of ascent is half the time of descent in covering the same distance, the coefficient of friction is
A
25
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B
35
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C
34
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D
58
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Solution
The correct option is B35 Up the plane: F=ma⇒(mgsinθ+f)=ma1 mgsinθ+μmgcosθ=ma1 a1=g(sinθ+μcosθ) s1=ut+12at2 (u is the initial velocity) s1=12g(sinθ+μcosθ)t21;
Down the plane: F=ma⇒(mgsinθ−f)=ma2 mgsinθ−μmgcosθ=ma2 a2=g(sinθ−μcosθ) s2=ut+12at2 ⇒s2=12g(sinθ−μcosθ)t22
Given: t1=t22;⇒12g(sinθ+μcosθ)t21=12g(sinθ−μcosθ)t22 ⇒(sinθ+μcosθ)=(sinθ−μcosθ)4⇒μ=35tanθ
Since θ=45∘ μ=35tan45∘=35