A block begins to slide down a rough inclined plane of angle 45∘ and moves 1m in 4√2s. What is the coefficient of friction μ, between the plane and the block? (Take g=10m/s2)
A
0.4
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B
0.5
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C
0.6
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D
0.8
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Solution
The correct option is D0.8 Let a be the acceleration of the block down the inclined plane. The given question can be depicted as shown below
The coefficient of friction between the block and the incline plane is μ Now, from the FBD we have mgsin45∘−f=ma......(1) and, N=mgcos45∘ Friction force, f=μN=μmgcos45∘ On putting the value of f in eq (1) we get mgsin45∘−μmgcos45∘=ma ⇒a=g(sin45∘−μcos45∘)=10(1−μ)√2m/s2
Using the equation of motion s=ut+12at2 ⇒1=0+12×10(1−μ)√2×(4√2)2⇒μ=0.8