A block begins to slide down a rough inclined plane of angle 45∘ and moves 1m in 4√2second. What is the coefficient of friction between the plane and the block? 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 The said condition can be shown as
From the figure we have N=mgcos45∘ f=μN=μmgcos45∘ So, from force equation we have, mgsin45∘−f=ma ⇒mgsin45∘−μmgcos45∘=ma ⇒a=g(sin45∘−μcos45∘)=10(1√2−μ√2)m/s2
Given, total length of the incline s=1m and time taken, t=4√2sec ∴s=ut+12at2 ⇒1=0+12×10(1√2−μ√2)(4√2)2 ⇒μ=0.8