a body takes twice as much time to slide down a rough inclined plane of inclination theta with the horizontal then its slider down over identical but a smooth inclined plane find the coefficient of kinetic energy
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The accelerations of the block sliding down a smooth and rough 45∘ inclined planes respectively are
a1=gsin45∘=g2–√
a2=g(sin45∘−μkcos45∘)=g2–√(1−μk)
Where μk is the coefficient of kinetic friction.
Now,we know that the square of the time of slide is inversely proportional to the acceleration.
Therefore t22t21=a1a2=11−μk
Since t2=2t1,we have
4=11−μk
μk=0.75
Hence the correct answer is 0.7
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