Question

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​

Answer 1

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