262work done total = change in mechanical energy
let height on total incline be h
hence work done by friction=μ .N.d =μ . (mg cosθ) (h/2. cosecθ)
change in mechanical energy = mgh
henced equating the two, we have μ =2tanθ
A body starts sliding down from the top of an inclined plane. The upper half of the inclined plane AB is smooth and the lower half BC is rough. The block just comes to rest at the bottom. Find the coefficient of friction between the block and lower half of the inclined plane?
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5 Answers
262
1oops sry.. even i was getting 2tanθ only.
another way of doing it is using v^2 - u^2 = 2as to find the acc. for half the incline and then equating to gsinθ-ugcosθ
