The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

Let m be mass of the block and L be length of the inclined plane.            According to work-energy theorem         W=ΔK=0         (∵Initial and final speeds are zero)       Work done by friction+ Work done by gravity= 0   −μmgcosθL2+mgsinθL=0 μ2cosθ=sinθ μ=2sinθcosθ=2tanθ       Alternate solution           For upper half smooth plane Acceleration of the block, a=gsinθHere, u = 0 (∵block starts from rest)a=sinθ,s=L2 Using, v2−u2=2as,we have     v2−0=2×gsinθ×L2 v=gLsinθ            −−−(i)      For lower half rough planeAcceleration of the block,a'=gsinθ−μgcosθwhere μis the coefficient of friction between the block and lower half of the planeHere, u=v=gLsinθ,v=0  ∵  block  comes  to  rest a=a'=gsinθ−μgcosθ,  s=L2 Again , using v2−u2=2as, we have 0−gLsinθ2=2×gsinθ−μgcosθ×L2 −gLsinθ=gsinθ−μgcosθL −sinθ=sinθ−μcosθ μcosθ=2sinθ μ=2tanθ