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

Question

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

Infinity Learn · Accepted Answer

Let m be mass of the block and L be length of the inclined plane.            According to $$work-energy$$ theorem         $$W=$$Δ$$K=0$$         $$($$∵Initial and final speeds are $$zero)$$       Work done by $$friction+$$ Work done by $$gravity=$$ $$0$$   −μmgcosθ$$L2+mgsin$$θ$$L=0$$ μ$$2cos$$θ$$=$$$$sin$$θ μ$$=2sin$$θ$$cos$$θ$$=2tan$$θ       Alternate solution           For upper half smooth plane Acceleration of the block, $$a=gsin$$θHere, u $$=$$ $$0$$ $$($$∵block starts from $$rest)a=$$$$sin$$θ,$$s=L2$$ Using, $$v2$$−$$u2=2as$$,we have     $$v2$$−$$0=2$$×gsinθ×$$L2$$ $$v=gLsin$$θ            −−−(i)      For lower half rough planeAcceleration of the block,a'$$=gsin$$θ−μgcosθwhere μis the coefficient of friction between the block and lower half of the planeHere, $$u=v=gLsin$$θ,$$v=0$$  ∵  block  comes  to  rest $$a=a$$'$$=gsin$$θ−μgcosθ,  $$s=L2$$ Again , using $$v2$$−$$u2=2as$$, we have $$0$$−gLsinθ$$2=2$$×gsinθ−μgcosθ×$$L2$$ −gLsinθ$$=gsin$$θ−μgcosθL −$$sin$$θ$$=$$$$sin$$θ−μ$$cos$$θ μ$$cos$$θ$$=2sin$$θ μ$$=2tan$$θ

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

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