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it takes a block twice as long to slide down a plane inclined at 45 to the horizontal as it would take for it to slide down the same plane had the plane had the plane been frictionless. Calculate the coefficient of friction between block and the inclined plane.
Let coefficient of friction be q then T=√2d/gsin 45 nT = √2d/gsin45- qgcos45 Equationg this we get q= tan 45( n*n -1/ n* n) On solving we get q= tan 45(2*2-1/2*2) (4-1/ 4) (3/4) Cheers!!!
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