A cubic block of mass m is sliding down on an inclined plane at 60° with an acceleration of g/2, the value of coefficient of kinetic friction is

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(3) √3-1. Idea: write the equation of motion down the slope and solve for μ. ma=mg sin θ-μ mg cos θ, so a=g(sin θ-μ cos θ). With θ=60°, sin 60°=(√3)/2, cos 60°=1/2 and a=g/2: 1/2=(√3)/2-μ/2. Multiplying through by 2, 1=√3-μ, so μ=√3-1≈0.73. Sanity check: μ must be less than tan 60°=√3≈1.73 or the block would not slide at all, and 0.73 comfortably is.

A cubic block of mass m is sliding down on an inclined plane at 60° with an acceleration of g/2, the value of coefficient of kinetic friction is

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