Consider a block kept on an inclined plane (inclined at 45°) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane (μ) is equal to

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(2) 0.33. Idea: both forces act along the incline, so write each one and set the ratio to 2. At 45° the sines and cosines are equal and cancel out. Pushing up, friction acts down the slope: Fᵤₚ=mg sin 45°+μ mg cos 45°=(mg)/(√2)(1+μ). Just preventing the slide, friction acts up the slope: F_down=mg sin 45°-μ mg cos 45°=(mg)/(√2)(1-μ). Setting Fᵤₚ=2F_down: 1+μ=2(1-μ) 1+μ=2-2μ 3μ=1, so μ=1/3≈0.33. Note the book prints 0.50 and 0.5 as two separate options; both are the same number and neither is the answer.

Consider a block kept on an inclined plane (inclined at 45°) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane (μ) is equal to

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Why the other options are wrong