A wooden crate is released from rest at the top of a ramp inclined at 37^ to the horizontal. When the ramp is perfectly smooth, the crate reaches the bottom with a speed v . When the ramp is rough, the crate reaches the

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Correct answer: B. (2)/(3)

Let the length of the ramp be L and the angle of inclination be θ = 37^ . For the smooth ramp, the acceleration of the crate is a_1 = g θ . Using the equation of motion, v^2 = 2a_1L = 2gL θ . For the rough ramp, the acceleration of the crate is a_2 = g θ - μ_k g θ . The velocity at the bottom is (v)/(3) , so ((v)/(3))^2 = 2a_2L = 2gL( θ - μ_k θ) . Dividing the second equation by the first, we get: (1)/(9) = ( θ - μ_k θ)/( θ) (1)/(9) = 1 - μ_k θ μ_k θ = 1 - (1)/(9) = (8)/(9) μ_k = (8)/(9) θ Given 37^ = (3)/(4) , substituting this value: μ_k = (8)/(9) × (3)/(4) = (2)/(3) Answer: (2)/(3)

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