Friction on a Ramp: Separating Static and Kinetic Coefficients
A group investigates friction using a 0.50 kg wooden block on a ramp. At a ramp angle of 20°, the block remains stationary after being placed gently. At 30°, the block initially stays still but slides after a light tap; a motion sensor records its velocity along the ramp. From t = 0.20 s to t = 0.80 s, the velocity data fit v = 0.85 + 1.6t, with v in m/s and t in s. The students use g = 9.8 m/s² and assume the ramp is uniform. One student writes, ‘At 30°, μk = tan 30° = 0.58, so friction is stronger than on the 20° trial.’ Another objects that the block was tapped, so the motion data must be ignored. The teacher asks for a quantitative decision about the coefficient of kinetic friction and whether the 20° observation is consistent with the 30° result. Air resistance is negligible, and the block moves straight down the slope.
1. Use the velocity fit to calculate the acceleration and the coefficient of kinetic friction. Show the force balance.
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The velocity fit is linear, so its slope is the acceleration: a = 1.6 m/s² down the slope. Along the slope, mg sin θ acts down and kinetic friction f_k = μ_k mg cos θ acts up. Newton's second law gives ma = mg sin θ − μ_k mg cos θ, so μ_k = (g sin θ − a)/(g cos θ). Substituting θ = 30°, g = 9.8, a = 1.6: μ_k = (4.9 − 1.6)/(9.8 × 0.866) = 3.3/8.49 ≈ 0.39. The intercept 0.85 m/s is the speed at t = 0 in the fit, not an acceleration; the block was already moving during the interval used. The result is a kinetic coefficient, valid only while sliding.
2. Evaluate the student's tan 30° claim and the objection that tapping invalidates the data. What can be concluded about μs at 20° and at 30°?
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tan 30° = 0.577 is not μ_k here because the block is accelerating; tan θ equals μ only when the block slides at constant velocity or is at the limit of static friction. The tap does not invalidate the slope: it supplies initial speed to break static friction, but once sliding, the acceleration is set by gravity and kinetic friction. The fact that the block initially stayed at rest at 30° is useful: static friction held, so μ_s ≥ tan 30° = 0.577. At 20°, μ_s ≥ tan 20° = 0.364. Both are consistent with μ_k ≈ 0.39 and with the usual μ_s > μ_k; the 30° trial gives the stronger lower bound on μ_s, not a value of μ_k.
3. Calculate the distance traveled between t = 0.20 s and t = 0.80 s. Is a constant-acceleration model reasonable for this interval?
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At t = 0.20 s, v = 0.85 + 1.6(0.20) = 1.17 m/s. At t = 0.80 s, v = 0.85 + 1.6(0.80) = 2.13 m/s. For constant acceleration, average velocity is (1.17 + 2.13)/2 = 1.65 m/s, so Δx = 1.65 × 0.60 = 0.99 m. A linear v-t fit is exactly a constant-acceleration model, and the forces are consistent with constant μ_k if friction is independent of speed. The stated neglect of air resistance supports that. The model should not be extrapolated far outside the fitted interval without evidence that μ_k remains constant.