Trolley on a Ramp: Is Friction Really Negligible?
A transition-year team is testing energy conservation with a 0.500 kg trolley on a 1.20 m ramp. The ramp's upper end is 0.168 m above the lower end, and a 0.050 m card on the trolley passes through a light gate at the bottom. Times from four releases are 0.032 s, 0.032 s, 0.033 s and 0.036 s. Using v = 0.050/t, the team gets 1.56 m/s, 1.56 m/s, 1.52 m/s and 1.39 m/s. They compare the kinetic energy at the gate with the loss in gravitational potential energy, find a shortfall, and conclude that friction is negligible but that the 0.036 s run is 'bad data'. The bench has a slight tilt, and the trolley's wheels feel stiff. They ask whether their conclusion is justified and how to report a defensible value.
1. Calculate the frictionless speed at the gate and compare it with the four measured speeds. Does the data support the claim that friction is negligible?
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Reasoning. For the ramp, sin θ = 0.168/1.20 = 0.140. The frictionless acceleration is a = g sin θ = 9.81 × 0.140 = 1.37 m s⁻². From v² = 2as, v = √(2 × 1.37 × 1.20) = 1.81 m s⁻¹. The four measured values are 1.56, 1.56, 1.52 and 1.39 m/s, all below 1.81 m/s. Conclusion. The data do not support 'friction is negligible': even the fastest run is 14% below the frictionless value. The card-average speed is close to the gate speed, so timing through the card cannot explain a 0.25 m/s shortfall. Friction (and possibly rotational energy of stiff wheels) is significant.
2. Should the 0.036 s reading be discarded? Outline a defensible procedure for deciding and for reporting the final speed.
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Do not discard automatically. With only four readings, a single low value may be a real sticking event or a mistrigger. The 0.036 s value is 9.7% below the median (1.54 m/s), which is large but not impossible. Check gate alignment, card edge, ramp joints and wheel stiffness; repeat at least six to ten releases. If the repeat distribution is tight around 1.54 m/s and the 1.39 m/s value recurs only when a wheel sticks, report the cause and exclude it. Otherwise quote the median with a range, e.g. 1.54 ± 0.03 m s⁻¹, and state that one reading was low. Selecting data only because it fits a conclusion is not acceptable.
3. Using the median speed, estimate the average friction force and coefficient of kinetic friction. State your assumptions.
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Using the median v = 1.54 m s⁻¹: E_k = ½(0.500)(1.54)² = 0.593 J. The potential-energy fall is mgh = 0.500 × 9.81 × 0.168 = 0.824 J. The missing 0.231 J is work done against friction over 1.20 m, so F_f = 0.231/1.20 = 0.193 N. The normal force is mg cos θ = 0.500 × 9.81 × 0.990 = 4.86 N, giving μ_k = 0.193/4.86 = 0.040. Assumptions: friction is constant, the card speed represents the gate speed, and wheel rotational energy is either included in the 'friction' term or negligible. The value is an effective coefficient, not a pure sliding value.