The Cart on the Carpet
A physics group releases a dynamics cart at 3.0 m/s along a carpeted track and measures how far it slides before stopping: 2.0 m. A student writes in the lab notebook: "Friction removes a fixed amount of energy per meter, so if we release the cart at 6.0 m/s it should travel 4.0 m - twice the speed, twice the distance." The group then runs the 6.0 m/s trial on the same track with the same cart and records a stopping distance of 8.6 m. The lab instructor asks the group to decide whether the model has been supported, and whether the difference between 8.0 m and 8.6 m matters.
1. Evaluate the student's reasoning step by step. Is the premise correct? Is the conclusion correct? Support your answer with a calculation.
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Premise: correct. Kinetic friction does a fixed amount of work per meter traveled (W = f·d, with f ≈ μmg constant). So the reasoning method is sound.
Conclusion: incorrect, because the student applied the fixed energy-per-meter idea without writing the energy balance.
½mv² = f·d → d = v² / (2μg). Distance scales with the square of speed, not the first power.
Check with the 3.0 m/s data: μ = v²/(2gd) = 9.0 / (2 × 9.8 × 2.0) ≈ 0.23.
Prediction at 6.0 m/s: d = 36 / (2 × 9.8 × 0.23) ≈ 8.0 m - four times the 2.0 m, not twice.
Physically: doubling speed quadruples kinetic energy, and since friction removes a constant amount per meter, four times the energy requires four times the distance.
The lesson: the student's qualitative statement ("fixed energy per meter") is right, and the error is purely in skipping the algebra that connects it to distance.
2. The group measures 8.6 m instead of 8.0 m, a 7.5% difference. Should they conclude the model failed? Describe how you would decide, including what the model assumes and what could produce a longer-than-predicted distance.
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Not on 7.5% alone - but it must be investigated, not waved away.
Assumptions the model needs: constant coefficient of friction along the whole path, level track, a cart that slides without veering or tumbling, negligible air resistance, and an accurately known release speed.
What could make the distance longer than predicted (i.e., less deceleration than modeled):
- The actual release speed was above 6.0 m/s (timing-gate spacing, cart pushed rather than released cleanly).
- The carpet varied - a slicker patch, or the cart rode up onto a seam.
- The mass distribution shifted so the effective μ dropped; a cart rocking onto two wheels has less contact friction.
- The 3.0 m/s trial was affected by an error, so the derived μ = 0.23 is too high - the comparison is only as good as the calibration trial.
What argues the model is fine: the discrepancy is systematic and small; a 7.5% error is within what a single un-repeated trial on a soft surface can produce.
How to decide: repeat each speed several times, measure the release speed independently (photogate), mark the track so both trials use the same segment, and check whether the stopping distance is reproducible. If repeated trials cluster at 8.0 ± 0.1 m, the model stands and the 8.6 m was experimental scatter. If they cluster at 8.6 m, the friction coefficient is not constant with speed and the model needs revising - but that is a conclusion from data, not from one trial.