Cart, Hanging Mass, and the Missing Acceleration
A group of students investigates motion using a 2.0 kg cart on a horizontal track, connected by a light string over a pulley to a hanging 0.50 kg mass. A motion sensor records the cart’s acceleration as 1.6 m/s² after release. The track is not perfectly frictionless, and the pulley has noticeable friction. The students expect the cart to accelerate at about 2.0 m/s² from a simple model. They repeat the trial with an extra 0.50 kg added to the cart and obtain 1.1 m/s². The string remains taut, and the hanging mass does not hit the floor during the 0.80 m run. The students disagree about whether the missing acceleration can be treated as ‘friction’ and whether mechanical energy is conserved. They also wonder whether adding mass to the cart should change the frictional force much. Use g = 9.8 m/s² and treat the string and pulley as ideal except for the stated friction.
1. Calculate the total resistive force implied by the first trial, treating it as a constant force opposing the cart’s motion. Show the force balance for the cart and for the hanging mass, and state any assumptions.
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For the hanging mass, taking downward as positive: m₂g − T = m₂a, so T = m₂(g − a) = 0.50(9.8 − 1.6) = 4.1 N. For the cart, T − f = m₁a, so f = T − m₁a = 4.1 − (2.0)(1.6) = 0.90 N. This f is an effective total resistance opposing the cart’s motion; it lumps track friction plus any pulley-bearing friction referred to the cart. Assumptions: the string is taut and massless, the pulley’s rotational inertia is negligible, g = 9.8 m/s², and f is approximately constant over the 0.80 m run. If the pulley has friction, the tension is not exactly the same on both sides; then 0.90 N is not a pure track-friction value but still the net resistive force needed to fit the cart’s acceleration. The calculation therefore supports the students’ idea qualitatively, but the label ‘friction’ should be treated as an effective resistance.
2. Using the first trial, test the students’ claim that mechanical energy is conserved. Calculate the change in gravitational potential energy, the final kinetic energy, and the energy not accounted for after the hanging mass has fallen 0.80 m.
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Use the first trial. The hanging mass falls h = 0.80 m, so ΔU = m₂gh = (0.50)(9.8)(0.80) = 3.92 J. The final speed from v² = 2ah = 2(1.6)(0.80) = 2.56 m²/s², so v = 1.6 m/s. Total kinetic energy is ½(m₁ + m₂)v² = ½(2.5)(2.56) = 3.20 J. Thus 3.92 − 3.20 = 0.72 J is missing. This equals fh = (0.90)(0.80) = 0.72 J, so the missing mechanical energy is consistent with work done against the effective resistance and converted to thermal energy. Mechanical energy of the cart–mass system is not conserved; the system’s total energy is conserved if the thermal energy of track, pulley, and surroundings is included. The result also shows that the simple ‘no friction’ prediction would overestimate both speed and kinetic energy.
3. The second trial gives 1.1 m/s². Use the same effective resistive force f = 0.9 N to predict the acceleration when the cart mass is 2.5 kg and hanging mass 0.50 kg. Compare with the observed value and comment on whether the constant-friction model is adequate.
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With m₁ = 2.5 kg and m₂ = 0.50 kg, and keeping f = 0.90 N: a = (m₂g − f)/(m₁ + m₂) = (4.9 − 0.90)/3.0 = 1.33 m/s². The observed value is 1.1 m/s², noticeably lower. If friction were proportional to the cart’s normal force, f = μ m₁g; from the first trial μ ≈ 0.90/(2.0×9.8) = 0.046, giving f ≈ 1.13 N for m₁ = 2.5 kg and a ≈ (4.9 − 1.13)/3.0 = 1.26 m/s². Both constant-f and proportional-f models predict more than 1.1 m/s², so the simple friction model is inadequate. Possible reasons: pulley friction and rotational inertia increase with load, the extra mass changes track contact or alignment, or the string is not exactly horizontal. The students should measure the resistive force directly, for example by letting the cart coast after a push, and repeat trials rather than relying on one run.