The trolley, the broken scale and the stopwatch
During a practical, a group pulls a 2.0 kg toy cart along a horizontal tiled floor with a spring balance held at a steady 3.0 N. One student starts a stopwatch as the cart passes a chalk mark and stops it when the cart passes a second mark 1.50 m away; the reading is 2.0 s. The stopwatch resolves 0.1 s, and the student's reaction time adds about ±0.2 s at each press. The metre scale has a broken zero end: 2.0 cm has snapped off, and the students placed the cart's front edge on the broken end and read the far mark from the scale markings. The group writes in their record: 'The cart moved at constant speed, so the floor is frictionless and no force opposed the motion.' Their teacher says the data support neither part of that sentence. Later the same cart is loaded with 2.0 kg of extra masses and pulled with the same 3.0 N on the same floor, and the group must predict the motion before trying it. Use g = 10 m/s².
1. Using the group's measurements, decide whether the floor is frictionless and estimate the friction force on the cart. State which measurement limits the precision of your answer.
Read the worked answer
Correct the length first. With 2.0 cm broken off the zero end, the true distance is 1.50 + 0.02 = 1.52 m.
Find acceleration from kinematics: s = ½at², so a = 2s/t² = (2 × 1.52)/(2.0)² = 0.76 m/s².
Find the net force: F_net = ma = 2.0 × 0.76 = 1.52 N. The pull is 3.0 N, so friction = 3.0 − 1.52 ≈ 1.5 N, and μ ≈ 1.5/(2.0 × 10) = 0.075.
Judge the group's statement: speed went from zero to about 1.5 m/s, so it was not constant, and a net force of only 1.52 N against a 3.0 N pull proves friction is not zero. Both parts of the record are wrong.
Precision: the timing error is ±0.2 s on 2.0 s, i.e. 10%, and a depends on 1/t², so a carries about 20% uncertainty. The friction force is therefore 1.5 ± 0.3 N. The 2 cm scale error is only 1.3% and is negligible. The honest report is 'about 1.5 N', never '1.48 N'.
2. Predict the motion of the cart when loaded to 4.0 kg total and pulled with the same 3.0 N on the same floor. Explain your reasoning.
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Friction depends on the normal force, so with the same surfaces μ is unchanged at about 0.075.
Loaded cart: normal force = 4.0 × 10 = 40 N, so limiting friction ≈ 0.075 × 40 = 3.0 N.
The pull is also 3.0 N, so the net force is essentially zero. Two judgements follow:
- Starting from rest: the pull must exceed the maximum static friction, which is a little larger than the sliding value. At 3.0 N it does not, so the cart most likely will not start at all.
- If already moving: kinetic friction and the pull cancel, so the cart moves with constant velocity, not with acceleration.
So the correct prediction is 'no acceleration; it either stays at rest or keeps a steady speed'. A student who simply doubles the mass and expects the acceleration to halve has forgotten that the opposing force also doubles.
3. The balance reads 3.0 N while the cart moves. A student says: 'The cart must pull back on the balance with a smaller force, otherwise it could not move forward.' Evaluate this statement.
Read the worked answer
The statement confuses an action–reaction pair with a pair of forces acting on the same body.
- By Newton's third law, the force of the balance on the cart and the force of the cart on the balance are equal in magnitude (3.0 N), opposite in direction, and act on different bodies. This is true whether the cart is at rest, speeding up or moving steadily.
- Whether the cart accelerates depends only on the forces on the cart: the 3.0 N pull and the roughly 1.5 N friction. Their resultant, not the reaction force, decides the motion.
- The reaction force acts on the balance and hence on the hand holding it; that is why the hand feels a backward pull.
A common wrong answer is 'the reaction is smaller so the cart can move'. The correct judgement is that the two forces of a third-law pair can never cancel each other because they never act on the same object.