Physics • Year 11 • Module 2 • Lesson 4
Newton’s Laws and Friction
Build HSC Band 5–6 extended-response technique on Newton’s Laws, multi-step friction calculations, and evaluation of experimental design.
1. Multi-step calculation — force at an angle (Band 5–6)
8 marks Band 5–6
Scenario. A 25 kg crate is pushed across a warehouse floor by a force of 150 N directed at 30° below the horizontal. The coefficient of kinetic friction between the crate and the floor is μk = 0.28. Use g = 9.8 m s−2, sin 30° = 0.500, cos 30° = 0.866.
Q1. Applying the Vector Protocol, calculate the acceleration of the crate. In your response you must:
- Define the positive direction and draw a labelled free body diagram (or describe one fully in words).
- Calculate the normal force, explaining why it is greater than mg.
- Calculate the kinetic friction force.
- Calculate the net horizontal force and the acceleration.
- State one assumption you made and assess whether it is reasonable.
2. Data + scenario: road surface testing for RMS (Band 5–6)
7 marks Band 5–6
Scenario. Roads and Maritime Services (NSW) tested three road surfaces to compare their wet-weather braking performance. A 1 600 kg car braked from 80 km h−1 on each surface. The table shows results from the tests.
| Surface type | Braking force (N) | Deceleration (m s−2) | Stopping distance (m) |
|---|---|---|---|
| Standard bitumen | 9 408 | 5.88 | 42.0 |
| Open-graded asphalt | 11 760 | 7.35 | 33.6 |
| Polished concrete | 5 880 | 3.68 | 67.2 |
Illustrative data. Initial speed: 80 km h−1 = 22.22 m s−1. All surfaces: wet conditions.
Q2. Analyse and evaluate the data above to assess the road safety performance of the three surfaces in wet conditions. In your response you must:
- Calculate μk for each surface and use these values to rank safety performance.
- Apply Newton’s Second Law to explain why the higher-friction surface produces a shorter stopping distance.
- Assess the claim: “Open-graded asphalt is always superior to standard bitumen and should replace it on all NSW roads.” Identify at least two limitations of this claim using the data and your physics knowledge.
- Propose one modification to the experimental design that would improve the validity of the comparison.
Q1 — Sample Band 6 response (8 marks), annotated
Step 1 — Positive direction + FBD: Define forward = positive, upward = positive. FBD: Weight W = 25×9.8 = 245 N downward; Fapplied = 150 N at 30° below horizontal (horizontal component forward, vertical component downward); FN upward; fk backward. [1 — positive direction defined and FBD described with all four forces correctly placed]
Step 2 — Normal force: The applied force has a downward vertical component (F sin30° = 150×0.500 = 75 N), which adds to the weight pressing the crate into the floor. Vertically: FN − W − F sin30° = 0 (no vertical acceleration). FN = mg + F sin30° = 245 + 75 = 320 N. FN > mg because the downward force component of the push presses the crate harder against the floor. [1 — correct FN with reason for it exceeding mg]
Step 3 — Kinetic friction: fk = μkFN = 0.28×320 = 89.6 N (backward, negative). [1]
Step 4 — Net horizontal force and acceleration: Horizontal component of applied force: F cos30° = 150×0.866 = 129.9 N (forward). Fnet = 129.9 − 89.6 = 40.3 N (forward). a = Fnet/m = 40.3/25 = 1.61 m s−2 (forward). [1 — correct horizontal Fnet] [1 — correct acceleration with direction]
Assumption and assessment: Key assumption: the floor is perfectly horizontal and rigid (no vertical acceleration of the crate other than from the applied force). This is reasonable for a flat warehouse floor. A secondary assumption: μk is constant regardless of speed; in reality μk can vary slightly with sliding speed on some surfaces, making this a minor limitation for high-speed scenarios. [1 — states one valid assumption; 1 — assesses reasonableness]
Marking criteria (8 marks): 1 = correct positive direction and FBD with all four forces; 1 = correct FN = 320 N with explanation of why it exceeds mg; 1 = correct fk = 89.6 N; 1 = correct horizontal Fnet = 40.3 N; 1 = correct a = 1.61 m s−2 with direction; 1 = states a valid assumption; 1 = assesses reasonableness of assumption; 1 = uses precise physics terminology throughout (FN, fk, net force, Vector Protocol steps).
Q2 — Sample Band 6 response (7 marks), annotated
Coefficients of kinetic friction: μk = a/g (since f = μmg and a = f/m = μg). Standard bitumen: μk = 5.88/9.8 = 0.60. Open-graded asphalt: μk = 7.35/9.8 = 0.75. Polished concrete: μk = 3.68/9.8 = 0.376. Safety ranking (highest to lowest): open-graded asphalt (0.75) > standard bitumen (0.60) > polished concrete (0.376). [1 — three μk values correct; 1 — ranking]
Newton’s Second Law explanation: Higher friction means a larger braking force (f = μmg) for the same mass. By Newton’s Second Law (a = Fnet/m = μg), a higher μk produces a greater deceleration. From kinematics (v2 = u2 + 2as), a larger deceleration (larger magnitude of a, same initial speed u) yields a shorter stopping distance s = u2/(2a). Open-graded asphalt has μk = 0.75 vs bitumen 0.60, giving 25% greater deceleration and proportionally shorter stopping distance (33.6 m vs 42.0 m). [1]
Evaluate the claim (two limitations): Limitation 1: The data is from wet conditions only. In dry conditions, polished concrete and standard bitumen may perform comparably to or better than open-graded asphalt; μk values differ between wet and dry. A claim about “all NSW roads” should consider both weather conditions. Limitation 2: The economic and maintenance cost of open-graded asphalt is higher than standard bitumen; it also has a shorter service life and is more prone to rutting in heavy traffic. A “superior on all roads” claim ignores these practical constraints that road engineers must balance with safety. Additional acceptable limitations: tested with one vehicle type/mass only; single tyre type; test was single trial (no repeat for reliability). [1 per limitation, max 2]
Design improvement: Repeat the test with at least three trials per surface under identical conditions (same vehicle, tyres, temperature, water depth) and calculate mean deceleration ± standard deviation to assess reliability and identify outliers. Alternatively, test with multiple vehicle types (truck, motorcycle, SUV) to improve the external validity of the comparison. [1]
Marking criteria (7 marks): 1 = correct μk for all three surfaces; 1 = correct safety ranking with justification; 1 = explains shorter stopping distance via Newton’s Second Law and kinematics; 1 = first limitation of the claim with reasoning; 1 = second limitation; 1 = states a valid design improvement; 1 = uses precise terminology throughout (μk, Fnet, deceleration, validity, reliability).