Physics • Year 11 • Module 2 • Lesson 4
Newton’s Laws and Friction
Apply Newton’s Laws and the friction equation to real scenarios, experimental data, and diagram analysis.
1. Interpret experimental data — braking distances at different loads
A driver tests the braking distance of a delivery van at the same initial speed (60 km h−1) with different loads. The table below records the results. 8 marks
| Trial | Mass of van (kg) | Braking force (N) | Deceleration (m s−2) | Braking distance (m) |
|---|---|---|---|---|
| 1 — Empty | 1 400 | 7 000 | calculate | 26.8 |
| 2 — Half load | 1 900 | 9 500 | calculate | 26.8 |
| 3 — Full load | 2 400 | 12 000 | calculate | 26.8 |
Illustrative data. Initial speed: 60 km h−1 = 16.67 m s−1.
1.1 Calculate the deceleration for each trial using a = Fnet/m. Show working. 3 marks (1 each)
1.2 Describe the trend in deceleration across the three trials. Explain why braking distance is the same despite the mass increasing. Refer to Newton’s Second Law in your answer. 3 marks
1.3 Calculate the coefficient of kinetic friction μk for the tyres and road surface, using Trial 1 data. 2 marks
2. Interpret graph — applied force vs acceleration for a 20 kg box
A student pushes a 20 kg box across a flat floor with progressively larger horizontal forces and measures the acceleration each time. The graph below shows the results. 7 marks
Figure 2.1. Acceleration of a 20 kg box vs applied horizontal force on a flat floor. Illustrative data; μk = 0.30.
2.1 Describe the trend in the graph from 0 N to 60 N, and from 60 N upward. Explain what is happening physically in each region. 3 marks
2.2 Use the graph to determine the kinetic friction force acting on the box. Explain your method. 2 marks
2.3 A student concludes: “Doubling the applied force doubles the acceleration.” Identify the flaw in this reasoning and write a corrected conclusion. 2 marks
3. Compare Newton’s First and Second Laws across five features
Complete the two-column table. For each feature, write a concise description that contrasts the two laws. 10 marks (1 per cell)
| Feature | Newton’s First Law | Newton’s Second Law |
|---|---|---|
| Net force condition | ||
| Object’s motion | ||
| Mathematical form | ||
| Delivery van scenario | ||
| Real-world example |
4. Predict and justify — pushing a heavy crate
A 25 kg crate sits stationary on a flat warehouse floor. The coefficient of static friction between the crate and floor is μs = 0.48 and the coefficient of kinetic friction is μk = 0.32. Use g = 9.8 m s−2. 5 marks
4.1 A horizontal force of 100 N is applied to the crate. Calculate the maximum static friction force (fs,max = μsFN). Predict whether the crate remains stationary or begins to slide. Justify your prediction by comparing the applied force with fs,max. Show all working. 3 marks
4.2 Once the crate is sliding, calculate the kinetic friction force and determine the net force and acceleration of the crate if the 100 N applied force continues. State which Newton’s Law applies now that the crate is in motion. 2 marks
Q1.1 — Decelerations
Trial 1: a = F/m = 7 000 / 1 400 = 5.0 m s−2. Trial 2: a = 9 500 / 1 900 = 5.0 m s−2. Trial 3: a = 12 000 / 2 400 = 5.0 m s−2.
Q1.2 — Trend and explanation
The deceleration is identical (5.0 m s−2) in all three trials. As mass increases, the braking force increases proportionally (both scale with the normal force via the same μk). By Newton’s Second Law, a = F/m = μmg/m = μg; mass cancels out. Because the deceleration is the same and the initial speed is the same, the braking distance (v²/2a) is also the same in all trials.
Q1.3 — Coefficient of kinetic friction
FN = mg = 1 400 × 9.8 = 13 720 N. μk = f / FN = 7 000 / 13 720 ≈ 0.51.
Q2.1 — Graph description
From 0 N to 60 N: the acceleration remains zero. The applied force is not yet sufficient to overcome the maximum static friction, so the box remains stationary. Above 60 N: the box begins to slide and the acceleration increases linearly with applied force. Once sliding, kinetic friction is constant (60 N), so each extra newton of applied force produces the same extra acceleration (0.05 m s−2 per N beyond 60 N).
Q2.2 — Kinetic friction from graph
The x-intercept of the linear region is 60 N. At that threshold, the net force is zero (a = 0), meaning the applied force exactly equals kinetic friction. Therefore fk = 60 N. This can also be verified: fk = μkmg = 0.30×20×9.8 ≈ 58.8 N (consistent with the graph to 2 significant figures).
Q2.3 — Flaw in student’s conclusion
The student is incorrect because F = ma uses the net force, not the applied force. Kinetic friction (60 N) is constant. If the applied force doubles from, say, 80 N to 160 N, the net force goes from 20 N to 100 N — a fivefold increase, not a doubling. Corrected conclusion: only the net force is proportional to acceleration; doubling the applied force increases the net force by more than double when friction is present.
Q3 — Compare and contrast table
Net force condition: N1 — Fnet = 0. N2 — Fnet ≠ 0 (unbalanced force). Object’s motion: N1 — at rest or constant velocity. N2 — accelerating (changing velocity). Mathematical form: N1 — Fnet = 0. N2 — Fnet = ma. Delivery van scenario: N1 — packages (no horizontal force) continue at original velocity. N2 — van and belted driver (friction/seatbelt force) decelerate. Real-world example: N1 — book resting on a table; car cruising at constant speed. N2 — car accelerating from traffic lights; box sliding and decelerating due to friction.
Q4.1 — Crate on flat floor (will it slide?)
FN = mg = 25×9.8 = 245 N. fs,max = μsFN = 0.48×245 = 117.6 N. The applied force is 100 N. Since 100 N < fs,max = 117.6 N, static friction can fully oppose the applied force. The crate remains stationary. (1 mark for correct fs,max; 1 mark for correct comparison; 1 mark for correct conclusion.)
Q4.2 — Kinetic friction once sliding
fk = μkFN = 0.32×245 = 78.4 N (backward). Fnet = Fapplied − fk = 100 − 78.4 = 21.6 N (forward). a = Fnet/m = 21.6/25 = 0.86 m s−2 (forward). Newton’s Second Law applies: a non-zero net force acts on the sliding crate, producing acceleration in the direction of the net force. (1 mark for correct fk and Fnet; 1 mark for correct acceleration and Newton’s Second Law.)