Physics • Year 11 • Module 2: Dynamics • Lesson 3
Inclined Planes
Apply the inclined plane force formulas to real data, worked calculations, and reasoning questions that match HSC Band 3–4 style.
1. Interpret experimental data — trolleys on airport ramps
An airport engineer tests five baggage ramps of different angles. A 20 kg test case is released from rest on each dry ramp (μ = 0.18). The table shows the ramp angle and the measured acceleration. 9 marks
| Ramp angle (θ) | Measured acceleration (m/s²) | Predicted acceleration using a = g(sin θ − μ cos θ) (m/s²) | Motion: slides / stays still |
|---|---|---|---|
| 5° | 0 (does not slide) | ||
| 15° | 0.81 | ||
| 25° | 2.48 | ||
| 35° | 3.97 | ||
| 45° | 5.62 |
Illustrative data. Use g = 9.8 m/s² and μ = 0.18 for all calculations. Round to 2 decimal places.
(a) Complete the “Predicted acceleration” column using the formula a = g(sin θ − μ cos θ). For θ = 5°, predict whether the object slides or stays still, and explain using the critical angle condition. (5 marks)
(b) Compare the measured and predicted accelerations for 25° and 45°. Identify one possible source of error that could explain any difference. (2 marks)
(c) Predict what the acceleration would be for the same ramp angles if the test case mass were doubled to 40 kg. Justify your answer. (2 marks)
2. Cause-and-effect chain — increasing the slope angle
Complete the cause-and-effect chain showing what happens to each quantity as a slope angle increases from 20° to 40°. Fill in the empty effect boxes. 5 marks (1 per correct effect)
Overall outcome:
3. Graph interpretation — acceleration vs slope angle
The graph below shows the theoretical acceleration of a block down a frictionless slope (solid line) and a slope with μ = 0.30 (dashed line) as slope angle varies from 0° to 60°. Use g = 9.8 m/s². 8 marks
(a) Describe the trend shown by the frictionless line and explain why the relationship is not linear (hint: think about what sin θ does as θ increases). (2 marks)
(b) Identify the approximate slope angle at which the dashed line first leaves zero. Explain what this angle represents physically. (2 marks)
(c) Estimate from the graph the acceleration of the block with μ = 0.30 at θ = 30°. Verify this estimate using the formula a = g(sin 30° − 0.30 × cos 30°). (2 marks)
(d) The vertical gap between the two lines represents the deceleration caused by friction. Describe what happens to this gap as the angle increases from 0° to 60° and explain why, using the friction formula f = μmg cos θ. (2 marks)
4. Predict and justify — icy road conditions
4 marks
Scenario. A 1200 kg car is parked on a sealed road that slopes at 12° to the horizontal. In dry conditions, the coefficient of static friction between the tyres and road is μs = 0.70. Overnight, the temperature drops and the road surface becomes icy, reducing the coefficient of static friction to μice = 0.09.
Predict whether the car will remain stationary or begin to slide in each condition. Justify your prediction by comparing W∥ with the maximum static friction force fmax in each case. Show calculations. Then state how the car’s mass affects your conclusion.
Q1 — Airport ramps data table
Predicted accelerations (a = 9.8(sin θ − 0.18 cos θ)):
- 5°: 9.8(0.0872 − 0.18×0.9962) = 9.8(0.0872 − 0.1793) = 9.8×(−0.0921) < 0. Negative result means the block does not slide. The critical angle is arctan(0.18) ≈ 10.2°; since 5° < 10.2°, the block stays still. Measured: 0. Consistent.
- 15°: 9.8(0.2588 − 0.18×0.9659) = 9.8(0.2588 − 0.1739) = 9.8×0.0849 ≈ 0.83 m/s². Measured: 0.81.
- 25°: 9.8(0.4226 − 0.18×0.9063) = 9.8(0.4226 − 0.1631) ≈ 2.54 m/s². Measured: 2.48.
- 35°: 9.8(0.5736 − 0.18×0.8192) = 9.8(0.5736 − 0.1475) ≈ 4.18 m/s². Measured: 3.97.
- 45°: 9.8(0.7071 − 0.18×0.7071) = 9.8×0.7071×0.82 ≈ 5.68 m/s². Measured: 5.62.
Marking: 1 mark for each of the five predicted values (with correct method shown) + 1 mark for the critical angle explanation at 5° = 5 marks total.
(b) Predicted values are slightly higher than measured. Possible sources: air resistance not included in the model; actual kinetic friction may be slightly higher than 0.18; the case may not start from a true zero-velocity rest position (systematic timing error). Award 1 mark for identifying a plausible source of error with a reason tied to the formula.
(c) The acceleration would be identical. Mass cancels from a = g(sin θ − μ cos θ). Both W∥ and f increase proportionally with m, so the net acceleration is unchanged. (1 mark correct prediction + 1 mark for explaining mass cancellation.)
Q2 — Cause-and-effect chain
Cause box 1 effect: W∥ = mg sin θ increases (larger downslope component of weight).
Cause box 2 effect: W∥ = mg sin θ increases.
Cause box 3 effect: N = mg cos θ decreases.
Cause box 4 effect: fmax = μN = μmg cos θ decreases.
Cause box 5 effect: Net force (Fnet = W∥ − f) increases significantly.
Overall outcome: The object accelerates more rapidly down the slope; it is less likely to remain stationary. Accept: “acceleration increases as θ increases.”
Q3 — Graph interpretation
(a) The frictionless line increases from 0 at 0° to approximately 8.5 m/s² at 60°. The relationship is not linear because a = g sin θ and sin θ is a non-linear (sinusoidal) function — the rate of increase of sin θ slows as θ approaches 90°. (1 mark describe trend; 1 mark explain non-linearity via sin.)
(b) The dashed line leaves zero at approximately 17°. This is the critical angle θc = arctan(0.30) ≈ 16.7°. Below this angle, maximum static friction exceeds W∥ so the object does not slide. (1 mark identify ~17°; 1 mark correct explanation using critical angle concept.)
(c) Graph estimate: approximately 2.6 m/s². Calculated: a = 9.8(sin 30° − 0.30×cos 30°) = 9.8(0.500 − 0.300×0.866) = 9.8(0.500 − 0.260) = 9.8×0.240 ≈ 2.35 m/s². (1 mark graph estimate consistent with graph; 1 mark correct calculation.)
(d) The gap decreases as angle increases. The friction deceleration is μg cos θ; as θ increases, cos θ decreases, so the friction contribution to deceleration decreases. At 90° the normal force (and therefore friction) becomes zero. (1 mark describe gap decreasing; 1 mark explain via cos θ decreasing in the friction formula.)
Q4 — Icy road prediction
Dry conditions (μ = 0.70):
W∥ = mg sin 12° = 1200×9.8×0.2079 ≈ 2445 N.
fmax = μmg cos 12° = 0.70×1200×9.8×0.9781 ≈ 8074 N.
Since fmax > W∥, the car stays still.
Icy conditions (μ = 0.09):
fmax = 0.09×1200×9.8×0.9781 ≈ 1038 N.
Since fmax < W∥ (1038 < 2445), the car slides.
Mass effect: Mass cancels when comparing W∥ and fmax. Both are proportional to mg, so whether the car slides is independent of mass. A heavier car would not stay still any more securely on the icy slope. (1 mark dry condition correct with working; 1 mark icy condition correct; 1 mark mass effect; 1 mark clear comparison/conclusion.)