Physics • Year 11 • Module 2: Dynamics • Lesson 6
Work and Kinetic Energy
Apply your understanding of work, kinetic energy, and the work-energy theorem to real data, multi-step scenarios, and a cause-and-effect chain.
1. Interpret data — speed and kinetic energy in road safety
The table below shows the speed and mass of five vehicles. 8 marks
| Vehicle | Mass (kg) | Speed (m/s) | KE (J) | KE relative to Vehicle A |
|---|---|---|---|---|
| A — Sedan at 30 km/h | 1200 | 8.33 | — (baseline) | |
| B — Sedan at 60 km/h | 1200 | 16.67 | ||
| C — Sedan at 90 km/h | 1200 | 25.0 | ||
| D — SUV at 60 km/h | 2400 | 16.67 | ||
| E — Motorbike at 60 km/h | 300 | 16.67 |
1.1 Calculate the kinetic energy (in J) and the KE relative to Vehicle A for each vehicle. Show your working for Vehicle A and B below the table. 5 marks
1.2 Compare vehicles B, D, and E (all travelling at 60 km/h). Explain how mass and speed each independently affect KE, using specific values from the table. 2 marks
1.3 Using your data, explain in one sentence why a speed limit reduction from 60 km/h to 30 km/h has a disproportionately large safety benefit. 1 mark
2. Interpret graph — braking distance and speed
The graph below shows the braking distance of a 1200 kg car on dry bitumen for different initial speeds. The braking force is constant at 9000 N. 7 marks
Figure 2. Braking distance vs initial speed for a 1200 kg car. Braking force = 9000 N (constant). Derived from Wnet = ΔKE: s = mv² / (2Fb). Dry bitumen, flat road. Illustrative data.
2.1 Describe the shape of the curve and explain why it is not a straight line. 2 marks
2.2 Use the graph to read off the braking distance at 20 m/s. Then use Wnet = ΔKE to verify this value algebraically. Show full working. 3 marks
2.3 The speed limit on a residential street is reduced from 50 km/h (13.9 m/s) to 40 km/h (11.1 m/s). Use the graph relationship s = mv²/(2Fb) to calculate the percentage reduction in braking distance. 2 marks
3. Compare: horizontal push vs angled pull
A 1000 kg cart starts from rest and is moved 50 m. In both scenarios F = 2000 N and μk = 0.2. 10 marks
| Quantity | Scenario A: horizontal push (θ = 0°) | Scenario B: rope at 20° above horizontal |
|---|---|---|
| Wapplied (J) — W = Fs cos θ | ||
| FN (N) — normal force | ||
| fk (N) — friction force | ||
| Wfriction (J) | ||
| Wnet (J) | ||
| vf (m/s) — from Wnet = ½mv² |
3.1 Which scenario produces the higher final speed? Explain why in terms of the work-energy theorem. 2 marks
3.2 Why does the angled pull reduce the normal force, and what effect does this have on friction? 2 marks
4. Predict and justify — cyclist on a hill
A 70 kg cyclist freewheels from rest at the top of a 12 m hill to the bottom. The road distance is 40 m. 5 marks
4.1 If there is no friction or air resistance, use ΔU = mgΔh and the work-energy theorem to find the cyclist’s speed at the bottom of the hill. 2 marks
4.2 Now a constant friction force of 80 N acts over the 40 m road distance. Predict (before calculating) whether the speed at the bottom will be more or less than the answer to 4.1. Then calculate the actual speed. 3 marks
Q1.1 — KE table
A: KE = ½ × 1200 × 8.33² = 600 × 69.4 = 41 667 J ≈ 41 700 J (baseline). B: KE = ½ × 1200 × 16.67² = 600 × 277.9 = 166 750 J ≈ 167 000 J (4× A). C: KE = ½ × 1200 × 25.0² = 600 × 625 = 375 000 J (9× A). D: KE = ½ × 2400 × 16.67² = 1200 × 277.9 = 333 500 J (8× A). E: KE = ½ × 300 × 16.67² = 150 × 277.9 = 41 700 J (≈ 1× A — same as baseline despite only ¼ the mass of B).
Q1.2 — Mass and speed comparison
Comparing B (1200 kg) and D (2400 kg) at the same speed shows doubling mass doubles KE (167 000 J vs 333 500 J). Comparing B (1200 kg) and E (300 kg) at the same speed shows KE is proportional to mass: E has ¼ the mass so ¼ the KE. Speed has a stronger effect than mass because KE ∝ v²: doubling speed quadruples KE while doubling mass only doubles KE.
Q1.3 — Speed limit benefit
Because KE ∝ v², halving speed from 60 km/h to 30 km/h reduces KE by a factor of four — meaning four times less energy must be dissipated in a collision, dramatically reducing injury severity and stopping distance.
Q2.1 — Curve shape
The curve is parabolic (concave up), not a straight line, because braking distance is proportional to v² (from s = mv²/(2Fb)). A straight line would imply s ∝ v, but the v² relationship means that each equal increment in speed produces an increasingly large increment in stopping distance.
Q2.2 — Algebraic verification at 20 m/s
Graph reading: approximately 26.7 m. Algebraic: Wnet = ΔKE → −Fb×s = 0 − ½mv² → s = mv²/(2Fb) = 1200 × 20² / (2 × 9000) = 1200 × 400 / 18 000 = 480 000 / 18 000 = 26.7 m. Consistent with graph reading.
Q2.3 — Percentage reduction in braking distance
s50 = 1200 × 13.9² / 18 000 = 1200 × 193.2 / 18 000 = 12.9 m. s40 = 1200 × 11.1² / 18 000 = 1200 × 123.2 / 18 000 = 8.2 m. Reduction = (12.9 − 8.2)/12.9 × 100 = 36%. Alternatively using the ratio: (40/50)² = 0.64, so 36% reduction. A 10 km/h speed reduction reduces stopping distance by 36%, significantly lowering the risk of fatal impact.
Q3 — Comparison table answers
Scenario A (θ = 0°): Wapplied = 2000 × 50 × 1 = 100 000 J. FN = 1000 × 9.8 = 9800 N. fk = 0.2 × 9800 = 1960 N. Wfriction = −1960 × 50 = −98 000 J. Wnet = 2000 J. vf = √(2 × 2000/1000) = 2.0 m/s.
Scenario B (θ = 20°): Wapplied = 2000 × 50 × cos20° = 2000 × 50 × 0.940 = 93 969 J. FN = 9800 − 2000×sin20° = 9800 − 684 = 9116 N. fk = 0.2 × 9116 = 1823 N. Wfriction = −1823 × 50 = −91 150 J. Wnet = 2819 J. vf = √(2 × 2819/1000) = 2.37 m/s.
Q3.1 — Higher speed explanation
Scenario B produces the higher final speed (2.37 vs 2.0 m/s). Although the angled rope does less work than a horizontal push (93 969 J vs 100 000 J), it also reduces the normal force, cutting friction significantly. The net work is higher in Scenario B (2819 J vs 2000 J), so by Wnet = ΔKE, more kinetic energy is gained.
Q3.2 — Angled pull and normal force
The upward component of the angled rope (F sin20° = 684 N) partially supports the cart’s weight, reducing the normal force from 9800 N to 9116 N. Since friction force = μk×FN, the reduced normal force means less friction (1823 N vs 1960 N), resulting in less negative work done by friction and a higher net work overall.
Q4.1 — No friction
Wgravity = mgΔh = 70 × 9.8 × 12 = 8232 J. Wnet = 8232 J (no friction). vf = √(2 × 8232 / 70) = √235.2 = 15.3 m/s.
Q4.2 — With friction
Prediction: speed will be less because friction does negative work, reducing Wnet. Wfriction = −80 × 40 = −3200 J. Wnet = 8232 − 3200 = 5032 J. vf = √(2 × 5032 / 70) = √143.8 = 12.0 m/s. Friction costs the cyclist 3.3 m/s — a 21.6% reduction in final speed.