Physics • Year 11 • Module 2: Dynamics • Lesson 9
Energy Synthesis
Apply your understanding of the six Phase 2 energy formulae to calculations, data interpretation, and scenario analysis.
1. Follow the synthesis chain — cyclist on a hill
A 70 kg cyclist starts from rest at the top of a frictionless 12 m hill and rides down to a flat road. On the flat road the drag force is 80 N and the cyclist’s power output is 400 W. 8 marks
1.1 Calculate the speed of the cyclist at the bottom of the hill. State which formula you used and why it is the appropriate choice here. 3 marks
1.2 When the cyclist reaches the flat road and maintains a constant speed, calculate that constant cruise speed. Show your reasoning using Newton’s first law before applying P = Fv. 2 marks
1.3 The cyclist cannot maintain the hill speed (from 1.1) on the flat road. Use your answers to explain quantitatively why this is the case. What power would be needed to maintain the hill speed against drag? 3 marks
2. Interpret experimental data — energy audit on a ramp
A student releases a 2.0 kg block from rest at the top of a 5.0 m high ramp. A sensor records the block’s speed at the bottom. The experiment is repeated three times. 8 marks
| Trial | Height (m) | Speed at bottom (m/s) | Actual KE at bottom (J) | Ideal KE (frictionless, J) | Energy lost (J) |
|---|---|---|---|---|---|
| 1 | 5.0 | 8.4 | |||
| 2 | 5.0 | 8.6 | |||
| 3 | 5.0 | 8.5 |
2.1 Calculate the ideal KE at the bottom (frictionless, from rest) and fill in all three rows of the table. Use g = 9.8 m/s². 4 marks (1 for ideal KE; 1 per trial for actual KE and energy lost)
2.2 A student concludes: “Conservation of mechanical energy does not apply here because energy was lost.” Identify the flaw in this reasoning and write a more precise conclusion with reference to which formula should be used and why. 2 marks
2.3 Identify one physical mechanism responsible for the energy loss and explain where that energy went. 2 marks
3. Compare the two work formulae
Complete the two-column comparison table below. 8 marks (1 per cell)
| Feature | W = Fs cosθ | Wnet = ΔKE |
|---|---|---|
| What it links | ||
| Derived from | ||
| When to use it | ||
| Common error |
4. Predict and justify — a loaded truck descending
A 3500 kg loaded truck travels at a constant speed of 25 m/s along a highway. The engine is cut off and the truck descends a hill (vertical height 8 m, road distance 100 m, μk = 0.05, engine off). 6 marks
4.1 Without calculating, predict qualitatively whether the truck will be faster or slower at the bottom of the hill compared to its highway speed. Justify your prediction using the energy categories from Card 1. 2 marks
4.2 Identify which Phase 2 formula you cannot use to find the speed at the bottom, and explain why. 2 marks
4.3 Outline the calculation steps you would follow (no full calculation required) to find the truck’s speed at the bottom. Name the formula used at each step and state the quantities that feed into each calculation. 2 marks
Q1.1 — Speed at bottom of hill (3 marks)
Formula: conservation of mechanical energy (KE1 + U1 = KE2 + U2). Appropriate because the hill is frictionless — only gravity does work.
v = √(2gh) = √(2 × 9.8 × 12) = √235.2 = 15.3 m/s [1 formula + reasoning; 1 substitution; 1 answer with units].
Marking criteria: 1 = states conservation of energy with justification (frictionless); 1 = correct substitution (mass cancels); 1 = v = 15.3 m/s with unit.
Q1.2 — Cruise speed on flat road (2 marks)
Constant speed → Newton’s first law: Fdrive = Fdrag = 80 N. Using P = Fv: v = P/F = 400/80 = 5.0 m/s.
Marking criteria: 1 = states Fdrive = Fdrag = 80 N via Newton 1; 1 = v = 5.0 m/s.
Q1.3 — Why hill speed cannot be maintained (3 marks)
At hill speed 15.3 m/s, power needed to overcome drag = Fdrag × v = 80 × 15.3 = 1224 W. Cyclist’s maximum power output is only 400 W — far less than the 1224 W required. Therefore the cyclist decelerates on the flat road until reaching 5.0 m/s (the speed where P needed = P available).
Marking criteria: 1 = calculates P needed at hill speed (80 × 15.3 = 1224 W); 1 = compares to available power (400 W < 1224 W); 1 = explains cyclist must slow down until P needed = P available.
Q2.1 — Energy audit table (4 marks)
Ideal KE (frictionless, from rest) = mgh = 2.0 × 9.8 × 5.0 = 98.0 J (same for all trials).
Trial 1: Actual KE = ½ × 2.0 × 8.4² = 70.6 J; Energy lost = 98.0 − 70.6 = 27.4 J.
Trial 2: Actual KE = ½ × 2.0 × 8.6² = 74.0 J; Energy lost = 98.0 − 74.0 = 24.0 J.
Trial 3: Actual KE = ½ × 2.0 × 8.5² = 72.3 J; Energy lost = 98.0 − 72.3 = 25.8 J.
Marking criteria: 1 = correct ideal KE 98.0 J; 1 per trial for both actual KE and energy lost correct (3 marks).
Q2.2 — Flaw in student’s reasoning (2 marks)
Flaw: The student confuses “conservation of mechanical energy does not hold” with “total energy is not conserved.” Total energy IS always conserved — the “lost” mechanical energy was converted to heat and sound by friction, not destroyed. The correct conclusion is: friction is present, so we cannot use KE1 + U1 = KE2 + U2; instead the work-energy theorem (Wnet = ΔKE) must be used, including the negative work done by friction.
Marking criteria: 1 = identifies flaw (total energy is still conserved; mechanical energy → heat); 1 = states Wnet = ΔKE is the correct formula to use (with friction present).
Q2.3 — Energy loss mechanism (2 marks)
Kinetic friction between the block and the ramp surface. The friction force acts over the length of the ramp and does negative work on the block (Wfriction = −fks). This work converts the block’s kinetic energy into thermal energy (heat) in the block and ramp surfaces. Accept also: air resistance (smaller contribution).
Marking criteria: 1 = names friction (kinetic) as the mechanism; 1 = states energy goes to heat/thermal energy in the surfaces.
Q3 — Compare and contrast table
What it links: W = Fs cosθ links force, displacement and direction → gives energy transferred. Wnet = ΔKE links all work done by all forces → gives the resulting change in speed/KE.
Derived from: W = Fs cosθ is the definition of work. Wnet = ΔKE is derived from Newton’s second law applied over a displacement.
When to use: W = Fs cosθ when you need the work done by a single specific force. Wnet = ΔKE when friction is present and you need the final speed or KE change after accounting for all forces.
Common error: W = Fs cosθ: forgetting to use θ between force and displacement (not just the slope angle). Wnet = ΔKE: including only the applied force, not all forces (forgetting friction or gravity component).
Q4.1 — Qualitative prediction (2 marks)
The truck will be faster at the bottom. Gravity converts PE (stored energy) to KE as the truck descends, increasing speed. Friction removes some KE (energy lost), but as long as the gravity work exceeds the friction work, net KE increases. Since the hill is relatively gentle (μk = 0.05 is low), the net effect is a speed increase.
Marking criteria: 1 = predicts faster with reasoning; 1 = correctly identifies PE → KE gain outweighs friction loss (or equivalent reasoning).
Q4.2 — Formula that cannot be used (2 marks)
Cannot use conservation of mechanical energy (KE1 + U1 = KE2 + U2). This formula requires that the system be frictionless and that only gravity does work. Here, friction is present (μk = 0.05), so mechanical energy is not conserved — it is reduced by the work done by friction.
Marking criteria: 1 = names conservation of mechanical energy; 1 = explains it cannot be used because friction is present.
Q4.3 — Calculation steps (2 marks)
Step 1: Find normal force on slope — FN = mg cosθ (where sinθ = h/s = 8/100). Step 2: Find friction force — fk = μk × FN. Step 3: Find work done by gravity — Wgrav = mgh (using vertical height, not slope distance). Step 4: Find work done by friction — Wfric = −fk × s (negative, opposes motion). Step 5: Apply Wnet = ΔKE: ½mv2² = ½mv1² + Wgrav + Wfric. Solve for v2.
Marking criteria: 1 = correct sequence identifying friction work and gravity work separately; 1 = names Wnet = ΔKE as the linking formula with v1 and v2 identified.