Physics • Year 11 • Module 2: Dynamics • Lesson 9

Energy Synthesis

Build HSC Band 5–6 extended-response technique on multi-step energy synthesis, evaluating student solutions, and constructing energy flow diagrams with full quantitative support.

Master · Extended Response

1. Full synthesis problem — car on a hill (Band 5–6)

10 marks Band 5–6

Scenario. A 1200 kg car starts from rest at the top of a frictionless 20 m hill and rolls to the flat road at the bottom. On the flat road the total resistance force is 900 N. The engine has a maximum output of 45 kW.

(a) Calculate the speed of the car at the bottom of the hill.
(b) Calculate the power needed to maintain this speed against resistance on the flat road.
(c) Calculate the reserve power available for acceleration.
(d) Using the reserve power, calculate the time taken to accelerate from the hill speed to 30 m/s on the flat road. Clearly state which formula you use at each step and why.
(e) Draw a labelled energy flow diagram for the complete scenario (hill + flat road acceleration). Label each energy transfer with the formula used.

Plan: (a) conservation (frictionless) → v = √(2gh); (b) P = Fv at that speed; (c) reserve = P_engine − P_needed; (d) ΔKE = ½m(v_f² − v_i²), then Δt = ΔKE/P_reserve; (e) label hill (PE → KE via conservation), flat (engine work → KE + heat via W_net = ΔKE).

2. Evaluate a student’s solution — spot the errors (Band 5–6)

8 marks Band 4–6

Question given to student. A 900 kg car cruises at constant speed on a flat highway with engine power 60 kW and resistance 1500 N. It then descends a 10 m hill (road distance 80 m, μ_k = 0.1, engine off). Find the speed at the bottom of the hill.

Student’s response (contains errors):

Step 1: v = P/F = 60000/1500 = 40 m/s  ← highway speed.
Step 2: Since the car descends a hill, use conservation of energy.
Step 3: KEbottom = KEtop + mgh = ½×900×1600 + 900×9.8×10
       = 720 000 + 88 200 = 808 200 J
Step 4: vbottom = √(2×808 200/900) = √1796 = 42.4 m/s

Q2. Evaluate the student’s solution. In your response you must:

Identify all errors in the student’s approach (there are at least two).
Explain why each is wrong with reference to the conditions for each formula.
Write a correct solution showing the proper calculation chain.
State the marking criteria you would use to award full marks for this question.

Errors: (1) used conservation when friction is present (μ_k = 0.1); (2) ignored friction work entirely. Correct: W_net = ΔKE with W_grav + W_fric. See Worked Example 2 in the lesson for the full method.

Answers — Do not peek before attempting

Q1 — Full synthesis solution and sample Band 6 response (10 marks)

(a) Speed at bottom of hill [2 marks]: Frictionless hill → use conservation of mechanical energy. KE_top + U_top = KE_bottom + U_bottom. Taking bottom as reference: mgh = ½mv², mass cancels. v = √(2gh) = √(2 × 9.8 × 20) = √392 = 19.8 m/s.

(b) Power needed to maintain this speed [2 marks]: Constant velocity → F_drive = F_resist = 900 N (Newton 1). P = Fv = 900 × 19.8 = 17 820 W = 17.8 kW.

(c) Reserve power [1 mark]: P_reserve = P_engine − P_needed = 45 000 − 17 820 = 27 180 W = 27.2 kW.

(d) Time to accelerate to 30 m/s [3 marks]: Use W_net = ΔKE (reserve power provides net work for acceleration). ΔKE = ½m(v_f² − v_i²) = ½ × 1200 × (900 − 392.04) = 600 × 507.96 = 304 776 J. Then P_reserve = ΔKE/Δt → Δt = 304 776/27 180 = 11.2 s.

(e) Energy flow diagram [2 marks]: Hill stage: “PE stored at top” → [ΔU = mgh; conservation KE₁+U₁=KE₂+U₂] → “KE at bottom (19.8 m/s)”. Flat stage: “Engine input (45 kW)” → [P = Fv; W_net = ΔKE] → “KE increase” + “heat lost to resistance (F×s)”. Award 1 mark for correct labels on each stage with formulae; 1 mark for showing energy lost to resistance as a separate branch.

Marking criteria summary (10 marks): 1 = conservation used for frictionless hill with justification; 1 = v = 19.8 m/s with unit; 1 = Newton 1 applied to get F_drive = 900 N; 1 = P_needed = 17.8 kW; 1 = P_reserve = 27.2 kW; 1 = ΔKE calculated correctly; 1 = Δt = 11.2 s; 1 = formula named at each step with condition stated; 1 = energy flow diagram — hill section correct; 1 = energy flow diagram — flat section with heat loss shown.

Q2 — Error evaluation with correct solution (8 marks)

Error 1 (Steps 2–4): Used conservation of mechanical energy when friction is present. The student wrote “use conservation of energy” but μ_k = 0.1 means kinetic friction acts on the hill. Conservation of mechanical energy (KE₁+U₁=KE₂+U₂) only applies when the system is frictionless. With friction, mechanical energy is not conserved — some is converted to heat. The correct formula is W_net = ΔKE.

Error 2: Ignored the work done by friction. Even if the student had used W_net = ΔKE, they included only W_gravity on the right-hand side. The friction force does negative work (W_fric = −f_k × s) which must be included in W_net.

Correct solution:

Step 1: Highway cruise speed. Constant v → F_drive = 1500 N. v = P/F = 60 000/1500 = 40 m/s (v₁ for hill).

Step 2: Geometry. sinθ = h/s = 10/80 = 0.125; cosθ = √(1−0.125²) ≈ 0.992. F_N = mg cosθ = 900 × 9.8 × 0.992 = 8745 N.

Step 3: Friction. f_k = μ_k × F_N = 0.1 × 8745 = 874.5 N (up slope).

Step 4: W_gravity = mgh = 900 × 9.8 × 10 = 88 200 J (positive — gravity aids descent).

Step 5: W_friction = −f_k × s = −874.5 × 80 = −69 960 J (negative — opposes motion).

Step 6: W_net = 88 200 − 69 960 = 18 240 J.

Step 7: ½mv₂² = ½mv₁² + W_net = ½×900×1600 + 18 240 = 738 240 J. v₂ = √(2×738 240/900) = √1640.5 = 40.5 m/s. (Car speeds up slightly as gravity work > friction loss.)

Marking criteria summary (8 marks): 1 = identifies Error 1 (conservation used when friction present); 1 = explains why (condition for conservation is frictionless); 1 = identifies Error 2 (friction work omitted from W_net); 1 = explains why (W_net must include ALL forces); 1 = correct W_grav = 88 200 J; 1 = correct W_fric = −69 960 J; 1 = W_net = ΔKE applied correctly; 1 = v₂ = 40.5 m/s with unit.