Physics • Year 11 • Module 2 • Lesson 10

Phase 2 Consolidation

Build HSC Band 5–6 technique on multi-step energy problems, algebraic derivations, and extended justification of formula choice and energy reasoning.

Master · Extended Response

1. Multi-step problem — rough slope with applied rope force (Band 4–5)

8 marks   Band 4–5

Scenario. A 15 kg box starts from rest on a flat floor. A rope applies a force of 90 N at 25° above the horizontal. The coefficient of kinetic friction between the box and the floor is μk = 0.3. The box is pulled 12 m.

Q1. Find the final speed of the box. In your response you must:

  • State your positive direction and sign convention before starting.
  • Calculate the work done by the applied force.
  • Calculate the normal force, accounting for the vertical component of the applied force.
  • Calculate the friction force and the work done by friction.
  • Apply Wnet = ΔKE to find the final speed.
  • Explain in one sentence why conservation of mechanical energy cannot be used here.
Stuck? Plan: state positive direction → Wapplied = 90 × 12 × cos25° → FN = mg − F sin25° → fk = μkFN → Wfriction = −fk × 12 → Wnet = Wapplied + Wfriction → v = √(2Wnet/m).

2. Algebraic derivation — power and speed (Band 5–6)

7 marks   Band 5–6

Context. A vehicle travels at constant speed against a constant resistance force F on a flat road. Use this situation to derive expressions and then explore the real-world implications of the power–speed relationship.

Q2. In your response you must:

  • State Newton’s First Law argument to identify the driving force at constant speed.
  • Derive the algebraic expression P = Fv for minimum power required at speed v.
  • Use your expression to show algebraically that doubling speed doubles power (when F is constant).
  • Explain why in practice, doubling speed requires more than double the power, referring to the nature of air resistance at high speeds.
  • A car requires 22.5 kW at 25 m/s. Calculate the resistance force and then predict the power required at 50 m/s if resistance were constant. Comment on whether this is realistic.
Stuck? Newton 1 at const v: Fdrive = Fresist = F. P = W/t = (F × s)/t = F × (s/t) = Fv. Double speed: Pnew = F × 2v = 2Fv = 2P. In reality air drag ∝ v², so F doubles and P ∝ v³ (increases by factor 8 at double speed).
Answers — Do not peek before attempting

Q1 — Sample Band 5–6 response (8 marks), annotated

Sign convention: Positive direction = forward (direction of motion). Negative work is done by friction because it acts backward. [1 mark]

Work done by applied force: Wapplied = F cosθ × s = 90 × cos25° × 12 = 90 × 0.9063 × 12 = 978.8 J. [1 mark]

Normal force: The rope has a vertical component (F sin25°) that partially supports the box, so FN = mg − F sin25° = 15 × 9.8 − 90 × sin25° = 147 − 38.0 = 109.0 N. [1 mark — must account for vertical component]

Friction force and work: fk = μk × FN = 0.3 × 109.0 = 32.7 N. Wfriction = −32.7 × 12 = −392.4 J. [1 mark]

Net work: Wnet = 978.8 + (−392.4) = 586.4 J. [1 mark]

Apply Wnet = ΔKE: ½mv² − 0 = 586.4. v = √(2 × 586.4/15) = √78.19 = 8.84 m/s. [1 mark]

Why conservation of mechanical energy cannot be used: Friction is a non-conservative force; it converts mechanical energy to thermal energy, so total mechanical energy at the bottom is less than at the start. Conservation of Emech would give a speed that is too high. [1 mark]

Marking criteria (8 marks): 1 sign convention; 1 Wapplied = 978.8 J; 1 FN with vertical force component deducted (common error: using FN = mg); 1 Wfriction = −392.4 J; 1 Wnet = 586.4 J; 1 correct final speed 8.84 m/s; 1 explicit explanation why conservation not used; 1 precise use of terminology (Wnet, non-conservative, ΔKE).

Q2 — Sample Band 6 response (7 marks), annotated

Newton’s First Law argument: At constant speed, acceleration = 0. By Newton’s First Law, the net force is zero. Therefore the driving force Fdrive = Fresistance = F. [1 mark]

Derivation of P = Fv: P = W/Δt. At constant speed v, work done over displacement s is W = F × s. So P = Fs/Δt = F × (s/Δt) = F × v. Therefore P = Fv. [1 mark — must show derivation, not just state]

Doubling speed: If v doubles to 2v: Pnew = F × (2v) = 2Fv = 2P. Therefore doubling speed doubles the required power, provided F remains constant. [1 mark]

Real-world explanation: Air resistance (drag) is not constant — it scales approximately with v² (drag ∝ v²). So doubling speed roughly quadruples the drag force. The required power P = Fdrag × v ∝ v² × v = v³. Doubling speed therefore increases power demand by a factor of 2³ = 8, not 2. This is why fuel consumption rises so sharply at motorway speeds. [1 mark]

Calculation: Fresist = P/v = 22 500/25 = 900 N. If F is constant at 50 m/s: P = 900 × 50 = 45 000 W = 45 kW. [1 mark]

Comment on realism: In reality, at 50 m/s, drag approximately quadruples (v doubles, Fdrag ∝ v² → increases by factor 4), giving Fresist ≈ 3600 N and P ≈ 180 kW — four times the “constant resistance” prediction. The constant-resistance model significantly underestimates power demand at high speed. [1 mark]

Marking criteria (7 marks): 1 Newton 1 argument giving Fdrive = Fresistance; 1 clear algebraic derivation P = Fv from P = W/t; 1 algebraic doubling proof; 1 real-world explanation referencing drag ∝ v² and P ∝ v³; 1 correct Fresist = 900 N; 1 correct Pnew = 45 kW; 1 evaluative comment on realism (drag not constant).

HSCScience • Physics • Year 11 • Module 2 • Lesson 10 • Worksheet 3 (Master)