Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Before you begin this lesson, take a moment to think about what you already know about this topic. Jot down your ideas — you will revisit them at the end.
Understand the core concepts covered in this lesson.
Apply your knowledge to solve problems and explain phenomena.
Evaluate and analyse scientific information and data.
No new content. Six formulae. Twelve common errors. Ten mixed practice questions. Three timed extended responses. This is your Phase 2 exam preparation.
Wrong: Friction always opposes motion and is always unwanted.
Right: Friction can be static and enable motion (walking, driving); it is essential for many everyday activities.
📐 Phase 2 Formula Sprint
🔬 Common Error Clinic
🔢 Mixed Practice
1. A 200 N force is applied at 35° above the horizontal to drag a box 15 m across a flat floor. Calculate the work done by this force.
Show full working.
Show full working in your book.
2. Calculate the kinetic energy of a 0.5 kg ball travelling at 12 m/s. Then state what happens to the kinetic energy if the speed doubles to 24 m/s — calculate to confirm.
3. State the condition required for conservation of mechanical energy to apply. Give one example of a scenario where it applies and one where it does not.
4. A 600 kg car decelerates from 25 m/s to rest on a flat road over a distance of 40 m. Using the work-energy theorem, calculate the average braking force. Show your sign convention clearly.
5. An object is dropped from rest at a height of 8 m above the ground on a frictionless surface. Find the speed of the object at a point 3 m above the ground. Use the full conservation equation and show all working.
6. A car travels at constant 40 m/s with engine power output of 80 kW. (a) Find the total resistance force. (b) If the speed increases to 50 m/s with the same resistance force, what power output is now required? (c) By what percentage has power demand increased?
7. A 15 kg box starts from rest and is pulled 12 m across a flat floor by a rope at 25° above horizontal. Applied force = 90 N, μk = 0.3. Find the final speed of the box. Apply the Vector Protocol.
8. Derive an algebraic expression for the minimum power P required to move a vehicle at constant speed v against a resistance force F. Use your expression to show that doubling the speed doubles the required power (assuming F stays constant). Explain why in practice, doubling speed requires more than double the power.
9. A 200 kg motorbike starts from rest at the top of a frictionless 20 m hill. At the bottom the engine cuts out and the bike travels along a flat road against a resistance force of 300 N. How far along the flat road does the bike travel before stopping? Show your reasoning in two clearly labelled stages.
10. A 1000 kg car cruises at constant speed with engine power 60 kW and resistance 1500 N on a flat road. It then climbs a hill (height 10 m, road distance 50 m, μk = 0.12, engine still on at same 60 kW). Find the speed at the top of the hill. Show all steps clearly.
⏱️ Timed Exam Block
A 400 kg cyclist and bike system start from rest at the top of a 12 m frictionless hill. At the bottom they travel along a flat road at constant speed. The cyclist's power output is 240 W and the resistance force on the flat road is 60 N.
Type your working below — attempt under timed conditions first.
A 900 kg car brakes from 28 m/s to rest on a flat road. The braking distance is 56 m.
An 800 kg object slides from rest down a rough slope. The vertical height is 18 m, the road distance is 60 m, and the coefficient of kinetic friction is μk = 0.2.
Q1: W = Fs cosθ = 200 × 15 × cos35° = 200 × 15 × 0.819 = 2457 J
Q2: KE = ½ × 0.5 × 144 = 36 J. At 24 m/s: KE = ½ × 0.5 × 576 = 144 J. Ratio = 144/36 = 4. Doubling speed quadruples KE.
Q3: Condition — no friction or air resistance; only conservative forces act. Applies: ball on frictionless rollercoaster. Does not apply: ball rolling down rough ramp (friction converts ME to heat).
Q4: Positive direction = forward. W_net = ΔKE: −F_b × 40 = 0 − ½ × 600 × 625 = −187 500. F_b = 187 500 / 40 = 4687.5 N. Negative sign confirms force opposes motion (backward).
Q5: g(h₁ − h₂) = ½v². v = √(2 × 9.8 × (8 − 3)) = √(2 × 9.8 × 5) = √98 = 9.90 m/s
Q6 (a) F = P/v = 80 000/40 = 2000 N. (b) P = 2000 × 50 = 100 000 W = 100 kW. (c) Increase = (100 − 80)/80 × 100 = 25%.
Q7: W_applied = 90 × 12 × cos25° = 978.3 J. F_N = 15×9.8 − 90×sin25° = 147 − 38.0 = 109 N. f_k = 0.3 × 109 = 32.7 N. W_friction = −32.7 × 12 = −392.4 J. W_net = 978.3 − 392.4 = 585.9 J. v = √(2 × 585.9 / 15) = √78.1 = 8.84 m/s
Q8 — Derivation: At constant v, Newton 1 → F_drive = F_resist = F. P = F_drive × v = Fv. If v doubles to 2v: P_new = F × 2v = 2Fv = 2P. Therefore doubling speed doubles power required when F is constant. In practice, air resistance force scales approximately with v² — so doubling speed roughly quadruples drag force, meaning power demand (P = F_drag × v ∝ v² × v = v³) actually increases by a factor of 8. This is why fuel consumption rises so sharply at highway speeds.
Q9: Stage 1 (frictionless hill): KE_bottom = mgh = 200 × 9.8 × 20 = 39 200 J. Stage 2 (flat road, engine off): W_net = ΔKE. −300 × s = 0 − 39 200. s = 39 200/300 = 130.7 m. The bike travels 130.7 m along the flat road before stopping.
Q10 — Setup note: Cruise speed: v₁ = P/F_resist = 60 000/1500 = 40 m/s. On the hill: sinθ = 10/50 = 0.2, cosθ = 0.980. F_N = 1000 × 9.8 × 0.980 = 9604 N. f_k = 0.12 × 9604 = 1152 N. W_gravity = −mgh = −1000 × 9.8 × 10 = −98 000 J (negative — gravity opposes upward motion). W_friction = −1152 × 50 = −57 600 J. For engine work, need time: assume average speed ≈ (v₁ + v₂)/2 but v₂ is unknown. Full solution requires iteration or treating as average: approximate t = 50/v_avg. For exam purposes: set up W_net = ΔKE and express that W_engine = P × t, then note this requires additional information (time or average speed on the hill). This question tests whether students can identify the setup even when the full numerical solution needs iteration.
Q11:
Q12:
(d) 176.4 kW is at the upper limit of a performance car engine — a standard car produces 80–120 kW. This reveals something important: braking forces are far larger than driving forces. A car can be stopped in 56 m from 28 m/s, but the engine cannot sustain 176 kW continuously. Brakes can dissipate energy much faster than engines can produce it — which is why emergency stops are possible even with modest engines.
Q13:
63.6% of the initial PE was converted to heat by friction — only 36.4% became kinetic energy. This is a very high μ_k = 0.2 for a slope, which explains the large loss. In real engineering, reducing μ_k (better bearings, lubrication) or reducing the normal force (lighter objects, shallower slopes) are the key strategies for improving energy efficiency.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
Put your knowledge of forces, work, energy and power to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–10.
Tick when you have finished all questions and checked your answers.