Physics • Year 11 • Module 2 • Lesson 10

Phase 2 Consolidation

Apply your understanding of work, energy and power to real calculations, graph interpretation and error correction across mixed Band 3–5 scenarios.

Apply · Data & Reasoning

1. Mixed calculations — select the right formula each time

Show all working. State the formula you are using before substituting. 16 marks (marks shown per question)

1.1 (2 marks) 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.

1.2 (2 marks) Calculate the kinetic energy of a 0.5 kg ball travelling at 12 m/s. Then state the kinetic energy when the speed doubles to 24 m/s — calculate to confirm.

1.3 (3 marks) A 600 kg car decelerates from 25 m/s to rest on a flat road over 40 m. Using the work-energy theorem, calculate the average braking force. State your sign convention.

1.4 (3 marks) An object is dropped from rest at height 8 m. Find its speed at 3 m above the ground. Use the full conservation equation (no friction).

1.5 (3 marks) A car travels at constant 40 m/s with engine power 80 kW. (a) Find the total resistance force. (b) If speed increases to 50 m/s with the same resistance, what power is now required? (c) By what percentage has power demand increased?

1.6 (3 marks) 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 rides along a flat road against a resistance force of 300 N. How far along the flat does the bike travel before stopping? Show two clearly labelled stages.

Stuck? Revisit Mixed Practice Q1–Q6 and Q9 in the lesson. Ask: is there friction? Is it asking for rate?

2. Interpret a kinetic energy vs speed graph

The graph below shows KE versus speed for a 2 kg object. 7 marks

Figure 2. KE versus speed for a 2 kg object. KE = ½mv².

2.1 (2 marks) Describe the shape of the curve and explain why it has this shape using the KE formula.

2.2 (2 marks) Use the graph to estimate the KE when v = 5 m/s, then verify algebraically using KE = ½mv².

2.3 (3 marks) A student claims: “The graph shows that going from 4 m/s to 8 m/s doubles the KE, because 8 is twice 4.” Identify the error in this reasoning and use the graph to show the correct statement.

Stuck? Revisit Error Clinic Card 2 (KE proportional to v, not v²) in the lesson.

3. Error diagnosis — find the mistake and fix it

Each student working below contains exactly one error. Identify it and write the corrected working. 9 marks (3 per error: 1 identify, 2 correct)

3.1 An object slides 80 m along a slope inclined at 30°. Student writes: “ΔU = mgΔh = 5 × 9.8 × 80 = 3920 J.”

Error:

Corrected working:

3.2 A box starts from rest. Applied force does 1200 J of work; friction does −400 J of work. Student writes: “W_net = ΔKE, so ½mv² = 1200 J.”

Error:

Corrected working:

3.3 A 1200 kg car travels at constant 25 m/s on a flat road. Student writes: “Driving force = mg = 1200 × 9.8 = 11 760 N. P = Fv = 11 760 × 25 = 294 kW.”

Error:

Corrected working:

Stuck? Revisit Error Clinic Cards 5, 6, and 4 in the lesson respectively.

Answers — Do not peek before attempting

Q1.1

W = Fs cosθ = 200 × 15 × cos35° = 200 × 15 × 0.8192 = 2457 J.

Marking: 1 mark correct formula with angle 35° (not 55°); 1 mark correct answer.

Q1.2

KE = ½ × 0.5 × 12² = 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.

Marking: 1 mark 36 J; 1 mark 144 J with ratio = 4 and correct conclusion.

Q1.3

Positive direction = forward. W_net = ΔKE: −F_b × 40 = 0 − ½ × 600 × 25² = −187 500 J. F_b = 187 500 / 40 = 4687.5 N. Negative sign confirms force opposes forward motion.

Marking: 1 sign convention stated; 1 correct KE calculation; 1 correct force.

Q1.4

Reference level: h = 0 at ground. KE₁ + U₁ = KE₂ + U₂: 0 + mgh₁ = ½mv² + mgh₂. Cancel m: g(h₁ − h₂) = ½v². v = √(2 × 9.8 × (8 − 3)) = √98 = 9.90 m/s.

Marking: 1 correct conservation equation written; 1 substitution; 1 answer 9.90 m/s.

Q1.5

(a) F = P/v = 80 000/40 = 2000 N. (b) P_new = 2000 × 50 = 100 000 W = 100 kW. (c) Increase = (100 − 80)/80 × 100 = 25%.

Marking: 1 per part.

Q1.6

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.

Marking: 1 correct KE at bottom; 1 correct W_net equation set-up; 1 answer 130.7 m.

Q2.1

The curve is parabolic (concave up), not a straight line. This is because KE = ½mv² — kinetic energy is proportional to vsquared. Each equal increase in speed produces a progressively larger increase in KE.

Q2.2

From graph, KE at v = 5 m/s ≈ 25 J. Algebraic check: KE = ½ × 2 × 25 = 25 J. Consistent.

Q2.3

Error: the student treated KE as proportional to v (linear reasoning), but KE ∝ v². Correct statement: At 4 m/s, KE = 16 J; at 8 m/s (double the speed), KE = 64 J. Ratio = 64/16 = 4. Going from 4 m/s to 8 m/s (doubling speed) quadruples KE, not doubles it.

Q3.1

Error: student used slope distance (80 m) as Δh instead of vertical height. Δh = 80 × sin30° = 80 × 0.5 = 40 m. Corrected: ΔU = 5 × 9.8 × 40 = 1960 J (exactly half the student’s answer, because sin30° = 0.5).

Q3.2

Error: student used only W_applied = 1200 J, not W_net. W_net = W_applied + W_friction = 1200 + (−400) = 800 J. Corrected: ½mv² = 800 J → v = √(2 × 800/m). (If m = 10 kg: v = √160 = 12.6 m/s, not 15.5 m/s.)

Q3.3

Error: weight acts vertically and does zero work on horizontal motion. On a flat road at constant velocity, F_drive = F_resistance (from Newton 1) — not mg. The driving force equals the resistance force stated in the problem. If resistance = 900 N: P = 900 × 25 = 22 500 W = 22.5 kW — far more realistic than 294 kW.