Work, Kinetic Energy, the Work-Energy Theorem, Gravitational PE, Conservation of Mechanical Energy, and Power — 45 minutes, 22 marks.
Phase 2 Quiz | Covers L06–L10 | No notes | Show all working
Section 1 — Multiple Choice
A 300 N horizontal force pushes a box 6 m along a flat floor. What is the work done by this force?
Which correctly gives the kinetic energy of a 4 kg object moving at 5 m/s?
A box slides horizontally across a flat floor. How much work does the normal force do on the box?
An 8 kg object is lifted 3.5 m vertically. What is the gain in gravitational potential energy?
A 2 kg ball is dropped from rest at a height of 10 m on a frictionless surface. What is its speed when it reaches a height of 4 m?
A 5 kg box starts from rest. The net work done on it is 360 J. What is the final speed of the box?
A crane lifts a 500 kg load through a height of 12 m in 30 s. What is the average power output of the crane?
A car travels at a constant speed of 20 m/s on a flat road. The total resistance force is 800 N. What is the engine's power output?
A cyclist rolls from rest down a frictionless 15 m hill, then travels along a flat road against an 80 N drag force. The cyclist's power output is 400 W. What is their constant cruise speed on the flat road?
A student uses KE₁ + U₁ = KE₂ + U₂ on a rough slope and calculates a final speed of 14 m/s. The actual speed of the object at the bottom will be:
Section 2 — Short Answer
A 60 kg person runs up a 4 m high staircase in 3 s. Calculate their average power output against gravity.
Show full working.
Show full working in your book.
A 1000 kg car decelerates from 30 m/s to 15 m/s over a distance of 90 m on a flat road. Using the work-energy theorem, calculate the average net braking force. State your sign convention clearly.
Show sign convention, formula setup, and full working.
Answer in your book — show sign convention.
A 20 kg object slides from rest down a slope. The vertical height is 10 m, the road distance is 50 m, and μk = 0.25. Calculate the speed at the bottom of the slope using the work-energy theorem.
Show all steps including slope geometry, normal force, friction, and net work.
Answer in your book — all steps required.
A 0.3 kg ball is thrown vertically upward at 15 m/s. Assuming no air resistance, use conservation of mechanical energy to find: (a) the maximum height reached, (b) the speed of the ball at half that maximum height on the way up.
Use the full conservation equation for both parts.
Answer in your book — use full conservation equation.
A car engine outputs 120 kW while the car travels at a constant 30 m/s on a flat road. (a) Calculate the total resistance force acting on the car. (b) The driver then applies full power while remaining at 30 m/s momentarily. Explain using Newton's Second Law what happens to the car's motion and why.
Part (b) requires a Newton's Second Law argument — not just a statement.
Answer in your book — include Newton's Second Law in part (b).
Section 3 — Extended Response
A 500 kg rollercoaster car starts from rest at the top of a 40 m drop. Due to friction, the car arrives at the bottom at 22 m/s rather than the theoretical maximum speed.
Show full working for all four parts.
Answer in your book — full working for all four parts.
A 700 kg car starts from rest on a flat road. The engine outputs a constant 35 kW. The resistance force is 500 N throughout.
Show full working for all four parts — part (c) must include a Newton's Second Law calculation.
Answer in your book — all four parts with full working.
Q1 — B (1800 J). W = Fs cosθ = 300 × 6 × cos0° = 1800 J. Force is horizontal, motion is horizontal — θ = 0°.
Q2 — B (50 J). KE = ½mv² = ½ × 4 × 25 = 50 J. Option A (20 J) uses v not v²; option D (40 J) forgets the ½.
Q3 — C. Normal force acts vertically upward; displacement is horizontal. W = Fs cos90° = 0. Perpendicular forces do zero work.
Q4 — B (274.4 J). ΔU = mgΔh = 8 × 9.8 × 3.5 = 274.4 J. Option A (28 J) multiplies mass × height only; option C halves the answer incorrectly.
Q5 — A (10.84 m/s). Conservation: g(h₁ − h₂) = ½v². v = √(2 × 9.8 × (10 − 4)) = √117.6 = 10.84 m/s. Option B (14.0) is the speed at h = 0, not h = 4. Option D is speed from 8 m.
Q6 — B (12 m/s). W_net = ΔKE: 360 = ½ × 5 × v². v² = 144. v = 12 m/s. Option A divides 360 by 5 without the ½; option C takes √(360/5) = 8.49 — forgets to multiply by 2 first.
Q7 — C (1960 W). P = mgh/Δt = 500 × 9.8 × 12 / 30 = 58 800 / 30 = 1960 W. Option A (200 W) divides by wrong values; option D (9800) forgets to divide by time.
Q8 — B (16 000 W). Constant v → F_drive = F_resist = 800 N. P = Fv = 800 × 20 = 16 000 W. Option D multiplies by 200 (wrong); option C uses only F.
Q9 — B (5.0 m/s). Hill speed from conservation: v = √(2 × 9.8 × 15) = 17.1 m/s — but this requires P = 80 × 17.1 = 1368 W, far more than 400 W. Cyclist decelerates on flat until P = F_drag × v: v = 400/80 = 5.0 m/s. Option A is the hill speed — this is the key synthesis trap.
Q10 — C. Friction converts mechanical energy to heat — E_mech decreases. Conservation assumes no energy loss and therefore overestimates final KE and speed. Actual speed < 14 m/s. Option D is wrong — we know for certain it will be less, we just don't know the exact value without μ_k.
Q11: W = mgh = 60 × 9.8 × 4 = 2352 J. P = W/Δt = 2352/3 = 784 W.
Q12: Positive direction = forward. W_net = ΔKE. −F_b × 90 = ½ × 1000 × 15² − ½ × 1000 × 30² = 112 500 − 450 000 = −337 500 J. F_b = 337 500/90 = 3750 N. The braking force acts backward (negative direction) with magnitude 3750 N.
Q13: sinθ = 10/50 = 0.200, cosθ = √(1 − 0.04) = 0.980. F_N = 20 × 9.8 × 0.980 = 192.1 N. f_k = 0.25 × 192.1 = 48.0 N. W_gravity = 20 × 9.8 × 10 = 1960 J. W_friction = −48.0 × 50 = −2400 J. W_net = 1960 − 2400 = −440 J. W_net is negative — the object does not reach the bottom under its own motion with this friction coefficient. The friction force (48 N) exceeds the gravitational driving component (mg sinθ = 20 × 9.8 × 0.2 = 39.2 N). The object decelerates and stops on the slope. (Note to teacher: a realistic μ_k for this slope would be ≤ 0.15.)
Q14: (a) At max height, v = 0: ½mv² = mgh_max → h_max = v²/(2g) = 225/19.6 = 11.48 m. (b) At h = h_max/2 = 5.74 m: KE₁ + U₁ = KE₂ + U₂. ½mv² + 0 = ½mv₂² + mgh/2. Cancel m: ½ × 225 = ½v₂² + 9.8 × 5.74. 112.5 = ½v₂² + 56.25. v₂² = 112.5. v₂ = 10.61 m/s.
Q15: (a) Constant v → P = F_resist × v → F_resist = 120 000/30 = 4000 N. (b) At 30 m/s with full engine power (120 kW): F_drive = P/v = 120 000/30 = 4000 N. But this equals the resistance — so F_net = 0 and the car maintains constant velocity. The car is already at the speed where it uses full power to overcome resistance. If "full power" means power above 120 kW, then F_drive > F_resist → F_net > 0 → Newton 2: a = F_net/m > 0 → car accelerates. (This question has an implicit ambiguity — the intended reading is that 120 kW is NOT the car's maximum power, so applying more power creates a net forward force.)
Q16:
Q17:
Note on Q17(a): This type of question — finding speed after a given distance when engine power is constant — requires the time taken, which in turn depends on the final speed. In the HSC, questions at this difficulty level usually provide the time, the final speed, or ask only for the maximum speed. If you set up the energy equation correctly and recognised the need to estimate or iterate, you have demonstrated strong conceptual understanding.
Tick when you have finished all questions and checked your answers.