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 = 0.15. 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; the engine's maximum output is 180 kW. (a) Calculate the total resistance force acting on the car. (b) The driver then applies full power while the car is momentarily still at 30 m/s. 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 and does 365 kJ of work over the first 200 m. 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 = 192.0 N. f_k = 0.15 × 192.0 = 28.8 N. Check it slides: driving component mg sinθ = 20 × 9.8 × 0.200 = 39.2 N > 28.8 N, so it does. W_gravity = 20 × 9.8 × 10 = 1960 J. W_friction = −28.8 × 50 = −1440 J. W_net = 1960 − 1440 = 520 J. W_net = ΔKE = ½mv², so v = √(2 × 520 / 20) = √52 = 7.2 m/s.
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) At constant velocity the net force is zero, so the driving force equals the resistance. P = Fv → F_resist = 120 000 / 30 = 4000 N. (b) At full power and still 30 m/s: F_drive = P/v = 180 000 / 30 = 6000 N. F_net = 6000 − 4000 = 2000 N forward. By Newton's Second Law a net forward force produces a forward acceleration, so the car speeds up. As v rises, F_drive = P/v falls, so F_net shrinks and the acceleration decreases; the car approaches a new, higher constant speed at which the driving force once again equals the resistance. You cannot put a number on that speed from this question: 4000 N is the resistance measured at 30 m/s, and resistance on a real car grows with speed, so the new balance point is reached below the 45 m/s that assuming a constant 4000 N would predict.
Q16:
Q17:
Note on Q17: The engine's work over the first 200 m is given so that you can answer with the work–energy theorem. Deriving it instead would mean integrating m dv/dt = P/v − F_resist, which is beyond this course — the driving force F = P/v is not constant, so you cannot pick one speed, multiply by 200 m and call it the work. The three figures here are mutually consistent: solving that equation exactly for this car gives v = 27.55 m/s, t = 10.45 s and W_engine = 365.7 kJ at the 200 m mark, which is what the rounded values above reproduce. If you check the numbers against each other, they will agree.
Tick when you have finished all questions and checked your answers.