What this is for. This diagnostic helps you decide which Unit 3 (Work, Energy, and Power) topics to revisit. It covers every topic, with extra questions on Topics 3.2 to 3.4. These are original Marlbridge practice questions, not past exam questions. The set is not calibrated against real exam results, so it does not give a predicted score. Treat each wrong answer as a pointer to one topic, not as a grade.
How to sit it. Allow about 30 minutes without notes. Use a calculator for arithmetic, roots and exponentials, but do the calculus by hand. Use g = 9.8 m/s². Quantities are in SI units; each coefficient carries the unit that makes its term correct.
Question 1 (multiple choice · 3.1)
Take +x forward. A 60 kg traveller walks forward at 1.5 m/s relative to a moving walkway. The walkway moves forward at 1.0 m/s relative to the airport floor. What is the traveller’s kinetic energy measured by someone standing on the floor?
- (A) 7.5 J
- (B) 68 J
- (C) 98 J
- (D) 190 J
Answer and explanation
Answer: (D). Relative to the floor, v = 1.5 + 1.0 = 2.5 m/s, so K = ½(60)(2.5)² = 187.5 J ≈ 190 J.
- (A) subtracts the speeds. Both velocities point forward, so they add.
- (B) is K in the walkway’s frame, 67.5 J. K depends on the observer.
- (C) adds two kinetic energies, 67.5 J + 30 J. Add the velocities, then square.
If you missed this: read “Kinetic energy depends on the observer” in the Topic 3.1 study guide.
Question 2 (multiple choice · 3.2)
Take +x along a level floor. A cable pulls a crate with a horizontal force F_x = 2.0 + 3.0x² as the crate moves from x = 0 to x = 2.0 m. How much work does the cable do?
- (A) 4.0 J
- (B) 10 J
- (C) 12 J
- (D) 16 J
Answer and explanation
Answer: (C). Integrate: W = ∫₀² (2.0 + 3.0x²) dx = [2.0x + x³]₀² = 4.0 + 8.0 = 12 J.
- (A) uses the starting force, 2.0 N, for the whole distance.
- (B) uses the midpoint force, 5.0 N. That works only for a linear F(x).
- (D) uses the mean of the end forces, 8.0 N. Also linear-only; this graph curves upward.
If you missed this: read “Variable forces: integrate along the path” in the Topic 3.2 study guide.
Question 3 (multiple choice · 3.2)
A crate sits on the flat bed of a truck. The truck speeds up along a level road, and the crate does not slip. In the road’s frame, which force does positive work on the crate?
- (A) No force, because static friction cannot do work when nothing slides
- (B) The static friction force from the truck bed
- (C) The normal force from the truck bed
- (D) None, because the crate is at rest on the truck
Answer and explanation
Answer: (B). The crate speeds up, so the net work on it is positive. The only horizontal force is static friction, forward, and in the road’s frame its point of application moves forward: W > 0.
- (A) “No sliding” is not “no displacement”. Crate and truck move together relative to the road.
- (C) The normal force is perpendicular to the displacement.
- (D) uses the truck’s frame. Work, like kinetic energy, depends on the frame.
If you missed this: read “The dot product and constant forces” and “Net work and the work–energy theorem” in the Topic 3.2 study guide.
Question 4 (multiple choice · 3.3)
A 0.50 kg ball falls from a shelf 2.0 m above the floor. Student P chooses U_g = 0 at the floor. Student Q chooses U_g = 0 at the shelf. Which statement is correct?
- (A) P: U_g = +9.8 J at the shelf. Q: U_g = −9.8 J at the floor. Both: ΔU_g = −9.8 J.
- (B) They find different ΔU_g but the same speed at the floor.
- (C) For Q, U_g = 0 at the shelf, so the ball cannot speed up.
- (D) Both must have U_g = 0 at the floor, where the ball stops.
Answer and explanation
Answer: (A). mgh = 0.50 × 9.8 × 2.0 = 9.8 J. The zero is a free choice, so the values of U_g differ, but every change is the same, ΔU_g = −9.8 J, and so is the speed at the floor, 6.3 m/s.
- (B) Only values of U depend on the zero; changes, forces and speeds do not.
- (C) What matters is ΔU, not the starting value. U_g can fall from 0 to −9.8 J.
- (D) The observer chooses the zero; the motion does not set it.
If you missed this: read “Defining U with an integral” (choosing the zero) in the Topic 3.3 study guide.
Question 5 (multiple choice · 3.3)
A planet has mass M and radius R. A probe of mass m is moved from the surface to a distance 3R from the planet’s centre. What is the change in the gravitational potential energy of the probe–planet system?
- (A) −2GMm/(3R)
- (B) +2GMm/(3R)
- (C) +8GMm/(9R)
- (D) +2GMm/R
Answer and explanation
Answer: (B). With U_g = −GMm/r: ΔU_g = −GMm/(3R) − (−GMm/R) = GMm(1 − ⅓)/R = +2GMm/(3R). Positive: the probe moves away against the attraction.
- (A) has the sign reversed. U_g rises towards zero as r grows.
- (C) uses 1/r², as for force. Potential energy goes as 1/r.
- (D) uses mgΔy with surface g over a height 2R. The field weakens with height, so this overestimates.
If you missed this: read “Gravitational potential energy” and Worked example 2 in the Topic 3.3 study guide.
Question 6 (multiple choice · 3.4)
A vertical toy launcher has an ideal spring with k = 500 N/m. A 0.20 kg ball sits on top of it, and the spring is compressed by 0.10 m. When released, the ball leaves the spring at its relaxed length. Ignoring air resistance, how high does the ball rise above its release point?
- (A) 0.64 m
- (B) 1.2 m
- (C) 1.3 m
- (D) 2.6 m
Answer and explanation
Answer: (C). System: ball + spring + Earth, with no external work or friction. K = 0 at release and at the top, so ½kx² = mgh. ½(500)(0.10)² = 2.5 J, so h = 2.5 ÷ (0.20 × 9.8) = 1.3 m (1.28 m).
- (A) drops the ½ from K only (½kx² = mv², then h = v²/2g).
- (B) is the height above the relaxed spring, not above the release point.
- (D) drops the ½ from the spring energy.
If you missed this: read “Mechanical energy and the energy rule” and “Comparing scenarios” in the Topic 3.4 study guide.
Question 7 (multiple choice · 3.5)
A constant net force speeds a 2.0 kg cart up from rest to 6.0 m/s in 3.0 s. What is the power of the net force at t = 3.0 s, and the average power over the 3.0 s?
- (A) 24 W at 3.0 s; 12 W average
- (B) 24 W at 3.0 s; 24 W average
- (C) 12 W at 3.0 s; 12 W average
- (D) 12 W at 3.0 s; 24 W average
Answer and explanation
Answer: (A). a = 6.0 ÷ 3.0 = 2.0 m/s², so F = 4.0 N. At 3.0 s, P = Fv = 4.0 × 6.0 = 24 W. Average: P_avg = ΔK/Δt = ½(2.0)(6.0)² ÷ 3.0 = 36 ÷ 3.0 = 12 W. With a constant force from rest, P rises linearly, so the average is half the final value.
- (B) treats the final power as the average, but Fv grows during the push.
- (C) uses the average for the instant too.
- (D) swaps the two values.
If you missed this: read “Average power” and “Instantaneous power: P = dW/dt = F·v” in the Topic 3.5 study guide.
Question 8 (short answer · 3.2)
Take +x along a level, frictionless track. A 0.60 kg cart passes x = 0 at +1.0 m/s. A fan on the cart gives a forward push that fades with distance: F_x = (3.0 N)e^(−x/2.0 m).
(a) Find the work done by the push from x = 0 to x = 4.0 m. (b) Find the cart’s speed at x = 4.0 m. (c) A student uses W = (3.0 N)(4.0 m) = 12 J. Explain the error.
Answer and explanation
(a) W = ∫₀⁴ 3.0e^(−x/2.0) dx = [−6.0e^(−x/2.0)]₀⁴ = 6.0(1 − e^(−2.0)) = 5.2 J (5.19 J).
(b) Only the push does work, so W = ΔK. K₀ = 0.30 J, so K = 5.49 J and v = √(2 × 5.49 ÷ 0.60) = 4.3 m/s.
(c) F × d needs a constant force. This force fades, so the work is the area under the F–x curve, far below the 12 J rectangle.
If you missed this: work through Worked example 1 in the Topic 3.2 study guide.
Question 9 (short answer · 3.4)
Take +x to the right, x > 0. A 0.25 kg glider moves without friction on an air track. Its system has U(x) = 0.50x² + 8.0/x.
(a) Find the equilibrium position and show whether it is stable. (b) The glider is released from rest at x = 1.0 m. Find the force on it there. (c) Find its speed as it passes the equilibrium position. (d) Find the other turning point.
Answer and explanation
(a) dU/dx = x − 8.0/x² = 0 gives x³ = 8.0, so x = 2.0 m. d²U/dx² = 1 + 16/x³ = 3.0 J/m² > 0: a minimum of U, so the equilibrium is stable. U(2.0) = 6.0 J.
(b) F_x = −dU/dx = −(1.0 − 8.0) = +7.0 N, towards lower U and the equilibrium.
(c) E = U(1.0) = 0.50 + 8.0 = 8.5 J. At x = 2.0 m, K = 8.5 − 6.0 = 2.5 J, so v = √(2 × 2.5 ÷ 0.25) = 4.5 m/s (4.47 m/s).
(d) Set U = E: 0.50x² + 8.0/x = 8.5, so x³ − 17x + 16 = 0. x = 1.0 (the release point) is one root; the other factor, x² + x − 16 = 0, gives x = 3.5 m (3.53 m).
If you missed this: for (c) and (d), read Worked example 1 in the Topic 3.4 study guide; for (a) and (b), “Equilibrium and stability on a U(x) graph” in the Topic 3.3 study guide.
Question 10 (short answer · 3.5)
An e-bike and rider have a total mass of 90 kg on a level road. Model the total resistive force as F = (0.30 N·s²/m²)v².
(a) Find the power the motor must supply to cruise at 8.0 m/s. (b) The motor’s greatest power is 500 W. Find the top speed. (c) At full power, find the acceleration at the moment the speed is 6.0 m/s.
Answer and explanation
(a) At constant speed the driving force balances resistance: F = 0.30 × 8.0² = 19.2 N. P = Fv = 19.2 × 8.0 = 150 W (154 W).
(b) P = 0.30v³ = 500, so v³ = 1667 and v = 12 m/s (11.9 m/s).
(c) Driving force P/v = 83 N; resistive force 0.30 × 6.0² = 10.8 N. a = (83.3 − 10.8) ÷ 90 = 0.81 m/s².
If you missed this: read “Instantaneous power: P = dW/dt = F·v” and “Net power and kinetic energy” in the Topic 3.5 study guide.
Your next step
Mark each question. Count a short answer as missed if any part went wrong.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 3.1 Translational Kinetic Energy | 1 | Topic 3.1 study guide |
| 3.2 Work | 2, 3, 8 | Topic 3.2 study guide |
| 3.3 Potential Energy | 4, 5, 9(a)–(b) | Topic 3.3 study guide |
| 3.4 Conservation of Energy | 6, 9(c)–(d) | Topic 3.4 study guide |
| 3.5 Power | 7, 10 | Topic 3.5 study guide |
How to use your result
- Read every explanation, even for questions you got right. A lucky guess is still a gap.
- Missed one question in a topic? Read the named section, then try that topic’s practice set.
- Missed two or more in a topic? Work through that topic’s whole study guide, then its practice set and checklist.
- Missed questions across several topics? Start with Topic 3.2; work integrals feed every other topic.
- Got everything right? Go to the mixed unit review, where each question combines topics.
Your result guides what to study next. It does not predict an exam score.