IB DP Physics: Work, energy, power and efficiency (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic A.3, Work, energy and power, in the IB Physics guide (first assessment 2025). SL and HL candidates answer the same set, because A.3 has no extra HL-only material in the guide. Take g = 9.81 m s⁻² where it is needed. Learn the content in the study guide and the revision notes. Other links: the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A rope pulls a crate with a constant force of 40 N. The rope is at 60° to the displacement and the crate moves 5.0 m. Calculate the work done by the rope. [2]

2. State what is meant by mechanical energy, and state the condition under which it is conserved. [3]

3. A ball of mass 0.060 kg has momentum 0.90 kg m s⁻¹. Determine its kinetic energy using the momentum, and its speed. [3]

4. A spring of spring constant 180 N m⁻¹ is compressed by 0.050 m. It then launches a 0.020 kg ball along a smooth horizontal surface.

(a) Calculate the energy stored in the spring. [2] (b) Calculate the launch speed of the ball. [2]

5. A crane lifts a 450 kg load at a constant speed of 0.80 m s⁻¹.

(a) Calculate the useful power output of the crane. [2] (b) The crane’s motor is 85% efficient. Calculate its input power. [2]

6. An electric motor has an input power of 2.5 kW. It delivers 1.9 kW of mechanical power and 0.45 kW is transferred as thermal energy in the windings. The rest is transferred as sound.

(a) Calculate the efficiency of the motor. [2] (b) A Sankey diagram is drawn with the input arrow 25 mm wide. Calculate the widths of the three output arrows. [3]

7. A car of mass 1200 kg is travelling at 18 m s⁻¹. It brakes to rest with a constant braking force over a distance of 36 m.

(a) Use the work done by the resultant force to calculate the braking force. [3] (b) The same car, with the same braking force, is travelling at 27 m s⁻¹. Calculate the stopping distance. [2]

8. A pendulum bob of mass 0.35 kg hangs from a string 1.2 m long. It is pulled aside until the string makes 40° with the vertical and released from rest.

(a) Show that the bob is 0.28 m above its lowest point at release. [2] (b) Calculate the speed at the lowest point if there are no resistive forces. [2] (c) The measured speed at the lowest point is 2.10 m s⁻¹. Calculate the work done against resistive forces. [2]

9. A block of mass 1.5 kg is released from rest on a rough slope at 25° to the horizontal. A spring of spring constant 400 N m⁻¹ lies along the slope with its free end 2.4 m from the block. A constant frictional force of 3.0 N acts on the block throughout. The block compresses the spring by a distance x before it stops momentarily. Determine x. [7]

10. A vehicle cruises at a constant 22 m s⁻¹. The total resistive force is 700 N. The engine is 25% efficient and the fuel has an energy density of 33 MJ per litre.

(a) Calculate the useful power output of the engine. [2] (b) Calculate the rate at which energy must be supplied by the fuel. [2] (c) Calculate the volume of fuel used per hour. [2] (d) Calculate the volume of fuel used to travel 100 km. [2]

11. An electric scooter and its rider have a combined mass of 95 kg. They climb a straight road at 3.0° to the horizontal at a constant speed of 5.0 m s⁻¹. The total resistive force is 22 N.

(a) Calculate the component of the weight along the road. [2] (b) Calculate the useful power the scooter’s motor delivers. [3] (c) The motor and battery system is 80% efficient. Calculate the electrical power drawn from the battery. [2] (d) The climb lasts 15 minutes. Determine the fraction of the motor’s useful work that is stored as gravitational potential energy. [3]

Answers

1. W = Fs cos θ = 40 × 5.0 × cos 60° [1] = 100 J [1] [2] Examiner insight: Writing the substitution before the answer protects the method if you slip on cos 60°; a bare number with no unit is easy to lose.

2. Mechanical energy is the sum of kinetic, gravitational potential and elastic potential energy [1]. It is conserved when no frictional or other resistive (non-conservative) forces act [1]. If such forces act, the loss in mechanical energy equals the work done against them, which appears as thermal energy [1]. [3] Examiner insight: An explanation worth three marks usually needs three separate points, so linking the loss to work done by resistive forces is worth adding after the condition.

3. Ek = p²/2m = 0.90² / (2 × 0.060) [1] = 6.75 J = 6.8 J [1]. v = p/m = 0.90 / 0.060 = 15 m s⁻¹ [1] [3] Examiner insight: If you use the momentum form directly you avoid finding the speed first, and the speed can then be used to cross-check Ek.

4. (a) Ee = ½k(Δx)² = ½ × 180 × 0.050² [1] = 0.225 J [1] (b) All of Ee becomes kinetic energy: ½ × 0.020 × v² = 0.225 [1], so v = √(2 × 0.225 / 0.020) = 4.7 m s⁻¹ [1] [4] Examiner insight: State the assumption that energy transfers fully from the spring to the ball; the smooth surface in the question is the reason you may do so.

5. (a) No acceleration, so the lifting force equals the weight: F = 450 × 9.81 = 4415 N [1]. P = Fv = 4415 × 0.80 = 3.5 kW (3532 W) [1] (b) η = Poutput/Pinput, so Pinput = 3532 / 0.85 [1] = 4.2 kW [1] [4] Examiner insight: If part (a) is wrong, a method that divides your own answer by 0.85 is more likely to gain credit if the division is shown.

6. (a) η = Poutput/Pinput = 1.9 / 2.5 [1] = 0.76 (76%) [1] (b) Widths are proportional to power, so 10 mm represents 1.0 kW. Useful: 19 mm [1]. Thermal: 4.5 mm [1]. Sound: 2.5 − 1.9 − 0.45 = 0.15 kW, so 1.5 mm; the three arrows total 25 mm [1] [5] Examiner insight: Show how the sound arrow is found by subtraction; you risk losing credit if the widths do not add up to the input width.

7. (a) Work done by the resultant force equals the change in kinetic energy: F × 36 = ½ × 1200 × 18² [1]. The change in kinetic energy is 194 400 J [1], so F = 194 400 / 36 = 5400 N [1] (b) F × d = ½ × 1200 × 27² = 437 400 J [1], so d = 437 400 / 5400 = 81 m [1] [5] Examiner insight: A quick check helps: speed rises by 1.5, kinetic energy by 1.5², so the distance should rise by 2.25 times, which gives 81 m.

8. (a) Δh = L(1 − cos 40°) = 1.2 × (1 − 0.766) [1] = 0.281 m, which rounds to 0.28 m [1] (b) mgΔh = ½mv², so v = √(2 × 9.81 × 0.2807) [1] = 2.35 m s⁻¹ [1] (c) Eg lost = 0.35 × 9.81 × 0.2807 = 0.964 J and Ek gained = ½ × 0.35 × 2.10² = 0.772 J [1]. Work against resistive forces = 0.964 − 0.772 = 0.19 J [1] [6] Examiner insight: In a “show that” part, quote the answer to more figures than the one you are asked to show, then round, so that your working supports the given value.

9. Total distance moved along the slope = 2.4 + x [1]. Gravitational component along the slope = 1.5 × 9.81 × sin 25° = 6.22 N [1]. Net force along the slope excluding the spring = 6.22 − 3.0 = 3.22 N [1]. Energy balance: gain of potential energy minus friction loss equals spring energy: 3.22 × (2.4 + x) = ½ × 400 × x² [1]. Rearranged: 200x² − 3.22x − 7.73 = 0 [1]. Quadratic formula: x = (3.22 + √(3.22² + 4 × 200 × 7.73)) / 400 [1]. x = 0.205 m (the negative root is rejected) [1] [7] Examiner insight: Take the spring compression into account in both the height lost and the friction distance; leaving x out of the 2.4 m term is a common way to reach an answer that looks tidy but is wrong.

10. (a) With no acceleration, the driving force balances the resistive force. Pout = Fv = 700 × 22 [1] = 15.4 kW [1] (b) Pin = 15.4 / 0.25 [1] = 61.6 kW [1] (c) Fuel energy per hour = 61 600 × 3600 = 2.22 × 10⁸ J [1]. Volume = 2.22 × 10⁸ / 3.3 × 10⁷ = 6.7 litres [1] (d) Journey time = 100 000 / 22 = 4545 s [1]. Volume = 6.72 × 4545 / 3600 = 8.5 litres [1] [8] Examiner insight: Work in joules and seconds, or use the rate in litres per hour consistently; mixing kW with hours in one line is where unit errors creep in.

11. (a) Component = mg sin θ = 95 × 9.81 × sin 3.0° [1] = 49 N [1] (b) Driving force = 48.8 + 22 = 70.8 N [1]. P = Fv = 70.8 × 5.0 [1] = 354 W [1] (c) η = Pout/Pin, so Pin = 354 / 0.80 [1] = 0.44 kW (442 W) [1] (d) Distance = 5.0 × 900 = 4500 m, so useful work = 70.8 × 4500 = 3.18 × 10⁵ J [1]. Gravitational energy gained = 48.8 × 4500 = 2.19 × 10⁵ J [1]. Fraction = 2.19 / 3.18 = 0.69 (69%) [1] [10] Examiner insight: The same ratio comes from 48.8 / 70.8, so you can state it without the distance; showing both routes checks your answer.

Where marks are usually lost

• Substituting the angle to the horizontal in W = Fs cos θ when the displacement is along a different line.
• Not squaring the speed in kinetic energy or the compression in elastic potential energy.
• Forgetting to include the compression x in the distance moved when a block keeps sliding as it compresses a spring.
• Applying conservation of mechanical energy when the question mentions friction or air resistance.
• Dividing output by input the wrong way round, giving an efficiency above 100%.
• Skipping the efficiency step in fuel problems, so the volume is too small.
• Leaving a quantity in kW, cm or hours while the rest of the equation is in SI units.
• Giving only the final number on a “show that” part, with no intermediate figure.
• Mixing up work done by a force with the change in the system’s energy when several forces act.

Next steps

• Read the revision notes and the study guide.
• Go to the IB DP Physics hub and the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class at /trial/.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topic A.3 Work, energy and power.