Mathematics 9709 · syllabus 4.5
Cambridge A Level Mathematics 9709 — Energy, work and power topical past papers
Work done by a force, kinetic and potential energy, the work–energy principle, and power as the rate of doing work, P = Fv, for moving vehicles.
- Questions
- 177
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 55
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Energy, work and power covers
Work is done when a force moves its point of application: a constant force F moving through a distance d at an angle θ to its line of action does work Fd cos θ. In Paper 4 (section 4.5) you use kinetic energy ½mv² and gravitational potential energy mgh, and the work–energy principle: the work done by the forces other than weight equals the gain in kinetic and potential energy. When no work is done against resistances, mechanical energy is conserved.
Energy methods need only the start and end of the motion, so they work even on curved paths, such as a particle on a smooth curved slide, where F = ma cannot be used directly. Power is the rate of doing work; for a driving force F at speed v, P = Fv. Typical questions involve a car or cyclist working at a constant power against a resistance, on the level or on a hill, and ask for a maximum speed or an acceleration at a given moment.
The section connects to Newton's laws, since P = Fv gives a driving force that goes into F = ma, and to momentum, where energy shows what a collision loses. The work done by a force at an angle uses its component along the motion, so the resolving of section 4.1 is needed here too, with g = 10 m s⁻² for weight and potential energy. Elastic potential energy and Hooke's law are not part of 9709.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 4.5 Energy, work and power
What you should be able to do
- Calculate the work done by a constant force, including one at an angle to the motion
- Calculate changes in kinetic and gravitational potential energy
- Apply the work–energy principle with resistances, on slopes and curved paths
- Use conservation of mechanical energy when no resistances act
- Use P = Fv to find driving forces, maximum speeds and accelerations
- Find average power as work done divided by time taken
Key ideas
- Work done = Fd cos θ, where θ is the angle between the force and the displacement
- Kinetic energy = ½mv²; gravitational potential energy = mgh
- Work done by driving forces − work done against resistances = gain in kinetic energy + gain in potential energy
- Power is the rate of doing work: P = Fv for a force in the direction of motion, and average power = work done ÷ time
- At maximum speed on a level road the driving force equals the resistance, so the acceleration is zero
How Energy, work and power has been examined
From 2022 to 2026, Energy, work and power was on Paper 4 · Mechanics in 11 of 11 exam series, worth about 24% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · Mechanics | 55 | 24% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 7.
Filed in Momentum under: Energy, Work & Power.
A worked example
A car of mass 1200 kg moves along a straight horizontal road with its engine working at 30 kW, against a constant resistance of 900 N. Find the car's acceleration when its speed is 20 m s⁻¹, and its greatest possible speed.
- Driving force at 20 m s⁻¹: F = P/v = 30 000 / 20 = 1500 N.
- Resultant force = 1500 − 900 = 600 N.
- Newton's second law: a = 600 / 1200 = 0.5 m s⁻².
- At the greatest speed the acceleration is zero, so the driving force equals the resistance: 30 000 / v = 900.
- v = 30 000 / 900 = 33.3 m s⁻¹.
Answer: 0.5 m s⁻²; greatest speed 33.3 m s⁻¹ (100/3)
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Counting the weight twice: include either the potential energy mgh or the work done by gravity, not both.
- Putting the resistance into P = Fv: P/v gives the engine's driving force, and the resultant is the driving force minus the resistance.
- Taking the work against friction on a slope as μmgd: the normal reaction on a slope at angle α is mg cos α, so the work is μmg cos α × d.
- Leaving power in kW: convert to W before using P = Fv with newtons and metres per second.
Exam technique
- Write the energy equation in words first, such as 'initial kinetic energy + work done by driving force = final kinetic energy + potential energy gained + work done against resistance', then substitute.
- For vehicles, find the driving force from F = P/v at the stated speed, then apply resultant force = ma at that instant.
- Choose an energy method when a question gives distances and speeds but no times, or when the path is curved.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.