Work Done and Energy Transfer Revision Notes | A Level AQA Physics
Work done is W = Fs cos θ, where F is force, s is displacement and θ is the angle between them. Free AQA notes with 2 worked examples and 14 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Work Done and Energy Transfer is part of Force, energy and momentum in A Level AQA Physics.
7 min
Reading time
14
Lessons in this topic
A Level AQA
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Reviewed by Revise Right
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What this lesson covers
Work done is W = Fs cos θ, where F is force, s is displacement and θ is the angle between them. Free AQA notes with 2 worked examples and 14 questions answered.
Calculate work done using $W = Fs\cos\theta$ and explain the significance of the angle between force and displacement
Describe the relationship between work done and energy transfer in mechanical systems
Interpret the area under a force–displacement graph to determine work done by a variable force
Apply the work equation to multi-step problems involving forces at angles
Key ideas to keep in view
Use this lesson to stay inside Force, energy and momentum while connecting the detail back to the wider Mechanics and materials topic.
Work done: The energy transferred to or from an object when a force causes displacement. Measured in joules (J). Equal to the product of the force component along the direction of displacement and the magnitude of that displacement.
Force–displacement graph: A graph plotting force (y-axis) against displacement (x-axis). The area enclosed between the curve and the x-axis equals the total work done, including for variable forces.
Variable force: A force whose magnitude changes with displacement, such as the force exerted by a spring. Work done by a variable force cannot be found by simple multiplication; the area under the force–displacement graph must be used instead.
Energy transfer: The process by which energy moves from one store to another. When work is done on an object, energy is transferred to that object's kinetic, gravitational potential, or elastic potential store.
Revision notes
Read the core explanations from Work Done and Energy Transfer before testing yourself from memory.
1. What Does 'Work Done' Actually Mean?
Push a book across a table and you transfer energy to it. That transfer has a precise name in physics: work done. The amount of energy transferred depends on how hard you push, how far the book moves, and — critically — the direction of your push relative to the motion.
This last point is where A-level physics parts company with GCSE. At GCSE, force and displacement were always assumed to be parallel. Here, they often are not — and that changes everything.
Notice that work done is a scalar quantity — it has magnitude but no direction. Even though force and displacement are both vectors, their combination through $\cos\theta$ produces a single number representing energy transferred.
2. The Equation $W = Fs\cos\theta$ — Unpacked
Picture someone pulling a suitcase along a flat floor using a handle angled at $\theta$ above the horizontal. The full force $F$ acts along the handle, but only the horizontal component — $F\cos\theta$ — actually moves the suitcase forward. The vertical component pushes into the floor and does no useful work on the suitcase's horizontal motion.
Two special cases are worth fixing in memory. When $\theta = 90°$, $\cos 90° = 0$, so $W = 0$ — a force perpendicular to motion does no work. When $\theta = 180°$, $\cos 180° = -1$, so work done is negative, meaning energy is transferred from the object (for example, friction opposing motion).
This result surprises many students, but the maths is unambiguous. Gravity only does work when there is a component of displacement in the vertical direction — which is why gravitational potential energy only changes when height changes.
3. Worked Calculation — Force at an Angle
Theory is clearest when tested against numbers. The following example mirrors the style of AQA calculation questions, where method marks are available even if arithmetic slips.
Worked example: Work Done by a Force at an Angle
A student drags a crate 12 m along a horizontal floor by applying a force of 80 N at an angle of 35° above the horizontal. Calculate the work done by the student on the crate. [3 marks]
$W = Fs\cos\theta$ — [M1] Selecting the correct formula earns the first method mark.
$F = 80 \text{ N}, \quad s = 12 \text{ m}, \quad \theta = 35°$ — State every value before substituting — this protects method marks if a slip occurs later.
$W = 80 \times 12 \times \cos 35°$ — [M1] Correct substitution earns the second method mark.
$W \approx 786 \text{ J}$ — [A1] The accuracy mark requires both the correct value and the correct unit (J).
Answer: $W \approx 786 \text{ J}$
4. Work Done and Energy Transfer — The Physical Link
Work done is not an abstract number — it represents a real physical change. When positive work is done on an object, energy is transferred to that object. When negative work is done (force opposing displacement), energy is transferred from the object.
Consider a ball thrown upward. Gravity does negative work on it as it rises (force downward, displacement upward), removing kinetic energy and storing it as gravitational potential energy. On the way down, gravity does positive work, converting that stored energy back into kinetic energy. The work–energy theorem formalises this: the net work done on an object equals its change in kinetic energy.
This connection between work and energy transfer is central to the AQA specification. Every time you calculate work done, you are calculating how much energy has moved between stores — whether that is kinetic, gravitational potential, elastic potential, or thermal.
5. Variable Forces and the Force–Displacement Graph
So far, the force has been constant throughout the displacement. Real situations are rarely that convenient. A spring being stretched exerts an increasing restoring force. A rocket burns fuel and becomes lighter, so the net force changes. For these cases, $W = Fs\cos\theta$ with a single value of $F$ no longer works.
Here is the key insight: if you plot force against displacement on a graph, the work done over any interval equals the area under the curve between those two displacement values. This works because each infinitesimally thin strip of the graph has area $F \times \delta s$, which is the tiny amount of work done over that tiny displacement — and summing all those strips gives the total work done.
For a linearly increasing force — such as the force in a spring obeying Hooke's law — the graph is a straight line through the origin, forming a triangle. The area of that triangle gives the elastic potential energy stored: $E = \frac{1}{2}Fx = \frac{1}{2}kx^2$, which you will recognise from the materials section of the course.
The graph above shows a spring with spring constant $k = 40 \text{ N m}^{-1}$ being extended from 0 to 0.5 m. The area under the line is a triangle with base 0.5 m and height 20 N, giving work done $= \frac{1}{2} \times 0.5 \times 20 = 5 \text{ J}$. That 5 J is the elastic potential energy stored in the spring.
Worked example: Finding Work Done from a Force–Displacement Graph
A force varies with displacement as follows: from 0 to 3 m, the force increases linearly from 0 N to 12 N. From 3 m to 5 m, the force remains constant at 12 N. Calculate the total work done over the full 5 m displacement. [4 marks]
Region 1 (0 to 3 m): triangle with base 3 m and height 12 N.
Region 2 (3 m to 5 m): rectangle with width 2 m and height 12 N. — Splitting the area into recognisable geometric shapes is the standard method for variable-force graphs.
$A_1 = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 12 = 18 \text{ J}$ — [M1] Correct use of triangle area formula. Verification: $\frac{1}{2} \times 3 = 1.5$, then $1.5 \times 12 = 18$.
$A_2 = width \times height = 2 \times 12 = 24 \text{ J}$ — [M1] Correct use of rectangle area formula.
$W_{total} = A_1 + A_2 = 18 + 24 = 42 \text{ J}$ — [A1] Total work done is the sum of all areas under the graph. Units must be stated as joules.
If the force had been constant at 12 N throughout: $W = 12 \times 5 = 60 \text{ J}$. Our answer of 42 J is less than 60 J, which makes sense because the force started at zero and built up gradually. — Order-of-magnitude checks like this catch arithmetic errors before they cost marks.
Answer: $W_{total} = 42 \text{ J}$
6. Quantitative Questions on Variable Forces
AQA explicitly states that quantitative questions may be set on variable forces. This means you need to be comfortable extracting work done from graphs with irregular shapes, not just triangles and rectangles.
When the curve is irregular, count the squares under the graph. Establish the value of one small square (width in metres multiplied by height in newtons gives joules per square), then count full squares and estimate partial ones. A common convention is to count a partial square as one full square if more than half is shaded, and ignore it if less than half is shaded.
Triangular region: $A = \frac{1}{2} \times base \times height$
Irregular curve: count grid squares, noting the joule value of each square
Known function (e.g. $F = kx$): integrate $\int F \, ds$ over the displacement range
The trapezoidal method is particularly useful when you have data points at regular intervals. If a force changes from 6 N to 14 N over a displacement of 4 m, the work done is $\frac{1}{2}(6 + 14) \times 4 = \frac{1}{2} \times 20 \times 4 = 40 \text{ J}$.
7. Exam Technique and Common Mistakes
Work done questions appear across all three AQA papers and range from single-step calculations to multi-part graph interpretation questions. The marks are genuinely accessible if you follow a consistent method — but there are predictable traps that cost students marks every year.
For extended-response questions asking you to 'explain' or 'evaluate' energy transfer, structure your answer as: state the relevant equation, identify the physical situation (direction of force relative to displacement), calculate or estimate the work done, and link explicitly to the energy store that changes. This chain of reasoning is what distinguishes a 5–6 mark response from a 2–3 mark one.
Every spec point in this lesson — the $W = Fs\cos\theta$ equation, variable force questions, and the significance of the force–displacement graph area — maps directly onto AQA mark schemes. Practise identifying which tool to reach for: constant force at an angle uses the equation; variable force uses the graph area. That distinction is the core of this topic.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
Work done is W = Fs cos θ, where F is the applied force in newtons, s is the displacement in metres and θ is the angle between the force and the direction of displacement. It is the energy transferred when a force causes a displacement, measured in joules, where 1 J = 1 N m. Work done is a scalar quantity.
What is the work-energy theorem?
The work–energy theorem states that the net work done on an object equals its change in kinetic energy. Positive work transfers energy to an object and negative work transfers energy from it: gravity does negative work on a ball thrown upward, storing energy as gravitational potential energy, then positive work as it falls back down.
What does the area under a force-displacement graph represent?
The area under a force–displacement graph equals the work done, and so the energy transferred, by the force. For a variable force, split the area into triangles and rectangles, count squares, or integrate. A force rising from 0 to 12 N over 3 m, then constant at 12 N for 2 m, does 18 + 24 = 42 J of work.
How do you calculate work done when a force is at an angle?
When a force acts at an angle, only the component F cos θ along the displacement does work, so W = Fs cos θ. If a student drags a crate 12 m with a force of 80 N at 35° to the horizontal, W = 80 × 12 × cos 35° = 960 × 0.8192 ≈ 786 J.
Why does a force perpendicular to motion do no work?
A force perpendicular to the motion does no work because θ = 90° and cos 90° = 0, so W = Fs cos θ = 0. Only the component of a force along the displacement transfers energy. At θ = 180°, as with friction opposing motion, cos 180° = -1 and the work done is negative, so energy is transferred from the object.
Is work done a scalar or a vector?
Work done is a scalar quantity: it has magnitude but no direction. Even though force and displacement are both vectors, combining them through W = Fs cos θ produces a single number representing the energy transferred.
What is one joule of work?
One joule is the work done, or energy transferred, when a force of 1 N causes a displacement of 1 m in the direction of the force, so 1 J = 1 N m. The area under a force–displacement graph also has units of N m, which is joules.
What does negative work done mean?
Negative work done means energy is transferred from the object, which happens when a force opposes the displacement. At θ = 180°, cos 180° = −1, so W is negative, as with friction opposing motion.
What work does gravity do on a ball thrown upwards?
Gravity does negative work on a ball thrown upwards as it rises, because the force is downward and the displacement upward, so kinetic energy is removed and stored as gravitational potential energy. On the way down gravity does positive work, converting that stored energy back into kinetic energy.
How do you find the work done stretching a spring from a graph?
The work done stretching a spring is the triangular area under its force–displacement line. For a spring with k = 40 N m⁻¹ extended by 0.5 m, the force reaches 20 N, so work done = ½ × 0.5 × 20 = 5 J, stored as elastic potential energy, E = ½Fx = ½kx².
How do you find work done from an irregular force–displacement graph?
For an irregular force–displacement curve, work done is found by counting squares under the graph. The value of one small square, width in metres times height in newtons, gives joules per square; full squares are counted, and a partial square counts as one if more than half is shaded and is ignored if less.
How do you use the trapezoidal method to find work done?
The trapezoidal method gives work done when a force changes between two values over a displacement: W = ½(F₁ + F₂) × Δs. For a force changing from 6 N to 14 N over 4 m, the work done is ½ × (6 + 14) × 4 = 40 J.
When do you use W = Fs cos θ and when do you use the graph area?
W = Fs cos θ is used for a constant force, including one at an angle, while the area under a force–displacement graph is used for a variable force. For a constant force the area is just a rectangle, W = F × s, so both methods agree.
How is work done found for a force that varies as a known function?
When force is a known function of displacement, such as F = kx, the work done is found by integrating F over the displacement range. Each thin strip under the graph has area F × δs, the tiny work done over that tiny displacement, and summing all the strips gives the total.