What is the definition of work in the context of Physics, specifically for Edexcel International A Levels?

In Physics, particularly in the Edexcel International A Levels curriculum, work is defined as the product of the force applied to an object and the distance over which that force is applied, in the direction of the force. Mathematically, it is expressed as W = F × d × cos(θ), where W is work, F is the force, d is the distance, and θ is the angle between the force and the direction of motion.

How is energy related to work in the study of Mechanics and Materials?

Energy is closely related to work in Mechanics and Materials. Energy is the capacity to do work. When work is done on an object, energy is transferred to or from that object. For example, when lifting a weight, work is done against gravity, and potential energy is stored in the object. This relationship is fundamental in understanding how forces cause changes in motion and energy states.

What are the different forms of energy covered in the Edexcel International A Levels Physics syllabus?

The Edexcel International A Levels Physics syllabus covers several forms of energy, including kinetic energy, potential energy (gravitational and elastic), thermal energy, chemical energy, and electrical energy. Each form of energy can be transformed into another, and understanding these transformations is crucial in solving problems related to energy conservation and transfer.

How is power defined and calculated in Physics, and why is it important?

Power in Physics is defined as the rate at which work is done or energy is transferred over time. It is calculated using the formula P = W/t, where P is power, W is work done, and t is the time taken. Power is measured in watts (W). Understanding power is important because it helps quantify how quickly energy is used or transferred, which is essential in applications ranging from engines to electrical circuits.

What is the principle of conservation of energy, and how does it apply to Mechanics and Materials?

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In Mechanics and Materials, this principle implies that the total energy within a closed system remains constant. For example, when a pendulum swings, its energy continuously transforms between kinetic and potential forms, but the total energy remains constant, assuming no energy is lost to friction or air resistance.

Why is "Multiplying the full force by the distance when the force acts at an angle to the motion" a common mistake in Work, Energy and Power?

Why it happens: Students learn W = F × d first and forget to resolve when the force is later applied at an angle. How to avoid it: Always use W = F s and take the **component of force along the displacement** (F). If the force is perpendicular to the motion, 90° = 0 and no work is done. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Forgetting to square the speed in $E_k = \tfrac12 mv^2$" a common mistake in Work, Energy and Power?

Why it happens: Rushing the substitution and treating the equation as if kinetic energy were linear in v. How to avoid it: Write the equation with v^2 explicitly and **square the speed before multiplying**. Remember kinetic energy scales with v^2 — doubling the speed quadruples the energy, not doubles it. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Using the distance along a slope as $\Delta h$ in $\Delta E_{grav} = mg\Delta h$" a common mistake in Work, Energy and Power?

Why it happens: The slope length is the distance travelled, so students substitute it without realising h must be vertical. How to avoid it: Use the **vertical** height change only. For a slope of length L at angle to the horizontal, h = L. Gravitational PE depends on height gained, not path length. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Saying mechanical energy is 'lost' or 'destroyed' when friction or drag acts" a common mistake in Work, Energy and Power?

Why it happens: Everyday language treats wasted energy as disappearing, contradicting conservation of energy. How to avoid it: Energy is **transferred to internal (thermal) energy** of the object and surroundings (and a little to sound). State that mechanical energy is **dissipated**, not destroyed — total energy is always conserved. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Using the weight (or the wrong force) instead of the driving force in $P = Fv$ for a vehicle" a common mistake in Work, Energy and Power?

Why it happens: Students see a vehicle and a speed but do not identify which force drives the motion. How to avoid it: In P = Fv, F is the **driving force**. At constant speed it equals the **total resistive force** (plus mg when climbing a slope), because the resultant force is zero — not the weight. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Inverting the efficiency fraction or quoting an efficiency above 100%" a common mistake in Work, Energy and Power?

Why it happens: Confusing which quantity is the useful output and which is the total input. How to avoid it: Always put **useful output on top and total input on the bottom**: efficiency = usefulinput × 100\%. A real machine can never exceed 100% — a value above 100% means the fraction is upside-down. Source: Edexcel IAL Physics Unit 1 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Work, Energy and Power

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Short Notes - Work, Energy and Power

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Detailed Study Notes

Detailed notes on Mechanics and Materials for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Work, Energy and Power — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 1)

Work done by a force ( $\Delta W = F\Delta s\cos\theta$ ), kinetic energy ( $E_k=\tfrac12 mv^2$ ) and gravitational potential energy ( $\Delta E_{grav}=mg\Delta h$ ), the principle of conservation of energy, power ( $P=\Delta W/\Delta t$ and $P=Fv$ ) and efficiency. Full mark-scheme working discipline with $g = 9.81\ \text{m s}^{-2}$ .

At a glance

Work done is energy transferred by a force: $\Delta W = F\Delta s\cos\theta$ — force × distance moved in the direction of the force. Measured in joules (J).
When the force is perpendicular to the motion, $\theta = 90°$ and $\cos 90° = 0$ , so no work is done (e.g. centripetal force, normal contact force on a level surface).
Kinetic energy $E_k = \tfrac12 mv^2$ — depends on $v^2$ , so doubling speed quadruples the kinetic energy.
Gravitational PE change $\Delta E_{grav} = mg\Delta h$ — use the vertical height change only.
Conservation of energy: energy is never created or destroyed, only transferred. In a frictionless drop, $mgh \rightarrow \tfrac12 mv^2$ , so $v=\sqrt{2gh}$ .
With friction or drag, some energy is transferred to thermal (internal) energy — it is not 'lost', it is dissipated.
Power is the rate of energy transfer: $P = \dfrac{\Delta W}{\Delta t} = \dfrac{\Delta E}{\Delta t}$ , measured in watts (W). For a constant force, $P = Fv$ .
Efficiency $= \dfrac{\text{useful output}}{\text{total input}} \times 100\%$ — always $< 100\%$ for a real machine because some energy is wasted (usually as heat).

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Use the equation $\Delta W = F\Delta s\cos\theta$ for the work done by a force, including resolving the force when it acts at an angle to the displacement.
Recall and use $E_k = \tfrac12 mv^2$ for the kinetic energy of a moving object.
Recall and use $\Delta E_{grav} = mg\Delta h$ for the change in gravitational potential energy near the Earth's surface.
Apply the principle of conservation of energy, including transfers between kinetic and gravitational potential energy and to internal (thermal) energy by friction.
Use the equations $P = \dfrac{\Delta W}{\Delta t}$ and $P = Fv$ for power, including vehicles moving at constant speed against resistive forces.
Use the relationship efficiency $= \dfrac{\text{useful energy (or power) output}}{\text{total energy (or power) input}}$ and interpret Sankey-style energy transfers.

Work done by a force

Work = force × distance moved in the direction of the force. Resolve when the force is at an angle.

Work is done whenever a force moves its point of application. The work done is the energy transferred, and it is calculated from:

$\Delta W = F\Delta s\cos\theta$

where $F$ is the force (N), $\Delta s$ is the displacement (m), and $\theta$ is the angle between the force and the displacement. Work is a scalar measured in joules (J): $1\ \text{J}$ is the work done when a force of $1\ \text{N}$ moves its point of application $1\ \text{m}$ in the direction of the force.

The $\cos\theta$ term is the key to this whole topic. It picks out the component of the force in the direction of motion:

When the force acts along the direction of motion, $\theta = 0°$ , $\cos 0° = 1$ , so $\Delta W = F\Delta s$ (the simple case).
When the force acts at an angle $\theta$ to the motion (e.g. pulling a sledge by a rope held above the ground), only the horizontal component $F\cos\theta$ does work along the ground.
When the force is perpendicular to the motion, $\theta = 90°$ , $\cos 90° = 0$ , so no work is done. This is why the normal contact force on a horizontal surface and the centripetal force in circular motion do zero work.

Only the component of force in the direction of motion does work. The horizontal component $F\cos\theta$ does work $F\Delta s\cos\theta$; the vertical component $F\sin\theta$ is perpendicular to the displacement and does no work.

A worked sense-check: a force of $50\ \text{N}$ pulls a crate $8.0\ \text{m}$ along the floor at $30°$ above the horizontal. Work done $= 50 \times 8.0 \times \cos 30° = 50 \times 8.0 \times 0.866 = 346\ \text{J}$ (3 s.f.). The vertical component does no work along the floor.

$\Delta W = F\Delta s\cos\theta$ — work is in joules (J).
$\theta$ is the angle between the force and the displacement.
Force perpendicular to motion ( $\theta = 90°$ ) does zero work.
Resolve an angled force: use the component along the displacement, $F\cos\theta$ .

See the full worked example for work, energy and power →

Kinetic energy and gravitational potential energy

$E_k = \tfrac12 mv^2$ depends on $v^2$ ; $\Delta E_{grav} = mg\Delta h$ uses the vertical height only.

Two energy stores dominate mechanics problems.

Kinetic energy is the energy of a moving object:

$E_k = \tfrac12 mv^2$

Because $E_k$ depends on $v^2$ , doubling the speed quadruples the kinetic energy, and tripling the speed gives nine times the kinetic energy. This is why stopping distances rise so steeply with speed.

Gravitational potential energy change near the Earth's surface is:

$\Delta E_{grav} = mg\Delta h$

where $\Delta h$ is the vertical height change and $g = 9.81\ \text{m s}^{-2}$ . On an inclined plane or staircase, you must use the vertical height gained, not the distance travelled along the slope.

To find the vertical height from a slope of length $L$ at angle $\alpha$ to the horizontal, use $\Delta h = L\sin\alpha$ .

On a slope, the gain in gravitational PE uses the **vertical** height $\Delta h = L\sin\alpha$, not the distance $L$ travelled along the slope.

Worked sense-check: a $2.0\ \text{kg}$ mass raised $1.5\ \text{m}$ vertically gains $\Delta E_{grav} = 2.0 \times 9.81 \times 1.5 = 29.4\ \text{J}$ . A $0.50\ \text{kg}$ ball moving at $6.0\ \text{m s}^{-1}$ has $E_k = \tfrac12 \times 0.50 \times 6.0^2 = 9.0\ \text{J}$ .

$E_k = \tfrac12 mv^2$ — quadratic in speed (double $v$ → ×4 energy).
$\Delta E_{grav} = mg\Delta h$ — uses vertical height change.
On a slope of length $L$ at angle $\alpha$ : $\Delta h = L\sin\alpha$ .
Both energies are scalars measured in joules (J).

See the full worked example for work, energy and power →

Conservation of energy

Energy is transferred, never created or destroyed. KE ⇌ GPE; friction dissipates energy as heat.

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one store to another or dissipated. The total energy of a closed system is constant.

In mechanics this gives a powerful shortcut. For an object falling freely (no air resistance) through a height $h$ , all the gravitational PE lost becomes kinetic energy:

$mgh = \tfrac12 mv^2 \quad\Rightarrow\quad v = \sqrt{2gh}$

The mass cancels, so a heavy and a light object reach the same speed after the same drop (Galileo's result). This $mgh \rightleftharpoons \tfrac12 mv^2$ exchange runs in both directions — a pendulum, a rollercoaster and a thrown ball all swap between the two stores continuously.

As the car descends, gravitational PE is converted into kinetic energy; climbing, the reverse occurs. Without friction the total mechanical energy $E = E_{grav} + E_k$ stays constant.

When friction or air resistance acts, mechanical energy is not conserved — but total energy still is. The "missing" mechanical energy is transferred to internal (thermal) energy in the object and surroundings (and a little to sound). The bookkeeping becomes:

$E_{grav\,lost} = E_k\,gained + E\,dissipated$

So a real object reaching the bottom of a slope arrives slower than $\sqrt{2gh}$ predicts. The difference $mgh - \tfrac12 mv^2$ equals the work done against friction.

Energy is conserved: transferred between stores, never destroyed.
Frictionless drop: $mgh = \tfrac12 mv^2 \Rightarrow v = \sqrt{2gh}$ (mass cancels).
With friction: $mgh = \tfrac12 mv^2 +$ energy dissipated as heat.
Energy 'lost' to friction is dissipated as internal (thermal) energy — not destroyed.

See the full worked example for work, energy and power →

Power — the rate of energy transfer

$P = \Delta W/\Delta t = \Delta E/\Delta t$ ; for a constant force, $P = Fv$ .

Power is the rate at which work is done, or equivalently the rate of energy transfer:

$P = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}$

Power is measured in watts (W), where $1\ \text{W} = 1\ \text{J s}^{-1}$ .

There is a second, hugely useful form. If a constant force $F$ moves an object a distance $\Delta s$ in time $\Delta t$ , then $P = \dfrac{F\Delta s}{\Delta t} = F \times \dfrac{\Delta s}{\Delta t}$ , and since $\dfrac{\Delta s}{\Delta t}$ is the velocity:

$P = Fv$

This is the equation for a vehicle's engine or a winch. At constant speed, the driving (engine) force exactly balances the total resistive force (drag + friction + the component of weight up a slope), so $P = F_{drive}\,v$ gives the power output needed to maintain that speed.

A car at constant speed up an incline: the driving force balances the resistive force plus the weight component $mg\sin\alpha$. The engine's useful output power is $P = Fv$.

Worked sense-check: a car maintains $25\ \text{m s}^{-1}$ against a total resistive force of $640\ \text{N}$ on level ground. Power output $= Fv = 640 \times 25 = 16\,000\ \text{W} = 16.0\ \text{kW}$ .

$P = \Delta W/\Delta t = \Delta E/\Delta t$ — watts (W), $1\ \text{W} = 1\ \text{J s}^{-1}$ .
For a constant force, $P = Fv$ .
At constant speed, driving force = total resistive force, so $P = F_{resist}\,v$ .
Up a slope at constant speed, $F =$ resistive force $+\ mg\sin\alpha$ .

See the full worked example for work, energy and power →

Efficiency and Sankey reasoning

Efficiency = useful output ÷ total input × 100%. Real machines waste energy, usually as heat.

No real machine transfers all of its input energy usefully — some is always dissipated, usually as heat to the surroundings. Efficiency measures the fraction transferred usefully:

$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%$

Because power is energy per second, you can equally write:

$\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\%$

Efficiency is a ratio, so it has no units; expressed as a percentage it is always less than 100% for a real device (a value of exactly 100% would violate the second law of thermodynamics).

A Sankey diagram shows this visually: the width of the arrow is proportional to the energy (or power). The input arrow splits into a useful output branch and one or more wasted branches, and the total width is conserved at every junction (conservation of energy).

A Sankey diagram: arrow width is proportional to energy. Of $100\ \text{J}$ in, $60\ \text{J}$ is transferred usefully and $40\ \text{J}$ is wasted as heat — an efficiency of $60\%$. The widths in equal the widths out (energy is conserved).

To find wasted energy from an efficiency, rearrange: useful output $=$ efficiency $\times$ input, and wasted $=$ input $-$ useful. For the motor above: useful $= 0.60 \times 100 = 60\ \text{J}$ , wasted $= 100 - 60 = 40\ \text{J}$ .

Efficiency $= \dfrac{\text{useful output}}{\text{total input}} \times 100\%$ — no units.
Works with energy or power (per second).
Real machines: efficiency $< 100\%$ ; waste is usually heat.
Sankey: arrow width ∝ energy; total width is conserved at each split.

See the full worked example for work, energy and power →

Quick recap

Work done $\Delta W = F\Delta s\cos\theta$ — include $\cos\theta$ when the force is at an angle; perpendicular force does zero work.
Kinetic energy $E_k = \tfrac12 mv^2$ scales with $v^2$ ; gravitational PE change $\Delta E_{grav} = mg\Delta h$ uses the vertical height.
Energy is conserved: in a frictionless drop $mgh = \tfrac12 mv^2$ so $v = \sqrt{2gh}$ .
Friction and drag dissipate energy as internal (thermal) energy — it is never destroyed.
Power $P = \Delta W/\Delta t = \Delta E/\Delta t$ , and for a constant force $P = Fv$ .
Efficiency $= \dfrac{\text{useful output}}{\text{total input}} \times 100\%$ , always below 100% for real machines.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Work done: $\Delta W = F\Delta s\cos\theta$ — measured in joules (J).
Kinetic energy: $E_k = \tfrac12 mv^2$ (depends on $v^2$ ).
Gravitational PE change: $\Delta E_{grav} = mg\Delta h$ (vertical height; $g = 9.81\ \text{m s}^{-2}$ ).
Frictionless fall: $mgh = \tfrac12 mv^2 \Rightarrow v = \sqrt{2gh}$ .
Power: $P = \dfrac{\Delta W}{\Delta t} = \dfrac{\Delta E}{\Delta t}$ ; for a constant force $P = Fv$ .
Efficiency $= \dfrac{\text{useful energy (or power) output}}{\text{total energy (or power) input}} \times 100\%$ .
Energy is conserved: transferred or dissipated (as heat), never created or destroyed.

How it’s examined

Work, energy and power is core Unit 1 (Mechanics and Materials) content, examined on the WPH11 paper. Expect: direct calculations of work, KE, GPE, power and efficiency (1–3 marks, AO2) — always show the equation, substitution with units and a 3 s.f. answer; 'Show that' questions (e.g. show that the power output is about 16 kW) where you must work through to the given value and quote one extra significant figure; conservation-of-energy problems with and without friction, where the high-value step is identifying that the discrepancy is energy dissipated as internal/thermal energy; $P = Fv$ vehicle questions (constant speed up an incline or against drag), where $F$ is the driving force balancing the resistive force plus $mg\sin\alpha$ ; efficiency/Sankey questions; and a levels-based 6-marker asking you to describe the energy transformations in a scenario (e.g. a bouncing ball, a pendulum or a hydroelectric scheme) — write linked continuous prose using conservation of energy, naming each store and the dissipation to heat.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 1 examiner reports; Edexcel IAL Physics Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — Work, Energy and Power

Step-by-step solutions to past-paper-style questions on work, energy and power, written exactly the way a tutor would explain them at the board.

Question type:

1Work done by a horizontal force (2 marks)
Getting startedDirect calculation• work, AO2, Unit 1
Question
A horizontal force of $35\ \text{N}$ pushes a box $4.0\ \text{m}$ across a level floor in the direction of the force. Calculate the work done. (2 marks)
Step-by-step solution
1. Step 1
  The force acts along the direction of motion, so $\theta = 0°$ and $\cos 0° = 1$ . Use $\Delta W = F\Delta s\cos\theta$ .
  $\Delta W = F\Delta s\cos\theta$
2. Step 2
  Substitute the values with units.
  $\Delta W = 35 \times 4.0 \times \cos 0° = 35 \times 4.0$
3. Step 3
  Evaluate to give the work done in joules.
  $\Delta W = 140\ \text{J}$
Answer
$\Delta W = 140\ \text{J}$
Examiner tip
Mark scheme: (1) correct substitution into $\Delta W = F\Delta s$ ; (2) answer $140\ \text{J}$ with the unit. The unit J is required for the accuracy mark.
2Kinetic energy of a moving object (2 marks)
Getting startedDirect calculation• kinetic energy, AO2, Unit 1
Question
A cyclist and bicycle have a combined mass of $80\ \text{kg}$ and move at $7.0\ \text{m s}^{-1}$ . Calculate their kinetic energy. (2 marks)
Step-by-step solution
1. Step 1
  Use the kinetic energy equation.
  $E_k = \tfrac12 mv^2$
2. Step 2
  Substitute mass and speed, remembering to square the speed.
  $E_k = \tfrac12 \times 80 \times 7.0^2 = \tfrac12 \times 80 \times 49$
3. Step 3
  Evaluate.
  $E_k = 1960\ \text{J} = 1.96\ \text{kJ}$
Answer
$E_k = 1960\ \text{J}$ (to 3 s.f., $1.96\ \text{kJ}$ )
Examiner tip
Mark scheme: (1) substitution with $v$ squared; (2) $1960\ \text{J}$ (or $1.96\ \text{kJ}$ ). The most common slip is forgetting to square $v$ , giving $280\ \text{J}$ .
3Gain in gravitational potential energy (3 marks)
Getting startedDirect calculation• gravitational PE, AO2, Unit 1
Question
A lift of mass $1200\ \text{kg}$ is raised vertically through $18\ \text{m}$ . Calculate the gain in gravitational potential energy. Take $g = 9.81\ \text{m s}^{-2}$ . (3 marks)
Step-by-step solution
1. Step 1
  Use the change in gravitational potential energy equation.
  $\Delta E_{grav} = mg\Delta h$
2. Step 2
  Substitute the mass, $g$ and the vertical height.
  $\Delta E_{grav} = 1200 \times 9.81 \times 18$
3. Step 3
  Evaluate and express to 3 significant figures.
  $\Delta E_{grav} = 211\,896\ \text{J} \approx 2.12 \times 10^5\ \text{J}$
Answer
$\Delta E_{grav} = 2.12 \times 10^5\ \text{J}$ (212 kJ)
Examiner tip
Mark scheme: (1) correct equation; (2) substitution with $g = 9.81$ ; (3) $2.12 \times 10^5\ \text{J}$ to 3 s.f. with unit. Using $g = 10$ loses the value here unless the question allows it.
4Work done by a force at an angle (3 marks)
Building confidenceDirect calculation• Adapted from Edexcel IAL Physics Unit 1• work, resolving, AO2, Unit 1
Question
A child pulls a sledge $12\ \text{m}$ across level snow using a rope that makes an angle of $25°$ with the horizontal. The tension in the rope is $90\ \text{N}$ . Calculate the work done by the tension. (3 marks)
Step-by-step solution
1. Step 1
  Only the horizontal component of the tension does work along the snow. Use $\Delta W = F\Delta s\cos\theta$ with $\theta = 25°$ .
  $\Delta W = F\Delta s\cos\theta$
2. Step 2
  Substitute the tension, distance and angle.
  $\Delta W = 90 \times 12 \times \cos 25°$
3. Step 3
  Evaluate $\cos 25° = 0.906$ and multiply.
  $\Delta W = 90 \times 12 \times 0.906 = 979\ \text{J}$
Answer
$\Delta W = 979\ \text{J}$ (3 s.f.)
Examiner tip
Mark scheme: (1) recognise $\Delta W = F\Delta s\cos\theta$ / resolve horizontally; (2) substitute $\cos 25°$ ; (3) $979\ \text{J}$ (allow 978–980). Omitting $\cos\theta$ (answer $1080\ \text{J}$ ) is the classic error and loses two marks.
5Speed of a falling object from energy (3 marks)
Building confidenceMulti-step problem• conservation of energy, AO2, Unit 1
Question
A ball of mass $0.20\ \text{kg}$ is dropped from rest and falls $2.5\ \text{m}$ . Assuming air resistance is negligible, calculate its speed just before it hits the ground. (3 marks)
Step-by-step solution
1. Step 1
  By conservation of energy, all the gravitational PE lost becomes kinetic energy. The mass cancels.
  $mg\Delta h = \tfrac12 mv^2$
2. Step 2
  Rearrange for $v$ .
  $v = \sqrt{2g\Delta h}$
3. Step 3
  Substitute $g = 9.81$ and $\Delta h = 2.5\ \text{m}$ and evaluate.
  $v = \sqrt{2 \times 9.81 \times 2.5} = \sqrt{49.05} = 7.00\ \text{m s}^{-1}$
Answer
$v = 7.00\ \text{m s}^{-1}$ (3 s.f.)
Examiner tip
Mark scheme: (1) equate $mg\Delta h = \tfrac12 mv^2$ (or use $v = \sqrt{2g\Delta h}$ ); (2) correct rearrangement; (3) $7.00\ \text{m s}^{-1}$ . Note the mass is not needed because it cancels — a deliberate distractor.
6Power from energy and time (3 marks)
Building confidenceDirect calculation• power, AO2, Unit 1
Question
A pump raises $450\ \text{kg}$ of water through a vertical height of $12\ \text{m}$ in $30\ \text{s}$ . Calculate the useful output power of the pump. Take $g = 9.81\ \text{m s}^{-2}$ . (3 marks)
Step-by-step solution
1. Step 1
  First find the useful energy transferred — the gain in gravitational PE of the water.
  $\Delta E = mg\Delta h = 450 \times 9.81 \times 12 = 52\,974\ \text{J}$
2. Step 2
  Power is energy transferred per unit time.
  $P = \frac{\Delta E}{\Delta t}$
3. Step 3
  Substitute the energy and the time.
  $P = \frac{52\,974}{30} = 1766\ \text{W} \approx 1.77\ \text{kW}$
Answer
$P = 1.77 \times 10^3\ \text{W}$ (1.77 kW, 3 s.f.)
Examiner tip
Mark scheme: (1) $\Delta E = mg\Delta h$ correct; (2) $P = \Delta E/\Delta t$ ; (3) $1.77\ \text{kW}$ . This is the useful output power; the input power would be higher because the pump is not 100% efficient.
7Engine power for a car up an incline (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 1• power, P=Fv, incline, AO2, Unit 1
Question
A car of mass $1100\ \text{kg}$ travels up a slope inclined at $6.0°$ to the horizontal at a constant $20\ \text{m s}^{-1}$ . The total resistive force (friction and drag) is $500\ \text{N}$ . Calculate the useful output power of the engine. Take $g = 9.81\ \text{m s}^{-2}$ . (4 marks)
Step-by-step solution
1. Step 1
  At constant speed the driving force balances the resistive force plus the weight component down the slope, $mg\sin\alpha$ .
  $F = F_{resist} + mg\sin\alpha$
2. Step 2
  Calculate the weight component along the slope.
  $mg\sin\alpha = 1100 \times 9.81 \times \sin 6.0° = 1100 \times 9.81 \times 0.1045 = 1128\ \text{N}$
3. Step 3
  Find the total driving force.
  $F = 500 + 1128 = 1628\ \text{N}$
4. Step 4
  Use $P = Fv$ with the driving force and the constant speed.
  $P = Fv = 1628 \times 20 = 32\,560\ \text{W} \approx 32.6\ \text{kW}$
Answer
$P = 3.26 \times 10^4\ \text{W}$ (32.6 kW, 3 s.f.)
Examiner tip
Mark scheme: (1) driving force = resistive + $mg\sin\alpha$ ; (2) $mg\sin\alpha = 1130\ \text{N}$ ; (3) total $F = 1630\ \text{N}$ ; (4) $P = Fv = 32.6\ \text{kW}$ . Omitting the $mg\sin\alpha$ term (giving $10\ \text{kW}$ ) is the standard error on incline questions.
8Energy dissipated by friction on a slope (4 marks)
StretchMulti-step problem• conservation of energy, friction, AO2, Unit 1
Question
A $3.0\ \text{kg}$ block slides from rest down a slope, dropping a vertical height of $1.6\ \text{m}$ . It reaches the bottom at $4.0\ \text{m s}^{-1}$ . Calculate the energy dissipated by friction. Take $g = 9.81\ \text{m s}^{-2}$ . (4 marks)
Step-by-step solution
1. Step 1
  Find the gravitational PE lost over the vertical drop.
  $\Delta E_{grav} = mg\Delta h = 3.0 \times 9.81 \times 1.6 = 47.1\ \text{J}$
2. Step 2
  Find the kinetic energy actually gained at the bottom.
  $E_k = \tfrac12 mv^2 = \tfrac12 \times 3.0 \times 4.0^2 = 24.0\ \text{J}$
3. Step 3
  By conservation of energy, the energy dissipated equals the PE lost minus the KE gained.
  $E_{dissipated} = \Delta E_{grav} - E_k$
4. Step 4
  Subtract.
  $E_{dissipated} = 47.1 - 24.0 = 23.1\ \text{J}$
Answer
$E_{dissipated} = 23.1\ \text{J}$ (transferred to internal/thermal energy)
Examiner tip
Mark scheme: (1) $\Delta E_{grav} = 47.1\ \text{J}$ ; (2) $E_k = 24.0\ \text{J}$ ; (3) energy dissipated = PE lost − KE gained; (4) $23.1\ \text{J}$ . State that this energy is transferred to internal (thermal) energy — saying it is 'lost' or 'destroyed' is penalised.
9Efficiency of a motor (3 marks)
StretchMulti-step problem• efficiency, power, AO2, Unit 1
Question
An electric winch motor draws $2.4\ \text{kW}$ of electrical power and lifts a load, gaining gravitational potential energy at a rate of $1.5\ \text{kW}$ . (a) Calculate the efficiency of the motor. (b) State what happens to the wasted energy. (3 marks)
Step-by-step solution
1. Step 1
  Efficiency is the useful power output divided by the total power input.
  $\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\%$
2. Step 2
  Substitute the useful (1.5 kW) and input (2.4 kW) powers.
  $\text{efficiency} = \frac{1.5}{2.4} \times 100\% = 62.5\%$
3. Step 3
  The wasted power ( $2.4 - 1.5 = 0.9\ \text{kW}$ ) is dissipated as internal (thermal) energy — heating the motor and surroundings (plus a little sound).
Answer
(a) efficiency $= 62.5\%$ . (b) The wasted $0.90\ \text{kW}$ is transferred to internal (thermal) energy in the motor and surroundings.
Examiner tip
Mark scheme: (1) efficiency = useful ÷ input; (2) $62.5\%$ (accept $0.625$ if not asked for a percentage); (3) wasted energy → heat/internal energy. Putting input over useful gives $160\%$ , which is impossible and signals the error.
10'Show that' — power output of a turbine (4 marks)
StretchShow that / prove• Adapted from Edexcel IAL Physics Unit 1• power, show that, AO2, Unit 1
Question
In a small hydroelectric scheme, $640\ \text{kg}$ of water falls through a vertical height of $22\ \text{m}$ every second onto a turbine. Show that the maximum power available to the turbine is about $140\ \text{kW}$ . Take $g = 9.81\ \text{m s}^{-2}$ . (4 marks)
Step-by-step solution
1. Step 1
  Each second, the gravitational PE lost by the falling water is the energy delivered.
  $\Delta E = mg\Delta h = 640 \times 9.81 \times 22$
2. Step 2
  Evaluate the energy transferred per second.
  $\Delta E = 138\,124.8\ \text{J}$
3. Step 3
  Since this energy is delivered every second ( $\Delta t = 1\ \text{s}$ ), the power is $P = \Delta E/\Delta t$ .
  $P = \frac{138\,124.8}{1} = 138\,125\ \text{W}$
4. Step 4
  Convert to kilowatts; to one extra significant figure this is $138\ \text{kW}$ , which is about $140\ \text{kW}$ as required.
  $P = 138\ \text{kW} \approx 140\ \text{kW}$
Answer
$P \approx 138\ \text{kW} \approx 140\ \text{kW}$ — shown.
Examiner tip
Mark scheme: (1) $\Delta E = mg\Delta h$ ; (2) energy per second $\approx 1.38 \times 10^5\ \text{J}$ ; (3) $P = \Delta E/\Delta t$ with $\Delta t = 1\ \text{s}$ ; (4) $138\ \text{kW} \approx 140\ \text{kW}$ . For a 'show that', work through to a more precise value ( $138\ \text{kW}$ ) and then state it rounds to the given $140\ \text{kW}$ — never just quote $140\ \text{kW}$ .

Model Answers — Work, Energy and Power

High-scoring sample answers for work, energy and power on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the work done by a force, and state the unit in which it is measured.
Model answer
Work done is the energy transferred when a force moves its point of application in the direction of the force. It is calculated from work $=$ force $\times$ distance moved in the direction of the force, that is $\Delta W = F\Delta s\cos\theta$ .

The unit of work is the joule (J), where one joule is the work done when a force of one newton moves its point of application one metre in the direction of the force.
Why this scores
Why this scores 2/2. (1) work = force × distance moved in the direction of the force (or energy transferred); (2) unit is the joule (J). Including the defining equation is creditworthy supporting detail.
Question 2
3 marks
Q (3 marks). A satellite moves in a circular orbit at constant speed. Explain why the gravitational force on the satellite does no work on it, even though the force is not zero.
Model answer
The work done by a force is given by $\Delta W = F\Delta s\cos\theta$ , where $\theta$ is the angle between the force and the direction of motion (displacement).

For a satellite in a circular orbit, the gravitational force always acts towards the centre of the circle (it is the centripetal force), while the satellite's velocity is always directed along the tangent to the circle. The force is therefore perpendicular to the direction of motion, so $\theta = 90°$ .

Since $\cos 90° = 0$ , the work done is $\Delta W = F\Delta s \times 0 = 0$ . No work is done, so the kinetic energy — and hence the speed — of the satellite stays constant.
Why this scores
Why this scores 3/3. (1) work depends on the component of force along the motion, $\Delta W = F\Delta s\cos\theta$ ; (2) gravitational force is perpendicular to the velocity ( $\theta = 90°$ ); (3) $\cos 90° = 0$ so no work is done / KE constant. The key is linking 'perpendicular' to $\cos 90° = 0$ .
Question 3
3 marks
Q (3 marks). The speed of a car increases from $10\ \text{m s}^{-1}$ to $30\ \text{m s}^{-1}$ . By what factor does its kinetic energy increase? Explain your answer.
Model answer
Kinetic energy is given by $E_k = \tfrac12 mv^2$ , so for a fixed mass the kinetic energy is proportional to the square of the speed, $E_k \propto v^2$ .

The speed increases by a factor of $\dfrac{30}{10} = 3$ . Because kinetic energy depends on $v^2$ , it increases by a factor of $3^2 = 9$ .

So the kinetic energy becomes nine times larger. (Checking with the numbers: $\tfrac12 m(30)^2 \div \tfrac12 m(10)^2 = 900/100 = 9$ .)
Why this scores
Why this scores 3/3. (1) $E_k = \tfrac12 mv^2$ so $E_k \propto v^2$ ; (2) speed ratio = 3; (3) energy factor = $3^2 = 9$ . A common wrong answer is 'three times', from treating the relationship as linear.
Question 4
4 marks
Q (4 marks). A pendulum bob is released from rest at a height of $0.25\ \text{m}$ above its lowest point. Assuming air resistance is negligible, calculate the speed of the bob at its lowest point and explain why your answer does not depend on the mass of the bob. Take $g = 9.81\ \text{m s}^{-2}$ . (4 marks)
Model answer
As the bob swings down, gravitational potential energy is converted into kinetic energy. By conservation of energy, with no air resistance, all the GPE lost becomes KE:

$mg\Delta h = \tfrac12 mv^2$

The mass $m$ appears on both sides and cancels, giving:

$v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.81 \times 0.25} = \sqrt{4.905} = 2.21\ \text{m s}^{-1}$

The speed is therefore $2.21\ \text{m s}^{-1}$ (3 s.f.). It does not depend on the mass because the mass cancels from the energy equation: a more massive bob has more GPE to convert, but it also requires more KE to reach a given speed, and these effects exactly balance.
Why this scores
Why this scores 4/4. (1) state $mg\Delta h = \tfrac12 mv^2$ / conservation of energy; (2) rearrange to $v = \sqrt{2g\Delta h}$ ; (3) $v = 2.21\ \text{m s}^{-1}$ ; (4) explain the mass cancels. The explanatory mark needs the reasoning (mass cancels because it appears on both sides), not just 'mass doesn't matter'.
Question 5
4 marks
Q (4 marks). A car travels along a level road at a constant speed of $30\ \text{m s}^{-1}$ . The engine provides a useful output power of $24\ \text{kW}$ . (a) Calculate the total resistive force acting on the car. (b) Explain why the car does not accelerate even though the engine is doing work. (4 marks)
Model answer
(a) For a constant force moving at speed $v$ , power is $P = Fv$ , so the driving force is:

$F = \frac{P}{v} = \frac{24\,000}{30} = 800\ \text{N}$

At constant speed the driving force balances the resistive force, so the total resistive force is $800\ \text{N}$ .

(b) The car moves at constant velocity, so by Newton's first law the resultant force on it is zero. The forward driving force of $800\ \text{N}$ is exactly balanced by the $800\ \text{N}$ of resistive forces (friction and air resistance). The engine still does work, but this energy is transferred to overcome the resistive forces — it is dissipated as internal (thermal) energy in the surroundings rather than increasing the car's kinetic energy. With no resultant force, there is no acceleration.
Why this scores
Why this scores 4/4. (a) 2 marks: $F = P/v$ and $F = 800\ \text{N}$ . (b) 2 marks: resultant force is zero (forces balanced) → constant velocity / no acceleration, and the work done goes to overcoming resistance (heat). Linking 'balanced forces' to Newton's first law is the high-value point.
Question 6
5 marks
Q (5 marks). A crane lifts a load of mass $250\ \text{kg}$ vertically through $15\ \text{m}$ in $8.0\ \text{s}$ . The crane's motor has an input power of $6.0\ \text{kW}$ . (a) Calculate the useful output power. (b) Calculate the efficiency of the crane. (c) State the form in which the wasted energy is transferred. Take $g = 9.81\ \text{m s}^{-2}$ . (5 marks)
Model answer
(a) The useful energy transferred is the gain in gravitational PE of the load:

$\Delta E_{grav} = mg\Delta h = 250 \times 9.81 \times 15 = 36\,787.5\ \text{J}$

The useful output power is this energy divided by the time:

$P_{out} = \frac{\Delta E}{\Delta t} = \frac{36\,787.5}{8.0} = 4598\ \text{W} \approx 4.60\ \text{kW}$

(b) Efficiency is the useful output power divided by the total input power:

$\text{efficiency} = \frac{P_{out}}{P_{in}} \times 100\% = \frac{4598}{6000} \times 100\% = 76.6\%$

(c) The wasted energy (about $1.4\ \text{kW}$ ) is transferred to internal (thermal) energy in the motor and surroundings, through friction in the moving parts and electrical resistance, with a small amount transferred as sound.
Why this scores
Why this scores 5/5. (a) 2 marks: $\Delta E_{grav} = mg\Delta h$ and $P_{out} = 4.60\ \text{kW}$ . (b) 2 marks: efficiency = useful ÷ input and $76.6\%$ . (c) 1 mark: wasted energy → heat/internal energy (allow sound). Carrying the unrounded $4598\ \text{W}$ into the efficiency calculation avoids rounding error.
Question 7
Adapted from Edexcel IAL Physics Unit 16 marks
Q (6 marks). A train of mass $4.5 \times 10^5\ \text{kg}$ accelerates from rest to $40\ \text{m s}^{-1}$ along a level track in $90\ \text{s}$ . The average resistive force during this time is $3.0 \times 10^4\ \text{N}$ and the train travels $1800\ \text{m}$ . (a) Calculate the kinetic energy gained by the train. (b) Calculate the work done against the resistive forces. (c) Hence calculate the average useful output power of the train's engines during the acceleration. (6 marks)
Model answer
(a) The train starts from rest, so the kinetic energy gained equals its final kinetic energy:

$E_k = \tfrac12 mv^2 = \tfrac12 \times 4.5 \times 10^5 \times 40^2 = \tfrac12 \times 4.5 \times 10^5 \times 1600 = 3.6 \times 10^8\ \text{J}$

(b) Work done against resistance is the resistive force times the distance travelled:

$W_{resist} = F\Delta s = 3.0 \times 10^4 \times 1800 = 5.4 \times 10^7\ \text{J}$

(c) The total useful energy output of the engines must supply both the kinetic energy gained and the work done against resistance:

$E_{total} = E_k + W_{resist} = 3.6 \times 10^8 + 5.4 \times 10^7 = 4.14 \times 10^8\ \text{J}$

The average useful output power is this energy divided by the time:

$P = \frac{E_{total}}{\Delta t} = \frac{4.14 \times 10^8}{90} = 4.6 \times 10^6\ \text{W} = 4.6\ \text{MW}$
Why this scores
Why this scores 6/6. (a) 2 marks: $E_k = \tfrac12 mv^2 = 3.6 \times 10^8\ \text{J}$ . (b) 2 marks: $W = F\Delta s = 5.4 \times 10^7\ \text{J}$ . (c) 2 marks: add KE gained and work done against resistance, then divide by time → $4.6\ \text{MW}$ . The trap is forgetting to include the work against resistance in the total energy — power must account for all the useful energy supplied.
Question 8
6 marks
Q (6 marks). A rollercoaster car starts from rest at the top of the first hill and travels along the track, which has several lower hills and dips, before coming to rest at the end. Describe, in terms of energy transfers, what happens to the car's energy as it travels from the top of the first hill to the end of the ride. Use the principle of conservation of energy in your answer. (This question is marked using levels.)
Model answer
At the top of the first hill the car is at rest, so it has its maximum gravitational potential energy and zero kinetic energy. The first hill must be the highest point of the ride because energy cannot be created.

As the car descends, gravitational potential energy is converted into kinetic energy, so it speeds up and reaches its greatest speed (maximum kinetic energy) at the lowest point of the track. As it climbs the next hill, the reverse happens: kinetic energy is converted back into gravitational potential energy, so it slows down. This continuous exchange between GPE and KE repeats over each hill and dip.

By the principle of conservation of energy, the total energy is constant — energy is transferred between stores, never created or destroyed. However, the car cannot rise as high on later hills as on the first, because resistive forces (friction in the wheels and bearings, and air resistance) do work against the motion. This work transfers some of the mechanical energy to internal (thermal) energy in the track, wheels and surrounding air, plus a small amount to sound.

As more and more mechanical energy is dissipated as heat, less kinetic and potential energy remains, so each successive hill the car can climb is lower than the last. Eventually all of the original gravitational potential energy has been transferred to internal energy and sound, the car has no kinetic energy left, and it comes to rest at the end of the ride. The total energy of the car and its surroundings is unchanged throughout — it has simply been transferred from a useful mechanical store into thermal energy.
Why this scores
Why this reaches the top level (5–6/6). Level 3 needs sustained, linked reasoning using conservation of energy: max GPE/zero KE at the start; GPE ⇌ KE exchange (fastest at the lowest points); resistive forces do work, dissipating mechanical energy as internal (thermal) energy and sound; this is why later hills are lower and the car finally stops; total energy is conserved throughout. Continuous prose naming each store and the dissipation — not a bare list — is required for the top band. Saying energy is 'lost' or 'used up' rather than 'transferred to thermal energy' caps the level.
Question 9
Adapted from Edexcel IAL Physics Unit 16 marks
Q (6 marks). A box of mass $5.0\ \text{kg}$ is pushed at constant speed up a frictionless ramp that rises a vertical height of $1.2\ \text{m}$ over a slope length of $3.0\ \text{m}$ . (a) Calculate the gain in gravitational potential energy. (b) Show that the force needed to push the box up the smooth ramp at constant speed is about $20\ \text{N}$ . (c) The same box is then pushed up a real (rough) ramp of identical dimensions, and the actual push force needed is $26\ \text{N}$ . Calculate the work done against friction over the $3.0\ \text{m}$ slope. Take $g = 9.81\ \text{m s}^{-2}$ . (6 marks)
Model answer
(a) Using the vertical height of $1.2\ \text{m}$ :

$\Delta E_{grav} = mg\Delta h = 5.0 \times 9.81 \times 1.2 = 58.86\ \text{J} \approx 58.9\ \text{J}$

(b) On a frictionless ramp at constant speed, the work done by the push force along the slope equals the gain in gravitational PE (conservation of energy, no energy lost to friction):

$F \times \Delta s = \Delta E_{grav} \quad\Rightarrow\quad F = \frac{\Delta E_{grav}}{\Delta s} = \frac{58.86}{3.0} = 19.6\ \text{N} \approx 20\ \text{N}$

So the force needed is about $20\ \text{N}$ , as required.

(c) On the rough ramp the push force is $26\ \text{N}$ , so the total work done by the push over the $3.0\ \text{m}$ slope is:

$W_{push} = F\Delta s = 26 \times 3.0 = 78\ \text{J}$

Of this, $58.86\ \text{J}$ raises the box (the gain in GPE is the same, since the height is unchanged). The remainder is the work done against friction:

$W_{friction} = W_{push} - \Delta E_{grav} = 78 - 58.86 = 19.1\ \text{J}$
Why this scores
Why this scores 6/6. (a) 2 marks: use the vertical height → $58.9\ \text{J}$ . (b) 2 marks: $F = \Delta E_{grav}/\Delta s$ worked to $19.6\ \text{N}$ then shown to round to $20\ \text{N}$ (a 'show that' needs the extra s.f.). (c) 2 marks: total push work $78\ \text{J}$ minus GPE gain $58.86\ \text{J}$ = $19.1\ \text{J}$ against friction. The common error in (a) is using the $3.0\ \text{m}$ slope length instead of the $1.2\ \text{m}$ vertical height.

Key Formulae — Work, Energy and Power

The formulae you need to memorise for work, energy and power on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Work done by a force
$\Delta W = F\Delta s\cos\theta$
$\Delta W$
work done / energy transferred (J)
$F$
force (N)
$\Delta s$
displacement (m)
$\theta$
angle between the force and the displacement
When to use
Whenever a force moves its point of application. Include $\cos\theta$ when the force acts at an angle to the motion; $\theta = 0°$ gives the simple $\Delta W = F\Delta s$ .
Example
A $90\ \text{N}$ tension in a rope at $25°$ over $12\ \text{m}$ : $\Delta W = 90 \times 12 \times \cos 25° = 979\ \text{J}$ .
Kinetic energy
$E_k = \tfrac12 mv^2$
$E_k$
kinetic energy (J)
$m$
mass (kg)
$v$
speed (m s⁻¹)
When to use
To find the energy of a moving object. Remember it depends on $v^2$ , so doubling the speed quadruples the kinetic energy. In the Edexcel formulae booklet (Appendix 7).
Example
$80\ \text{kg}$ at $7.0\ \text{m s}^{-1}$ : $E_k = \tfrac12 \times 80 \times 7.0^2 = 1960\ \text{J}$ .
Change in gravitational potential energy
$\Delta E_{grav} = mg\Delta h$
$\Delta E_{grav}$
change in gravitational PE (J)
$m$
mass (kg)
$g$
gravitational field strength, $9.81\ \text{m s}^{-2}$
$\Delta h$
vertical height change (m)
When to use
For changes in height near the Earth's surface. Use the vertical height; on a slope of length $L$ at angle $\alpha$ , $\Delta h = L\sin\alpha$ . In the formulae booklet (Appendix 7).
Example
$1200\ \text{kg}$ raised $18\ \text{m}$ : $\Delta E_{grav} = 1200 \times 9.81 \times 18 = 2.12 \times 10^5\ \text{J}$ .
Power (rate of energy transfer)
$P = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}$
$P$
power (W)
$\Delta W,\ \Delta E$
work done / energy transferred (J)
$\Delta t$
time taken (s)
When to use
When energy is transferred over a measured time. $1\ \text{W} = 1\ \text{J s}^{-1}$ . In the formulae booklet (Appendix 7).
Example
$52\,974\ \text{J}$ in $30\ \text{s}$ : $P = 52\,974 / 30 = 1.77\ \text{kW}$ .
Power of a constant force
$P = Fv$
$P$
power (W)
$F$
force (N) — the driving force
$v$
speed (m s⁻¹)
When to use
For a constant force moving at speed $v$ — especially vehicles at constant speed, where the driving force equals the total resistive force (plus $mg\sin\alpha$ up a slope). In the formulae booklet (Appendix 7).
Example
Driving force $1628\ \text{N}$ at $20\ \text{m s}^{-1}$ : $P = 1628 \times 20 = 32.6\ \text{kW}$ .
Efficiency
$\text{efficiency} = \frac{\text{useful energy (or power) output}}{\text{total energy (or power) input}} \times 100\%$
$useful output$
energy or power transferred usefully (J or W)
$total input$
total energy or power supplied (J or W)
When to use
To quantify how much input energy is transferred usefully. It is a ratio (no units); always $< 100\%$ for a real device. In the formulae booklet (Appendix 7).
Example
Useful $1.5\ \text{kW}$ from $2.4\ \text{kW}$ input: efficiency $= 1.5/2.4 \times 100\% = 62.5\%$ .

Key Definitions and Keywords — Work, Energy and Power

Definitions to memorise and the exact keywords mark schemes credit for work, energy and power answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Work done
Examiner keyword
The energy transferred when a force moves its point of application in the direction of the force: $\Delta W = F\Delta s\cos\theta$ , where $\theta$ is the angle between the force and the displacement. Measured in joules (J).
Joule (J)
Examiner keyword
The SI unit of energy and work. One joule is the work done when a force of one newton moves its point of application one metre in the direction of the force ( $1\ \text{J} = 1\ \text{N m}$ ).
Energy
Examiner keyword
The capacity to do work, measured in joules (J). It can be transferred between stores (e.g. kinetic, gravitational potential, thermal) but cannot be created or destroyed.
Kinetic energy
Examiner keyword
The energy an object has because of its motion, $E_k = \tfrac12 mv^2$ . It depends on the square of the speed, so doubling the speed quadruples the kinetic energy.
Gravitational potential energy (change)
Examiner keyword
The energy stored when an object is raised in a gravitational field. Near the Earth's surface the change is $\Delta E_{grav} = mg\Delta h$ , using the vertical height change $\Delta h$ .
Principle of conservation of energy
Examiner keyword
Energy cannot be created or destroyed, only transferred from one store to another or dissipated. The total energy of a closed system is constant.
Mechanical energy
Examiner keyword
The sum of an object's kinetic and gravitational potential energy. It is conserved only when no resistive (friction or drag) forces act.
Dissipation of energy
Examiner keyword
The transfer of useful energy to less useful forms — usually internal (thermal) energy — when work is done against friction or air resistance. The energy is spread to the surroundings, not destroyed.
Internal (thermal) energy
Energy associated with the random motion and arrangement of particles. Work done against friction and drag is transferred to internal energy, raising the temperature of the object and surroundings.
Power
Examiner keyword
The rate of doing work or transferring energy, $P = \dfrac{\Delta W}{\Delta t} = \dfrac{\Delta E}{\Delta t}$ . Measured in watts (W), where $1\ \text{W} = 1\ \text{J s}^{-1}$ .
Watt (W)
Examiner keyword
The SI unit of power. One watt is a rate of energy transfer of one joule per second ( $1\ \text{W} = 1\ \text{J s}^{-1}$ ).
P = Fv
Examiner keyword
The power developed by a constant force $F$ moving at speed $v$ . For a vehicle at constant speed, $F$ is the driving force, which balances the total resistive force (and any weight component up a slope).
Driving force
The forward force provided by an engine or motor. At constant velocity it equals the total resistive force (plus $mg\sin\alpha$ when moving up a slope), giving zero resultant force.
Efficiency
Examiner keyword
The fraction of input energy (or power) transferred usefully: efficiency $= \dfrac{\text{useful output}}{\text{total input}} \times 100\%$ . It is a ratio with no units and is always less than 100% for a real device.
Sankey diagram
An energy-flow diagram in which the width of each arrow is proportional to the energy (or power). The input arrow splits into useful and wasted branches, and the total width is conserved at every junction.

Common Mistakes and Misconceptions — Work, Energy and Power

The traps other students keep falling into on work, energy and power questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Multiplying the full force by the distance when the force acts at an angle to the motion
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students learn $W = F \times d$ first and forget to resolve when the force is later applied at an angle.
How to avoid it
Always use $\Delta W = F\Delta s\cos\theta$ and take the component of force along the displacement ( $F\cos\theta$ ). If the force is perpendicular to the motion, $\cos 90° = 0$ and no work is done.
✕Forgetting to square the speed in $E_k = \tfrac12 mv^2$
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Rushing the substitution and treating the equation as if kinetic energy were linear in $v$ .
How to avoid it
Write the equation with $v^2$ explicitly and square the speed before multiplying. Remember kinetic energy scales with $v^2$ — doubling the speed quadruples the energy, not doubles it.
✕Using the distance along a slope as $\Delta h$ in $\Delta E_{grav} = mg\Delta h$
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
The slope length is the distance travelled, so students substitute it without realising $\Delta h$ must be vertical.
How to avoid it
Use the vertical height change only. For a slope of length $L$ at angle $\alpha$ to the horizontal, $\Delta h = L\sin\alpha$ . Gravitational PE depends on height gained, not path length.
✕Saying mechanical energy is 'lost' or 'destroyed' when friction or drag acts
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Everyday language treats wasted energy as disappearing, contradicting conservation of energy.
How to avoid it
Energy is transferred to internal (thermal) energy of the object and surroundings (and a little to sound). State that mechanical energy is dissipated, not destroyed — total energy is always conserved.
✕Using the weight (or the wrong force) instead of the driving force in $P = Fv$ for a vehicle
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students see a vehicle and a speed but do not identify which force drives the motion.
How to avoid it
In $P = Fv$ , $F$ is the driving force. At constant speed it equals the total resistive force (plus $mg\sin\alpha$ when climbing a slope), because the resultant force is zero — not the weight.
✕Inverting the efficiency fraction or quoting an efficiency above 100%
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Confusing which quantity is the useful output and which is the total input.
How to avoid it
Always put useful output on top and total input on the bottom: efficiency $= \dfrac{\text{useful}}{\text{input}} \times 100\%$ . A real machine can never exceed 100% — a value above 100% means the fraction is upside-down.

Work, Energy and Power — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Work, Energy and Power

Short Notes - Work, Energy and Power

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Work done by a horizontal force (2 marks)

2Kinetic energy of a moving object (2 marks)

3Gain in gravitational potential energy (3 marks)

4Work done by a force at an angle (3 marks)

5Speed of a falling object from energy (3 marks)

6Power from energy and time (3 marks)

7Engine power for a car up an incline (4 marks)

8Energy dissipated by friction on a slope (4 marks)

9Efficiency of a motor (3 marks)

10'Show that' — power output of a turbine (4 marks)

Work done by a force

Kinetic energy

Change in gravitational potential energy

Power (rate of energy transfer)

Power of a constant force

Efficiency

Work done

Joule (J)

Energy

Kinetic energy

Gravitational potential energy (change)

Principle of conservation of energy

Mechanical energy

Dissipation of energy

Internal (thermal) energy

Power

Watt (W)

P = Fv

Driving force

Efficiency

Sankey diagram

✕Multiplying the full force by the distance when the force acts at an angle to the motion

✕Forgetting to square the speed in Ek=12mv2E_k = \tfrac12 mv^2Ek​=21​mv2

✕Using the distance along a slope as Δh\Delta hΔh in ΔEgrav=mgΔh\Delta E_{grav} = mg\Delta hΔEgrav​=mgΔh

✕Saying mechanical energy is 'lost' or 'destroyed' when friction or drag acts

✕Using the weight (or the wrong force) instead of the driving force in P=FvP = FvP=Fv for a vehicle

✕Inverting the efficiency fraction or quoting an efficiency above 100%

Work, Energy and Power — frequently asked questions

✕Forgetting to square the speed in $E_k = \tfrac12 mv^2$

✕Using the distance along a slope as $\Delta h$ in $\Delta E_{grav} = mg\Delta h$

✕Using the weight (or the wrong force) instead of the driving force in $P = Fv$ for a vehicle