# Energy, Work and Power

> CIE IGCSE Physics · CIE 0625 2026-2028
> Source: https://www.owlsprep.com/study/cie-0625-u1-energy-work-and-power/

This guide covers all Core and Extended content for CIE IGCSE Physics 0625 Unit 1.7 Energy, Work and Power, including formula applications, energy transfer rules, and exam-standard worked problems aligned to the 2026-2028 syllabus.

**Prerequisites:** [Force and mass calculations (CIE IGCSE 0625 Unit 1.4)](https://www.owlsprep.com/study/cie-0625-u1-forces-mass-weight/); [Units and prefixes (CIE IGCSE 0625 Unit 1.1)](https://www.owlsprep.com/study/cie-0625-u1-units-measurement/)

## Learning objectives

- Define energy, work done and power with correct SI units
- Calculate work done as force × distance moved in direction of force
- Calculate power as work done / time or energy transferred / time
- Apply conservation of energy to Core exam problems
- Calculate efficiency for Extended exam scenarios
- Recall and use $E_k = \tfrac{1}{2}mv^2$ and $\Delta E_p = mg\Delta h$ for Extended exam scenarios

## Core: Definitions of Energy and Work Done

**Energy** — The capacity to do work, measured in joules (J). Energy cannot be created or destroyed, only transferred between stores or converted between forms. Total energy in a closed system is always conserved.

*Example:* A raised ball has gravitational potential energy, which transfers to kinetic energy as it falls, with no energy lost if air resistance is negligible.

Work done is equal to the total energy transferred between stores when a force acts on an object. It is only calculated for distance moved in the exact direction of the applied force.

**Work Done** — Work done equals energy transferred when a force moves an object through a distance in the direction of the force.

*Notation:* W

*Example:* Pushing a box across a floor requires work done against friction, transferring chemical energy from your body to thermal energy in the floor and box.

$$W = F \times d$$

**Worked example:** A student pushes a box across a flat floor with a constant force of 40 N. The box moves 3.5 m in the direction of the force. Calculate the work done by the student.

1. Use the work done formula:
2. $$W = F \times d$$
3. Substitute given SI values: F = 40 N, d = 3.5 m
4. $$W = 40 \times 3.5$$
5. Calculate result and add correct units:
6. $$W = 140 \text{ J}$$

> **Exam tip:** Always confirm distance is in the same direction as the applied force; work done by gravity on a horizontally moving object is 0.

## Core: Definition and Calculation of Power

**Power** — The rate of energy transfer, or the rate of doing work. Measured in watts (W), where 1 W equals 1 joule of energy transferred per second.

*Notation:* P

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

Higher power values mean work is completed faster. For example, a 2 kW kettle boils water faster than a 1 kW kettle because it transfers more thermal energy per second to the water.

**Worked example:** A lift transfers 12000 J of gravitational potential energy to raise a group of people in 8 seconds. Calculate the power of the lift motor.

1. Use the power formula with energy transferred E = 12000 J, time t = 8 s:
2. $$P = \frac{E}{t}$$
3. Substitute values:
4. $$P = \frac{12000}{8}$$
5. Calculate result and add units:
6. $$P = 1500 \text{ W} (or 1.5 \text{ kW})$$

> **Exam tip:** Remember 1 kilowatt (kW) = 1000 W, this unit conversion is tested frequently in both Core and Extended papers.

## Core: Conservation of Energy Principle

The conservation of energy principle states that energy cannot be created or destroyed, only converted between forms or transferred between objects. The total energy in a closed, isolated system never changes.

**Worked example:** A ball is dropped from a height of 2 m, with 30 J of gravitational potential energy at the point it is released. Assuming no air resistance, calculate the kinetic energy of the ball just before it hits the ground.

1. Apply conservation of energy: all initial gravitational potential energy (GPE) transfers to kinetic energy (KE) when no energy is lost to air resistance.
2. Total initial energy = Total final energy, so initial GPE = final KE
3. Final kinetic energy = 30 J

> **Exam tip:** If energy losses (friction, air resistance, sound) are stated, subtract the wasted energy from the initial total to find the final useful energy.

## Extended Only: Efficiency Calculations

> **Extended Only Content**
>
> This section is only required for candidates taking Extended papers (Paper 2/4). Core candidates (Paper 1/3) may skip this section.

**Efficiency** — The fraction of total input energy (or power) that is converted to useful output energy (or power), often expressed as a percentage.

*Notation:* \eta

*Example:* A 60% efficient lightbulb converts 60 J of every 100 J of electrical input energy to useful light energy, with 40 J wasted as heat.

$$\text{Efficiency (\%)} = \frac{\text{Useful Output}}{\text{Total Input}} \times 100$$

**Worked example:** An electric motor has a total power input of 2000 W. It produces 1200 W of useful mechanical power to lift a load. Calculate the percentage efficiency of the motor.

1. Use the extended efficiency formula for power:
2. $$\text{Efficiency (\%)} = \frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100$$
3. Substitute values:
4. $$\text{Efficiency (\%)} = \frac{1200}{2000} \times 100$$
5. Calculate result and add % symbol:
6. $$\text{Efficiency} = 60\%$$

> **Exam tip:** Efficiency can be calculated using either energy values or power values, as the time term cancels out for both input and output.

## Extended Only: Kinetic and Gravitational Potential Energy

> **Extended Only Content**
>
> These two equations are Supplement (Extended) content for Paper 2/4. Core (Paper 1/3) candidates are not required to calculate with them.

**Kinetic energy** — The energy an object has because of its motion, measured in joules (J). $m$ = mass (kg), $v$ = speed (m/s).

*Notation:* E_k = \frac{1}{2}mv^2

*Example:* A 2 kg ball moving at 3 m/s has $E_k = \tfrac{1}{2} \times 2 \times 3^2 = 9$ J.

**Change in gravitational potential energy** — The energy transferred to or from an object's gravitational store when its height changes, measured in joules (J). $m$ = mass (kg), $g$ = gravitational field strength (N/kg), $\Delta h$ = change in height (m).

*Notation:* \Delta E_p = mg\Delta h

*Example:* Lifting a 2 kg book 1.5 m raises its GPE by $\Delta E_p = 2 \times 9.8 \times 1.5 = 29.4$ J.

**Worked example:** A 0.5 kg ball is dropped from rest and falls through a height of 1.8 m. Take g = 9.8 N/kg and ignore air resistance. Calculate (a) the loss in gravitational potential energy, and (b) the speed of the ball just before it lands.

1. Step (a): Calculate the loss in GPE using $\Delta E_p = mg\Delta h$:

   $$\Delta E_p = 0.5 \times 9.8 \times 1.8 = 8.82 \text{ J}$$
2. Step (b): By conservation of energy, all the lost GPE becomes kinetic energy, so $E_k = 8.82$ J. Rearrange $E_k = \tfrac{1}{2}mv^2$ to make $v$ the subject:

   $$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 8.82}{0.5}} = \sqrt{35.28} = 5.9 \text{ m/s}$$

> **Exam tip:** Both $E_k = \tfrac{1}{2}mv^2$ and $\Delta E_p = mg\Delta h$ give energy in joules only when mass is in kg, speed in m/s and height in m, so convert units first.

## Common pitfalls

- **Wrong:** Calculating work done using distance perpendicular to the applied force
  - Why it fails: Work is only done when distance is aligned with the force; e.g. carrying a box horizontally does no work against gravity.
  - Correct: Use only the component of distance that matches the force direction, or 0 if force and distance are perpendicular.
- **Wrong:** Forgetting to convert units (cm to m, kW to W) before calculations
  - Why it fails: All formulae require SI units to give correct results in joules or watts.
  - Correct: Convert all given values to SI units (N, m, s, J, W) before substituting into formulae.
- **Wrong:** Assuming all initial energy transfers to useful energy without accounting for losses
  - Why it fails: Most real systems lose energy to heat, sound, or friction, so total energy is conserved but useful energy is reduced.
  - Correct: Subtract wasted energy from total input energy to find useful output energy if losses are stated.
- **Wrong:** Writing efficiency with units or as a decimal when percentage is requested
  - Why it fails: Efficiency is a unitless ratio; percentage efficiency requires multiplication by 100 and a % symbol.
  - Correct: Multiply the ratio by 100 and add % if the question asks for percentage efficiency.
- **Wrong:** Confusing total work done with power
  - Why it fails: Power is the rate of doing work, not the total amount of work completed over time.
  - Correct: Use W = F×d for total work, P = W/t for the speed at which work is done.

## Cheatsheet

| Quantity | Symbol | Formula | Units | Tier |
| --- | --- | --- | --- | --- |
| Work Done | W | W = F × d | Joules (J) | Core |
| Power | P | P = W/t = E/t | Watts (W) | Core |
| Energy Transfer | E | Total Input = Useful Output + Wasted Energy | Joules (J) | Core |
| Kinetic energy | E_k | E_k = ½mv² | Joules (J) | Extended |
| Change in GPE | ΔE_p | ΔE_p = mgΔh | Joules (J) | Extended |
| Efficiency | η | (Useful Output / Total Input) × 100% | % / unitless | Extended |

## What's next

Now you have mastered energy, work and power for CIE IGCSE Physics 0625, you can move on to related topics in the Motion, Forces and Energy unit. Core candidates should practice basic calculation questions to build speed and accuracy with unit conversions, while Extended candidates can tackle multi-step problems linking work, power and efficiency. Be sure to review the conservation of energy principle regularly, as it underpins all energy-related topics in the syllabus.

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