# Work and Power

> Edexcel International GCSE Physics · 4PH1 2017
> Source: https://www.owlsprep.com/study/edexcel-igcse-physics-s4-work-and-power/

This guide covers core work, energy and power calculations for Edexcel IGCSE Physics (4PH1) Papers 1 and 2, including energy conservation rules and common exam pitfalls to avoid.

**Prerequisites:** [SI units and prefixes for Edexcel IGCSE Physics](https://www.owlsprep.com/study/edexcel-igcse-physics-s1-units-measurements/); [Energy stores and transfers](https://www.owlsprep.com/study/edexcel-igcse-physics-s4-energy-stores-transfers/)

## Learning objectives

- Recall and apply the work done formula $W=Fd$, and relate work done to energy transferred
- Calculate gravitational potential energy using $GPE=mgh$ with $g=10\text{ N/kg}$
- Calculate kinetic energy using $KE=\frac{1}{2}mv^2$
- Apply conservation of energy to link GPE, KE and work done for moving/falling objects
- Define power as rate of energy transfer, and apply $P=W/t$ for power calculations

## Work Done and Energy Transferred

**Work done** — Energy transferred when a force moves an object a distance in the direction of the applied force

*Notation:* $W = F \times d$

*Example:* Lifting a 10N box 2m upwards does 20J of work, transferring 20J to the box's GPE store

1 joule (J) equals 1 newton-metre (N·m). Only distance moved *in the same direction as the force* counts towards work done: if you carry an object horizontally, the upward lifting force does zero work as there is no vertical movement.

**Worked example:** A student pushes a trolley with a constant horizontal force of 35N across a flat floor for a distance of 12m. Calculate the work done by the student on the trolley.

1. Recall the work done formula: $W = F \times d$
2. Substitute given values: $F = 35 \text{ N}$, $d = 12 \text{ m}$
3. $$W = 35 \times 12 = 420$$
4. State answer with correct units: $420 \text{ J}$

> **Exam tip:** Always include units in your final answer, and show every step of your calculation to access method marks even if your final answer is wrong.

*Calculator:* allowed

## Gravitational Potential Energy (GPE)

**Gravitational Potential Energy** — Energy stored in an object due to its position above ground level

*Notation:* $GPE = m \times g \times h$

*Example:* A 2kg mass lifted 3m gains $2 \times 10 \times 3 = 60 \text{ J}$ of GPE

GPE increases when an object is lifted higher, and decreases when it falls. Use $g = 10 \text{ N/kg}$ for all calculations unless the question specifies a different value.

**Worked example:** A crane lifts a 450kg pallet of bricks to a height of 8m above the ground. Calculate the gain in GPE of the pallet.

1. Recall GPE formula: $GPE = mgh$
2. Substitute values: $m=450 \text{ kg}$, $g=10 \text{ N/kg}$, $h=8 \text{ m}$
3. $$GPE = 450 \times 10 \times 8 = 36000$$
4. Final answer: $36000 \text{ J}$ or $36 \text{ kJ}$

*Calculator:* allowed

## Kinetic Energy (KE)

**Kinetic Energy** — Energy stored in any moving object

*Notation:* $KE = \frac{1}{2} \times m \times v^2$

*Example:* A 4kg dog running at 3m/s has $\frac{1}{2} \times 4 \times 3^2 = 18 \text{ J}$ of KE

> **warning**
>
> Always square the speed *before* multiplying by mass and $\frac{1}{2}$. Forgetting to square speed or omitting the $\frac{1}{2}$ are the two most common errors on this topic.

**Worked example:** Calculate the kinetic energy of a 1200kg car travelling at a constant speed of 15m/s.

1. Recall KE formula: $KE = \frac{1}{2}mv^2$
2. First calculate the square of the speed: $v^2 = 15^2 = 225 \text{ m}^2/\text{s}^2$
3. Substitute values: $m=1200 \text{ kg}$, $v^2=225$
4. $$KE = 0.5 \times 1200 \times 225 = 135000$$
5. Final answer: $135000 \text{ J}$ or $135 \text{ kJ}$

*Calculator:* allowed

## Conservation of Energy Linking Work, GPE and KE

Energy cannot be created or destroyed, only transferred between stores. For falling objects or objects moving down ramps with no friction, all GPE lost is transferred to KE gained: $mgh = \frac{1}{2}mv^2$. If friction or air resistance is present, some energy is transferred to thermal stores of the surroundings, so work done against friction equals the difference between GPE lost and KE gained.

**Worked example:** A 0.5kg ball is dropped from a height of 1.8m. Assume no air resistance. Calculate the speed of the ball just before it hits the ground.

1. Apply conservation of energy: GPE lost = KE gained, so $mgh = \frac{1}{2}mv^2$
2. Mass $m$ appears on both sides and cancels out: $gh = \frac{1}{2}v^2$
3. Rearrange to solve for $v$: $v = \sqrt{2gh}$
4. Substitute values: $g=10 \text{ N/kg}$, $h=1.8 \text{ m}$
5. $$v = \sqrt{2 \times 10 \times 1.8} = \sqrt{36} = 6$$
6. Final answer: $6 \text{ m/s}$

> **Exam tip:** If a question mentions friction or air resistance, you cannot directly equate GPE lost to KE gained. You must state the assumption of no energy loss to access marks when cancelling mass.

*Calculator:* allowed

## Power Calculations

**Power** — Rate of doing work or rate of energy transfer, measured in watts (W) where 1W = 1J/s

*Notation:* $P = \frac{W}{t}$

*Example:* A machine that does 100J of work in 2s has a power output of 50W

Two devices can do the same amount of work, but the device that completes the work faster has a higher power output. This formula must be recalled, as it is not provided on the exam formula sheet.

**Worked example:** A runner does 42000J of work in 60 seconds while sprinting. Calculate the runner's average power output.

1. Recall power formula: $P = W/t$
2. Substitute values: $W=42000 \text{ J}$, $t=60 \text{ s}$
3. $$P = 42000 \div 60 = 700$$
4. Final answer: $700 \text{ W}$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using distance perpendicular to the force in $W=Fd$ calculations
  - Why it fails: Work is only done when the force causes movement in its own direction
  - Correct: Only use the component of distance parallel to the applied force in work done calculations
- **Wrong:** Forgetting to square speed or omitting the $\frac{1}{2}$ in $KE=\frac{1}{2}mv^2$
  - Why it fails: Kinetic energy scales with the square of speed, and the $\frac{1}{2}$ is a required constant in the relationship
  - Correct: Always calculate $v^2$ first, then multiply by $\frac{1}{2}$ and mass
- **Wrong:** Using $g=9.81 \text{ N/kg}$ instead of $10 \text{ N/kg}$
  - Why it fails: Edexcel IGCSE Physics explicitly specifies using $g=10 \text{ N/kg}$ unless stated otherwise, so using 9.81 will lead to incorrect answers
  - Correct: Use $g=10 \text{ N/kg}$ for all calculations unless the question gives a different value
- **Wrong:** Relying on the formula sheet for the $P=W/t$ power formula
  - Why it fails: Edexcel IGCSE Physics (4PH1) provides no formula sheet at all — every equation, including $P=VI$, must be memorised
  - Correct: Memorise $P=W/t$ as part of your core formula revision
- **Wrong:** Equating GPE lost to KE gained when friction/air resistance is present
  - Why it fails: Friction does work against the moving object, transferring energy to thermal stores, so not all GPE becomes KE
  - Correct: Subtract work done against friction from GPE lost to find KE gained, or state the assumption of no friction if you equate GPE and KE

## Cheatsheet

| Formula | Variables | Units | Must Recall? |
| --- | --- | --- | --- |
| $W=F \times d$ | W = work done, F = force, d = distance in direction of force | W: J, F: N, d: m | Yes |
| $GPE=mgh$ | GPE = gravitational potential energy, m = mass, g = g-field strength, h = height | GPE: J, m: kg, g: N/kg, h: m | Yes |
| $KE=\frac{1}{2}mv^2$ | KE = kinetic energy, v = speed | KE: J, v: m/s | Yes |
| $P=W/t$ | P = power, t = time taken | P: W, t: s | Yes |
| Conservation of energy (no friction) | $mgh = \frac{1}{2}mv^2$ | All units as above | Yes |

## What's next

Now you have mastered work, energy and power calculations, you can move on to related energy topics in Edexcel IGCSE Physics. Next, learn about energy resources, efficiency and Sankey diagrams to understand how energy is converted for human use, and how to calculate the efficiency of energy transfer processes. You can also practice applying these work and power rules to exam-style questions on motion, such as calculating the power of accelerating vehicles or the work done by braking forces. Make sure to memorise all required formulas, as they are not provided on the exam formula sheet, and practice calculation questions to avoid common errors like forgetting to square speed in KE calculations.

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