# Conservation of Energy

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u5-conservation-of-energy/

This module covers the principle of conservation of energy, its application to mechanical systems, and how to account for energy transfers and losses in real-world scenarios. You will learn to solve problems for falling objects, pendulums, and powered vehicles.

**Prerequisites:** [Work done by forces](https://www.owlsprep.com/study/cie-9702-u5-work-done/); [Kinetic and gravitational potential energy](https://www.owlsprep.com/study/cie-9702-u5-kinetic-potential-energy/)

## Learning objectives

- State the principle of conservation of energy
- Apply conservation of energy to solve mechanical problems
- Account for energy losses in real systems
- Calculate efficiency of energy transfer processes

## The Principle of Conservation of Energy

**Principle of Conservation of Energy** — Energy cannot be created or destroyed. It can only be converted from one form to another, or transferred between bodies. The total energy in a closed system remains constant.

*Example:* A ball falling through air converts gravitational potential energy to kinetic energy, with a small amount lost as heat to air resistance; total energy is unchanged.

A closed system is defined as one where no energy enters or leaves the system. In open systems, energy can transfer between the system and its surroundings. For A-level problems, you will almost always work with closed systems, even when energy is lost to the surroundings from the working system — you just account for the lost energy in your total calculation.

**Worked example:** A 2.0 kg ball is dropped from rest at a height of 5.0 m above the ground. Assuming no air resistance, calculate the speed of the ball just before impact.

1. Apply conservation of energy: Loss of gravitational potential energy = Gain in kinetic energy
2. $$mgh = \frac{1}{2}mv^2$$
3. Mass $m$ is a common factor on both sides, so it cancels out:
4. $$gh = \frac{1}{2}v^2$$
5. Substitute $g = 9.81 \text{ ms}^{-2}$ and $h = 5.0$:
6. $$v = \sqrt{2 \times 9.81 \times 5.0} = \sqrt{98.1} \approx 9.9 \text{ ms}^{-1}$$

> **Exam tip:** Mass almost always cancels out in problems without friction or air resistance, so you do not need the mass of the object to calculate final speed.

## Applications to Mechanical Systems

Conservation of energy can be applied to almost any mechanical problem, including pendulums, roller coasters, and objects moving on slopes. When non-conservative forces like friction or air resistance act, some energy is transferred to the surroundings as heat (and sound), so total mechanical energy (kinetic + potential) does not stay constant.

**Worked example:** A 0.5 kg pendulum is pulled sideways so its centre of mass is 10 cm higher than at its lowest point. It is released from rest, and has a speed of 1.2 m/s when it passes through the lowest point. Calculate the energy lost to heat due to air resistance.

1. Calculate initial gravitational potential energy relative to the lowest point:
2. $$E_p = mgh = 0.5 \times 9.81 \times 0.10 = 0.4905 \text{ J}$$
3. Calculate final kinetic energy at the lowest point:
4. $$E_k = \frac{1}{2}mv^2 = 0.5 \times 0.5 \times (1.2)^2 = 0.36 \text{ J}$$
5. By conservation of energy, energy lost equals the difference between initial and final mechanical energy:
6. $$\text{Energy lost} = 0.4905 - 0.36 \approx 0.13 \text{ J}$$

> **tip**
>
> Set your reference level for gravitational potential energy to the lowest point in the problem to avoid negative potential energy values and simplify calculations.

## Efficiency of Energy Transfer

The principle of conservation of energy applies to all energy transfers, not just mechanical ones. Common examples include chemical energy converted to kinetic energy in car engines, and electrical energy converted to light in bulbs. For all real processes, some energy is lost as wasted heat, so we calculate efficiency to quantify useful energy output.

**Efficiency** — The ratio of useful energy output to total energy input, expressed as a decimal or percentage. It is always less than 1 (100%) for real systems.

*Notation:* \eta

*Example:* A car engine typically has an efficiency of ~20-30%, meaning 70-80% of fuel energy is lost as heat.

**Worked example:** A car engine has an efficiency of 25%. The car travels 1 km at constant speed against an average resistive force of 500 N. Calculate the total chemical energy used by the engine.

1. Calculate useful work done (useful energy output) to overcome resistance:
2. $$W = Fd = 500 \times 1000 = 500\,000 \text{ J}$$
3. Use the efficiency formula, rearrange to find total energy input:
4. $$\eta = \frac{\text{Useful output}}{\text{Total input}} \implies 0.25 = \frac{500\,000}{E_{\text{total}}}$$
5. $$E_{\text{total}} = \frac{500\,000}{0.25} = 2.0 \times 10^6 \text{ J}$$

**Check your understanding**

Check your understanding:

1. The total energy of an open system is always constant. True or false?

   - True
   - False

   *Answer:* False

   *Why:* False. Energy can enter or leave an open system, so the system's total energy can change. Conservation of energy applies to the entire system plus surroundings, not the open system alone.

## Common pitfalls

- **Wrong:** Forgetting to account for energy losses when friction/air resistance is mentioned
  - Why it fails: Assuming all initial potential energy converts to kinetic energy, leading to overestimated final speed
  - Correct: Always check for non-conservative forces in the question; add energy lost as heat to the right-hand side of your conservation equation
- **Wrong:** Cancelling mass when it is not a common factor on all terms
  - Why it fails: When energy losses do not depend on mass or one side of the equation does not include mass, cancelling mass leads to incorrect results
  - Correct: Only cancel mass if it appears in every term of the energy conservation equation
- **Wrong:** Using absolute values of potential energy instead of changes
  - Why it fails: Adds unnecessary constant terms that introduce calculation errors
  - Correct: Always calculate the change in potential energy between two points; constant terms always cancel out
- **Wrong:** Flipping the efficiency ratio to input over output
  - Why it fails: Gives efficiency values greater than 100% which is impossible for real systems
  - Correct: Remember: $\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}}$

## Cheatsheet

| Concept | Key Relationship |
| --- | --- |
| Conservation of energy (closed system) | Total initial energy = Total final energy |
| With friction/air resistance | Initial energy = Final mechanical energy + Energy lost as heat |
| Efficiency | $\eta = \frac{\text{Useful output}}{\text{Total input}} \times 100\%$ |
| Falling object (no resistance) | $v = \sqrt{2gh}$ |

## What's next

Conservation of energy is one of the most fundamental laws in physics, underpinning all topics from classical mechanics to quantum physics and thermodynamics. Mastering this sub-topic gives you a core problem-solving tool that you will use for every subsequent topic in A-level Physics. Next, you will learn how to relate energy transfer to time via power, and calculate efficiency for more complex systems. This principle also forms the basis for understanding conservation of momentum in collisions and energy transfers in thermodynamics.

- [Power](https://www.owlsprep.com/study/cie-9702-u5-power/)
- [Energy efficiency](https://www.owlsprep.com/study/cie-9702-u5-energy-efficiency/)
- [Deformation of solids](https://www.owlsprep.com/study/cie-9702-u6-overview/)

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