Energy

CORE VS EXTENDED GUIDE

• Core students study only the unlabelled sections.
• Extended students must study everything, including Extended Extended points.
• Extended = Core + Supplement.

The Seven Energy Stores

Every physical system stores energy in one or more recognised forms. Energy is not a substance that flows — it is a quantity that describes the capacity of a system to cause changes. The seven examinable energy stores are:

The internal (thermal) store increases whenever the temperature of an object rises, because particles move faster and therefore possess greater kinetic energy.

MisconceptionStudents sometimes say "heat energy" as though heat is a store. Heat is not a store — it is a transfer pathway (by heating). The store is the internal (thermal) store. Exam cue: always write "internal (thermal) store," never "heat energy store."

Energy Transfers Between Stores

Energy is transferred from one store to another during events and processes. Energy transfer pathways describe the mechanism by which energy moves between stores. The four examinable transfer pathways are:

When describing any energy transfer, always state three things:

• the store energy is transferred from
• the store energy is transferred to
• the pathway

For example: "Energy is transferred from the chemical store of the fuel to the kinetic store of the car mechanically (by forces)."

Examiner InsightCIE mark schemes require the phrasing "energy is transferred from the [X] store to the [Y] store." Writing "energy is converted from X to Y" loses marks on recent papers. Exam cue: always use "transferred from… to…" with named stores.

Exam TipArrows must be labelled with the pathway, not just drawn as plain connectors.

Conservation of Energy and Flow Diagrams

The principle of conservation of energy states that energy cannot be created or destroyed — it can only be transferred from one store to another. The total energy of a closed system remains constant.

Principle of conservation of energy: energy cannot be created or destroyed; it can only be transferred from one store to another.

This principle means that in any process, the total energy before equals the total energy after. No energy is lost — but energy is often transferred to stores that are not useful, such as the internal (thermal) store of the surroundings. This energy is sometimes described as "wasted" or "dissipated," but it still exists.

A simple flow diagram (also called an energy flow diagram) represents the conservation of energy visually. Boxes represent energy stores before and after an event, and arrows show the transfer pathways. The sum of all output energies equals the input energy.

For example, consider a light bulb supplied with 100 J of energy each second:

The total output (10 J + 90 J = 100 J) equals the total input (100 J), which confirms conservation of energy.

Examiner InsightWhen interpreting flow diagrams, CIE often asks students to calculate the wasted energy. Use: wasted energy = total input energy − useful output energy. Exam cue: always check that all outputs sum to the input.

Extended Calculating Kinetic Energy

Key Equations

Kinetic energy:

$${E}_{k}=\frac{1}{2}m{v}^{2}$$

Variables:

• ${E}_{k}$ = kinetic energy, in J (joules)
• $m$ = mass of the object, in kg (kilograms)
• $v$ = speed of the object, in m s⁻¹ (metres per second)

SI unit: J

Rearrangements:

Starting from ${E}_{k}=\frac{1}{2}m{v}^{2}$:

Rearranging for mass:

$$m=\frac{2{E}_{k}}{{v}^{2}}$$

Rearranging for speed:

$$v=\sqrt{\frac{2{E}_{k}}{m}}$$

Proportionality

• Kinetic energy is directly proportional to mass when speed is constant. Doubling the mass doubles the kinetic energy.
• Kinetic energy is proportional to the square of speed when mass is constant. Doubling the speed quadruples the kinetic energy.

Kinetic energy is the energy stored by an object because of its motion. Any object with mass that moves possesses kinetic energy. Because kinetic energy depends on ${v}^{2}$, a small increase in speed causes a large increase in kinetic energy — this is why high-speed collisions are far more destructive.

Representation note — subscripts: The symbol ${E}_{k}$ uses a subscript "k" to identify this as kinetic energy, distinguishing it from other forms of energy. Subscripts appear slightly below and smaller than the main letter.

MisconceptionStudents often forget that speed is squared in ${E}_{k}=\frac{1}{2}m{v}^{2}$. If speed doubles, kinetic energy quadruples — not doubles. Exam cue: always square the speed before multiplying by mass.

Worked Example

A cyclist has a mass of 60 kg and travels at 8.0 m s⁻¹. Calculate her kinetic energy.

Finding the kinetic energy

Equation used

$${E}_{k}=\frac{1}{2}m{v}^{2}$$

Given

$$m=60\text{ kg}$$

$$v=8.0{\text{ m s}}^{-1}$$

Substituting:

$${E}_{k}=\frac{1}{2}\times 60\times (8.0{)}^{2}$$

$${E}_{k}=\frac{1}{2}\times 60\times 64$$

$${E}_{k}=1920\text{ J}$$

The kinetic energy is 1920 J.

Extended Calculating Gravitational Potential Energy

Key Equations

Change in gravitational potential energy:

$$\Delta {E}_{p}=mg\Delta h$$

Variables:

• $\Delta {E}_{p}$ = change in gravitational potential energy, in J (joules)
• $m$ = mass of the object, in kg (kilograms)
• $g$ = gravitational field strength, in N kg⁻¹ (newtons per kilogram); on Earth, $g\approx 9.8$ N kg⁻¹ (use 10 N kg⁻¹ where the question permits)
• $\Delta h$ = change in height, in m (metres)

SI unit: J

Rearrangements:

Starting from $\Delta {E}_{p}=mg\Delta h$:

Rearranging for mass:

$$m=\frac{\Delta {E}_{p}}{g\Delta h}$$

Rearranging for change in height:

$$\Delta h=\frac{\Delta {E}_{p}}{mg}$$

Proportionality

• Gravitational potential energy is directly proportional to mass when $g$ and $\Delta h$ are constant. Doubling the mass doubles the gravitational potential energy.
• Gravitational potential energy is directly proportional to the change in height when mass and $g$ are constant. Doubling the height doubles the gravitational potential energy.

Gravitational potential energy is the energy stored by an object because of its position in a gravitational field. When an object is raised, energy is transferred from another store (e.g. chemical) to the gravitational potential store. The Δ symbol means "change in," so $\Delta {E}_{p}$ refers to the change in gravitational potential energy and $\Delta h$ refers to the change in height — not the total height above the ground.

Representation note — Δ notation: The Greek letter Δ (delta) always means "change in" the quantity that follows it. So $\Delta h={h}_{\text{final}}-{h}_{\text{initial}}$.

Worked Example

A book of mass 2.0 kg is lifted from the floor onto a shelf 1.5 m above the floor. The gravitational field strength is 9.8 N kg⁻¹. Calculate the change in gravitational potential energy.

Finding the change in gravitational potential energy

Equation used

$$\Delta {E}_{p}=mg\Delta h$$

Given

$$m=2.0\text{ kg}$$

$$g=9.8{\text{ N kg}}^{-1}$$

$$\Delta h=1.5\text{ m}$$

Substituting:

$$\Delta {E}_{p}=2.0\times 9.8\times 1.5$$

$$\Delta {E}_{p}=29.4\text{ J}$$

$$\Delta {E}_{p}\approx 29\text{ J (2 s.f.)}$$

The change in gravitational potential energy is 29 J.

Extended Conservation of Energy in Multi-Stage Problems and Sankey Diagrams

The principle of conservation of energy applies to complex, multi-stage processes just as it does to simple ones. In a multi-stage problem, the output energy from one stage becomes the input energy for the next stage. At every stage, total energy in equals total energy out.

A Sankey diagram is a scaled arrow diagram that represents energy transfers quantitatively. The width of each arrow is proportional to the amount of energy it represents. The input arrow enters from the left. Useful output arrows continue to the right, and wasted energy arrows branch off (usually downward). Because the diagram is drawn to scale, the total width of all output arrows equals the width of the input arrow — this visually confirms conservation of energy.

Representation note — Sankey diagrams: Read a Sankey diagram by comparing arrow widths. A wider arrow represents more energy. The useful output arrow is typically drawn continuing horizontally to the right. Wasted energy arrows branch off at an angle. To find the wasted energy, subtract the useful output from the total input, or measure the width of the waste arrow relative to the scale.

Multi-stage example: A ball is dropped from rest at a height of 5.0 m. Assuming no air resistance, the gravitational potential energy lost equals the kinetic energy gained. This links the two equations:

$$mg\Delta h=\frac{1}{2}m{v}^{2}$$

The mass $m$ cancels:

$$g\Delta h=\frac{1}{2}{v}^{2}$$

Rearranging for speed:

$$v=\sqrt{2g\Delta h}$$

This is a powerful result: the speed of a freely falling object depends only on the height fallen and $g$, not on the mass.

Worked Example — Multi-stage with Sankey diagram interpretation

A coal-fired power station receives 500 MJ of chemical energy each second. The Sankey diagram shows that 200 MJ is transferred as useful electrical energy and the rest is wasted. Of the wasted energy, 250 MJ is transferred to the internal (thermal) store of cooling water and the remainder is transferred to the internal (thermal) store of waste gases.

Step 1: Finding the total wasted energy

$${E}_{\text{wasted}}={E}_{\text{input}}-{E}_{\text{useful}}$$

$${E}_{\text{wasted}}=500-200$$

$${E}_{\text{wasted}}=300\text{ MJ}$$

Step 2: Finding the energy transferred to waste gases

$${E}_{\text{gases}}={E}_{\text{wasted}}-{E}_{\text{cooling water}}$$

$${E}_{\text{gases}}=300-250$$

$${E}_{\text{gases}}=50\text{ MJ}$$

The energy transferred to the waste gases is 50 MJ each second.

Worked Example — Conservation of energy linking $\Delta {E}_{p}$ and ${E}_{k}$

A stone of mass 0.20 kg is dropped from rest at a height of 3.0 m. Assuming no air resistance and using g = 9.8 N kg⁻¹, calculate the speed of the stone as it reaches the ground.

Step 1: Finding the gravitational potential energy lost

Equation used

$$\Delta {E}_{p}=mg\Delta h$$

Given

$$m=0.20\text{ kg}$$

$$g=9.8{\text{ N kg}}^{-1}$$

$$\Delta h=3.0\text{ m}$$

Substituting:

$$\Delta {E}_{p}=0.20\times 9.8\times 3.0$$

$$\Delta {E}_{p}=5.88\text{ J}$$

By conservation of energy (no air resistance), the kinetic energy gained equals the gravitational potential energy lost:

$${E}_{k}=5.88\text{ J}$$

Step 2: Finding the speed

Equation used

$${E}_{k}=\frac{1}{2}m{v}^{2}$$

Rearranging for speed:

$$v=\sqrt{\frac{2{E}_{k}}{m}}$$

Substituting:

$$v=\sqrt{\frac{2\times 5.88}{0.20}}$$

$$v=\sqrt{\frac{11.76}{0.20}}$$

$$v=\sqrt{58.8}$$

$$v = 7.668\ldots \text{ m s}^{-1}$$

$$v\approx 7.7{\text{ m s}}^{-1}\text{ (2 s.f.)}$$

The speed of the stone as it reaches the ground is 7.7 m s⁻¹.

Examiner InsightMulti-stage conservation questions are very common on Paper 4. CIE expects students to explicitly state the conservation of energy link (e.g. "gravitational potential energy lost = kinetic energy gained") before equating the two expressions. Exam cue: always write the conservation statement as a separate line before substituting.