Calculating energy changes
Kinetic energy
Anything that moves has energy in its store. You must recall this equation:
kinetic energy = ½ × mass × speed², or Eₖ = ½mv²
| Quantity | Symbol | Unit |
|---|---|---|
| Kinetic energy | Eₖ | joules (J) |
| Mass | m | kilograms (kg) |
| Speed | v | metres per second (m/s) |
Example: a car of mass 1200 kg travels at 20 m/s. Find its kinetic energy.
- Eₖ = ½mv²
- Eₖ = ½ × 1200 × 20²
- Eₖ = 0.5 × 1200 × 400 = 240 000 J (240 kJ)
Common mistake
Square the speed only, before you halve or multiply by the mass. Doubling the speed makes the kinetic energy four times bigger.
Key terms
Gravitational potential energy
Lifting an object raises the energy in its store. You must recall this equation:
, or Eₚ = mgh
| Quantity | Symbol | Unit |
|---|---|---|
| Gravitational potential energy | Eₚ | joules (J) |
| Mass | m | kilograms (kg) |
| Gravitational field strength | g | newtons per kilogram (N/kg) |
| Height | h | metres (m) |
On Earth, g = 9.8 N/kg.
Example: a 50 kg student climbs stairs 12 m high. How much does their g.p.e. increase?
- Eₚ = mgh
- Eₚ = 50 × 9.8 × 12
- Eₚ = 5880 J
Key terms
Elastic potential energy
Stretching or squashing a spring fills its store. This equation is on the equation sheet:
elastic potential energy = ½ × spring constant × extension², or Eₑ = ½ke²
| Quantity | Symbol | Unit |
|---|---|---|
| Elastic potential energy | Eₑ | joules (J) |
| Spring constant | k | newtons per metre (N/m) |
| Extension | e | metres (m) |
It only works if the spring is not stretched past its .
Example: a spring with k = 40 N/m is stretched by 5 cm.
- Convert: 5 cm = 0.05 m
- Eₑ = ½ke² = 0.5 × 40 × 0.05²
- Eₑ = 0.5 × 40 × 0.0025 = 0.05 J
Exam tip
Convert to SI units first: centimetres to metres, grams to kilograms, kilojoules to joules.
Key terms
Linking the stores
When a ball falls, ignoring air resistance, all the energy that leaves its gravitational potential store goes into its kinetic store:
Eₚ lost = Eₖ gained
Example: a 0.5 kg ball is dropped from 5 m. How fast is it going just before it lands?
- Eₚ = mgh = 0.5 × 9.8 × 5 = 24.5 J
- So Eₖ = 24.5 J
- Rearrange Eₖ = ½mv² to get v² = 2Eₖ ÷ m = (2 × 24.5) ÷ 0.5 = 98
- v = √98 = 9.9 m/s
The mass cancels out in this kind of question, so a heavy ball and a light ball land at the same speed if there's no air resistance.