4.1.1.2 Changes in Energy — Calculating Energy Transfers — Physics with Kate

AQA GCSE Physics · Topic 4.1 Energy

Changes in Energy: Calculating Energy Transfers

In this lesson you'll learn to: use E_p = mgh, E_k = ½mv², E_e = ½ke² and W = Fs; convert into the right units before you substitute; and follow the energy through a transfer to find a speed or a distance.

▶ Prefer to watch? Scroll to the bottom for the video on this topic

▶

Start here — have a play

Drag the height slider and watch the numbers. The working on the right recalculates every time — so it's a worked example that answers whatever question you set it. There are four scenarios: a dive, a loop, a braking bike and a bow and arrow. Make the ramp too short and the cart comes off the track; brake harder and the bike stops sooner.

Mass of diver Board height Loop radius

Energy stores

Gravitational potential0 J

Elastic potential0 J

Kinetic0 J

Thermal0 J

The working

▲ Change the mass on the diving board — the speed doesn't budge, and that's not a bug. Watch where the energy ends up once she stops in the water.

Now learn to do it on paper

The notes below give you both equations, the unit conversions that catch people out, and three worked examples laid out exactly the way you should set yours out in the exam.

Read the notes ↓

✎

Revision notes

The two equations

Both are on your equation sheet, but you still need to know what every letter means and what unit it has to be in. That's where most marks are lost.

E_p = m g h

gravitational potential energy = mass × gravitational field strength × height

Symbol	Quantity	Unit
E_p	gravitational potential energy	J (joules)
m	mass	kg
g	gravitational field strength	N/kg — use 9.8
h	height it rises or falls	m

E_k = ½ m v²

kinetic energy = ½ × mass × (speed)²

Symbol	Quantity	Unit
E_k	kinetic energy	J (joules)
m	mass	kg
v	speed	m/s

💡 Exam tip — it's v², not 2v

Square the speed first, then multiply. If the speed doubles, the kinetic energy goes up four times — not twice. This is the single most common slip in the whole topic.

Getting the units right before you substitute

The equations only work in kilograms, metres and metres per second. Exam questions deliberately give you grams, centimetres or kilometres to see whether you notice. Convert first, substitute second.

Prefix	Means	To convert to the base unit	Example
M (mega)	× 1 000 000	MJ → J: × 1 000 000	2 MJ = 2 000 000 J
k (kilo)	× 1000	kJ → J: × 1000	4.5 kJ = 4500 J
c (centi)	÷ 100	cm → m: ÷ 100	80 cm = 0.80 m
m (milli)	÷ 1000	mm → m: ÷ 1000	250 mm = 0.25 m

And the three that catch people out most often:

grams → kilograms: divide by 1000 (250 g = 0.25 kg)
centimetres → metres: divide by 100 (80 cm = 0.80 m)
tonnes → kilograms: multiply by 1000 (1.2 t = 1200 kg)

⭐ Key fact — answers come out in joules

Put kg, m and m/s in, and the answer is always in J. If the number is big, you can write it with a prefix: 2940 J = 2.94 kJ. Both are correct — but always write the unit, because a number with no unit scores no marks.

Transferring GPE into kinetic energy

When something falls, its gravitational potential store empties into its kinetic store. If we ignore air resistance, none of it is wasted, so:

E_p lost = E_k gained

m g h = ½ m v²

The m appears on both sides, so it cancels:

v = √(2 g h)

the speed after falling a height h — mass makes no difference at all

⭐ Key fact — why mass cancels

A heavy diver and a light diver jumping from the same board hit the water at the same speed. The heavy one has more energy in both stores, and the two effects cancel exactly. Try it on the animation above.

Elastic potential energy

Stretch or squash a spring, an elastic band or a bow and you store energy in its elastic potential store. How much depends on how stiff it is and how far you pull it.

E_e = ½ k e²

elastic potential energy = ½ × spring constant × (extension)²

Symbol	Quantity	Unit
E_e	elastic potential energy	J (joules)
k	spring constant (how stiff it is)	N/m
e	extension — how far it stretches	m

💡 Exam tip — e is the extension, not the length

e is how much longer the spring has got, not its total length. If a 10 cm spring is stretched to 25 cm, then e = 15 cm = 0.15 m, not 0.25 m. And like v in E_k, the extension is squared — pull it twice as far and you store four times the energy.

Work done

Whenever a force moves something, energy is transferred. We call that transfer work done, and it is measured in joules just like every other energy.

W = F s

work done = force × distance moved along the line of the force

Symbol	Quantity	Unit
W	work done	J (joules)
F	force	N
s	distance moved	m

⭐ Key fact — 1 joule = 1 newton-metre

Doing 1 J of work means moving something 1 m against a force of 1 N. That's what a joule is.

Using work done to find a stopping distance

This is the classic exam question. A vehicle braking has to lose all the energy in its kinetic store, and the brakes do that by doing work against friction:

½ m v² = F × d

kinetic energy at the start = work done by the brakes

So the braking distance is d = ½mv² ÷ F. All of that energy ends up in the thermal store of the brakes — which is exactly why brakes get hot.

⭐ Key fact — double the speed, four times the distance

Because E_k depends on v², a car at 30 m/s needs four times the braking distance of the same car at 15 m/s — not twice. Doubling the mass only doubles it. This is a favourite exam question, and the reason speed limits matter.

🔢

Worked examples

Set your working out like this every time: equation → convert → substitute with units → answer to 2 significant figures. You get marks for the working even if the final number is wrong.

EXAMPLE 1A diver of mass 60 kg stands on a board 5.0 m above the water. Calculate the energy in her gravitational potential store, and how fast she is moving as she reaches the water.

Equation E_p = m g h

Substitute — with units E_p = 60 kg × 9.8 N/kg × 5.0 m = 2940 J

All of it becomes kinetic, so rearrange E_k = ½mv² v = √(2E_k ÷ m) = √(2 × 2940 ÷ 60) = √98 = 9.899… m/s

E_p = 2900 J (2.9 kJ) · v = 9.9 m/s Both given to 2 significant figures, both with units.

EXAMPLE 2A ball of mass 250 g is dropped from a height of 80 cm. Calculate its speed just before it hits the floor.

Convert first — the numbers are in the wrong units 250 g ÷ 1000 = 0.25 kg 80 cm ÷ 100 = 0.80 m

Energy in the GPE store E_p = 0.25 kg × 9.8 N/kg × 0.80 m = 1.96 J

Speed at the floor — mass cancels, so use v = √(2gh) v = √(2 × 9.8 × 0.80) = √15.68 = 3.959… m/s

E_p = 2.0 J · v = 4.0 m/s Had you substituted 250 and 80 straight in, you'd have been out by a factor of 100 000.

EXAMPLE 3A car of mass 1200 kg is travelling at 15 m/s. It freewheels up a hill. Ignoring friction, how high does it rise?

Energy in the kinetic store — square the speed first E_k = ½ × 1200 kg × 15² = ½ × 1200 × 225 = 135 000 J

In kilojoules, if you prefer 135 000 J ÷ 1000 = 135 kJ

All of it becomes GPE, so rearrange E_p = mgh for h h = E_p ÷ (m g) = 135 000 ÷ (1200 × 9.8) = 135 000 ÷ 11 760 = 11.47… m

h = 11 m Use the unrounded 135 000 J here, not the rounded answer — see the tip below.

EXAMPLE 4An archer draws a bow with a spring constant of 600 N/m back by 0.50 m. The arrow has a mass of 30 g. Calculate the speed of the arrow as it leaves the bow.

Convert the mass — it's in grams 30 g ÷ 1000 = 0.030 kg

Energy stored in the drawn bow — square the extension E_e = ½ k e² = ½ × 600 N/m × 0.50² = ½ × 600 × 0.25 = 75 J

All of it is transferred to the arrow's kinetic store E_k = E_e = 75 J

Rearrange E_k = ½mv² for v v = √(2E_k ÷ m) = √(2 × 75 ÷ 0.030) = √5000 = 70.7… m/s

v = 71 m/s Forget the g → kg conversion and you'd get 1.3 m/s — a very slow arrow.

EXAMPLE 5A cyclist and bike have a total mass of 80 kg and are travelling at 10 m/s. The brakes apply a constant force of 400 N. Calculate the braking distance.

Energy in the kinetic store E_k = ½ × 80 kg × 10² = ½ × 80 × 100 = 4000 J

The brakes must do that much work to stop it W = E_k = 4000 J

Rearrange W = Fs for the distance s = W ÷ F = 4000 J ÷ 400 N = 10 m

s = 10 m All 4000 J is now in the thermal store of the brakes and the surroundings.

💡 Exam tip — round only at the very end

In Example 3, rounding E_k to 140 kJ before the last step gives h = 12 m instead of 11 m. Carry the full number through your calculator and round once, on the final answer.

💡 Exam tip — when do you actually need 2 significant figures?

The mark for rounding to 2 s.f. is only given when the question specifically asks for it — "give your answer to 2 significant figures".

If the question doesn't ask, you can leave your answer to any number of decimal places, as long as it is to 2 s.f. or more. So with no instruction, 9.9 m/s, 9.90 m/s and 9.899 m/s would all be accepted — but 10 m/s would not, because that has been rounded too far.

Two things that are always true: never round to fewer than 2 s.f., and always write the unit. And when you do count significant figures, start from the first non-zero digit — so 0.0396 to 2 s.f. is 0.040.

📘 Now do it in your workbook

Exam-style calculation practice for this spec point, with every step of the working shown in the answers so you can see exactly where marks are given.

AQA GCSE Physics Workbook · ENERGY · pages 5–8

⬇ Free: pages 5–8 (PDF) Get the full workbook on Amazon

The full workbook covers the whole of Topic 1 Energy with exam-style questions and worked answers.

▶

Prefer to watch? Here's the whole thing

Every calculation on this page worked through out loud, including the unit conversions.

▲ In GHL you can also use the lesson's built-in video field instead of this embed.

Now put it into practice

Calculations only stick once you've done them yourself. Pages 5–8 of the workbook are free and get you started, and the full book has exam-style questions with worked answers for every spec point in Topic 1 Energy.

⬇ Free: pages 5–8 (PDF) Full workbook on Amazon