Energy: stores, transfers and work

From a braking car to a law of nature

Physics · beta (2A) · 21 September 2026

A car brakes hard on a dry road

At about 50 km/h (30 mph) it stops in about 14 m. Brakes only: the driver's reaction time is not included.

And at about 100 km/h (60 mph)?

about 28 m
about 40 m
about 55 m

Source: The Highway Code (UK), typical braking distances: 14 m at 30 mph, 55 m at 60 mph. Speeds in km/h are rounded.

7 min. THE HOOK: nothing to do with energy so far, don't name it. Ask: "who has been in a car that brakes hard?" Take the vote by raised hands and click the buttons (the count is kept and shown on the "Back to the car" tally slide). Write the three counts on the board too: a public commitment. DO NOT SOLVE NOW. If someone asks: "we'll see at the end of the lesson, the work is the price of the answer." Expected: most vote 28 m ("double speed, double distance"). That is the linear intuition we want to break. ANSWER (for you): about 55 m, four times the distance, not twice. The Highway Code table: 30 mph → 14 m, 60 mph → 55 m. Wording: these are typical values for a dry road and good brakes; the real distance depends on tyres, road and driver. Thinking distance (9 m → 18 m) is NOT included: it does double, because it grows linearly with speed.

Stores and transfers of energy

Energy stores (nouns)

kinetic
gravitational potential
elastic (strain)
chemical
thermal (internal)
nuclear

Energy transfers (verbs)

mechanically, by a force
electrically, by a current
by heating
by waves (light, sound)

A store holds energy. A transfer moves it from one store to another.

A flow diagram: the braking car

Where does the energy of the car come from, and where does it end up?

Chemical storefuel in the tank

→engine, mechanical transfer

Kinetic storethe moving car

→friction in the brakes

Thermal storebrakes, tyres, road, air

The car's kinetic energy is not "used up": it becomes thermal energy in the brakes. The brakes get hot.

Your turn: draw the flow diagram

A bouncing ball that comes back lower each time

Gravitational potentialball held high

→

Kineticjust before impact

→

Elasticball squashed

→

Kinetic, then gravitationalit goes up again, a bit lower

At each impact a small share goes to the thermal store (ball, floor, air) and to sound.

A hydroelectric dam

Gravitational potentialwater in the reservoir

→

Kineticwater falling

→

Kineticturbine spinning

→

Electricaloutput of the generator

Friction and heating of the wires: a share always goes to the thermal store.

The law of conservation of energy

Law

Energy cannot be created or destroyed. It can only be transferred from one store to another.

Total energy before = total energy after
"Wasted" energy is not gone: it is in the thermal store of the surroundings, spread out and hard to use again
We measure energy in joules (J)

Before the first formula

A car goes at 50 km/h. What is that in m/s?
What is the weight of a 2 kg mass, with g = 10 N/kg?

Kinetic energy

Definition

The energy stored in a moving object.

\htmlClass{c-Ek}{E_k}\;=\;\tfrac{1}{2}\,\htmlClass{c-m}{m}\,\htmlClass{c-v}{v}^{2}

E_kkinetic energy, in joules (J)

mmass of the object, in kilograms (kg)

vspeed of the object, in metres per second (m/s), squared

Try the law

What does the square do?

Worked example: the braking car

A car of mass 1000 kg goes at 50 km/h, then at 100 km/h. What is its kinetic energy in each case?

v_1=\dfrac{50}{3.6}=13.9\ \mathrm{m/s}\qquad v_2=\dfrac{100}{3.6}=27.8\ \mathrm{m/s}

first the units: speed must be in m/s

E_{k1}=\tfrac12\times1000\times13.9^2\approx 96\,600\ \mathrm{J}\approx 97\ \mathrm{kJ}

then the law

E_{k2}=\tfrac12\times1000\times27.8^2\approx 386\,000\ \mathrm{J}\approx 386\ \mathrm{kJ}

same calculation, other speed

\dfrac{E_{k2}}{E_{k1}}\approx 4

double the speed: the brakes must remove four times the energy

Three typical mistakes

Speed in km/h. \tfrac12\times1000\times50^2=1\,250\,000\ \mathrm{J}: about 13 times too big, and it looks reasonable
Forgetting the square. \tfrac12\times1000\times13.9=6950\ \mathrm{J}: it only grows in proportion to v, not to v²
Forgetting the half. 1000\times13.9^2=193\,000\ \mathrm{J}: exactly double the right answer

Check every answer: is the unit joule, and is the speed in m/s?

Quick check

The speed of a cyclist goes from 4 m/s to 12 m/s

Same mass. The kinetic energy becomes…

3 times bigger
6 times bigger
9 times bigger

The speed is multiplied by 3, so the energy is multiplied by 3², which is 9. Check: at 4 m/s Ek is 8m, at 12 m/s it is 72m, and 72/8 = 9.

Your turn 1: kinetic energy

Write: data, unit conversion, formula, substitution, result with unit.

(a) A ball of mass 0.50 kg moves at 8.0 m/s. Calculate its kinetic energy.
(b) The ball is now thrown at 16 m/s. What is its kinetic energy? How does it compare with (a)?
(c) A cyclist and bike, total mass 75 kg, travel at 36 km/h. Calculate the kinetic energy.

Gravitational potential energy

Definition

Lifting an object increases the energy stored in its gravitational field.

\htmlClass{c-dEp}{\Delta E_p}\;=\;\htmlClass{c-m}{m}\,\htmlClass{c-g}{g}\,\htmlClass{c-dh}{\Delta h}

\Delta E_pchange in gravitational potential energy (J)

mmass (kg)

ggravitational field strength: 10 N/kg on Earth

\Delta hchange in height (m)

Only the change in height counts

A 2.0 kg book goes from the table (0.75 m above the floor) to a shelf (1.95 m above the floor).

Calculate ΔE_p.

\Delta h = 1.95-0.75 = 1.20\ \mathrm{m}

\Delta E_p = 2.0\times10\times1.20 = 24\ \mathrm{J}

Not 2.0 × 10 × 1.95 = 39 J: the floor is not where the book started.

Conservation at work: a falling ball

A ball of mass 0.20 kg is dropped from 1.8 m. How fast is it going just before it hits the floor? (Ignore air resistance.)

\Delta E_p = 0.20\times10\times1.8 = 3.6\ \mathrm{J}

the gravitational store goes down by 3.6 J

E_k = 3.6\ \mathrm{J}

conservation: the kinetic store goes up by the same amount

\tfrac12\times0.20\times v^2=3.6\ \Rightarrow\ v^2=36\ \Rightarrow\ v=6.0\ \mathrm{m/s}

solve for the speed: about 22 km/h

What if the ball were twice as heavy? Same speed: the mass cancels.

Your turn 2: potential energy and conservation

(a) A climber of mass 60 kg goes up a mountain path and gains 200 m in height. Calculate the increase in gravitational potential energy.
(b) A 0.10 kg stone is thrown straight up at 12 m/s. How high above the hand does it go? (Ignore air resistance.)

How does a force move energy? Work

Definition

When a force moves an object, it transfers energy to it. The energy transferred is the work done.

\htmlClass{c-W}{W}\;=\;\htmlClass{c-F}{F}\,\htmlClass{c-d}{d}

Wwork done = energy transferred, in joules (1 J = 1 N × 1 m)

Fforce, in newtons (N)

ddistance moved in the direction of the force, in metres (m)

The potential energy formula was a work formula

Lift the 2.0 kg book from the table to the shelf, slowly and at constant speed: Δh = 1.20 m.

F = mg = 2.0\times10 = 20\ \mathrm{N}

constant speed: the forces balance, the lifting force equals the weight

W = F\,d = 20\times1.20 = 24\ \mathrm{J}

the hand transfers 24 J to the book

\Delta E_p = mg\,\Delta h = 24\ \mathrm{J}

the same number: the gravitational store gains exactly what the hand transfers

mg\,\Delta h is just F\,d with F=mg and d=\Delta h.

Back to the car

The brakes exert an average force of about 7000 N (dry road, good tyres). Work done by the brakes = kinetic energy removed.

F\,d = E_k\ \Rightarrow\ d=\dfrac{E_k}{F}

the brakes stop the car when all the kinetic energy has been transferred

50\ \mathrm{km/h}:\quad d=\dfrac{96\,600}{7000}\approx 14\ \mathrm{m}

100\ \mathrm{km/h}:\quad d=\dfrac{386\,000}{7000}\approx 55\ \mathrm{m}

Same force, four times the energy to remove: four times the distance.

6 min. CLOSING THE HOOK. The formula d = Ek / F is the same as W = F d with W = Ek (energy transferred = kinetic energy removed). Numbers checked: 96 605 J / 7000 N = 13.8 m; 386 420 J / 7000 N = 55.2 m; both agree with the Highway Code 14 m and 55 m. F ≈ 7000 N is about 0.7 of the car's weight (10 000 N): an average value chosen to match a dry road; it is DATA, not something they can compute yet. Then look back at the class prediction on the next slide. Say honestly: the force is assumed constant. A real stop has small variations, and the reaction (thinking) distance is not included. Thinking distance doubles (9 → 18 m in the Highway Code); braking distance quadruples. Expected question: "why isn't the force bigger at 100 km/h?" Because it is limited by the grip of the tyres, not by the speed.

Back to the class prediction

At about 100 km/h (60 mph) the braking distance is…

Not double: about four times. Kinetic energy grows with v^2.

Try it

Braking distance, speed and braking force

Your turn 3: work

(a) A crate of mass 40 kg is pushed 5.0 m along the floor by a horizontal force of 120 N. How much work is done by the force?
(b) You carry a 20 kg bag horizontally for 10 m at constant speed. How much work does your upward force do on the bag?
(c) A cyclist and bike, total mass 80 kg, travel at 6.0 m/s and stop with a braking force of 120 N. What distance do they need? And if they were going at 12 m/s?

7 min. (c) is the real-world question (the same problem as the hook, on a bicycle); (b) is the conceptual one. Cut (c) if short of time and use it as the first exercise of 2 October. SOLUTIONS: (a) W = F d = 120 × 5.0 = 600 J (the mass is not needed). (b) 0 J: the upward force is at right angles to the displacement, so the distance moved in the direction of the force is zero. (Your muscles still use energy, but that isn't work done on the bag.) (c) Ek = ½ × 80 × 6.0² = 1440 J; d = 1440 / 120 = 12 m. At 12 m/s: Ek = ½ × 80 × 12² = 5760 J; d = 5760 / 120 = 48 m: four times again. WRONG NUMBERS: (a) 40 × 5.0 × … = 200 J or 4000 J if they use the mass. (b) 200 N × 10 m = 2000 J (ignoring the direction). (c) d = 24 m at 12 m/s (linear thinking); Ek = ½×80×6 = 240 J (no square) → d = 2.0 m.

For 2 October: your climb

Measure your mass (bathroom scale) and your own weight in N (g = 10 N/kg).
Measure the height of three different steps of a staircase with a ruler, to the nearest half division. Write all three values.
Take the mean step height. Count the steps of one flight: Δh = number of steps × mean height.
Calculate ΔE_p for climbing that flight, and the work W = F d done by your leg force. Are they the same?

Bring a table with your own numbers, written by hand. Everyone gets a different answer.

In summary

Energy is stored in stores and moved by transfers: it is conserved, never lost
Kinetic energy: E_k=\tfrac12mv^2. Double the speed, four times the energy
Gravitational potential energy: \Delta E_p=mg\Delta h. Only the change in height counts
Work: W=Fd is the energy a force transfers. In the direction of the force
Braking: d=E_k/F. That is why the distance is four times longer