Kinetic energy and the work-energy theorem — Physics

Before this

This leads to

Twice as fast, four times as far

A car brakes hard from 20 m/s (72 km/h) and stops in 40 m. Now the same car, on the same road, brakes just as hard from 40 m/s. How far does it travel before it stops?

Most people answer 80 m. Twice the speed, twice the distance. It feels obvious. Make your own guess below before reading on.

Predict, then reveal — doubling the speed

From 20 m/s, the car stops in 40 m. From 40 m/s, with the same braking force, how far does it travel before it stops? Choose one.

The speed–time graph of the two stops. The car slows at the same rate, 5 m/s every second, in both.

The area under a speed–time graph is the distance. From 40 m/s, the stop takes twice as long (8 s instead of 4 s) and starts twice as high. Both sides of the triangle double, so its area is four times as large: 160 m instead of 40 m.

The answer is 160 m: four times as far, not twice. The graph shows one reason. Twice the speed means twice the time to stop and twice the average speed while stopping. Those two doublings multiply.

This lecture finds the deeper reason. It is hidden in a quantity called kinetic energy, and in the way that quantity depends on the square of the speed.

What the total work does

The last lecture ended with an unfinished thought. Whether an object speeds up depends on the total work done by all the forces on it. And it promised to show exactly how that total is connected to the speed. Here is the connection.

Take an object of mass $m$ moving in a straight line. A constant resultant force $F$ acts on it, along its motion, while it moves through a displacement $s$ . Its speed changes from $u$ to $v$ . Follow four steps.

Step 1. The resultant force gives a constant acceleration. Newton's second law says $F = ma$ .

Step 2. With constant acceleration, one of the SUVAT equations links the speeds to the displacement:

v^{2} = u^{2} + 2 a s

Step 3. Rearrange it to get $a s$ on its own: $a s = \frac{1}{2} v^{2} - \frac{1}{2} u^{2}$ .

Step 4. Multiply both sides by $m$ . On the left, $ma s = (ma) s = F s$ , which is the work done by the resultant force. On the right, each term gains an $m$ :

F s = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

Look at what has appeared. The quantity $\frac{1}{2} m \times (speed)^{2}$ is calculated once at the end and once at the start. The total work done is exactly the change in that quantity. This quantity is the energy the object has because it is moving. It is called kinetic energy.

Kinetic energy and the work-energy theorem

An object of mass $m$ moving at speed $v$ has kinetic energy $E_{k} = \frac{1}{2} m v^{2}$ , measured in joules.

The total work done by all the forces on an object equals the change in its kinetic energy: $W_{total} = Δ E_{k} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ .

(Some books write KE or $E_{K}$ . They mean the same thing.)

That is it. That is the whole idea. Positive total work raises the kinetic energy, so the object speeds up. Negative total work lowers it, so the object slows down. Zero total work leaves the speed unchanged, like the crate pushed at steady speed in the last lecture.

The derivation used a constant force, because SUVAT needs a constant acceleration. The theorem is more general than that. It holds for forces that change during the motion too, but proving that needs calculus. That generality is its power. You do not need to know how the force varied, only the total work it did.

Check the units. $\frac{1}{2} m v^{2}$ has units of kg × (m/s)², which is kg m²/s². A joule is a newton metre, and a newton is kg m/s². So a joule is (kg m/s²) × m = kg m²/s². The units match, as they must if work can turn into kinetic energy.

One word of care. Work and kinetic energy are measured in the same unit, but they are not the same thing. Kinetic energy is a store: something the object has. Work is a transfer: something a force does, which moves energy into or out of that store.

Why the speed is squared

Here is the reason behind the square, without any algebra. Think about the braking car again.

To stop a car, the brakes must remove all its kinetic energy. The braking force does negative work, $- F s$ , and that must equal the loss of kinetic energy. For a fixed braking force, the distance is proportional to the kinetic energy that must be removed.

Now double the speed. The car moving at 40 m/s has a velocity twice as large, so each kilogram of it needs twice as much speed removed. And while that speed is being removed, the car moves twice as fast on average, so it covers twice as much ground for every second of braking. Twice the speed to remove, at twice the average speed: four times the distance, and four times the energy.

The same pattern continues. Three times the speed means nine times the kinetic energy. Ten times the speed means a hundred times the kinetic energy. The graph below shows this for a 1200 kg car. The curve gets steeper and steeper, because each extra 10 m/s adds more energy than the one before.

Stop and think

A 1500 kg car speeds up from 10 m/s to 20 m/s. Later it speeds up from 20 m/s to 30 m/s. Both changes are "10 m/s more". Do they need the same total work?

Show the answer

No. From 10 to 20 m/s: $Δ E_{k} = \frac{1}{2} \times 1500 \times (2 0^{2} - 1 0^{2}) = 750 \times 300 = 225000$ J. From 20 to 30 m/s: $750 \times (3 0^{2} - 2 0^{2}) = 750 \times 500 = 375000$ J. The second change needs about 1.7 times as much work. Kinetic energy depends on the square of the speed, so the same gain in speed costs more energy when you are already moving fast. This is one reason fuel use climbs quickly at high speeds.

Explore the two quantities yourself. Choose a mass and a speed. The braking force is fixed at 6000 N.

Explore — mass, speed and stopping

Move the sliders. First double the mass and watch both bars. Then put the mass back and double the speed. Which change has the bigger effect, and by what factor?

Mass m = 1200 kg

Speed v = 20 m/s

Try:

Kinetic energy of a 1200 kg car at different speeds, and its braking distance with a 6000 N braking force.

E_{k} = \frac{1}{2} m v^{2}

for

m = 1200

kg. Doubling the speed from 20 to 40 m/s multiplies the kinetic energy by 4, from 240 kJ to 960 kJ. Doubling the mass instead would only double it.

Mass	Speed	$E_{k}$	Braking distance
1200 kg	20 m/s	240 kJ	40 m
2400 kg	20 m/s	480 kJ	80 m
1200 kg	40 m/s	960 kJ	160 m

What did you see? Doubling the mass doubled both bars. Doubling the speed made both bars four times as long. Mass appears once in $\frac{1}{2} m v^{2}$ , but speed appears twice, as $v \times v$ . That is why speed matters so much more than mass for stopping.

The real stopping table

This is not only a textbook result. The United Kingdom's Highway Code gives typical stopping distances for cars. Each one is a thinking distance (the car travels at full speed while the driver reacts) plus a braking distance. Here are its braking distances, with the speeds converted.

Speed	Braking distance	Compared with 20 mph	(speed ratio)²
20 mph ≈ 8.9 m/s	6 m	1	1
40 mph ≈ 17.9 m/s	24 m	4	4
60 mph ≈ 26.8 m/s	55 m	≈ 9.2	9
70 mph ≈ 31.3 m/s	75 m	12.5	12.25

The braking distances follow the square of the speed, to within rounding. The table assumes a deceleration of about 6.5 m/s². You can check this from $v^{2} = 2 a s$ . The thinking distances in the same table are 6 m, 12 m, 18 m and 21 m. They grow only in proportion to the speed, because the car simply travels at its speed for a fixed reaction time. So at high speed, braking becomes most of the stopping distance.

Worked examples

Worked Example 1 — Kinetic energy at two speeds

Find the kinetic energy of a 1200 kg car at 20 m/s and at 40 m/s.

At 20 m/s: $E_{k} = \frac{1}{2} \times 1200 \times 2 0^{2} = 600 \times 400 = 240000$ J $= 240$ kJ.

At 40 m/s: $E_{k} = \frac{1}{2} \times 1200 \times 4 0^{2} = 600 \times 1600 = 960000$ J $= 960$ kJ.

Check. $960 \div 240 = 4$ . Double the speed, four times the energy ✓. Square the speed before multiplying, never after.

Worked Example 2 — Braking distance, with clear signs

The same car brakes from 20 m/s to rest. The braking force is 6000 N. How far does it travel? Then repeat from 40 m/s.

Choose a positive direction. Take the direction of motion as positive. The braking force points backwards, so the work it does over a displacement $s$ is $W = - 6000 s$ . It is the only horizontal force, so it is the resultant force.

Apply the theorem. $W_{total} = Δ E_{k}$ , and the car ends at rest:

- 6000 s = 0 - 240000

Both sides are negative: the work removes energy, and the kinetic energy falls. Dividing gives $s = 240000 \div 6000 = 40$ m. The minus signs cancel because they describe the same thing. Keep them both, or drop them both; never drop just one.

From 40 m/s. $- 6000 s = 0 - 960000$ , so $s = 160$ m. Four times as far, as the predict question showed.

Check with SUVAT. $a = - 6000 \div 1200 = - 5$ m/s². From $v^{2} = u^{2} + 2 a s$ : $0 = 400 + 2 (- 5) s$ , so $s = 40$ m ✓.

Worked Example 3 — The force of a kick

A 0.50 kg football is kicked from rest and leaves the foot at 12 m/s. The foot is in contact with the ball for 0.15 m. Find the average force of the foot on the ball.

The ball's kinetic energy rises from 0 to $\frac{1}{2} \times 0.50 \times 1 2^{2} = 36$ J. During contact, the foot's push is far larger than any other force, so take the foot's work as the total work: $F \times 0.15 = 36$ , which gives $F = 240$ N.

Does it make sense? The ball's weight is $0.50 \times 9.8 = 4.9$ N. The average kick is about 49 times the ball's weight. That is large, but a kick is a short, violent push, so a large force is reasonable. Notice what the theorem let you skip: the force certainly varied during the kick, but only its total work mattered.

Worked Example 4 — Turning 400 kJ into a speed

In the last lecture, a car travelled 500 m with a driving force of 1200 N and resistive forces of 400 N. The total work done on it was 400 kJ, and that lecture promised to turn this into a speed. Suppose the car's mass is 1000 kg and it started at 10 m/s. How fast is it moving at the end?

Starting energy. $\frac{1}{2} \times 1000 \times 1 0^{2} = 50000$ J $= 50$ kJ.

Final energy. The total work adds 400 kJ, so $E_{k} = 50 + 400 = 450$ kJ.

Speed. $\frac{1}{2} \times 1000 \times v^{2} = 450000$ , so $v^{2} = 900$ and $v = 30$ m/s.

Check. The resultant force is $1200 - 400 = 800$ N, so $a = 0.8$ m/s². Then $v^{2} = 1 0^{2} + 2 \times 0.8 \times 500 = 100 + 800 = 900$ ✓. Notice that the speed went from 10 to 30 m/s, three times as large, while the kinetic energy went from 50 to 450 kJ, nine times as large.

Three classic mistakes

Trap 1 — Squaring the wrong thing

$\frac{1}{2} m v^{2}$ means half of $m$ times $v^{2}$ . Only the speed is squared. A common slip is to work out $\frac{1}{2} m v$ and then square the whole thing. For the car at 20 m/s that gives $1200 0^{2} = 144$ million J, not 240 000 J. Square the speed first, then multiply.

Trap 2 — Using one force instead of the total

The theorem uses the total work done by all the forces. In Worked Example 4, using only the driving force's 600 kJ would give a final speed of about 36 m/s, too high. The resistive forces' negative work must be included. If the object moves at a steady speed, the total work is zero, even though individual forces do work.

Trap 3 — Speed in the wrong unit

The joule is built from kg, m and s, so the speed must be in m/s. A speed in km/h put straight into $\frac{1}{2} m v^{2}$ gives an answer $3. 6^{2} \approx 13$ times too large. Divide km/h by 3.6 first.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Take $g = 9.8$ m/s².

A 60 kg runner moves at 8.0 m/s. Find her kinetic energy. How fast would a 0.16 kg cricket ball have to travel to have the same kinetic energy?

Hint 1
Use $E_{k} = \frac{1}{2} m v^{2}$ for the runner first.

Hint 2
For the ball, $\frac{1}{2} \times 0.16 \times v^{2}$ must equal the runner's energy. Solve for $v^{2}$ , then take the square root.

Full answer
Runner: $\frac{1}{2} \times 60 \times 8. 0^{2} = 30 \times 64 = 1920$ J. Ball: $0.08 v^{2} = 1920$ , so $v^{2} = 24000$ and $v \approx 155$ m/s, about 560 km/h. The ball has $\frac{1}{375}$ of the mass, so it needs $375 \approx 19$ times the speed. Check: $0.08 \times 154. 9^{2} \approx 1920$ J ✓.
A 0.010 kg bullet travelling at 400 m/s enters a wooden block and stops after 0.12 m. Find the average force of the wood on the bullet.

Hint 1
All the bullet's kinetic energy is removed by the wood's force.

Hint 2
Take the bullet's motion as positive. Then $- F \times 0.12 = 0 - \frac{1}{2} \times 0.010 \times 40 0^{2}$ .

Full answer
Kinetic energy: $\frac{1}{2} \times 0.010 \times 160000 = 800$ J. So $- F \times 0.12 = - 800$ , which gives $F \approx 6670$ N, about 6.7 kN, against the motion. That is more than 68 000 times the bullet's weight of 0.098 N. A small mass can carry a lot of energy when its speed is high.
A car's braking distance from 15 m/s is 18 m. Using only the idea that braking distance is proportional to $v^{2}$ , predict the braking distance from 30 m/s and from 25 m/s. Then find the braking force if the car's mass is 1100 kg.

Hint 1
30 m/s is twice 15 m/s. 25 m/s is $\frac{5}{3}$ of 15 m/s.

Hint 2
Multiply 18 m by the square of each speed ratio. For the force, the braking force's work removes all of $\frac{1}{2} m v^{2}$ over 18 m.

Full answer
From 30 m/s: $18 \times 2^{2} = 72$ m. From 25 m/s: $18 \times (\frac{5}{3})^{2} = 18 \times \frac{25}{9} = 50$ m. Force: $E_{k} = \frac{1}{2} \times 1100 \times 1 5^{2} = 123750$ J, so $F = 123750 \div 18 = 6875$ N, about 6.9 kN. Check at 30 m/s: $\frac{1}{2} \times 1100 \times 900 \div 6875 = 495000 \div 6875 = 72$ m ✓.
Amara pulls a 20 kg sledge from rest across level snow. Her rope pulls with 80 N at 25° above the horizontal, and friction is 60 N. After 30 m, how fast is the sledge moving?

Hint 1
Find the work done by each force. Only the rope's horizontal component does work along the motion; friction does negative work; weight and the normal force do none.

Hint 2
Total work $= 80 cos 2 5^{\circ} \times 30 - 60 \times 30$ . This equals $\frac{1}{2} \times 20 \times v^{2}$ .

Full answer
Rope: $80 \times 30 \times cos 2 5^{\circ} \approx 2175$ J. Friction: $- 60 \times 30 = - 1800$ J. Weight and normal force: 0 J each, because they are at 90° to the motion. Total work $\approx 375$ J. So $\frac{1}{2} \times 20 \times v^{2} = 375$ , giving $v^{2} \approx 37.5$ and $v \approx 6.1$ m/s. Check with forces: the resultant is $80 cos 2 5^{\circ} - 60 \approx 12.5$ N, so $a \approx 0.63$ m/s², and $v^{2} = 2 \times 0.63 \times 30 \approx 37.5$ ✓. Most of the rope's work went into heating the snow; only about a sixth became kinetic energy.

The bigger idea

The work-energy theorem changes the question you ask. Newton's second law asks about forces at each moment. The theorem asks only about the start and the end: how much work was done in total, and how much kinetic energy changed. You can find a final speed without ever finding the acceleration or the time. In the kick and the bullet, the force changed in ways nobody measured, and it did not matter.

It also raises a new question. In the last lecture, Ravi lifted a box at steady speed. He did 78.4 J of work, but the box's kinetic energy did not change, because gravity did −78.4 J. So where did Ravi's energy go? It did not disappear. It was stored in the box's height, ready to come back as kinetic energy if the box falls. That store is gravitational potential energy, the subject of the next lecture. Kinetic energy is the first store of energy you can calculate. Soon, energy will become an account that always balances.

Key concepts in this lesson

Kinetic energy is the energy an object has because it is moving, $E_{k} = \frac{1}{2} m v^{2}$ , measured in joules.
The work-energy theorem says that the total work done by all the forces on an object equals the change in its kinetic energy.
The theorem follows from $F = ma$ and $v^{2} = u^{2} + 2 a s$ , and it also holds when the force varies.
Because kinetic energy depends on $v^{2}$ , doubling the speed gives four times the kinetic energy and, for the same braking force, four times the braking distance.
Work is a transfer of energy and kinetic energy is a store, so positive total work speeds an object up and negative total work slows it down.