Conservation of energy — the big idea — Physics

Before this

Gravitational potential energy
Kinetic energy and the work-energy theorem
Temperature vs heat · coming soon

This leads to

Power — energy per second
The zeroth and first law of thermodynamics · coming soon
Carnot cycle and efficiency · coming soon

A ball that does not come back

Drop a 0.50 kg ball from a height of 2.0 m. Take the floor as the reference level. At the start, its gravitational potential energy is $E_{p} = 0.50 \times 9.8 \times 2.0 = 9.8$ J. It hits the floor, bounces, and rises to only 1.4 m. At the top of the bounce, $E_{p} = 0.50 \times 9.8 \times 1.4 = 6.86$ J. (These numbers are invented, but a real ball behaves like this.)

The last lecture showed a thrown ball trading potential energy for kinetic energy, with the total fixed at 9.6 J the whole way. Here the total seems to have fallen from 9.8 J to 6.86 J. Where did 2.94 J go? The last lecture promised that this lecture would turn the "fixed total" into a principle, and add what happens when friction or air resistance takes some energy away. Here it is.

This is the disruption. The trade between $E_{p}$ and $E_{k}$ is real, but it is not the whole story. If you look only at those two, energy seems to disappear. If you look everywhere, it never does.

Stores, transfers and one rule

Energy is not a substance, but it behaves like money in a set of accounts. It sits in stores: kinetic (motion), gravitational potential (height), elastic (stretch or squash), thermal, chemical (in fuels, food and batteries) and nuclear. The thermal store is the random jiggling of particles, which we feel as temperature. It moves between stores by transfers: a force doing work, heating, an electric current, and radiation such as light and sound.

Now look at the ball again. When it hits the floor, it squashes. The squash is not perfectly springy. Some of the energy makes the particles of the ball and the floor jiggle faster, so both become very slightly warmer. Some leaves as the sound of the bounce, and that sound also ends up as thermal energy in the air and walls. Measure everything, and the missing 2.94 J is all there.

The bounce, as energy accounts (invented numbers). Before the drop, all 9.8 J is potential energy. At the top of the bounce, 6.86 J is potential energy again, and 2.94 J is now thermal energy and sound. The two bars are the same height.

The principle of conservation of energy

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

So in a closed system, one that no energy enters or leaves, the total energy stays the same.

The word "closed" matters. Your phone's battery does not obey "total energy stays the same" on its own, because energy flows into it from the charger. Draw the boundary wide enough, around the power station, the cable and the phone, and the total is constant again. Choosing the system is the first step in every energy problem, just as choosing the positive direction is the first step in every motion problem.

When energy spreads out into the surroundings as thermal energy, we say it is dissipated. It is not lost from the universe. It is lost to us, because a little warmth spread through a room cannot easily be gathered up again to do a job. Physicists call this energy "wasted", never "destroyed".

Where friction's energy goes

Now put a block on a rough ramp. As it slides down, gravity speeds it up and friction slows it down. The two energy lectures before this one already contain everything needed to write the energy accounts exactly. Step through the argument below. Each line uses one result you have met before.

Step through — energy accounts on a rough ramp

Press Next to add one step. Read the reason before you go on.

Each step gives one line and the result it uses.

Step 1. W(total) = ΔEₖ. The work-energy theorem: the work done by all the forces equals the change in kinetic energy.

Step 2. W(gravity) + W(friction) = ΔEₖ. Only gravity and friction do work. The normal force does none, because it is at 90° to the motion.

Step 3. W(gravity) = −ΔEₚ. From the last lecture: gravity's work equals the decrease in potential energy.

Step 4. −ΔEₚ + W(friction) = ΔEₖ. Substitute step 3 into step 2.

Step 5. ΔEₖ + ΔEₚ = W(friction). Move ΔEₚ to the other side.

Step 6. W(friction) = −Fs. Friction points against the motion (θ = 180°), so its work is negative, from the work lecture.

Result: ΔEₖ + ΔEₚ = −Fs. Kinetic and potential energy together fall by exactly Fs, the work done against friction.

That is it. That is the whole idea, in one line. On a rough surface, kinetic and potential energy together are not conserved. They fall by exactly $F s$ , the work done against friction. And $F s$ is precisely the thermal energy that appears in the block and the ramp. The total of all three never changes:

E_{p} lost = E_{k} gained + work done against friction

Use the explorer to watch the accounts. The ramp is 5.0 m long and 2.5 m high. The block's mass is 0.40 kg, so it starts with $E_{p} = 0.40 \times 9.8 \times 2.5 = 9.8$ J above the bottom.

Explore — the energy accounts of a sliding block

The 0.40 kg block slides from rest down the 5.0 m ramp, which is 2.5 m high. Change the friction force. Compare the two bars: the energy at the top, and where it has gone at the bottom.

Presets:

Friction force =

The accounts for the 0.40 kg block on the 5.0 m ramp, 2.5 m high. At the top, Eₚ = 9.8 J in every row.

Friction (N)	Thermal energy, F × 5.0 m (J)	Eₖ at the bottom (J)	Speed (m/s)
0	0	9.8	7.0
0.5	2.5	7.3	≈ 6.0
1.0	5.0	4.8	≈ 4.9
1.5	7.5	2.3	≈ 3.4
1.9	9.5	0.3	≈ 1.2

In every row, thermal energy + Eₖ = 9.8 J.

What should you have seen? The bar at the bottom is always exactly as tall as the bar at the top. Friction never removes energy from the total. It only changes which store the energy ends up in. More friction means more thermal energy and less kinetic energy, so a slower block. At 1.9 N, almost everything becomes thermal energy, and the block crawls to the bottom.

Stop and think

The last lecture showed that, without friction, a block reaches the same speed at the bottom of a steep ramp and a gentle ramp of the same height. Now give both ramps the same friction force of 1.0 N. Which block is faster at the bottom?

Show the answer

The block on the steep ramp. Both start with the same 9.8 J. But the gentle ramp is longer, so friction acts over a longer distance and does more work. For a 2.5 m height, a 60° ramp is about 2.89 m long and a 20° ramp about 7.31 m. The work done against friction is about 2.89 J and about 7.31 J. So the steep block keeps about 6.91 J of kinetic energy (about 5.9 m/s) and the gentle one only about 2.49 J (about 3.5 m/s). "Same height, same speed" is true only when nothing is dissipated.

Useful energy and efficiency

Every real machine moves energy from one store to another, and every real machine dissipates some on the way. A motor gets warm. A bicycle chain squeaks. So we ask how much of the energy went where we wanted it. That is the efficiency. Both energies are measured in joules:

efficiency = (useful energy output \div total energy input) \times 100%

Efficiency can also be written as a decimal, such as 0.82 instead of 82%. Conservation of energy sets a hard limit. The useful output is part of the input, so efficiency can never be more than 100%. Any claim of more than 100% means that some input has been left out of the accounts.

A Sankey diagram draws the accounts as a picture. The input enters as a wide band on the left. It splits into a useful band that carries on to the right and a wasted band that turns away. The width of each band is proportional to the energy it carries, so the widths always add up.

A filament bulb. The US Department of Energy states that about 90% of the energy used by incandescent (filament) bulbs is released as heat and only 10% as light. The useful band is one tenth of the input band.

This is why filament bulbs have mostly been replaced. The same source says that an LED needing only about 9 W can be as bright as a 75 W filament bulb. The LED wastes far less, so its Sankey diagram has a much wider useful band. It still has a wasted band, because LEDs also get warm.

Worked examples

Take $g = 9.8$ N/kg throughout.

Worked Example 1 — A pulley system

Grace lifts a 5.0 kg load through 3.0 m with a pulley system. She pulls the rope with a force of 20 N through a distance of 9.0 m. Find the efficiency, and say where the rest of the energy goes.

System and input. The energy input is the work Grace does on the rope: $W = F s = 20 \times 9.0 = 180$ J.

Useful output. The load gains $Δ E_{p} = m g Δ h = 5.0 \times 9.8 \times 3.0 = 147$ J.

Efficiency. $\frac{147}{180} \times 100% \approx 81.7%$ .

The rest. $180 - 147 = 33$ J is dissipated, mainly by friction in the pulley wheels and the rope. It warms them slightly.

Check. Notice that the pulley lets Grace pull with less force than the load's weight (20 N instead of 49 N), but over three times the distance. A machine can trade force for distance. It can never give out more energy than it takes in, and here 147 J is less than 180 J ✓.

Worked Example 2 — The ramp, with and without friction

A 0.40 kg block starts from rest at the top of the 5.0 m ramp, 2.5 m above the bottom. Find its speed at the bottom (a) with no friction, (b) with a friction force of 1.0 N.

(a) All the lost $E_{p}$ becomes $E_{k}$ : $m g h = \frac{1}{2} m v^{2}$ . The mass cancels, so $v^{2} = 2 g h = 2 \times 9.8 \times 2.5 = 49$ , and $v = 7.0$ m/s.

(b) Work done against friction $= F s = 1.0 \times 5.0 = 5.0$ J. So $E_{k} = 9.8 - 5.0 = 4.8$ J. Then $\frac{1}{2} \times 0.40 \times v^{2} = 4.8$ , so $v^{2} = 24$ and $v \approx 4.9$ m/s.

Check. Accounts at the bottom: $4.8 + 5.0 = 9.8$ J, the same as at the top ✓. The speed is lower with friction, as it must be.

Notice something about (b). The mass no longer cancels, because the 5.0 J taken by friction does not depend on the mass. A 4.0 kg block losing the same 5.0 J would keep 93 J of its 98 J, and reach about 6.8 m/s.

Worked Example 3 — A motor lifting a box

An electric motor uses 500 J of electrical energy and lifts a box, giving it 350 J of potential energy. Find the efficiency, and describe its Sankey diagram.

Efficiency $= \frac{350}{500} \times 100% = 70%$ .

Sankey diagram. An input band of 500 J splits into a useful band of 350 J (the box's $E_{p}$ ) and a wasted band of 150 J. The wasted energy is thermal energy in the motor's wires and bearings, and a little sound.

Check. $350 + 150 = 500$ ✓. Nothing has been created or destroyed. Every joule is in a store you can name.

Worked Example 4 — Where does a car's fuel go?

The US government's fuel-economy website estimates that, in a typical petrol car in mixed city and highway driving, only about 16% to 25% of the fuel's energy reaches the wheels. About 68% to 72% is lost in the engine itself, mainly as thermal energy.

Efficiency. Taking "moving the car" as the useful output, the efficiency is about 16% to 25%.

Follow the useful part further. On a level road, the energy that reaches the wheels pushes the car against air resistance and rolling friction. All of it is dissipated as thermal energy in the air and the road. When the driver brakes to stop, the car's kinetic energy becomes thermal energy in the brakes. So at the end of a trip on level ground, every joule of the fuel's chemical energy has become thermal energy. Conservation of energy is still true: the total has not changed. The energy has simply spread out.

Three traps

Trap 1 — "Friction destroys energy"

Friction transfers energy to the thermal store; it destroys nothing. Write "transferred to thermal energy" or "dissipated", not "lost". In an exam answer, always say which store the energy went to.

Trap 2 — Using the ramp length as the height

$Δ E_{p} = m g Δ h$ needs the vertical height (2.5 m), not the length along the slope (5.0 m). The slope length belongs in the work done against friction, $F s$ . One problem often needs both numbers, each in its own place.

Trap 3 — Efficiency upside down

Efficiency is useful output ÷ total input, so it is always 1 or less (100% or less). If your answer is more than 100%, you have divided the wrong way round, or left an input out.

Try it yourself

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

A crane's motor uses 12 250 J of electrical energy to lift a 250 kg crate through 4.0 m. Find the efficiency, and the energy wasted.

Hint 1
The useful output is the crate's gain in potential energy.

Hint 2
$Δ E_{p} = 250 \times 9.8 \times 4.0$ .

Full answer
Useful output $= 250 \times 9.8 \times 4.0 = 9800$ J. Efficiency $= \frac{9800}{12 250} \times 100% = 80%$ . Wasted energy $= 12250 - 9800 = 2450$ J, dissipated as thermal energy in the motor and the cables. Check: $9800 + 2450 = 12250$ ✓.
A 0.060 kg ball is dropped from 1.5 m and bounces to 0.90 m. How much energy is dissipated in the bounce? If each bounce keeps the same fraction of the energy, how high is the second bounce?

Hint 1
Compare $E_{p}$ at 1.5 m and at 0.90 m.

Hint 2
$E_{p}$ is proportional to height, so the fraction kept is $0.90 \div 1.5$ .

Full answer
$E_{p}$ at 1.5 m $= 0.060 \times 9.8 \times 1.5 \approx 0.882$ J. At 0.90 m, $\approx 0.529$ J. Dissipated $\approx 0.353$ J (40%). The fraction kept is $0.90 \div 1.5 = 0.60$ . The second bounce keeps 60% of the 0.90 m height: $0.60 \times 0.90 = 0.54$ m. Check: the heights 1.5, 0.90, 0.54 each fall by the same factor, 0.6 ✓.
Mei and her sledge have a total mass of 30 kg. They start from rest at the top of a hill 12 m high and reach the bottom at 12 m/s, after sliding 60 m along the slope. How much energy was dissipated? What was the average resistive force (friction and air resistance together)?

Hint 1
Energy dissipated = $E_{p}$ lost − $E_{k}$ gained.

Hint 2
The dissipated energy is the work done against the resistive force over the 60 m: $F \times 60$ .

Full answer
$E_{p}$ lost $= 30 \times 9.8 \times 12 = 3528$ J. $E_{k}$ gained $= \frac{1}{2} \times 30 \times 1 2^{2} = 2160$ J. Dissipated $= 3528 - 2160 = 1368$ J. Average force: $F \times 60 = 1368$ , so $F = 22.8$ N. Check: without resistance, $v = 2 \times 9.8 \times 12 \approx 15.3$ m/s, so a lower 12 m/s makes sense ✓.
A 2.0 kg block slides from rest down a smooth (frictionless) curved slide, dropping 1.25 m. At the bottom it slides onto a rough floor, where friction is 4.9 N. How far along the floor does it slide before stopping? Then show that, if the friction is $F = μ m g$ with the same $μ$ , the distance does not depend on the mass at all.

Hint 1
The shape of the smooth slide does not matter: only the drop in height does. All the $E_{p}$ becomes $E_{k}$ at the bottom.

Hint 2
On the floor, all that kinetic energy becomes thermal energy: $F \times d = E_{k}$ . For the second part, $μ = 4.9 \div (2.0 \times 9.8)$ .

Full answer
$E_{p}$ lost $= 2.0 \times 9.8 \times 1.25 = 24.5$ J, so the block reaches the floor with 24.5 J of kinetic energy. On the floor, $4.9 \times d = 24.5$ , so $d = 5.0$ m. For the general case, $m g h = μ m g \times d$ . The $m g$ cancels, so $d = \frac{h}{μ}$ . Here $μ = \frac{4.9}{19.6} = 0.25$ , and $d = \frac{1.25}{0.25} = 5.0$ m ✓. A heavier block has more energy, but it is also pressed harder onto the floor, and the two effects cancel exactly.

The bigger idea

Conservation of energy has been tested more than almost any other law in science, and no experiment has ever found an exception. It once looked as if one had been found. Early in the twentieth century, measurements of beta decay, a kind of radioactivity, showed electrons leaving nuclei with a whole range of energies. The energy accounts said they should all have the same energy. Some energy seemed to vanish. Niels Bohr was ready to consider that energy might not be conserved in the nucleus. In a letter dated 4 December 1930, Wolfgang Pauli proposed what he called a "desperate remedy": a tiny, neutral, invisible particle that carried the missing energy away. Physicists trusted the law so much that they accepted a particle nobody had seen. It is now called the neutrino. Clyde Cowan and Frederick Reines finally detected it in 1956, beside a nuclear reactor in South Carolina, about 26 years after Pauli's letter.

The same law explains why a perpetual motion machine, one that runs for ever and does work with no energy input, can never be built. Every real machine dissipates some energy, so with no input its stores must run down. Conservation tells you that the total never changes. A deeper law, met later in the course, explains why the energy always spreads out and never gathers itself back up. That is why the warm air around a filament bulb never turns back into electrical energy. The next lecture adds time to the accounts: not how much energy is transferred, but how fast.

Key concepts in this lesson

Energy cannot be created or destroyed; it is only transferred between stores, so the total energy of a closed system stays the same.
When energy seems to disappear, it has usually been dissipated as thermal energy (and sound) in the surroundings.
On a rough surface, the loss in potential energy equals the gain in kinetic energy plus the work done against friction, $F s$ .
Efficiency is useful energy output ÷ total energy input (× 100%), and it can never be more than 100%.
A Sankey diagram shows the energy transfers as bands whose widths are proportional to the energy, so the output bands always add up to the input band.