Module 5 · the conservation laws, continued

Energy, in all its disguises

Module 2 said energy is a ledger, not a substance: Dennis has 28 blocks, his mother counts them every evening, and when the count comes up short she hunts for where the missing ones went rather than give up the number. This module is the audit. Five situations where the total is fixed and something is hiding — a pendulum, a lever, a loop, a bungee cord, and a bucket of water — and, at the end, the one line that says even mass is an entry in the book.

60 minutesfive benches, three checkpoints
VoicesLewin's pendulum, Feynman's reversible machine, Joule's paddle wheel
You needModules 1, 2, 3
Bench 1

Lewin puts his life on the line

A 15 kg steel ball hangs from the ceiling on a long cable. Walter Lewin stands against the wall, pulls the ball back until it touches his chin, and lets go. It swings across the lecture hall and comes back — toward his face. He does not flinch. He has done this every year for thirty years.

W = F·s  (force along the motion × distance moved)    Wnet = ½mv² − ½mu²F is the force on the object and s the distance it moves along that force. The work-energy theorem, and the joule (1 J = 1 N·m) it is measured in. Module 3 got it from the third equation of motion: multiply v² = u² + 2as through by ½m. It is the bridge between force and energy, and every column in every ledger below is a special case of it.

What is the one thing he must not do when releasing the ball?

Bench 1 · The pendulum and the chinthe two bars are the ledger: kinetic and potential
Release with zero push first. Then give it 0.5 m/s and watch where it returns to.

The ledger has two columns. The gravitational column holds mgh: largest at the top of the swing, zero at the bottom. The kinetic column holds ½mv²: zero at the top, largest at the bottom. As one empties the other fills, and their total never moves. So the ball can only ever climb back to the height at which its kinetic energy was last zero — the height of release, if he released it from rest. His chin is safe for one reason only: he gave the ball no kinetic energy at the start, so there is none left over to carry it past his chin on the way back.

mghchin + 0 = ½mv²bottom + 0 = mghreturn + 0The mass cancels from every term, which is why it does not matter whether the ball is 15 kg or 15 tonnes — and why Lewin can do the demonstration with any ball that happens to be in the room.

Notice what the ledger is silent about. It says nothing about the path: how long the cable is, how the ball speeds up and slows, what the tension is at the bottom. Force-based analysis would need all of that. Energy skips it. That is the practical reason physicists reach for energy first — it answers "how high" and "how fast" without asking "how, exactly."

Bench 2

Feynman's reversible machine: where mgh comes from

Every textbook hands you mgh. Feynman refuses to. In Chapter 4 he derives it from a single assumption — perpetual motion is impossible: no machine can go on lifting weights forever without being fed anything. (The Earth orbits forever, but it lifts nothing.) That one assumption, and nothing else, gives the whole of energy — and the derivation is the cleverest thing in the first hundred pages.

Imagine a lifting machine: a lever, a pulley, anything. You lower a 1 kg weight by some height and, in doing so, raise a 3 kg weight. The machine is reversible if you can run it backward — lower the 3 kg and lift the 1 kg back — with no friction and no other change. The question: if the 1 kg drops one metre, how high does a reversible machine raise the 3 kg?

Suppose some reversible machine could raise the 3 kg by more than one-third of a metre — say by 0.4 m. What follows?

So no reversible machine can raise 3 kg by more than ⅓ m per metre of 1 kg dropped. And none can raise it by less either: if one only managed 0.3 m, run it backwards to lower the 3 kg by 0.3 m and lift the 1 kg back to 1 m, then drop that 1 kg through a machine that does give the full ⅓ m — the 3 kg ends 0.033 m higher than before, from nothing. Free lifting again, which the impossibility already ruled out. Every reversible machine therefore gives exactly ⅓ m. The ratio ⅓ is 1 kg / 3 kg, and the same argument for any two weights gives the same shape of answer: dropping mass m₁ by h₁ can lift mass m₂ by h₂ only if m₁h₁ = m₂h₂. The quantity weight × height is what a reversible machine conserves.

Σ mi g hi = constant   (for any reversible rearrangement of weights)Σ is the Greek capital sigma; it means "add up over every weight in the machine — the first, the second, and so on". What the argument itself fixes is mass × height. Multiply that by g, the weight per kilogram at Earth's surface (about 9.8 N/kg), and mass × height becomes weight × height — force × distance — which is work in joules. That is mgh: gravitational potential energy, derived from "you can't get something for nothing" and nothing else.

Real machines are not reversible: friction means they lift less than ⅓ m, never more. That is not a violation — the missing height is exactly accounted for by the heat in the bearings, which is Bench 5. Feynman's point is this. Every real machine falls short of ⅓ m, and each falls short by a different amount depending on how well it is oiled. But no machine ever exceeds it. So ⅓ m is not a property of any particular machine — it is a ceiling the world imposes on all of them, and that ceiling is what we name and measure. That is what we mean when we say the falling 1 kg has a certain energy: it is the most that fall could ever buy.

Why this is the right way round

Notice the direction of the logic. We did not start from "energy is conserved" and deduce that perpetual motion is impossible. We started from the impossibility of perpetual motion — an experimental fact, tested by every failed inventor in history — and deduced what quantity must be conserved. The law was discovered by people asking "what is it that a machine can never increase?" The same question, asked about heat engines in the 1820s, produced thermodynamics; asked about chemical reactions, produced chemical energy; asked about nuclear reactions, produced E = mc², where E is the total energy the mass would release if it disappeared entirely. It is the most productive question in physics, and it is Dennis's mother's question: "where did the blocks go?"

Bench 3

The loop-the-loop: energy meets the banked bend

A cart rolls down a track from height h into a vertical circular loop of radius r. Frictionless. How high must the start be for the cart to get round?

At the top of the loop the cart is upside down — the track is above it. A track can push a cart, but it cannot hold on and pull. So at the top, gravity (with any push the track adds) has to bend the cart's path into the circle. Module 1's Bench 4 gave the price of that turn: going round a circle of radius r at speed v needs an inward acceleration of v²/r. If v²/r is less than g, gravity alone is already bending the path more sharply than the circle needs, and the cart curves inside the track — it leaves the rails and falls. So the cart has to be moving fast enough at the top to keep the turn tight. Work out that minimum speed, and then let energy tell you the start height that gives it.

Loop radius 1 m. The minimum starting height for a frictionless cart to complete the loop is:

Bench 3 · Loop-the-looploop radius 1 m
at the top:  N + mg = mv²r,  where N is the normal push of the rail on the cart, and N ≥ 0  ⟹  v²top ≥ gr
energy:  mgh = mg(2r) + ½mv²top  ⟹  h ≥ 2r + ½r = 2.5r Two laws in one problem: energy tells you the speed at any height, and Newton's second law in the radial direction tells you whether that speed is enough. Neither alone can answer the question.

In the wild

Real roller-coaster loops are not circles — they are teardrops, tighter at the top and wider at the bottom. A circular loop entered fast enough to survive the top gives a punishing 6g at the bottom; the teardrop's larger bottom radius eases that to about 4g while the tighter top keeps the cart pressed to the rails. The shape is called a clothoid, and it was worked out in the 1970s by an engineer who understood this bench. Every modern loop is one.

Bench 4

The bungee cord: a third column in the ledger

A stretched spring stores energy. Hooke's law says its force is kx; the work to stretch it is the area under that line, a triangle: ½kx². That is a third column, and it lets the ledger handle a bungee jump.

You stand on a platform H metres above the river (where H is the platform's height above the water), tied to a cord of natural length L₀ — L₀ is that slack length, before any stretch. You fall L₀ metres freely — the cord is slack — and then it begins to stretch. At the lowest point you are momentarily at rest. All the potential energy you have lost is now in the cord.

Bench 4 · The jumpfour columns now: gravitational, kinetic, elastic, thermal
Jump at 60 kg and note the lowest point. Now double your mass to 120 kg — does the lowest point double? Then halve the stiffness k instead and see which of the two matters more.
mg(L₀ + x) = ½kx²  ⟹  x = mg + √(m²g² + 2kmgL₀)kx is the extension — how far the cord has stretched beyond L₀. A quadratic, because the elastic column goes as x². The bungee company's entire job is to make sure L₀ + x stays comfortably less than H for every customer. Double the jumper's mass and x does not double — the square root in the formula sees to that — which is why they weigh you first.

Look at what the ledger gives you for free. The lowest point does not depend on how you fell — head-first, spread-eagled, screaming — only on the numbers in the equation. And the maximum force on you, kx at the bottom, comes out of the same calculation: for a 60 kg jumper on the default cord it is roughly three times your weight. The elastic column also explains why the ride ends: on every bounce, a little energy leaks into the cord's internal friction as heat, the amplitude shrinks, and you end up dangling. Where did the energy go? Bench 5.

Bench 5

Joule's bucket, a brake disc, and the Sun

Until 1843 heat and motion were thought to be different things. James Joule, a brewer's son who did physics in his spare time, hung weights from a pulley, let them fall, and used the falling to turn a paddle wheel inside an insulated bucket of water. The water warmed. He measured how much, and found that a fixed quantity of weight-times-height always produced a fixed rise in temperature. Mechanical energy was becoming heat at a fixed exchange rate. The blocks had been found in the bathwater.

Insulated is the word doing the work in that description. No heat could cross the bucket's wall from outside — the only way energy got in at all was the work the falling weight did on the paddle. Whatever left the gravitational column of the ledger therefore had to arrive somewhere, and the thermometer found it as a rise in the water's internal energy. Write that as a rule for any system with a boundary, not just this one bucket, and it is the first law of thermodynamics:

ΔU = Q + WU is the system's internal energy — the sum of the microscopic kinetic and potential energies of its molecules — so ΔU is how much of it changed. The change equals the heat added to it plus the work done on it. Joule's bucket is the special case Q = 0 — insulated, so no heat crosses the boundary at all — and every joule the falling weight's work delivers shows up as ΔU, exactly the temperature rise his thermometer caught. Module 8's engines and fridges are the general case, where Q and W are both doing something at once, in opposite directions round a cycle.
Bench 5 · Exchange ratesthree audits, one currency
Run Joule's drop first and read the temperature rise — it is tiny. Now brake the car from 100 km/h and read the energy dumped into the discs. Ask yourself how many of Joule's 10 kg drops that one stop is worth.

Joule's number — about 4,186 joules to raise a kilogram of water by one degree — is why heat is so hard to notice and so easy to underestimate. Dropping 10 kg through 2 m barely warms a litre. But a car's kinetic energy, dumped into four brake discs in a few seconds, warms them by fifty degrees in one stop — and by several hundred down a long alpine descent, which is why racing discs glow. The exchange rate is fixed; only the amount changes.

One warning about the money picture, because this is where it stops being honest. Rupees change into dollars, and dollars back into rupees, all day. The thermal column is not like that. Motion turns into heat completely and by itself — the brake disc heats up on its own — but heat does not turn back into motion by itself, and never all of it. The ledger balances; the conversions are not all two-way. Why that is so is Module 8.

Before reading the summary table below, one gloss on the thermal row. mcΔT means: m is the mass being warmed, ΔT is how many degrees it warms by, and c is that substance's own number — the joules needed to raise one kilogram of it by one degree. For water c is Joule's 4,186; for steel it is about fifteen times smaller, which is why the brake disc glows and the water does not.

ColumnFormulaWhere you saw the blocks hide
kinetic½mv²the pendulum at the bottom of its swing
gravitationalmghthe pendulum at the top; the reversible machine
elastic½kx²the bungee cord at its lowest point
thermalmcΔTJoule's water; the brake disc; the bungee cord after ten bounces
chemicalbond energiesthe skater's muscles in Module 2; petrol; your breakfast
massmc²the Sun, below

Every arrow drawn in this module points the same way: ordered energy — height, speed, stretch — turns into the thermal column, and never all the way back on its own. The ledger says the total is fixed; it says nothing about which column the total is in. That second question — what gets used up when nothing is lost — is the second law of thermodynamics, and Module 8 will run its audit.

The last column

In 1905 Einstein added one more entry, and it is the strangest. Mass itself is a form of energy, at the exchange rate c², which is about 9 × 10¹⁶ joules per kilogram. The Sun radiates 3.8 × 10²⁶ watts — a watt is one joule every second, so that is 3.8 × 10²⁶ joules of light every second. Divide by 9 × 10¹⁶ and each second the Sun is losing four million tonnes of its own mass, converted to light by nuclear fusion in its core — and has been for four and a half billion years, and has barely noticed. The blocks were in the mass all along. A gram of anything, fully converted, is about 25 gigawatt-hours — roughly ten hours of everything Chennai plugs in.

Is the ledger ever wrong?

Not so far, but it has had close calls. In 1930 radioactive beta decay appeared to lose energy: the electrons came out with less than the books said they should. Niels Bohr was prepared to abandon conservation of energy. Wolfgang Pauli instead proposed an invisible particle carrying the missing amount — a "desperate remedy," he called it — and the neutrino was detected 26 years later. Every time the ledger has failed to balance, the resolution has been a new column, never a broken law. Whether that is a fact about the universe or about how physicists prefer to think is a question worth carrying around for a decade or two.

Bench 5, continued

The ledger, written as one rule

Five benches, one pattern each time: something falls while something else changes, and a total stays fixed. Write the pattern down properly — the way Bench 2's reversible-machine argument earns it, rather than just asserting it — and it reads:

ΣEi = constant   (kinetic + gravitational + elastic + thermal + chemical + … , for an isolated system)"Isolated" is carrying all the weight in that sentence: no work done on the system from outside, no heat crossing its boundary. The pendulum's ledger only balances because the cable, the bearing and the air are (nearly) not taking energy out. Let heat leak across the boundary, or let something outside do work on the system, and the sum genuinely changes — by exactly the amount that crossed, which is the first law again, generalised past Joule's one bucket to any system at all.

A worked example, by hand rather than by slider, to show the method and not just an answer:

A 2 kg ball is dropped from 5 m. How fast is it moving when it hits the ground?

  1. Name the system, and check it is isolated. Ball plus Earth's gravity; no air resistance is mentioned, so treat the pair as isolated — nothing does work on it from outside, and no heat crosses in or out.
  2. List every column at the start. At the top, released from rest: kinetic energy is 0; gravitational potential energy is mgh = 2 × 9.8 × 5 = 98 J. Nothing else is present, so those two columns hold the entire 98 J.
  3. List every column at the end. Just before impact, height is 0, so gravitational potential energy is 0. By the rule above, the 98 J has not gone anywhere — it must now all be kinetic energy, ½mv².
  4. Set the two sums equal and solve. ½ × 2 × v² = 98  ⟹  v² = 98  ⟹  v ≈ 9.9 m/s.

Check it the kinematics way, as a cross-check rather than a second method: v² = u² + 2as = 0 + 2(9.8)(5) = 98 — the same 98, and therefore the same 9.9 m/s. The energy method never needed the word "acceleration", the time of flight, or even the path taken, which is exactly why Bench 1 reached for it first.

Checkpoint

Three questions

1. Two balls of equal mass are thrown from a cliff with the same speed, one straight up and one straight down. Ignoring air, which hits the ground with more kinetic energy?

2. A block slides down a rough ramp and arrives at the bottom moving slower than energy conservation with mgh alone would predict. The kinetic energy that is 'missing' has:

3. You hold a 10 kg box stationary at arm's length for a full minute, and your arms burn with effort. How much work, in the physics sense, have you done on the box?

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