The floor that does no work

The observation that kinetic energy depends on the observer finds that kinetic energy is a number about the observer: a train has energy in one frame and none in another, and what the conservation laws constrain is not how much there is but how much changes.

There is a second thing the accounting does not mean, and it is nearer home. Consider a person standing on a floor who jumps.

The jumper leaves the ground with kinetic energy. Something supplied it. The only external forces acting are gravity, which is pulling downward and therefore taking energy away, and the floor’s normal force, which is pushing upward.

So the floor supplied it. That is the natural conclusion, it is what most people say, and it is wrong.

Work is a force times the displacement of its own point of application #

Work is defined at the point where the force acts. If a force $\mathbf{F}$ is applied at a point, and that point moves through $d\mathbf{r}$, the work is $\mathbf{F}\cdot d\mathbf{r}$ — with $d\mathbf{r}$ being the displacement of the point the force is applied to, not of anything else.

The floor’s force is applied to the soles of the feet. The soles of the feet are in contact with the floor and they do not move while the contact lasts. They move at the instant the contact ends, by which time the force is zero.

So the floor’s force acts through zero displacement, and its work is exactly zero. No energy crosses the boundary between the floor and the jumper. The floor is not a source of energy, it is not warmed or cooled by the jump, and a jump on a perfectly rigid floor leaves the floor exactly as it was.

What supplied the kinetic energy is chemical energy stored in the muscles. It was inside the jumper before the jump and is inside the jumper’s kinetic energy afterwards, and it never crossed a boundary at all.

The integral that is not work, and is still correct #

That leaves an arithmetic puzzle. Newton’s second law is $\mathbf{F}_{\text{net}} = M\mathbf{a}_{\text{cm}}$ for any body whatever, rigid or not — the centre of mass moving as though it were the whole thing. Dot both sides with the centre of mass’s displacement and integrate, and out comes

$$\Delta\left(\tfrac12 M v_{\text{cm}}^2\right) = \int \mathbf{F}_{\text{net}}\cdot d\mathbf{r}_{\text{cm}}.$$

That is a true equation, it is exact, and its right-hand side for the jump is not zero. The floor’s force integrated over the centre of mass’s rise gives 293 joules, which is exactly the kinetic energy at take-off. The arithmetic works.

It works because it is not a statement about energy. The equation is Newton’s second law with both sides multiplied by a displacement; the right-hand side is a force dotted with the wrong displacement — the centre of mass’s rather than the point of application’s — and the quantity it computes is sometimes called pseudowork, or the centre-of-mass work, to keep it apart from the real thing.

Two equations, both exact, that happen to look alike:

• The centre-of-mass equation, $\Delta KE_{\text{cm}} = \int \mathbf{F}_{\text{net}}\cdot d\mathbf{r}_{\text{cm}}$, which follows from Newton’s second law and says nothing about where energy went.
• The first law of thermodynamics, $\Delta U + \Delta KE = W + Q$, which is conservation of energy and accounts for every joule.

For a particle they are the same equation, because a particle has no interior, no shape, and one displacement. For anything that deforms they are different, and the difference between them is the change in internal energy.

For the jump: the first says 293 joules of centre-of-mass work produced 293 joules of kinetic energy. The second says zero external work plus zero heat produced 293 joules of kinetic energy and therefore minus 293 joules of internal energy — the chemical energy spent. Both are right and only the second is about energy.

Where it matters most: a collision #

The cleanest case is one where the external force is zero altogether, so there is nothing to argue about.

Here the centre-of-mass equation is trivially satisfied: no external force, so no change in the centre of mass’s velocity, so no change in its kinetic energy. And that is true — the centre of mass of the pair travels at one metre a second before and after.

The kinetic energy of the system is not the kinetic energy of the centre of mass. The system had two carts with different velocities, and the energy of the relative motion is what vanished. The centre-of-mass equation never knew about it, because the centre-of-mass equation only ever tracks one degree of freedom.

Where the joule went is not mysterious and it is not accounted for by any equation of motion: it is in the deformation of the coupling, the sound of the impact and the warmth of both carts — warmth being a change of internal energy and not necessarily of temperature. It is internal energy, it is what entropy counts, and the equation that keeps track of it is the first law.

Friction’s two works, which do not cancel #

The place the distinction does the most useful work is the one place everybody has to get it right eventually: how much heat sliding friction makes.

Put a block on a table and let it slide to a stop. Friction acts backwards on the block through the block’s displacement $d$, so it does $-fd$ of work on the block, and the block loses $fd$ of kinetic energy. The heat generated is also $fd$. The two agree, and it looks as though the work done by friction is the heat.

It is not, and the case that shows it is two blocks sliding on each other with both free to move. Friction acts backwards on the upper block through its displacement $d_1$, and forwards on the lower one through its displacement $d_2$. The work done on the upper is $-fd_1$; on the lower, $+fd_2$. Those do not cancel. Their sum is

$$-f(d_1 - d_2) = -f\,d_{\text{rel}},$$

and the heat generated is $f d_{\text{rel}}$ — the friction force times the relative sliding, not the displacement of either body.

Put numbers on it. A one-kilogram block at three metres a second sliding on a two-kilogram block that is free to move, with a coefficient of 0.3 between them and a frictionless floor beneath: they end at one metre a second together, the kinetic energy falls from 4.5 joules to 1.5, and three joules become heat. The upper block travelled further than the lower, by exactly enough that $f d_{\text{rel}}$ is three joules. Neither block’s own displacement gives that number.

The general statement is worth keeping. A force between two bodies that move differently does two different amounts of work, the difference is what is dissipated, and no single displacement gives it. A force between two bodies that move together — a normal force, a rolling contact, a rigid joint — does two works that cancel exactly, dissipates nothing, and can be ignored in an energy ledger for that reason.

That is also why an inelastic collision, a slipping clutch and a brake all have the same arithmetic underneath: in each case two surfaces move relative to one another while a force acts between them, and the product is the heat.

The force that accelerates everything and transfers nothing #

The same distinction settles a question that comes up whenever a body pushes off the world.

A car accelerates because the road pushes it forward. The road’s push is static friction at the contact patch of each driven tyre, and a rolling tyre’s contact patch is instantaneously at rest — that is what rolling without slipping means. A force applied at a point that is not moving does no work.

So the road supplies no energy to the car. The fuel does. The road supplies the external force without which the momentum could not change at all, and the two requirements are separate and both binding:

To change the momentum, an external force is needed. Nothing internal can do it — which is why a collision is easier to analyse than the forces inside it — that is the content of conservation of momentum, and it is why a car in space cannot accelerate by spinning its wheels however much fuel it burns.

To change the kinetic energy, an energy source is needed. The external force need not be the source, and in every case here it is not.

The same reading covers a sprinter, a person walking, a rocket pushing on nothing at all, and the crossover at which a car stops being grip-limited and starts being power-limited — which is exactly the crossover between the first requirement binding and the second.

Why the two equations were separated so late #

The distinction has a literature and it is surprisingly recent — the term pseudowork dates from the late 1970s, and papers arguing that introductory textbooks conflate the two equations were still being published decades later.

The reason for the delay is that for most of what mechanics was used on, the two agree. A planet, a projectile, a pendulum bob, a block on an incline: all of them can be treated as particles or as rigid bodies whose contact points do not slip, and for every one of those the centre-of-mass equation and the work-energy theorem are the same statement. The cases where they differ are bodies that change shape while forces act on them — a jumper, a car, a colliding pair, a person walking — and those entered physics teaching much later than the planets did.

There is also a naming problem that keeps the confusion alive. The centre-of-mass equation is often written with the same symbol $W$ on the right, and it is often called the work-energy theorem, because for a particle it is one. A student who learns it that way and then meets a jumper has been handed an equation that gives the right number and a name for it that gives the wrong account.

The clean way to hold it is to notice which quantity each equation is about. Newton’s second law is about momentum; integrating it gives a statement about the centre of mass’s kinetic energy, which is one number derived from the momentum and is not the system’s energy. The first law is about energy, it has a term for what is stored inside, and it is the only one of the two that can say where anything came from.

That is the same separation the magnetic force makes in a different setting: a force that changes a momentum without transferring any energy is not a contradiction, it is what a force perpendicular to a velocity does, and the bookkeeping has room for it.

The skater, the wall and the rocket #

Three more cases, quickly, because together they show that the pattern is general rather than a curiosity about floors.

A skater pushes off a wall and glides away with kinetic energy. The wall’s force acts on hands that are in contact with the wall and not moving, so the wall does no work; the energy is the skater’s own. A wall is a perfect supplier of momentum and a perfect non-supplier of energy, which is exactly what a rigid immovable object is for.

A person walking on level ground is doing no net work against anything, and is tired at the end of it. Nothing external moves, no height is gained, nothing is lifted — and the metabolic cost of walking a mile is real and measurable. It goes into repeatedly accelerating and decelerating the limbs and into the elastic losses of muscle and tendon doing that, all of it internal, none of it work in the mechanical sense. The centre-of-mass equation returns nearly zero for a mile of level walking and is right, and it is the wrong equation for the question.

A rocket has no external body to push on at all, so neither equation has a contact to argue about. Its momentum changes because it throws mass backwards, and its kinetic energy comes from the chemistry in the tank. Here the two requirements — an external force for momentum and an internal source for energy — are met by the same act, which is why a rocket seems to escape the argument and why it is cleanest to treat as pure momentum bookkeeping.

What the three have in common is that the body supplying the force and the body gaining the energy are different bodies, and nothing in the equations of motion requires them to be the same.

A contact patch creeps and a floor deflects #

A real contact patch is not perfectly at rest. A tyre creeps: the rubber deforms and the contact slips by a per cent or two of the vehicle speed even when it is described as rolling without slipping, so the friction force does a small negative work and that work is dissipated as heat in the tyre. The ideal statement is that the work is zero; the honest one is that it is small and negative, and it is one of the several losses that make a real vehicle less efficient than its engine — the same creep that makes a tyre’s grip a response rather than a coefficient.

The floor is not rigid either. A real floor deflects under a jumper by a fraction of a millimetre, so the point of application does move and the floor does do a small amount of work — on a trampoline, a great deal of it, which is the whole point of a trampoline. The statement that the floor does no work is exact for a rigid floor and is a very good approximation for concrete.

And the “centre of mass” is doing more work than it looks. For a rotating body the centre-of-mass equation tracks the translation and says nothing about the rotation, which has its own energy and its own equation. A wheel rolling down a slope has to be accounted for twice, and using the centre-of-mass equation alone gives the wrong answer for the speed at the bottom by the fraction of the energy that went into spinning.

Which displacement, and where the boundary is drawn #

They cannot show the point of application. Every figure here plots a force against a time or a distance, and the whole argument turns on which distance — the centre of mass’s or the contact’s — and the two are different curves for the same force. A plot of force against the centre of mass’s displacement looks exactly like a plot of force against the point of application’s, and the area under one is 293 joules while the area under the other is nothing.

Nor can they show the internal energy going. The jumper’s chemical energy falls by rather more than 293 joules, because muscle is around a quarter efficient and the rest is heat — so the honest ledger has about twelve hundred joules of chemistry becoming three hundred of motion and nine hundred of warmth. None of that is visible in a force-time curve.

And they cannot show that the ledger is a choice of boundary. Whether the floor does work on the jumper depends on where the boundary between them is drawn, and drawing it at the skin gives zero. Drawing it around the jumper and the floor makes the floor’s force internal and the question does not arise. That freedom is not a weakness; it is how the first law is always used, and the only requirement is to say where the boundary is before counting what crosses it.

Still open: how much of a step is elastic return #

A running human stores energy in the tendons on landing and returns it on the next push, and how much is genuinely elastic rather than being re-supplied by muscle is measured indirectly and argued about. The Achilles tendon is the largest store; estimates of the fraction of the stride’s energy it returns run from a third to over a half, and they depend on the method — on force plates and motion capture inferring what a tendon did, rather than on measuring the tendon.

It matters here because an elastic return is a case where the boundary is genuinely ambiguous. A tendon is internal to the runner, so its energy never crosses any boundary and does no work by the definition used above; and functionally it behaves exactly like a spring in the ground would. Whether to model it as part of the body or as part of the environment is a decision about the accounting rather than about the physics, and different laboratories make it differently, which is one reason the published fractions differ as much as they do.

The habit worth carrying away is the one about equations that look alike. Two exact equations can have the same shape and answer different questions, and the way to tell them apart is to ask what each displacement belongs to. Dot a force with the centre of mass’s motion and get a statement about the centre of mass; dot it with its own point of application and get a statement about energy. Neither is wrong and only one of them is about where the energy came from.