Work, Energy and Power

Beginner

In everyday speech, holding a heavy suitcase for ten minutes is hard work. In physics it is no work at all, because work has a precise meaning: a force does work when it moves something along the direction it pushes or pulls. Work = force × distance, measured in joules (J). Push a trolley with 10 N for 3 m and you do 30 J of work.

Doing work on something gives it energy, the ability to do work. Two kinds matter here. A raised object has height energy (gravitational potential energy): weight × height. Lift a 20 N bag of flour 1.5 m onto a shelf and it stores 30 J. A moving object has motion energy (kinetic energy), $\frac{1}{2}m{v}^2$. Because the speed is squared, a car at twice the speed carries four times the energy.

Energy is never created or destroyed; it only changes form. Let a ball fall: it loses height energy and gains exactly as much motion energy. When it lands, the motion energy squashes the ball and the floor for an instant, and part of it becomes heat and sound, so the ball never bounces back to where it started.

Power tells you how fast work is done, in watts: one watt is one joule per second. Two people carrying the same box up the same stairs do the same work; the faster one develops more power.

Check yourself: you carry a box across a level room at a steady speed. How much work does your upward push do on it? Answer: none, because the box moves sideways, at right angles to your push.

In the animation, the blue "height" bar shrinks and the gold "motion" bar grows as the ball falls, while the green total stays the same.

Intermediate

Work: the force along the displacement

Holding a heavy suitcase still is tiring, yet the hand holding it does no work on it. In physics a force does work only if the object moves while the force acts, and only the part of the force along the motion counts. For a constant force $F$ and a straight displacement $d$ at an angle $\theta$ to the force,
$W=Fd\cos \theta$

Work is measured in joules, $1\mathrm{J}=1\mathrm{N}\cdot \mathrm{m}$, about the work needed to lift a 100 g apple by one metre. It is a scalar: it has no direction, but it has a sign.

The angle between the force and the displacement sets that sign:

• Along the motion, $\theta =0^{\circ }$ and $W=Fd$: the force feeds energy into the object.
• At right angles, $\theta =90^{\circ }$ and the force does no work: the floor under a sliding box, or a string whirling a stone in a circle.
• Against the motion, $\theta =180^{\circ }$ and $W=-Fd$: friction and brakes take energy out of the motion.

When several forces act, each does its own work, and the net work is their sum.

Kinetic energy and the work–energy theorem

Push a cart of mass $m$ with a constant net force $F$ over a distance $d$. With $a=F/m$, the kinematics relation $v^2=v_0^2+2ad$ multiplied by $m/2$ becomes $Fd=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2$. What the work changes is the kinetic energy,
$KE=\frac{1}{2}m{v}^2$
and the net work done by all the forces equals its change. This is the work–energy theorem:
$W_{\mathrm{net}}=\Delta KE$

Positive net work speeds an object up and negative net work slows it down. Kinetic energy is never negative, ignores the direction of motion and grows with the square of the speed.

Potential energy and the conservation of energy

When a ball of mass $m$ goes from height $h_1$ down to $h_2$, its weight $mg$ does work $mg(h_1-h_2)$ whatever the path, since only the vertical part of each small step lines up with the weight. Work that depends only on the end points can be stored and given back; this stored energy is the potential energy:
$PE=mgh$

The height $h$ is measured from a level you choose, because only changes of $PE$ matter. Gravity's work is positive going down, where $PE$ falls, and negative going up.

If gravity, or an ideal spring, is the only force doing work, every joule that leaves $PE$ turns up in $KE$, and the mechanical energy stays the same:
$KE+PE=\text{constant}\text{(no friction or drag)}$

So an object dropped from a height $h$ lands at $v=\sqrt{2gh}$, from $mgh=\frac{1}{2}m{v}^2$. The mass cancels, and on a frictionless slide the shape of the track does not matter either: the track pushes at right angles to the motion and does no work.

Worked example: the falling ball in the animation

The 2 kg ball is released from rest at $h_0=18\mathrm{m}$. With the ground as $h=0$, all of its energy is potential at the start:
$E=mg{h}_0=2\cdot 9.81\cdot 18\approx 353\mathrm{J}$

Halfway down, at $h=9\mathrm{m}$, half of it has changed form: $PE\approx 177\mathrm{J}$ and $KE\approx 177\mathrm{J}$, so $v=\sqrt{2KE/m}\approx 13.3m/s$. Just before landing $PE=0$ and $KE=E$:
$v=\sqrt{2g{h}_0}=\sqrt{2\cdot 9.81\cdot 18}\approx 18.8m/s$

These are the values on the $PE$, $KE$ and $v$ readouts, and the final frame, the instant just before landing, shows $KE=E$ and $PE=0$. All the way down, the blue $PE$ bar and the gold $KE$ bar add up to the green $E$ bar, while the velocity arrow $v$ grows.

The fall lasts $\sqrt{2h_0/g}\approx 1.92\mathrm{s}$, yet the halfway height is passed only at 1.35 s. At half the time, 0.96 s, the ball has dropped just 4.5 m, and only a quarter of the energy, 88 J, has changed form. Energy is shared out by height, not by time, and the ball covers more height every second: its faint images, 0.4 s apart, spread ever further apart.

Worked example: pulling a loaded sled

A child pulls a 25 kg sled from rest for 20 m across level snow. The rope is at $30^{\circ }$ to the ground with a tension of 60 N, and friction on the runners is 30 N. How fast is the sled going at the end?

The rope does $60\cdot 20\cdot \cos 30^{\circ }\approx 1039\mathrm{J}$ of work, and friction, pointing backwards, does $30\cdot 20\cdot \cos 180^{\circ }=-600\mathrm{J}$. The weight and the push of the snow are vertical, at right angles to the motion, so they do no work. The net work is 439 J, and from rest the work–energy theorem gives
$v=\sqrt{\frac{2W_{\mathrm{net}}}{m}}=\sqrt{\frac{2\cdot 439}{25}}\approx 5.9m/s$

Newton's laws agree: the forward part of the pull, $60\cos 30^{\circ }\approx 52\mathrm{N}$, leaves a net force of 22 N, so $a\approx 0.88m/{\mathrm{s}}^2$ and $v^2=2\cdot 0.88\cdot 20\approx 35{\mathrm{m}}^2/{\mathrm{s}}^2$. The energy route needed neither the acceleration nor the time.

Power, and reading energy graphs

Power is the rate at which work is done, measured in watts, $1\mathrm{W}=1J/s$:
$P=\frac{W}{t}$
For a force along the motion at speed $v$, the work in a short time $\Delta t$ is $Fv\Delta t$, so $P=Fv$. At the end of the pull the rope delivers $52\cdot 5.93\approx 308\mathrm{W}$, but only 154 W on average over the 6.7 s, since the sled started from rest.

On a graph of force against distance, the area under the line is the work: a rectangle $Fd$ for a constant force, a triangle $\frac{1}{2}k{x}^2$ for a spring whose force grows with the stretch, $F=kx$. On a graph of energy against time, the slope is the power. Gravity's power on the falling ball, $mgv$, rises from zero to $2\cdot 9.81\cdot 18.8\approx 369\mathrm{W}$ at landing, so the gold bar fills slowly at first and then ever faster.

Common mistakes

• Work is not effort. Holding a 20 kg suitcase still for ten minutes does no work on it; your muscles burn chemical energy just to stay tense.
• Kinetic energy is not proportional to speed. From 15 m/s to 30 m/s a 1200 kg car's $KE$ goes from 135 kJ to 540 kJ, so with the same braking force it needs four times the braking distance.
• Work can be negative. Friction did $-600\mathrm{J}$ on the sled: that energy left the motion and reappeared as heat in the runners and the snow.
• A single value of $PE$ means nothing without its zero level. The ball has 353 J relative to the ground and 0 J relative to its starting point, yet the fall releases 353 J either way.
• Heavier objects do not land faster when only gravity works. A 4 kg ball dropped from 18 m carries 706 J, twice as much, but still lands at 18.8 m/s: the mass cancels.

Where you meet it

Roller-coaster designers start from $KE+PE$: a car released from rest 40 m above the lowest point can reach at most $\sqrt{2\cdot 9.81\cdot 40}\approx 28m/s$, about 100 km/h. Hydroelectric plants turn the potential energy of water into electricity, about 0.98 MJ for every tonne that falls 100 m.

Later lessons build on the same energy balance: Simple Machines: Levers, Pulleys and Gears (no machine gives out more work than it takes in), Elasticity: Hooke's Law, Stress and Strain (the energy in a stretched spring), Simple Harmonic Motion, Bernoulli's Principle, Electric Field and Potential, and First Law of Thermodynamics and Heat Engines, which adds heat to the account.

Advanced

Work as a line integral

The rule $W=Fd\cos \theta$ needs a constant force and a straight path. Real forces change from place to place and real paths bend, so we need a definition of work that copes with both. Everything else in this lesson is built on it.

Legend

$A$	Start of the path.
$B$	End of the path.
$d\vec{r}$	One short step along the path, of length $dr$.	$\mathrm{m}$
$\vec{F}$	Force on the particle at the start of the step.	$\mathrm{N}$
$\theta$	Angle between $\vec{F}$ and the step.	$\mathrm{rad}$
$F\cos \theta$	Component of $\vec{F}$ along the step: the only part that does work.	$\mathrm{N}$

Figure 1 shows a particle moving along a curved path from $A$ to $B$ while the force on it changes from point to point. The idea is to cut the path into steps so short that each one is practically straight and the force hardly changes along it. Over one such step the simple rule holds again.

Look at the enlarged step $d\vec{r}$: an arrow of length $dr$ pointing along the path. Where it starts, the force is $\vec{F}$, at the angle $\theta$ to the step. The small dots along the curve mark where other steps begin; each has its own force and its own angle.

Only the part of $\vec{F}$ along the step pushes the particle along it; the part at right angles only pushes it sideways. To find the part along the step, drop the dashed perpendicular from the tip of $\vec{F}$ onto the line of the step: the green bar it cuts off has the length $F\cos \theta$. The work done in this one step is that component times the length of the step:
$dW=F\cos \theta dr=\vec{F}\cdot d\vec{r}$

The dot product appears because it is exactly this: the component of one vector along the other, times the length of the other. The total work from $A$ to $B$ is the sum over all the steps. As the steps get ever shorter, the sum becomes the line integral
$W={\int }_A^B\vec{F}\cdot d\vec{r}={\int }_A^B\left(F_{x}dx+F_{y}dy+F_{z}dz\right)$
The second form writes the dot product out in components, which is how the integral is usually worked out.

Two results follow at once. For a constant force on a straight path, $\vec{F}$ comes out of the integral and the steps add up to the displacement, which gives back $W=Fd\cos \theta$. And dividing $dW$ by the time $dt$ the step takes, with $d\vec{r}/dt=\vec{v}$, gives the power the force delivers at each instant, $P=\vec{F}\cdot \vec{v}$.

Work is not something a body has. It describes a process: an amount of energy passed from whatever exerts the force to the body it acts on.

Deriving the work–energy theorem

The line integral becomes useful once it is tied to motion. Newton's second law makes the link, and the result, the work–energy theorem, says that the net work done on a particle equals the change in its kinetic energy $KE=\frac{1}{2}m{v}^2$.

Start with how $KE$ changes in time. The speed squared is the dot product of the velocity with itself, $v^2=\vec{v}\cdot \vec{v}$, and by the product rule its rate of change is $2\vec{v}\cdot d\vec{v}/dt$. For a particle of constant mass $m$, therefore,
$\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)=m\vec{v}\cdot \frac{d\vec{v}}{dt}={\vec{F}}_{\mathrm{net}}\cdot \vec{v}$
The last step is Newton's second law, $md\vec{v}/dt={\vec{F}}_{\mathrm{net}}$: the rate of change of $KE$ is the power of the net force.

Now integrate over time from start to finish. In a time $dt$ the particle takes the step $d\vec{r}=\vec{v}dt$, so the time integral of ${\vec{F}}_{\mathrm{net}}\cdot \vec{v}$ becomes a line integral along the path:
$\Delta KE=\int {\vec{F}}_{\mathrm{net}}\cdot d\vec{r}=W_{\mathrm{net}}$

Nothing in this derivation assumed anything about the force or the path. The theorem therefore holds for the net force on a particle whether or not that force has a potential energy.

Forces that do no work

Some forces act on a body all the time yet never change its speed. Spotting them simplifies energy problems, because they drop out of the balance.

Legend

$h$	Starting height above the floor: how far the ball drops.	$\mathrm{m}$
$m\vec{g}$	Weight of the ball, straight down.	$\mathrm{N}$
$\vec{N}$	Push of the track on the ball, at right angles to the track.	$\mathrm{N}$
$\vec{v}$	Velocity of the ball, along the track.	$m/s$

Figure 2 shows a ball that slides from rest down a frictionless slide of any shape, starting at the height $h$ above the floor. Two forces act on it: its weight $m\vec{g}$, straight down, and the push $\vec{N}$ of the track. A smooth track can only push at right angles to its own surface. The ball moves along that surface, so its velocity $\vec{v}$ points along the track. The small square marks the right angle between $\vec{N}$ and $\vec{v}$: at every instant $\vec{N}\cdot \vec{v}=0$, so the power of $\vec{N}$ is zero, and the push does no work however the track bends.

The same holds in general. Split any force into a part along the velocity and a part across it. Only the part along $\vec{v}$ changes the speed; the part across it only bends the path. The push of a frictionless track, the tension in a pendulum string and the magnetic force on a moving charge are all at right angles to $\vec{v}$, so none of them does work. In Figure 2 that leaves the weight alone to change the kinetic energy.

To find the work of the weight, let $y$ be the ball's height above the floor at any moment: $y=h$ at the start and $y=0$ at the end. The weight points straight down, so only the vertical part of $\vec{v}$, which is $dy/dt$, counts:
$m\vec{g}\cdot \vec{v}=-mg\frac{dy}{dt}=-\frac{d}{dt}\left(mgy\right)$
The minus sign appears because the weight points down while $y$ is measured upwards. Over the whole slide the weight therefore does the work $mgh$, which depends only on the drop, not on the shape of the track. The theorem then gives $\frac{1}{2}m{v}^2=mgh$: a frictionless slide of any shape delivers the same speed at the bottom, $v=\sqrt{2gh}$.

Read the equation for the power the other way round: the rate at which $KE$ grows is exactly the rate at which $mgy$ falls. With no other force doing work, the sum $\frac{1}{2}m{v}^2+mgy$ therefore stays constant along any path. Conservation of energy is not an extra assumption here: it follows from Newton's second law together with the form of the force.

What the energy graph shows

The Advanced view of the animation plots these energies against time for the lesson's falling ball, so the bookkeeping can be watched as it happens.

The ball, of mass $m=2\mathrm{kg}$, is released from rest at the height $h_0=18\mathrm{m}$ and falls freely. After a time $t$ its height is $h=h_0-\frac{1}{2}g{t}^2$ and its speed is $v=gt$. Putting these into $KE=\frac{1}{2}m{v}^2$ and $PE=mgh$ gives
$KE(t)=\frac{1}{2}m{g}^2{t}^2,PE(t)=mg{h}_0-\frac{1}{2}m{g}^2{t}^2$
The coefficient is $\frac{1}{2}m{g}^2\approx 96.2J/{\mathrm{s}}^2$. The gold $KE$ parabola opens upward and the blue $PE$ parabola is its mirror image. In the sum the $t^2$ terms cancel, so the green total stays flat at $mg{h}_0\approx 353\mathrm{J}$: $dE/dt=0$.

The slope of the $KE$ curve is the power that gravity delivers, $dKE/dt=m{g}^{2}t=mgv$, which is $\vec{F}\cdot \vec{v}$ for the weight. It is zero at release, about 184 W half-way through the time of the fall and $2\cdot 9.81\cdot 18.8\approx 369\mathrm{W}$ at landing. The fall lasts $\sqrt{2h_0/g}\approx 1.92\mathrm{s}$, so the average power is $353/1.92\approx 184\mathrm{W}$: half the final value, because $v$ grows in proportion to time. The slope of $PE(t)$ is exactly the opposite, which is how the graph says $W=-\Delta PE$.

Plotted against height instead of time, both energies would be straight lines, $PE=mgh$ and $KE=mg(h_0-h)$. The curves in time only reflect that the ball covers more height per second as it speeds up.

Conservative forces and potential energy

The work of the weight depended only on where the ball started and where it ended. Forces with this property are special: their work can be stored and recovered, and that store is potential energy.

A force is conservative if its work from a point $A$ to a point $B$ is the same along every path between them. Equivalently, its work around every closed loop is zero: go out along one path and back along another, and the two contributions cancel. Gravity, ideal springs and electrostatic forces are conservative. Because their work depends only on the end points, it can be written as the drop of a function of position, the potential energy $U$:
$W_{A\to B}=-(U_B-U_A)$
Only differences of $U$ appear, so $U$ is defined up to an additive constant: we are free to choose where it is zero.

For gravity there is also an argument, with no dynamics at all, for why weight times height is the right quantity to keep track of. Picture ideal machines (levers, pulleys) that lower one weight to raise another, and call a machine reversible if running it backwards undoes exactly what it did. Suppose some machine could lift a load higher than a reversible one for the same descent of the driving weight. Run it forwards and the reversible one backwards: the driving weight ends up back where it started, but the load ends up higher, for nothing. Repeated, that would be perpetual motion, which is never observed.

So no machine beats a reversible one, and all reversible machines do exactly the same. A balanced lever trades weight for height in exact proportion, so what they all keep unchanged is the total of weight times height.

Reading a potential energy graph

A graph of $U$ against position shows at a glance where a body can go, where it turns back and which way it is pushed, without solving any equation of motion.

Legend

$x$	Position of the body along the line.	$\mathrm{m}$
$U$	Potential energy of the body at position $x$.	$\mathrm{J}$
$E$	Total energy $KE+U$: constant, drawn as the level line.	$\mathrm{J}$
$KE$	Kinetic energy: the gap between the level $E$ and the curve $U$.	$\mathrm{J}$
$F$	Force $-dU/dx$ on the body at the marked point: it points downhill and is larger where the slope is steeper. Zero at the two points marked $F=0$.	$\mathrm{N}$
$x_1$	Left turning point, where $U=E$.	$\mathrm{m}$
$x_2$	Right turning point, where $U=E$.	$\mathrm{m}$

Figure 3 plots the potential energy $U(x)$ of a body that moves along a line, where $x$ is its position. The horizontal line is its total energy $E=KE+U$, which stays constant while only conservative forces do work. Since $KE=E-U$, the kinetic energy at each $x$ is the height of the gold gap between the level $E$ and the curve below it.

Kinetic energy cannot be negative, so the body can only be where the curve lies on or below $E$. The gap closes at the turning points $x_1$ and $x_2$, where $U=E$: there the body stops for an instant and turns back, so it shuttles between them like a marble in a bowl. The hill on the right rises above $E$, so a body with this energy cannot get over it.

The graph also gives the force. When the body moves by $dx$, the force does the work $Fdx$, and for a conservative force that work equals the drop in potential energy, $-dU$. Setting $Fdx=-dU$ gives
$F=-\frac{dU}{dx}$
and in three dimensions $\vec{F}=-\nabla U$. Because of the minus sign, on any graph of $U(x)$ the force points downhill. In Figure 3 the arrow $F$, drawn on the slope between the bottom of the well and $x_2$, points back towards the bottom; where the curve is steeper, the force is larger. Where the slope is zero the force vanishes, at the two points marked $F=0$.

Those two points are equilibria, but of different kinds. The bottom of the well is a minimum of $U$: a small displacement either way meets a force back towards it, so the equilibrium is stable. The top of the hill is a maximum: a small displacement meets a force that pushes the body further away, so the equilibrium is unstable. As a check, the spring potential $U=\frac{1}{2}k{x}^2$ gives back Hooke's law, $F=-kx$.

Friction, heat and the full energy balance

Not every force is conservative. Friction is the everyday exception, and including it completes the energy account.

The work of kinetic friction grows with the length of the path, and it is negative whichever way the body slides, because friction always opposes the sliding. A round trip therefore leaves negative work, not zero, so no potential energy can describe it. Split the net work into the part done by conservative forces, which is $-\Delta U$, and the rest, $W_{\mathrm{nc}}$ (non-conservative). The work–energy theorem, $\Delta KE=-\Delta U+W_{\mathrm{nc}}$, then becomes
$\Delta KE+\Delta U=W_{\mathrm{nc}}$
$W_{\mathrm{nc}}$ collects friction and drag, which make it negative, and also positive contributions such as the push of an engine or of muscles.

Here is the effect in numbers. A 2 kg block released from the animation's 18 m height slides down a straight ramp at the angle $\alpha$ to the horizontal, with the coefficient of kinetic friction $\mu =0.2$. The normal force is $mg\cos \alpha$, so friction is $\mu mg\cos \alpha$, and it acts along the whole ramp, of length $h_0/\sin \alpha$. Their product is the work of friction:
$W_{\mathrm{nc}}=-\mu mg\cos \alpha \cdot \frac{h_0}{\sin \alpha }=-\frac{\mu mg{h}_0}{\tan \alpha }$
On a $30^{\circ }$ ramp this is $-122\mathrm{J}$, which leaves $353-122=231\mathrm{J}$ of kinetic energy and a speed of 15.2 m/s at the bottom. On a $60^{\circ }$ ramp it is only $-41\mathrm{J}$, and the block arrives at 17.7 m/s. Same drop, different paths, different work: the signature of a non-conservative force.

The energy is not destroyed. The rubbing sets the atoms of both surfaces jiggling, and that random motion is thermal energy. At the level of atoms the forces are conservative; friction only turns ordered motion into disordered motion. A block sliding a distance $d$ on a level floor, pressed on it by the normal force $N$, produces $\mu Nd$ of thermal energy. Counting it, energy is conserved exactly: this is the first law of thermodynamics seen from mechanics. The heat never gathers back into the block's motion by itself; the second law of thermodynamics describes this one-way change.

Worked problem: a ball landing on a spring

This problem puts gravity and a spring into one energy balance, and shows how large a force it takes to stop a falling body over a short distance.

Legend

$m$	Mass of the ball.	$\mathrm{kg}$
$k$	Stiffness of the spring.	$N/m$
$h_0$	Height of the release point above the top of the spring.	$\mathrm{m}$
$x$	Compression of the spring at the lowest point of the fall.	$\mathrm{m}$
$h_0+x$	Total distance fallen from release to the lowest point.	$\mathrm{m}$

Figure 4 sets it up. The animation's ball, of mass $m=2\mathrm{kg}$, is released from rest a height $h_0=18\mathrm{m}$ above a light spring of stiffness $k=5000N/m$, which stands upright with its top at ground level. How far, $x$, is the spring compressed when the ball stops? On the right of the figure, the ball falls $h_0$ to reach the spring and then a further $x$ while it squeezes it, so it falls $h_0+x$ in all. The figure is not to scale: the real $x$ is about fifty times smaller than $h_0$.

The ball starts and ends at rest, so $\Delta KE=0$. With no losses, all the work the weight does over the whole fall must end up stored in the spring: the original 353 J plus the extra $mgx$ released as the ball sinks below ground level. So
$mg(h_0+x)=\frac{1}{2}k{x}^2$
With the numbers this is a quadratic in $x$, and its positive root is the answer:
$2500x^2-19.62x-353.16=0,x\approx 0.380\mathrm{m}$

The extra term changes the answer by only 1 %: without it, $x=\sqrt{2\cdot 353/5000}\approx 0.376\mathrm{m}$. The peak force on the ball, $kx\approx 1900\mathrm{N}$, is about 97 times its weight. Crash mats and crumple zones work the other way: they absorb the energy over a longer distance, and twice the distance means half the average force.

The same problem can be read on a potential energy graph. Write the potential energy as a function of the compression $x$ at any moment while the ball touches the spring (the answer above is the largest value of $x$), with $U=0$ at the top of the spring:
$U(x)=\frac{1}{2}k{x}^2-mgx$
The spring term grows as the spring is squeezed; the gravity term falls as the ball sinks.

The minimum of $U$, where $dU/dx=kx-mg=0$, lies at $x=mg/k\approx 3.9\mathrm{mm}$: there the spring force balances the weight, and the ball could rest on the spring. The lowest point of the fall is not this minimum but the turning point where $U=E$, as in Figure 3. Without losses the ball would bounce, leave the spring at 18.8 m/s and climb back to 18 m.

Limits of validity

Both energy formulas of this lesson are approximations, and the theorem itself is about particles. Here is where each one stops working.

$mgh$ is the near-surface form of the gravitational potential energy $U=-GMm/r$, where $M$ is the Earth's mass and $r$ the distance from its centre. Exactly, raising a mass from the surface, at the radius $R\approx 6371\mathrm{km}$, by a height $h$ takes $mghR/(R+h)$. So $mgh$ is too high by a fraction of about $h/R$: $3\times 10^{-6}$ for the 18 m drop, but 6 % for a lift to the 400 km height of the space station. Letting $h$ grow without limit gives the energy needed to escape from the Earth altogether, $mgR$, about 62.5 MJ per kilogram, which corresponds to an escape speed of 11.2 km/s.

$\frac{1}{2}m{v}^2$ is the low-speed form of the relativistic kinetic energy $(\gamma -1)m{c}^2$, where $c$ is the speed of light and $\gamma =1/\sqrt{1-v^2/c^2}$. The first correction adds a fraction $\frac{3}{4}{v}^2/c^2$ to the energy: about $3\times 10^{-15}$ for the ball at 18.8 m/s, while at half the speed of light the classical formula falls 19 % short. Energy also has mass: the ball's 353 J corresponds to $353/c^2\approx 4\times 10^{-15}\mathrm{kg}$.

The theorem is a statement about a particle. For a body that spins, deforms or has its own energy stores, the integral $\int {\vec{F}}_{\mathrm{net}}\cdot d{\vec{r}}_{\mathrm{cm}}$, taken along the path of the centre of mass, gives only the change in the kinetic energy of the centre-of-mass motion. When you jump, the floor pushes you up, but the point where it pushes does not move, so the floor does no work: your kinetic energy comes from chemical energy in your muscles. A solid ball rolling without slipping from 18 m puts two-sevenths of its kinetic energy into spin and arrives at 15.9 m/s instead of 18.8 m/s.

At the scale of atoms quantum mechanics takes over: energy is still conserved, but bound energies come in discrete levels, and a particle can even be found where $U>E$, which classical mechanics forbids.

Where this leads

Emmy Noether showed in 1918 that energy conservation follows from a symmetry: the laws of physics are the same today as tomorrow. Within mechanics, the same energy balance appears in other lessons: Momentum and Collisions sorts collisions by whether kinetic energy survives, Rotational Dynamics and Torque adds the energy of spin, and Gravitation and Orbits uses $U=-GMm/r$ for orbits and escape speeds. Elasticity: Hooke's Law, Stress and Strain and Simple Harmonic Motion build on $U=\frac{1}{2}k{x}^2$.

Beyond mechanics, Electric Field and Potential turns $U$ into energy per unit charge, Bernoulli's Principle applies the energy balance to a flowing fluid, and First Law of Thermodynamics and Heat Engines makes heat a full member of the account. Special Relativity replaces $\frac{1}{2}m{v}^2$ with $(\gamma -1)m{c}^2$, Fission, Fusion and E = mc² weighs the energy stored in nuclei, and Schrödinger's Equation and Tunnelling shows what happens where classical motion would need a negative kinetic energy.