Power

In 1782 James Watt had a marketing problem. His improved steam engine was replacing horses at English breweries, and a brewer naturally asked the killer question: how many horses does one engine replace? Watt measured a strong dray horse turning a mill wheel, multiplied force by distance, divided by time, and rounded up generously: 33,000 foot-pounds per minute. The horsepower was born — and with it the realization that what industry buys and sells is not energy but the rate of delivering it. That rate is power. Two engines can do the same total work — lift the same water, grind the same grain — but the one that does it faster is the more powerful, and the distinction runs the modern world: your laptop charger is rated in watts, a cyclist's fitness is measured in watts, a power station is nothing but its wattage. In mechanics the concept sharpens into a single dot product, $P = \mathbf{F}\cdot\mathbf{v}$ : only the component of force along the velocity delivers energy, which is why a force can be enormous yet powerless — the normal force under a sliding block transfers nothing, and gravity does no work on a satellite in circular orbit however hard it pulls. Power is also where energy methods become differential: net power equals the rate of change of kinetic energy, the heartbeat form of the work-energy theorem. Master it and you can size a motor, explain why a car's acceleration dies at speed, and read a cycling power meter like a physicist.

Key takeaways

Power is the rate at which work is done or energy is transferred: $P = \mathrm{d}W/\mathrm{d}t$ , measured in watts ( $1\ \mathrm{W} = 1\ \mathrm{J/s}$ ; $1\ \mathrm{hp} \approx 746\ \mathrm{W}$ ).
For a force $\mathbf{F}$ on a particle moving with velocity $\mathbf{v}$ , the instantaneous power is the dot product $P = \mathbf{F}\cdot\mathbf{v} = Fv\cos\theta$ — only the velocity-aligned component of force counts.
The work-energy theorem in rate form: $P_{\text{net}} = \mathrm{d}T/\mathrm{d}t$ — net power feeds kinetic energy directly.
Average power is energy over elapsed time, $\bar P = W/\Delta t$ ; instantaneous power is its limit.
At constant power the driving force falls as $F = P/v$ : acceleration is fierce from low speed and starves at high speed, giving $v \propto \sqrt{t}$ from rest.
Power is frame-dependent (velocity is), even though Newtonian forces are not — two observers disagree on the rate of energy transfer.
Key idea: power is force dotted into velocity, $P=\mathbf{F}\cdot\mathbf{v}$ — equivalently the time rate of work. Only the velocity-aligned part of a force delivers power, and net power equals $\mathrm{d}T/\mathrm{d}t$ .

Where this lives today

The watt is the axis of the energy transition. An electric car's spec sheet is a small physics exam: a 150 kW motor, regenerative braking that captures deceleration power back into the battery, and a charging curve measured in kilowatts that tapers exactly because charging power, like mechanical power, is a rate the chemistry can only sustain briefly at its peak. Grid operators live inside equation (6): a country's demand is a power curve, storage is its integral, and the daily scramble is matching the two in real time. At human scale, power became a consumer product. Cycling power meters — strain gauges in the crank measuring pedal force, multiplied by crank angular velocity, literally $P = \mathbf F\cdot\mathbf v$ at 60 samples per second — turned training into applied mechanics; a rider's "functional threshold power" of, say, 250 W is the sustained $P$ their physiology can pay. The same dot product runs in reverse in wind turbines, where the power available scales as the cube of wind speed (the same $cv^3$ law as the car's drag), which is why a site with 10 percent more wind is worth 33 percent more energy — and why turbine siting is a business of decimal points.

The physical picture

Push a stalled car and you feel the difference between force and power in your muscles. Straining against a car that barely creeps, you exert a huge force but deliver almost no power — force times a trickle of velocity is a trickle of energy. Once the car rolls at walking pace, the same push pours energy into it ten times faster. Power measures how fast you are spending energy, never how hard you are pushing. The dot product is the whole geometry of the concept. Carry a suitcase horizontally at constant speed and you do no work against gravity: the force is vertical, the motion horizontal, the angle between them $90°$ , the power zero. Between the extremes, only the shadow of the force along the velocity earns anything:

Figure

Only the along-track component of a force delivers power

The third picture to internalize is the car on a hill at constant speed. Its kinetic energy never changes, so the net power is zero — yet the engine is working furiously. The engine's positive power exactly cancels the negative power of gravity and drag. Power accounting tracks each force's contribution separately even when the total cancels; that is what makes it an accounting system rather than a single number.

The core equations

Start from the definition. Work accumulates over time, and power is its instantaneous rate:

P \equiv \frac{\mathrm{d}W}{\mathrm{d}t} \qquad (1)

where: $P$ — instantaneous power (watts); $W$ — work done (joules). A joule per second is a watt: a phone charges at $\sim 20$ W, a kettle draws $\sim 2000$ W, a strong cyclist sustains $\sim 300$ W. To connect the rate of work to the motion, feed the work element $\mathrm{d}W = \mathbf{F}\cdot\mathrm{d}\mathbf{r}$ into (1). Dividing the displacement by $\mathrm{d}t$ turns it into the velocity, and the central formula of this article falls out:

P = \mathbf{F}\cdot\mathbf{v} \qquad (2)

where: $\mathbf{v}$ — velocity of the point where the force is applied. Read it geometrically: the force's projection onto the velocity, times the speed. A force with no shadow along the motion — normal force, centripetal force, the magnetic force on a moving charge — delivers exactly zero power no matter its size. For hand computation the same statement in magnitudes and the angle is often quicker:

P = Fv\cos\theta \qquad (3)

where: $\theta$ — the angle between $\mathbf{F}$ and $\mathbf{v}$ . The three regimes of the hero figure are $\theta = 0$ (full feed), $\theta = 90°$ (steering, zero power), $\theta = 180°$ (braking, negative power). Apply (2) to the net force and Newton's second law turns it into the differential heartbeat of energy methods:

P_{\text{net}} = \frac{\mathrm{d}T}{\mathrm{d}t}, \qquad T = \tfrac{1}{2}mv^2 \qquad (4)

where: $T$ — kinetic energy. Net power in, kinetic energy rises; net power negative, it falls; net power zero, speed constant (though the direction may still turn, as in circular orbit). Integrate (4) over time and the work-energy theorem $W_{\text{net}} = \Delta T$ reappears. When only totals matter — a stair climb, a full battery charge — the average is the honest number:

\bar{P} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t} \qquad (5)

where: $\bar P$ — average power over the interval $\Delta t$ . Instantaneous power (1) is the limit of (5) as the interval shrinks; a sprinter's peak power can be triple their one-hour average, which is why both numbers appear on a cycling power report. Conversely, energy is the time-integral of power — the fact your electricity meter lives on (a kilowatt-hour is $3.6\times10^6$ J of energy, not power):

W = \int_{t_1}^{t_2} P\,\mathrm{d}t \qquad (6)

where: the integral runs over the operating interval. Constant $P$ makes it $W = P\,\Delta t$ ; varying $P$ (a real drive cycle) needs the full integral. Finally, a subtlety worth owning early: power is frame-dependent. Newtonian forces are unchanged by a Galilean boost, but velocities transform as $\mathbf{v}' = \mathbf{v} - \mathbf{V}$ , so

P' = \mathbf{F}\cdot(\mathbf{v}-\mathbf{V}) = P - \mathbf{F}\cdot\mathbf{V} \qquad (7)

where: $\mathbf{V}$ — the boost velocity. The ground observer and the train passenger disagree about how fast a given force feeds energy to a given body — legitimately, because kinetic energy itself is frame-dependent. Total energy balance holds separately in each frame.

Which formula answers which question

you want	use	notes
rate a force feeds energy	$P = \mathbf F\cdot\mathbf v$ , eq. (2)	zero for perpendicular forces
quick hand computation	$P = Fv\cos\theta$ , eq. (3)	watch the sign of $\cos\theta$
how fast speed builds	$P_{\text{net}} = \mathrm dT/\mathrm dt$ , eq. (4)	net power only
motor/engine sizing from a task	$\bar P = W/\Delta t$ , eq. (5)	totals, not instants
energy from a power rating	$W = \int P\,\mathrm dt$ , eq. (6)	kWh is energy, kW is power
driving force available at speed $v$	$F = P/v$	constant-power engine's force curve

Derivation ARate form of the work-energy theorem

We are deriving equation (4): the net power on a particle equals the time derivative of its kinetic energy. This is the differential version of the work-energy theorem, and the derivation is the cleanest two-line argument in energy methods — but it earns its keep, because the rate form is what you integrate in real problems where force and speed both change (a car accelerating under constant engine power, Derivation B). The route: start from the power of the net force, insert Newton's second law, and recognize a total time derivative. Watch the single hypothesis: mass must be constant, or the derivative picks up extra terms.

Step 1. The power delivered by the net force is $P_{\text{net}} = \mathbf{F}_{\text{net}}\cdot\mathbf{v}$ . Why: powers add linearly — the dot product distributes over the sum of forces, so the sum of each force's power is the power of the resultant.

Step 2. Insert Newton's second law $\mathbf{F}_{\text{net}} = m\,\mathrm{d}\mathbf{v}/\mathrm{d}t$ :

P_{\text{net}} = m\,\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\cdot\mathbf{v}

Step 3. Recognize the right side as a total derivative using the product rule on $\mathbf{v}\cdot\mathbf{v} = v^2$ :

\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\cdot\mathbf{v} = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{v}\cdot\mathbf{v}\right) = \frac{1}{2}\frac{\mathrm{d}(v^2)}{\mathrm{d}t}

Why this is the crux: differentiating the square of the speed produces exactly the factor of two that cancels the one-half in kinetic energy — the same algebra that fixes the $\tfrac12$ in $T = \tfrac12 mv^2$ in the first place.

Step 4. Assemble:

P_{\text{net}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\tfrac{1}{2}mv^2\right) = \frac{\mathrm{d}T}{\mathrm{d}t}

Integrating both sides over any time interval recovers the integrated theorem $W_{\text{net}} = \Delta T$ . Verified: matches Morin ch. 5 and Taylor ch. 4; dimensional check, $\mathrm{W} = \mathrm{J/s}$ on both sides; for uniform circular motion $\mathbf F \perp \mathbf v$ gives $P_{\text{net}} = 0$ and indeed $T$ is constant.

Go deeper · Derivation B · 1 minMotion under constant power

Derivation BMotion under constant power

Here we answer a deceptively simple question with real engineering consequences: how does a body accelerate when the engine delivers constant power rather than constant force? The result — speed growing only as $\sqrt{t}$ and distance as $t^{3/2}$ — explains why a car's push melts away as it gathers speed. The strategy is to treat the rate form (4) as a differential equation for $v(t)$ and separate variables. Watch the low-speed end: constant power implies a force $F = P/v$ that formally diverges as $v \to 0$ , so the idealization must break down at launch — real drivetrains are traction-limited off the line.

Step 1. Write the rate form with constant $P$ along a line:

P = \frac{\mathrm{d}}{\mathrm{d}t}\left(\tfrac{1}{2}mv^2\right) = mv\,\frac{\mathrm{d}v}{\mathrm{d}t}

Why: with no other forces, all the engine's power feeds kinetic energy; friction and drag would appear as additional negative-power terms.

Step 2. Separate variables and integrate from the initial speed $v_0$ :

\int_{v_0}^{v} m v'\,\mathrm{d}v' = \int_0^t P\,\mathrm{d}t' \quad\Longrightarrow\quad \tfrac{1}{2}m\left(v^2 - v_0^2\right) = Pt

Why the left side is just the kinetic energy gained: energy delivered equals energy stored — this line is the work-energy theorem wearing integration limits.

Step 3. Solve for the speed:

v(t) = \sqrt{v_0^2 + \frac{2P}{m}\,t}

The speed grows without bound but ever more slowly — each additional m/s costs more energy than the last, because kinetic energy is quadratic in speed.

Step 4. From rest ( $v_0 = 0$ ), integrate once more for the distance:

x(t) = \int_0^t \sqrt{\frac{2P}{m}}\,t'^{1/2}\,\mathrm{d}t' = \frac{2}{3}\sqrt{\frac{2P}{m}}\,t^{3/2}

Why the exponent: velocity goes as $t^{1/2}$ , and integrating raises the power by one — the constant-power analogue of the constant-force $t^2/2$ . *Verified: differentiate to check — $\dot x = \sqrt{2P/m}\,t^{1/2} = v(t)$ and $mv\dot v = m \cdot \sqrt{2Pt/m}\cdot \tfrac12\sqrt{2P/(mt)} = P$ , constant as required; matches the standard result quoted in Morin ch. 5 problems.*

The two curves of Derivation B's world — the force an engine can still supply, and the force drag demands — tell a car's whole speed story in one frame:

Figure

Constant power meets quadratic drag — where the curves cross is the top speed

Limiting cases

Case	Result	Meaning
force parallel to velocity	$P = Fv$	maximum transfer — thrust during acceleration
force perpendicular to velocity	$P = 0$	normal, centripetal, magnetic forces steer without feeding energy
force antiparallel to velocity	$P = -Fv$	friction, drag, braking — the body is bled
constant power from rest	$v = \sqrt{2Pt/m}$ , $x \propto t^{3/2}$	acceleration dies as $1/v$ ; the launch instant is traction-limited, not power-limited
net power zero	$T$ constant	speed fixed, direction free — uniform circular motion
boosted frame	$P' = P - \mathbf F\cdot\mathbf V$	power is frame-dependent; each frame's books balance separately

Worked examples

1 · The stair-climb watt (confidence builder)

A person of mass $70\ \mathrm{kg}$ runs up a flight of stairs of vertical height $3.0\ \mathrm{m}$ in $5.0\ \mathrm{s}$ . Find the average mechanical power against gravity, in watts and horsepower. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Work done against gravity: $W = mgh = 70 \times 9.8 \times 3.0 = 2058\ \mathrm{J}$ .

Step 2. Average power from eq. (5): $\bar P = W/\Delta t = 2058/5.0 = \boxed{412\ \mathrm{W}}$ (to three figures; about $410$ W).

Step 3. In Watt's own unit: $412/746 = \boxed{0.55\ \mathrm{hp}}$ — half a horse, sustainable by a fit human for seconds, not minutes. (A pro cyclist's hour record is about $440$ W; your body also spends roughly four metabolic joules per mechanical joule, so this climb costs $\sim 8$ kJ of food energy.)

Step 4. Sanity check via eq. (2): average vertical speed $\bar v = 3.0/5.0 = 0.60\ \mathrm{m/s}$ ; the supporting force is about $mg = 686$ N, and $P = Fv = 686\times0.60 = 412$ W ✓ — the same number from the force-velocity side.

2 · Top speed and hill climb at constant power (exam level)

A car of mass $1500\ \mathrm{kg}$ has a drivetrain that delivers $P = 90\ \mathrm{kW}$ to the wheels. Aerodynamic drag is $F_d = cv^2$ with $c = 0.75\ \mathrm{kg/m}$ (rolling resistance folded in). Find the top speed on level ground. Then find the power needed to hold $25\ \mathrm{m/s}$ up a 6 percent grade.

Step 1. Top speed: at $v_{\max}$ acceleration ceases, so the entire engine power feeds drag: $P = F_d\,v_{\max} = c\,v_{\max}^3$ .

Step 2. Solve: $v_{\max} = (P/c)^{1/3} = (90{,}000/0.75)^{1/3} = (120{,}000)^{1/3} = \boxed{49.3\ \mathrm{m/s} \approx 178\ \mathrm{km/h}}$ .

Step 3. Note the cube-root law: doubling the power multiplies top speed by $2^{1/3} = 1.26$ — only 26 percent faster for twice the engine. That scaling is why top-speed bragging rights are so expensive.

Step 4. On the grade at constant $v = 25\ \mathrm{m/s}$ : gravity's power drain is $P_g = mgv\sin\theta \approx 1500 \times 9.8 \times 25 \times 0.06 = 22{,}050\ \mathrm{W} \approx 22.1\ \mathrm{kW}$ (for small grades $\sin\theta \approx \tan\theta = 0.06$ ).

Step 5. Drag at $25\ \mathrm{m/s}$ : $P_d = cv^3 = 0.75 \times 25^3 = 0.75\times15{,}625 = 11{,}719\ \mathrm{W} \approx 11.7\ \mathrm{kW}$ .

Step 6. Total: $P = 22.1 + 11.7 = \boxed{33.8\ \mathrm{kW}}$ — barely a third of the car's 90 kW. The engine's headroom, $56$ kW, is what remains for acceleration; via eq. (4) the car could still gain speed at $a = P_{\text{spare}}/(mv) = 56{,}200/(1500\times25) \approx 1.5\ \mathrm{m/s^2}$ on that hill.

Origins · 1 minJames Watt and how this was discovered

James Watt

1736–1819 · Glasgow and Birmingham

Watt was a university instrument maker in Glasgow when, in 1765, he was handed a broken model of a Newcomen steam engine and saw in it a thermodynamic scandal: the cylinder was heated and chilled on every stroke, throwing away most of the coal's energy. His fix — a separate condenser, patented in 1769 — multiplied the engine's output per bushel of coal and effectively launched the Industrial Revolution. Watt turned out to be as shrewd a marketer as an engineer: to sell engines to brewers and mine owners who thought in horses, he defined the horsepower from measurements of dray horses at work, deliberately overestimating so no customer could complain of being short-changed. He also gave engineering the flyball governor (automatic feedback control), the copying press, and the term he and partner Matthew Boulton billed by — duty, work delivered per fuel consumed, the ancestor of efficiency. The SI unit of power was named for him in 1882: a fitting close of the ledger for the man who taught the world to price energy by the second.

History

1712

— Thomas Newcomen's atmospheric engine pumps water from Cornish mines: the first machine sold, in effect, by its power.

1769

— James Watt patents the separate condenser, multiplying steam-engine output per ton of coal.

1783

— Watt defines the horsepower as 33,000 foot-pounds per minute, so factories can compare engines with the animals they replace.

1882

— The British Association adopts the watt as the unit of power, honoring Watt alongside the volt and ampere.

1960

— The watt is enshrined in the SI as one joule per second, the rate at which all energy flows are now metered.

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 5 (Energy) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., ch. 1 · Landau & Lifshitz, Mechanics, §6-§7 · Taylor, Classical Mechanics, ch. 4.

Provenance. Definitions and formulas from Morin Ch. 5, Taylor Ch. 4, Goldstein Ch. 1, Landau-Lifshitz §6-7. A verifier should recompute: (i) $P=\mathbf F\cdot\mathbf v$ from $\mathrm dW=\mathbf F\cdot\mathrm d\mathbf r$ ; (ii) the rate-form work-energy theorem $P_{\text{net}}=\mathrm dT/\mathrm dt$ ; (iii) constant-power kinematics $v=\sqrt{v_0^2+2Pt/m}$ and $x=\tfrac23\sqrt{2P/m}\,t^{3/2}$ ; (iv) numerical check of the stair-climb power ( $\approx410\,\mathrm W$ ) and the $30\,\mathrm W$ dot-product example; (v) the conversion $1\,\mathrm{hp}\approx746\,\mathrm W$ (mechanical horsepower). Worked-example numbers on this page ( $412$ W and $0.55$ hp for the stair climb; $v_{\max} = 49.3$ m/s, hill-climb budget $33.8$ kW and spare acceleration $1.5\ \mathrm{m/s^2}$ for the 90 kW car) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Velocity and Acceleration · Newton's Second Law · Vectors in Kinematics and Coordinate Systems · Relative Motion and Galilean Transformations. Leads to: Circular Motion and Centripetal Acceleration · Normal Force.

Is a kilowatt-hour a unit of power or energy?

Energy — and the name says so once you parse it. A kilowatt-hour is a power (one kilowatt) multiplied by a time (one hour), which by equation (6) is an energy: $1\ \mathrm{kWh} = 1000\ \mathrm{W} \times 3600\ \mathrm{s} = 3.6\times10^6\ \mathrm{J}$ . Your electricity bill charges for energy; the circuit breaker limits power. The confusion cuts both ways: "my house used 20 kWh yesterday" is an energy statement, "this shower heater draws 8 kW" is a rate statement. When headlines mix them up ("the battery stores 100 MW"), the physics has already gone wrong — storage is joules or kWh, delivery is watts.

If the net power on my car is zero at top speed, why does the engine still burn fuel furiously?

Because net power being zero means the accounting cancels, not that nothing flows. At top speed the engine pours $P = cv^3$ into the car while drag drains exactly the same amount into the air; kinetic energy holds steady per equation (4). The engine's fuel pays for a real, continuous energy current — car to air to heat — even though the speedometer freezes. This is the constant-speed hill climb of the physical picture: individual powers are large and opposite, the sum is zero. "Zero net power" and "engine off" are as different as a bathtub with taps and drain matched versus a bathtub empty.

How can power be frame-dependent? Surely the engine burns the same fuel for every observer

It does — but the body the energy lands in has different kinetic energy in different frames, so the delivery rate differs too. Equation (7) makes it explicit: $P' = P - \mathbf F\cdot\mathbf V$ . A locomotive pulling with force $F$ at 30 m/s delivers $30F$ watts to the train in the ground frame, and zero in the train's own frame, where the coupling force acts on a stationary object. No paradox survives careful bookkeeping: in each frame, all energy flows (fuel chemistry, kinetic energy, heat, work done on the rails and Earth) balance internally. What is frame-independent is the dissipated power — the friction heating — because it depends on relative sliding velocities only.

Why does acceleration die at high speed even before drag matters?

Kinetic energy is quadratic in speed, so each additional metre per second costs more joules than the last. Going 10 to 11 m/s costs $\tfrac12 m(121-100) = 10.5m$ joules; 40 to 41 m/s costs $40.5m$ — four times more for the same speed increment. A constant-power engine supplies joules at a fixed rate, so the increments must slow down: that is the $v = \sqrt{2Pt/m}$ of Derivation B. Equivalently, force is $F = P/v$ , falling as speed rises, while $a = P/(mv)$ . Drag then piles on top, stealing a growing share $cv^3$ of the fixed budget until, at $v_{\max}$ , nothing is left to accelerate with.

The formula F equals P over v predicts infinite force at zero speed. What actually happens when a car launches?

The idealization "constant power at all speeds" fails at launch, in two honest ways. First, real drivetrains cannot deliver rated power at zero engine or motor speed — combustion engines idle and need gearing or a slipping clutch; electric motors are torque-limited (hence force-limited) from zero up to a base speed, and only above it hold constant power. Second, even a drivetrain that could push harder would spin the tires: static friction caps the transmissible force at $\mu_s N$ , roughly $mg$ for good tires, i.e. about $1g$ of acceleration. So a launch is traction-limited at first (constant force, $v \propto t$ ), then becomes power-limited ( $v \propto \sqrt{t}$ ) once $P/v$ falls below the traction ceiling. Drag racers live exactly on that crossover.

My electric kettle is rated 2 kW. Is that the power it consumes or the power it delivers to the water?

Rated power is what it draws from the wall; delivery to the water is that times an efficiency, and for kettles the two are unusually close — resistive heating "wastes" energy only as heat, which is the product, so $\eta \gtrsim 0.85$ with losses limited to warming the kettle body and escaping steam. Check it with equation (5): heating 1.0 L of water from 20 °C to 100 °C stores $mc\Delta T = 1.0 \times 4186 \times 80 \approx 3.3\times10^5$ J; at 2 kW that predicts $t = W/P \approx 170$ s, and real kettles take about three minutes. Doing the same audit on a motor or an engine gives much lower $\eta$ — there the heat is not the product but the leak.

Human power: what can a person actually sustain?

Useful anchors, all straight from equation (5). Sustained aerobic output: a healthy adult manages 75–100 W for hours (that is why gym bikes feel cruel at 200 W); a trained amateur cyclist holds 200–300 W for an hour; the elite hour record sits near 440 W. Peak output: a sprinter's legs can spike above 1500 W for a few seconds — which is why stair-sprinting (Example 1's 412 W) is easy briefly and impossible for long. For engineering history's sake: Watt's horse, 746 W, is a fair sustained figure for a working horse, roughly triple a laboring human — so "half a horsepower" for the stair climb means precisely what it sounds like.