Power

Two lifts carry identical loads to the tenth floor. Both do exactly the same work — the same $mgh$ , banked joule for joule. But one arrives in twenty seconds and the other in sixty, and every passenger can feel that the two machines are not equals. Work is blind to the difference; the quantity that sees it is power: the rate at which energy is transferred. The fast lift is a three-times-more-powerful machine doing the identical job. Power is where this unit's accounting meets the clock — and the market. Your electricity bill, your car's engine rating, a sprinter's burst, a pump's delivery: all quoted in watts and kilowatts, all priced per kilowatt-hour. The two working equations are short: $P = W/t$ for the average rate, and the sharper instantaneous form $P = Fv$ — force times speed — which quietly runs half of JEE's mechanics-with-machines problems. That second form carries the module's best physical insight: at fixed power, force and speed trade off against each other, which is why your bicycle has gears, why a car's top speed on a hill is lower than on the flat, and why a tugboat can pull hardest precisely when it moves slowest. The formulas are two lines; the exam craft is in choosing snapshots, converting stubborn units (kilowatt-hours are energy, horsepower is power, and marks die on the difference), and reading "constant speed" as the code phrase it is: driving force equals resistance, so $P = fv$ .

Key takeaways

Power is the rate of energy transfer: $P_{\rm avg} = \dfrac{W}{t}$ , in watts ( $1$ W $= 1$ J/s). Same work in less time = more power.
Instantaneous form: $P = Fv$ (force along the velocity; $P = Fv\cos\theta$ in general) — the machine's live output at one moment.
At constant speed the driving force just balances resistance, so engine power $= f v$ : the code phrase "maximum/steady speed" means exactly this equation.
At fixed power, force and speed are enemies: $F = P/v$ . Slow crawl → huge force available; top speed → only enough force to match drag. That trade-off is gearing.
Lifting steadily: $P = \dfrac{mgh}{t} = mgv$ . Pumps ejecting water add a kinetic term: $P = \lambda gh + \tfrac12 \lambda v^2$ with $\lambda$ the kg lifted per second.
The kilowatt-hour is energy, not power: $1$ kWh $= 3.6 \times 10^6$ J — the unit your electricity bill counts.
Efficiency works with powers exactly as with energies: $\eta = P_{\rm useful}/P_{\rm input}$ .
Key idea: power is energy per second — $P = W/t$ on average, $P = Fv$ in the moment — and at constant power the product $Fv$ is a fixed budget that force and speed must share.

The core insight

Power measures how fast energy moves — the average $W/t$ over an interval, the instantaneous $Fv$ at a moment — so a machine's rating fixes the product of the force it can supply and the speed at which it supplies it, making "constant cruising speed" the statement $P = (\text{total resistance}) \times v$ and gearing the art of spending one budget as either force or speed.

Where this lives today

Power ratings are the grammar of the energy transition. Electric-vehicle spec sheets quote motor power (a family hatchback: ~150 kW peak) and charging power — a 350 kW charger delivers a kilowatt-hour every ten seconds, which is the only reason road-trip charging is tolerable. Grid operators live on the demand curve, a national $P(t)$ that must be matched second by second; the infamous evening ramp, when millions of kettles and cookers switch on together, is a gigawatt-scale $P = W/t$ event. Data centres are now planned around power before land — a large AI cluster draws over 100 MW continuously, a small city's worth. And human power still calibrates it all: you idle at ~100 W, a Tour de France rider holds ~400 W for hours, and an elite sprinter's two-second burst tops 2 kW — the wattage of your kitchen kettle, which is why the kettle is the physicist's favourite unit of athletic awe.

The physical picture

Watch a single flight of stairs judge two athletes. Both haul the same body weight up the same 3 m: the same $\sim 2000$ J of work against gravity. The sprinter does it in 2 s, the walker in 8 s. Nothing in the work ledger distinguishes them — but the sprinter's muscles delivered $\sim 1000$ J per second while the walker's managed 250, and it is the per-second figure that decides what a machine (or a body) can and cannot do. Power is capability; work is just the total delivered. A wristwatch battery holds respectable energy but microscopic power; a camera flash capacitor holds modest energy and ferocious power. Never confuse the tank with the tap.

Figure

Hero — same work, different power: the staircase as a wattmeter

The second picture to install is the trade-off at fixed power. A machine rated at $P$ watts can push with force $F = P/v$ — enormous when barely moving, feeble at speed. A cyclist in low gear crawls uphill exerting a huge drive force at tiny wheel speed; the same 250-watt legs in top gear on the flat supply only a small force, but at high speed. Tugboats, tractors, and electric motors all live on this hyperbola. Its exam signature: a vehicle's maximum speed is reached when the engine's available force $P/v$ has fallen to exactly match the total resistance — no leftover force, no acceleration, just balance.

The math: the average, the instant, and the trade-off

The average, over any interval:

P_{\rm avg} = \frac{W}{t} \qquad (1)

where: $W$ — energy transferred (work done) in time $t$ ; unit the watt, $1$ W $= 1$ J/s. Read it with a concrete pair: a crane doing $3 \times 10^4$ J of lifting in 20 s runs at 1.5 kW; your body idles at roughly 100 W — a human-shaped filament bulb — and a trained cyclist holds 300–400 W for an hour. The instantaneous form — the rate right now:

P = \vec F \cdot \vec v = Fv\cos\theta \qquad (2)

where: $F$ — the force the machine applies; $v$ — the speed of the point where it applies it; $\theta$ — the angle between them. Derivation A shows this is just (1) applied to a vanishingly short interval. Two special readings pay rent: force along motion gives $P = Fv$ (engines, pumps, climbing); force perpendicular gives zero power (the centripetal family transfers no energy — now stated in power language). The constant-speed workhorse. "The car cruises at its top speed $v_{\max}$ " encodes zero acceleration, so the drive force equals the total resistance $f$ :

P_{\rm engine} = f\,v_{\max} \qquad (3)

where: $f$ collects rolling resistance and drag (given, at this level). On a hill, gravity's along-slope component joins the resistance: $P = (f + mg\sin\alpha)\,v$ — which is why the same engine tops out slower uphill, worked example 4's whole business. Steady lifting and pumping, the boards' favourite:

P = \frac{mgh}{t} = \lambda g h \quad\text{and, with ejection speed } v:\quad P = \lambda\left(gh + \tfrac12 v^2\right) \qquad (4)

where: $\lambda = m/t$ — the kilograms handled per second (for water, numerically the litres per second). The first term banks height; the second buys the stream's kinetic energy — forgetting it is the single most-harvested error in pump problems.

the problem says	write
"does $W$ joules in $t$ seconds"	$P = W/t$
"moving at $v$ under applied force $F$ "	$P = Fv\cos\theta$
"maximum / constant / steady speed"	$P = (\text{total resistance})\times v$
"pump raises $\lambda$ kg/s through $h$ , ejects at $v$ "	$P = \lambda(gh + \tfrac12 v^2)$
"consumes $E$ kWh"	energy $= E \times 3.6\times10^6$ J — then divide by time if power is wanted

Derivation AFrom the average to the instant — why P = Fv

We derive the instantaneous form (2) from the average form (1). The strategy: apply $W/t$ to an interval so short that the force and velocity have no room to change, and watch the displacement collapse into speed-times-time. The one subtlety: only the force's along-motion component transfers energy — the cosine rides along from the work formula automatically.

Step 1. Over a very short interval $\Delta t$ , a force $F$ at angle $\theta$ to the motion does work

\Delta W = F\,\Delta s\,\cos\theta

where $\Delta s$ is the small distance moved.

Step 2. Divide by the interval, per definition (1):

P = \frac{\Delta W}{\Delta t} = F\cos\theta \cdot \frac{\Delta s}{\Delta t}

Step 3. In the short-interval limit, $\Delta s/\Delta t$ is the instantaneous speed $v$ (the same slope-of-tangent idea kinematics used for velocity):

P = Fv\cos\theta

force along motion: $P = Fv$ ; force perpendicular: $P = 0$ — steering is free, energizing is not.

Step 4. Consistency check backward: if $F$ and $v$ are constant, the average over any interval is $Fv\cos\theta \cdot$ (anything)/(the same anything) — the average and the instant agree, as they must for steady motion. Why the exam cares about the distinction: accelerating objects have a CHANGING $P = Fv$ even under constant force — the power grows as the object speeds up. "Work done in the third second" (average) and "power at $t = 3$ s" (instant) are different numbers; the constant-speed problems where they coincide are the safe harbour. *Verified: a 2000 N drive force at the moment the car passes 15 m/s delivers $P = 30$ kW; the same force at 25 m/s delivers 50 kW — same push, higher bill, which is why top speed exists at all.*

Go deeper · Derivation B · 1 minThe pump's two invoices — height and stream speed

Derivation BThe pump's two invoices — height and stream speed

We derive the pump formula (4), the JEE-Main/NEET staple, from the energy chain over one second. The strategy: audit what happens to the mass handled per second — it gains height AND it gains speed, and the pump pays both invoices. The classic omission to watch: the ejected stream's kinetic energy, which examiners hide behind the innocent phrase "delivers water at speed $v$ ".

Step 1. Let the pump handle $\lambda$ kilograms per second, raising each kilogram through height $h$ and ejecting it at speed $v$ . Consider the parcel of water handled in one second: mass $\lambda$ .

Step 2. Its potential-energy invoice: $\lambda g h$ joules (banked as height).

Step 3. Its kinetic-energy invoice: the water started at rest in the well and leaves the nozzle at $v$ : $\tfrac12 \lambda v^2$ joules (carried away by the stream).

Step 4. The pump's useful power is the total energy delivered per second:

P_{\rm useful} = \lambda gh + \tfrac12 \lambda v^2

With pump efficiency $\eta$ , the input (electrical) power is $P_{\rm in} = P_{\rm useful}/\eta$ — the efficiency module's grade, now in watts. Why $\lambda$ and not $m$ : power problems live per-second, so the natural mass variable is the flow rate. For water, 1 litre = 1 kg makes $\lambda$ numerically the litres per second — a free unit check. *Verified: $\lambda = 2$ kg/s, $h = 10$ m, $v = 10$ m/s, $g = 10$ : $P = 2(100) + \tfrac12(2)(100) = 200 + 100 = 300$ W — the kinetic invoice is HALF the size of the height invoice here, far too large to drop.*

Limiting cases

case	result	meaning
$t \to$ small at fixed $W$	$P \to$ large	a firecracker beats a battery on power, not on energy
$v \to 0$ at fixed $P$	$F = P/v \to$ huge	the tugboat/low-gear end of the hyperbola
$v \to v_{\max}$	$F = P/v$ falls to $f$	no spare force, no acceleration — top speed defined
$\theta = 90°$	$P = 0$	centripetal forces energize nothing — circular motion at constant speed is power-free
hill steepens at fixed $P$	$v_{\max}$ falls	resistance gains $mg\sin\alpha$ ; balance arrives sooner
ejection speed $v \to 0$	pump formula → $\lambda gh$	pure lifting; the kinetic invoice politely vanishes

Common trapA kilowatt-hour is not a kilowatt

The electricity meter's unit — the kilowatt-hour — is ENERGY: one kilowatt of power flowing for one hour, $1000 \times 3600 = 3.6 \times 10^6$ joules. It is not a rate, not a power, and not interchangeable with the kilowatt. The exam versions of the confusion: "a 2 kW heater runs for 3 hours — how much energy?" (answer $6$ kWh $= 2.16\times10^7$ J, multiply by time); and "a house consumes 12 kWh in a day — average power?" (answer $500$ W, divide by time in the right units). The same trap wears a mechanics coat as horsepower: hp is a POWER unit ( $1$ hp $\approx 746$ W), so "a 5 hp motor" tells you the rate, never the total. Say it once before every power problem: watts are per-second; joules and kilowatt-hours are totals.

Worked examples

1 · Board level — the crane's rating (CBSE/ICSE pattern, 3 marks)

A crane lifts a $400$ kg load through $30$ m at constant speed in $60$ s. ( $g = 10$ m/s².) Find (a) the work done on the load, (b) the crane's useful power output, (c) its input power if it is 75% efficient.

Step 1. (a) Constant speed → the cable force equals the weight; work banked as height: $W = mgh = 400 \times 10 \times 30 = \boxed{1.2 \times 10^5\ \mathrm{J}}$ .

Step 2. (b) $P_{\rm useful} = W/t = \dfrac{1.2\times10^5}{60} = \boxed{2.0\ \mathrm{kW}}$ . (Equivalently $P = mgv$ with $v = 0.5$ m/s ✓.)

Step 3. (c) $\eta = P_{\rm useful}/P_{\rm in} \Rightarrow P_{\rm in} = \dfrac{2000}{0.75} \approx \boxed{2.7\ \mathrm{kW}}$ — input exceeds output, always.

2 · Board level — reading the meter (IGCSE/ICSE pattern)

A household runs a $2.0$ kW water heater for 90 minutes and a $100$ W fan for 10 hours. Electricity costs ₹8 per kWh (use the local unit symbol as given). Find (a) the total energy used in kWh and in joules, (b) the cost, (c) the average power over the whole day (24 h).

Step 1. (a) Heater: $2.0\ \mathrm{kW} \times 1.5\ \mathrm{h} = 3.0$ kWh. Fan: $0.10 \times 10 = 1.0$ kWh. Total $\boxed{4.0\ \mathrm{kWh}} = 4.0 \times 3.6\times10^6 = \boxed{1.44 \times 10^7\ \mathrm{J}}$ .

Step 2. (b) Cost $= 4.0 \times 8 = \boxed{₹32}$ .

Step 3. (c) $P_{\rm avg} = \dfrac{4.0\ \mathrm{kWh}}{24\ \mathrm{h}} = 0.167$ kW $\approx \boxed{170\ \mathrm{W}}$ — note the kWh/h division needs no joule conversion at all.

3 · Competitive level — the pump with a nozzle (JEE-Main/NEET pattern, ~3 minutes)

A pump lifts water from a well $20$ m deep and ejects it through a nozzle at $10$ m/s, delivering $5.0$ kg of water per second. The pump is 80% efficient. ( $g = 10$ m/s².) Find (a) the useful power, (b) the electrical input power.

Step 1. The per-second audit (derivation B): height invoice $\lambda gh = 5.0 \times 10 \times 20 = 1000$ W.

Step 2. Stream invoice: $\tfrac12 \lambda v^2 = \tfrac12 \times 5.0 \times 100 = 250$ W. Useful total: $P = \boxed{1250\ \mathrm{W}}$ .

Step 3. (b) $P_{\rm in} = \dfrac{1250}{0.80} \approx \boxed{1.6\ \mathrm{kW}}$ .

Step 4. Sanity: the kinetic invoice was a quarter of the total — dropping it (the designed error, answer 1000 W useful / 1250 W input) is a 20% miss, comfortably outside every rounding tolerance. Timing note: ~3 min. The mark-scheme skeleton is exactly the two invoices + the efficiency division. If the problem gives litres per minute, convert to kg/s FIRST ( $60$ L/min $= 1$ kg/s) — flow-rate units are where this problem actually kills.

4 · Competitive level — top speed, flat versus hill (JEE-Main pattern, ~3 minutes)

A car of mass $1000$ kg has a maximum engine output of $40$ kW. Total resistance to motion (rolling + drag, assumed constant) is $1000$ N. ( $g = 10$ m/s².) Find its top speed (a) on a level road, (b) up a slope of $\sin\alpha = 0.05$ .

Step 1. (a) Top speed = zero acceleration = drive force exactly balances resistance, with the engine flat out: $P = f\,v_{\max}$ , so $v_{\max} = \dfrac{40\,000}{1000} = \boxed{40\ \mathrm{m/s}}$ (144 km/h).

Step 2. (b) Uphill the resistance gains gravity's along-slope component: $f' = f + mg\sin\alpha = 1000 + 1000 \times 10 \times 0.05 = 1500$ N.

Step 3. $v'_{\max} = \dfrac{40\,000}{1500} = \boxed{27\ \mathrm{m/s}}$ (about 96 km/h) — same engine, one-third slower, purely because the fixed $Fv$ budget must now fund a bigger force.

Step 4. Structure worth banking: at fixed power, $v_{\max} \propto 1/(\text{total resistance})$ . Any option scaling top speed with $\sqrt{P}$ or with mass alone (on the flat) misreads the balance. Timing note: ~3 min. The phrase "maximum speed" IS the equation $P = f v$ — no kinematics, no energy snapshots. The uphill designed error is forgetting $mg\sin\alpha$ or, subtler, multiplying the resistance by $\cos\alpha$ too (only do that if the problem's $f$ was specified as $\mu N$ -based — here it was given as a flat 1000 N).

Exam traps

Energy units posing as power (and back). kWh, joule, calorie = energy; watt, kilowatt, horsepower = power. The pitfall box's two conversions ( $\times$ time, $\div$ time) cover every version. $1$ hp $\approx 746$ W; $1$ kWh $= 3.6\times10^6$ J — both worth memorizing cold.

Average versus instantaneous. Under constant force an accelerating body's $P = Fv$ changes every moment. "Power at the instant..." wants (2) with that instant's $v$ ; "average power over the trip" wants total work over total time. They agree only for steady motion — check which the question asks.

"Maximum speed" left uncashed. The phrase is not scenery: it means $a = 0$ , drive $=$ resistance, $P = fv_{\max}$ . Students who try energy snapshots or kinematics here dissolve into unknowns.

The hill's tax forgotten (or double-charged). Uphill at steady speed: resistance $= f + mg\sin\alpha$ . Forgetting the gravity term is the common miss; the subtle overcharge is also billing $mgh/t$ separately — $mg\sin\alpha \times v$ is $mgh/t$ , one invoice, not two.

The pump's stream ignored. "Ejects at $v$ " adds $\tfrac12\lambda v^2$ . Its partner error: using the well depth for $h$ but measuring ejection height wrongly when the nozzle stands above ground — read the geometry sentence twice.

$P = Fv$ with the wrong angle. The $v$ belongs to the point of application, along the force's direction ( $\cos\theta$ in general). A force perpendicular to motion — tension in the whirling string, the normal force on a track — transfers zero power no matter how large.

Efficiency applied to the wrong side. Input $=$ useful $/\eta$ (bigger); useful $=$ input $\times \eta$ (smaller). If your "input" came out below your "useful," the fraction flipped.

Flow-rate unit stumbles. Litres per minute → kg/s for water: divide by 60. Tonnes, minutes, km/h — power problems are unit minefields; convert everything to SI in line one.

Origins · 1 minJames Watt and how this was discovered

James Watt

1736–1819 · Glasgow & Birmingham

Watt was an instrument maker at Glasgow University when, repairing a model Newcomen engine in 1764, he saw that it wasted most of its steam reheating the cylinder — and invented the separate condenser, the single improvement that launched the industrial age. To sell engines he needed a number buyers understood, so he measured what a brewery horse could sustain and coined horsepower; the metric unit that replaced it now carries his own name. Watt's marketing arithmetic — rate of work, priced and compared — is precisely this module's $P = W/t$ , and every engine nameplate, light bulb and electricity bill still speaks his language.

History

1712

— Newcomen's atmospheric engine pumps water from Cornish mines: useful power, dreadful efficiency

1765

— Watt conceives the separate condenser; steam power becomes economic

1783

— Watt defines the horsepower (~746 W) to price engines against horses

1889

— The British Association adopts the watt as the unit of power, honouring him

1948

— The SI locks the watt as one joule per second, the definition this module uses

Going further

Texts. NCERT Physics Class 11, Work, Energy and Power chapter (power section) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 7 · CIE IGCSE 0625 §1.7.3 (energy resources & power calculations).

Provenance. All results are standard. A verifier should recompute: the stairs pair (250 W vs 1000 W); the crane (1.2×10⁵ J, 2.0 kW, 2.7 kW); the meter reading (4.0 kWh, ₹32, ~170 W); the pump (1000 + 250 = 1250 W, 1.6 kW input); the car (40 m/s flat, 27 m/s on the 0.05 slope); the derivation-B check (300 W with the kinetic invoice a full third of the height invoice).

Prerequisites. Work Done by a Force · Kinetic Energy & the Work–Energy Theorem. Leads to. ARCHETYPE: Energy Accounting on Rough Inclines · Pumps, Motors & Engine Power (archetype, ahead in this unit) · Power (UG deep dive).

Is power just "work divided by time," or is there more to it?

That ratio is the definition, but two refinements matter. First, average versus instantaneous: $W/t$ over a long interval smears out variation, while $P = Fv$ reads the rate at one moment — an accelerating car's instantaneous power climbs even under constant force. Second, power applies to every energy transfer, not only mechanical work: your heater's 2 kW is heat per second, a lamp's 10 W is light (and heat) per second. The watt is the universal per-second currency unit; this module trains it on mechanics first.

Why do cars have gears if the engine's power is fixed?

Because power fixes only the product $Fv$ , and different situations need different splits. Pulling away or climbing, you need force — low gear trades wheel speed away to multiply force ( $F = P/v$ is large when $v$ is small). Cruising, you need speed — top gear accepts a small force (just enough to match drag) at large $v$ . The gearbox is a force-speed exchange bureau operating at (nearly) constant power. Electric motors deliver strong force across a wide speed range, which is why many EVs manage with a single gear.

If a porter holds a suitcase stationary, work is zero — is power zero too? He's clearly "putting out effort".

Mechanical power delivered to the suitcase: zero — no displacement, no work, no rate of work. The porter's metabolic power is very much nonzero (muscles burn fuel to sustain tension, as the work module's FAQ explained), but it all leaves as body heat, not as energy given to the suitcase. Physics bills the suitcase's account; biology bills the porter's lunch. Exams ask about the suitcase.

Why does P = Fv use the speed of the point of application? What else would it use?

Consider a winch dragging a sledge by a rope: the engine's power is the rope tension times the rope's speed — the speed at which the point where the force acts actually moves. If the sledge (and rope end) moves at 2 m/s under 500 N, the winch delivers 1 kW, whatever the winch drum's internals do. The distinction bites in pulley systems, where a movable pulley makes the rope end move at twice the load's speed: same power, half the force at double the speed — the trade-off again, now inside the machine.

How can the SAME engine have a maximum power AND a maximum force? Which limits what?

They rule different regimes. At low speed the limit is usually force (grip, gearbox, motor torque): you cannot use full power because $F = P/v$ would exceed what the wheels can transmit. Past a crossover speed, power takes over: force available falls as $1/v$ until it just matches resistance at $v_{\max}$ . Exam problems live almost entirely in the power-limited regime and say so with phrases like "maximum power output" — take the phrase, write $P = fv$ , and don't import torque complications the problem never raised.

Does maximum power mean maximum efficiency?

No — the two are independent gradings and often pull opposite ways. A motor near stall delivers great force, terrible efficiency (most input becomes heat); the same motor lightly loaded runs efficiently but delivers little useful power. Real machines have an operating sweet spot chosen by engineers. In exam terms: efficiency converts input to useful power at some stated percentage; maximum power is a separate ceiling. Use each number only for the sentence that quotes it.

My phone charger says 20 W, my house connection says 5 kW, a power station says 500 MW — how do these compare to mechanics problems?

Same unit, same meaning: joules per second. The charger fills a ~ $4\times10^4$ J battery in about half an hour ( $20$ W $\times$ 2000 s). The house ceiling, 5 kW, is why you cannot run three heaters and a kettle together. The station's 500 MW is a hundred thousand houses' worth. Keeping such ladders in your head — 100 W human, 1 kW kettle, 100 kW car, 1 GW power station — is genuinely useful exam equipment: it flags absurd answers (a crane rated 30 W, a pump at 2 GW) before the examiner does.