Power

Two workers can lift the identical stack of bricks to a rooftop — same force, same height, same total work — yet if one finishes in ten minutes and the other takes an hour, everyone calls the first one more "powerful." That everyday instinct is exactly what physics means by power: not how much work gets done, but how fast it gets done. A tiny motor and a massive crane can, in principle, do the same amount of work; what separates them is the rate. This matters because almost every real machine — a pump, a car engine, a lift motor — is rated by its power, not by the work it can do (which is unlimited given enough time). Power lets us compare machines fairly, predict how long a job will take at a given rating, and connect the force a machine exerts to the speed at which it moves things. This article builds the two working forms of the power formula — average power from work and time, and instantaneous power from force and velocity — and applies them to pumps and engines, the two classic ISC numerical settings.

The core insight

Power is the rate of doing work, $P = W/t$ on average or $P = \vec F \cdot \vec v$ instantaneously, and it is what separates "how much work" from "how fast."

The core idea

Average power. Power is defined as work done per unit time:

P = \frac{W}{t}

where: $W$ is the work done (or energy transferred) in a time interval $t$ . This is an average over the interval — it says nothing about whether the work was done steadily or in bursts within that time. The SI unit is the watt ( $1\ \text{W} = 1\ \text{J/s}$ ); larger machines are often rated in kilowatts ( $1\ \text{kW} = 1000\ \text{W}$ ) or, historically, horsepower.

Instantaneous power. For a force $\vec F$ acting on a body moving with velocity $\vec v$ at some instant, the power delivered right then is:

P = \vec F \cdot \vec v = Fv\cos\theta

where: $\theta$ is the angle between the force and the velocity vectors. This follows directly from $P = dW/dt = (\vec F \cdot d\vec S)/dt = \vec F \cdot \vec v$ , since $d\vec S/dt = \vec v$ by definition of velocity. When the force acts along the direction of motion ( $\theta = 0°$ , the everyday case of an engine driving a car forward), this simplifies to $P = Fv$ . Two ideas worth holding onto side by side:

$W = Pt$ . Rearranging $P = W/t$ tells you the work (or energy) delivered by a machine running at roughly constant power for a known time — this is exactly how electricity or fuel consumption is estimated for a steadily running device.
Average versus instantaneous need not match. If force or velocity changes during an interval, the instantaneous power $Fv\cos\theta$ at any one moment can differ from the average power $W/t$ over the whole interval — a car accelerating from rest delivers very different power at the start than once it is cruising, even though its average over the trip is a single number.

Figure

Two same-height weight lifts side by side, one finished quickly and one finished slowly, showing equal work but different power

Derivation AInstantaneous power from work

The goal is to show that the force-and-velocity formula $P = Fv\cos\theta$ is not a separate rule to memorise but falls directly out of the definition $P = W/t$ once that ratio is shrunk to an instant, so that the two "forms" of power a student meets in this chapter are really one idea viewed at two different time scales.

Step 1. Start from the definition of work done by a force $\vec F$ over a small displacement $d\vec S$ : $dW = \vec F \cdot d\vec S$ . This holds instant by instant, whatever the body's speed or the angle between force and motion.

Step 2. Divide both sides by the tiny time $dt$ over which that displacement happens: $\dfrac{dW}{dt} = \vec F \cdot \dfrac{d\vec S}{dt}$ . The left side is, by definition, the instantaneous power $P$ .

Step 3. Recognise $d\vec S/dt$ as the instantaneous velocity $\vec v$ of the body, so $P = \vec F \cdot \vec v$ . Writing the dot product out with the angle $\theta$ between $\vec F$ and $\vec v$ gives the usable scalar form $P = Fv\cos\theta$ , which reduces to the familiar $P = Fv$ whenever the force acts along the direction of travel.

Step 4. Check the special cases. If the force is perpendicular to the velocity ( $\theta = 90°$ ), $\cos\theta = 0$ and the force delivers zero power at that instant — it changes direction but not speed, as with a satellite's centripetal force. If the force opposes the velocity ( $\theta = 180°$ ), $P$ is negative — the force removes energy from the body, as friction does while a block decelerates. These limits confirm the formula behaves exactly as physical intuition demands.

Worked example

A pump lifts $300\ \text{kg}$ of water through a height of $10\ \text{m}$ in $50\ \text{s}$ (take $g = 9.8\ \text{m/s}^2$ ). Find the power of the pump.

Step 1. Work done against gravity in lifting the water: $W = mgh = (300)(9.8)(10) = 29{,}400\ \text{J}$ .

Step 2. Power is work divided by the time taken: $P = \dfrac{W}{t} = \dfrac{29{,}400}{50} = 588\ \text{W}$ , or about $0.588\ \text{kW}$ .

The pump delivers 588 W of power.

Competitive stretch

A car engine exerts a constant driving force of $2000\ \text{N}$ on a car moving at a constant $25\ \text{m/s}$ along a level road. Find the power developed, and the maximum speed the car could sustain at this same driving force if the engine's output is capped at $40\ \text{kW}$ .

Step 1. Force and velocity are aligned ( $\theta = 0°$ ), so $P = Fv = (2000)(25) = 50{,}000\ \text{W} = 50\ \text{kW}$ at $25\ \text{m/s}$ .

Step 2. At the power-limited maximum, $P_{max} = Fv_{max}$ , so $v_{max} = \dfrac{P_{max}}{F} = \dfrac{40{,}000}{2000} = 20\ \text{m/s}$ .

With the power capped at 40 kW, the car's top speed at this driving force is 20 m/s.

Common traps

Using $P = W/t$ (an average) where the instantaneous $P = Fv\cos\theta$ is needed, or vice versa, when force or velocity is changing during the interval.
Dropping the $\cos\theta$ factor when the force is not aligned with the velocity — a force applied at an angle delivers less power than the same force applied along the motion.
Confusing units: work and energy are in joules ( $\text{J}$ ), power is in watts ( $\text{J/s}$ ) — a "powerful" machine does work quickly, not necessarily a lot of it.
Treating a device's power rating as the power it always delivers — a rating is typically a maximum or nominal value; actual instantaneous power varies with load and speed.

Ask the tutor: Want to try a few more power problems involving lifts, pumps, or accelerating vehicles?

Why do we need two different formulas for power instead of just one?

Because they answer different questions. $P = W/t$ tells you the average rate over a stretch of time — useful when you know total work and total time, like a pump running for a fixed duration. $P = Fv\cos\theta$ tells you the power at one instant, given the force and speed right then — useful for a car whose speed is changing. They are the same physical quantity, just evaluated over an interval versus at a point; for constant force and constant speed, the two agree exactly.

If a car engine can develop more power, does that mean it can go faster forever?

No — power alone does not fix a top speed; the driving force at that speed matters too, and other forces like air resistance and friction eventually balance the engine's force. What a power limit does tell you is the maximum sustainable speed for a given driving force, as in the worked example: with force fixed, $v_{max} = P_{max}/F$ . In a real car, as speed rises, resistive forces grow, so the engine cannot keep exerting the same force at every speed — that is a separate, more advanced complication beyond this simple picture.

Is horsepower the same as watts, and do I need to convert between them for ISC problems?

Horsepower and watts both measure power, but they are different units from different systems, related by $1\ \text{hp} \approx 746\ \text{W}$ . ISC numericals almost always work directly in watts or kilowatts using $P = W/t$ and $P = Fv$ , so you will rarely need the horsepower conversion — but recognise it if a question mentions an engine "rated in hp," since you would convert to watts before applying either formula.

Power

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads