In short
Power is the rate of doing work: P = \frac{dW}{dt}. For a force F acting on a body moving at velocity v, the instantaneous power is P = \vec{F} \cdot \vec{v}. The SI unit is the watt (1 W = 1 J/s). Average power over a time interval is \bar{P} = \frac{W}{t}. One horsepower equals 746 W.
Two autorickshaws sit at a traffic signal in Bangalore. When the light turns green, both need to reach 30 km/h. The first one is a classic two-stroke with a 7 kW engine. The second is a newer four-stroke putting out 11 kW. Both carry the same load — one passenger with a backpack. The 11 kW engine reaches 30 km/h in about 5 seconds. The 7 kW engine takes nearly 8 seconds to reach the same speed.
Both engines did the same total work — they gave the same kinetic energy to the same mass. But the stronger engine did it faster. That "faster" is what physics calls power.
Power is not about how much work you do. It is about how quickly you do it. A construction worker carrying bricks up to the second floor of a building does the same work whether they sprint up the stairs in 30 seconds or walk up in 2 minutes. The difference is power: the sprinting worker delivers work at a higher rate.
Defining power — the rate of doing work
Start with something you already know: work. When a force \vec{F} moves an object through a displacement d\vec{s}, the work done is dW = \vec{F} \cdot d\vec{s}. This tells you how much energy has been transferred. But it tells you nothing about how long the transfer took.
To capture the time dimension, define power as the rate at which work is done:
Why the derivative: work done is a cumulative quantity — it adds up over time. Taking its derivative with respect to time gives you the instantaneous rate at which energy is being transferred, which is exactly what you need to describe "how fast" the work is happening.
This is the general definition. If the force and velocity are both known at an instant, you can rewrite this in a more useful form.
From dW/dt to F · v
Since dW = \vec{F} \cdot d\vec{s}, divide both sides by dt:
Why this step works: the dot product is between the force vector and the displacement vector. Dividing the displacement by time gives velocity. The force does not depend on dt, so it passes through untouched.
But \frac{d\vec{s}}{dt} = \vec{v}, the instantaneous velocity. Therefore:
Why this is powerful: you no longer need to track the work done over an interval. If you know the force and velocity at any single instant, you know the power at that instant. This is the formula you will use most often.
Since this is a dot product, power depends on the angle \theta between the force and velocity:
When the force is along the direction of motion (\theta = 0°), P = Fv — the full magnitude of force times the full speed. When the force is perpendicular to the velocity (\theta = 90°), P = 0. This is why the normal force on a car driving on a flat road delivers zero power — it acts vertically while the car moves horizontally.
Instantaneous power
The instantaneous power delivered by a force \vec{F} to a body moving at velocity \vec{v} is
where \theta is the angle between the force and the velocity. When \vec{F} and \vec{v} are parallel, P = Fv.
Reading the definition. \vec{F} is the force acting on the body — not the net force necessarily, but the specific force whose power you want to measure. \vec{v} is the body's velocity at that instant. The dot product picks out the component of force along the direction of motion. If the force has no component along the velocity, it does no work and delivers no power.
Average power
You do not always need the instantaneous value. Often, you care about the total work done over a time interval divided by that interval. This is the average power:
Why this is useful: when your electricity bill says you used 150 units (kWh) in a month, the bill is measuring total energy, and the rate at which you consumed it is the average power. Most practical calculations — fuel consumption, engine ratings, energy costs — use average power.
The connection between instantaneous and average is the same as for velocity: the average power over an interval is the time-averaged value of the instantaneous power over that interval:
If the power is constant (a motor running at steady speed under constant load), average power equals instantaneous power. If the power varies (an engine accelerating from rest), the average is lower than the peak.
Units — the watt and horsepower
The watt
The SI unit of power is the watt (W):
Why these dimensions: power = work / time = (force × distance) / time = (kg·m/s²)(m) / s = kg·m²/s³. The watt bundles this into a single clean unit.
Some everyday benchmarks to anchor your intuition:
| Source | Approximate power |
|---|---|
| LED bulb | 9 W |
| Ceiling fan | 75 W |
| Human body at rest (basal metabolic rate) | 80 W |
| Human cycling at moderate effort | 75–150 W |
| Human climbing stairs briskly | 300–500 W |
| Washing machine motor | 500 W |
| Immersion water heater (Indian standard) | 1000–1500 W |
| Autorickshaw engine | 7,000–11,000 W (7–11 kW) |
| Small car (Maruti Alto) | ~35 kW (47 bhp) |
| Delhi Metro train (per car) | ~300 kW |
| ISRO PSLV first-stage solid motor | ~5,000 kW |
The kilowatt-hour — energy, not power
Your BSES or TPDDL electricity bill shows consumption in kilowatt-hours (kWh). Despite the name containing "watt," a kWh is a unit of energy, not power:
Why energy: power is the rate, energy is what you accumulate. Running a 1000 W heater for 1 hour uses 1 kWh of energy. Running it for 2 hours uses 2 kWh. The power is the same; the energy doubles.
If electricity costs ₹8 per unit (1 unit = 1 kWh), running that 1000 W heater for 3 hours costs:
Horsepower
The horsepower (hp) is a legacy unit still used for engine ratings in India and worldwide:
When a car's specification sheet says "82 bhp" (brake horsepower), that means the engine produces 82 \times 746 = 61{,}172 W \approx 61 kW at its peak RPM. The "brake" refers to the measurement method — a dynamometer brake measures the torque and rotation rate at the output shaft.
Why does horsepower persist? The number was chosen to sell steam engines: mine owners could compare the engines to their horses. The physics community moved to watts, but the automotive and appliance industries kept horsepower because customers are used to it.
The physics of power — what it tells you about motion
Power during constant-force acceleration
A car of mass m accelerates from rest under a constant net force F. By Newton's second law, a = F/m, so v = at = Ft/m. The instantaneous power is:
Why power grows linearly: the force is constant, but the speed is increasing. Since power = force × speed, and speed grows with time, power grows with time too. The engine must deliver more and more power as the car goes faster, even though the force hasn't changed.
This is the fundamental constraint on acceleration at high speeds. A car engine has a maximum power output P_{\max}. As speed increases, the force the engine can deliver falls:
At low speeds, the engine has force to spare and acceleration is strong. At high speeds, the available force drops, and acceleration tapers off. Eventually, the driving force equals the drag and friction forces, and the car reaches its top speed — the point where all the engine's power goes into fighting resistance.
Why engines are rated by power, not force
An autorickshaw engine produces about 7 kW. A Maruti Alto engine produces about 35 kW. Both can exert large forces at low speed (in first gear). But the Alto can sustain a higher speed on a highway because it can maintain more power at high velocity. The power rating tells you how much energy per second the engine can deliver — and that determines the speed-force trade-off.
This is also why the Delhi Metro uses regenerative braking. When a train decelerates, its motors run as generators, converting kinetic energy back into electrical energy and feeding it to the grid. A 300-ton train entering a station at 80 km/h carries roughly \frac{1}{2}(300{,}000)(22.2)^2 \approx 74 MJ of kinetic energy. If regenerative braking recovers 30% of this, it feeds about 22 MJ back — enough to power roughly 6,000 LED bulbs for an hour.
Worked examples
Example 1: Climbing the stairs at Rajiv Chowk
A Delhi Metro commuter (mass 70 kg, including bag) climbs a staircase of vertical height 8 m at Rajiv Chowk station. They take 20 seconds to reach the top. Find the average power they deliver against gravity.
Step 1. Find the work done against gravity.
Why: the only work that matters against gravity is the vertical displacement. Whether the staircase is steep or gentle, the work is the same — it depends only on the height gained and the weight.
Step 2. Compute the average power.
Why: dividing total work by total time gives average power. This is about 0.37 horsepower — roughly a third of a horse's sustained effort, which is reasonable for a person climbing briskly.
Step 3. Compare with the human body's capability.
A healthy adult can sustain about 75 W of mechanical power while cycling and can produce short bursts up to 500 W climbing stairs. This commuter's 274 W is well within the burst range — they are working hard but not at maximum effort.
Why this comparison matters: it connects the abstract number (274 W) to something the reader can feel in their body. Climbing stairs is hard precisely because you are delivering several hundred watts against gravity.
Step 4. Express as metabolic power.
The human body is about 25% efficient at converting food energy to mechanical work. So the total metabolic power while climbing is:
Why: three-quarters of the food energy goes to heat, not useful work. This is why you sweat when climbing stairs — most of your energy output is thermal.
Result: The commuter delivers an average mechanical power of 274 W (about 0.37 hp) against gravity. Their body actually consumes about 1.1 kW of metabolic power, with the rest dissipated as heat.
What this shows: Climbing stairs is a high-power activity for the human body. The average power formula \bar{P} = W/t directly tells you how demanding a physical task is — and why elevators exist in metro stations alongside staircases.
Example 2: An autorickshaw accelerating from a signal
An autorickshaw (total mass 400 kg including passengers) accelerates from rest to 36 km/h (10 m/s) in 8 seconds. The engine delivers a constant force during this acceleration. Find (a) the average power delivered by the engine, and (b) the instantaneous power at t = 8 s, just as the rickshaw reaches 36 km/h. Neglect friction and air resistance.
Step 1. Find the acceleration.
Why: constant force means constant acceleration. The rickshaw gains 1.25 m/s of speed every second.
Step 2. Find the driving force.
Why: Newton's second law directly. With friction neglected, the entire engine force goes into accelerating the mass.
Step 3 (a). Find the total work done, then the average power.
Total work equals the kinetic energy gained (by the work-energy theorem):
Why use the work-energy theorem: it gives the total work in one step without needing the displacement. Since the only force is the engine's, the work done equals the kinetic energy gained.
Step 3 (b). Find the instantaneous power at t = 8 s.
At t = 8 s, the velocity is 10 m/s. The force is still 500 N.
Why the instantaneous power is double the average: power increases linearly with time (since P = Fv and v = at). At t = 0, P = 0. At t = 8 s, P = 5 kW. The average of a quantity that grows linearly from 0 to 5 kW is \frac{0 + 5}{2} = 2.5 kW — exactly what you computed.
Step 4. Verify: average of a linear function.
Since P(t) = F \cdot at = (500)(1.25)t = 625t:
Why verify with integration: it confirms that the average-power formula W/\Delta t and the time-averaged instantaneous-power formula give the same answer, as they must.
Result: Average power = 2.5 kW (3.35 hp). Instantaneous power at the end = 5 kW (6.7 hp). The instantaneous power at the end is exactly twice the average — a direct consequence of power growing linearly from zero under constant acceleration.
What this shows: Even though the force is constant, the power demand increases with speed. An engine's peak power output limits how fast you can accelerate at high speed, even if the engine can provide enough force at low speed.
Common confusions
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"Power and energy are the same thing." They are not. Energy is the total amount of work done or stored. Power is the rate at which energy is transferred. A 100 Wh battery stores energy. It can deliver that energy at 100 W for 1 hour, or at 50 W for 2 hours. The power and duration are different; the energy is the same.
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"A powerful engine always means a fast car." Not necessarily. Power determines the maximum speed and the rate of acceleration, but the actual speed also depends on the car's mass, aerodynamic drag, and road friction. A 100 kW engine in a 2000 kg SUV will be slower than a 100 kW engine in a 800 kg hatchback.
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"If power is zero, nothing is happening." Zero power means no work is being done at that instant. But forces can still act. The normal force on a car driving on a flat road exerts zero power because it is perpendicular to the motion. Centripetal force on a satellite in circular orbit delivers zero power — the force is perpendicular to the velocity, so P = Fv\cos 90° = 0. The satellite moves fast and feels a large force, but its kinetic energy never changes.
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"Horsepower is bigger than watts." Horsepower is a larger unit — 1 hp = 746 W — but it measures the same physical quantity. Saying an engine produces "100 hp" is identical to saying it produces "74,600 W." The number is different because the unit is different, just as 1 km = 1000 m does not mean kilometres are "bigger" than metres.
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"Average power and instantaneous power give the same information." They do not. A sprinter running 100 m in 10 seconds might have an average power of 500 W, but their instantaneous power during the first 2 seconds (acceleration phase) could peak at 2000 W. Average power smooths over the peaks and troughs. For designing engines, you need the peak (instantaneous maximum). For calculating electricity bills, you need the average.
If you came here to understand power, use the formula P = \vec{F} \cdot \vec{v}, and solve problems, you have what you need. What follows is for readers preparing for JEE Advanced who want to see power in more general contexts: variable force, angular motion, and the formal connection between power and energy curves.
Power with variable force — the general integral
When force varies with time or position, the total work is:
Why this is the general form: P = dW/dt, so W = \int P\,dt. This replaces the simple W = Fd when force and speed are not constant.
For a force that depends on position (like a spring), you can still use this. A spring with F = -kx compressed by x_0 and released: the velocity at position x is v = \sqrt{(k/m)(x_0^2 - x^2)} (from energy conservation), and the power delivered by the spring at that instant is:
The sign carries physical meaning: power is positive when force and velocity point the same way (spring accelerating the block toward equilibrium), and negative when they oppose (spring decelerating the block as it overshoots past equilibrium).
Power in rotational motion
For a rotating body, the analogous formula replaces force with torque and linear velocity with angular velocity:
Why this is the rotational analogue: work in rotation is W = \tau \theta (torque times angular displacement), so P = dW/dt = \tau \cdot d\theta/dt = \tau \omega.
This is how engine power is actually measured. A dynamometer measures the torque \tau at the output shaft and the rotational speed \omega (in rad/s). The product gives power:
where N is the RPM. A car engine producing 120 N·m of torque at 4000 RPM delivers:
Power and the velocity-time curve
Here is a useful connection: if you plot P vs t, the area under that curve is the total work done:
This means you can read energy from a power-time graph the same way you read displacement from a velocity-time graph. For JEE problems, this allows you to solve "power delivered by a variable engine" problems graphically.
Terminal velocity under constant power
A classic JEE problem: a car engine delivers constant power P on a level road. Air resistance is f = bv^2. Find the terminal velocity.
At terminal velocity, the driving force equals the drag force, and the speed is constant. The driving force is F = P/v (since P = Fv). Setting F = f:
Why the cube root: at constant power, force falls as 1/v. Drag rises as v^2. They balance when 1/v = v^2 (up to constants), giving v^3 = \text{const}, hence v \propto P^{1/3}. Doubling the engine power increases terminal velocity by only 2^{1/3} \approx 1.26 — a 26% increase, not a doubling.
This explains why doubling a car's engine power does not double its top speed. Most of the extra power goes into fighting the steeply rising air resistance.
ISRO's rocket engines — thrust power
A rocket engine produces thrust F while the rocket moves at velocity v. The power delivered by the thrust is P = Fv. For ISRO's PSLV, the first-stage solid booster produces about 4,800 kN of thrust. At a velocity of 1 km/s (achieved about 60 seconds after launch), the thrust power is:
That is 4.8 gigawatts — roughly the output of four large nuclear power plants — concentrated in a single rocket stage for about two minutes.
Where this leads next
- Work Done by a Force — the foundational concept that power builds on: how forces transfer energy through displacement.
- Kinetic Energy and the Work-Energy Theorem — the theorem connecting net work to changes in kinetic energy, used directly in power calculations.
- Conservation of Mechanical Energy — when total mechanical energy stays constant, power analysis tells you how it flows between kinetic and potential forms.
- Work Done by Variable Forces — extending the power integral to forces that change with position or time.
- Circular Motion — where centripetal force delivers zero power, connecting back to why P = Fv\cos\theta matters when \theta = 90°.