Power

Two labourers carry identical fifty-kilogram sacks of cement up the same flight of stairs to the same twelve-metre rooftop. One jogs up in eight seconds flat; the other trudges up over two minutes, stopping twice to catch his breath. Both perform exactly the same work against gravity — the sack ends up at the same height, so the energy transferred is identical. Yet ask any contractor which labourer "worked harder," and the answer is instant: the fast one. Something about the rate of doing work matters, separately from the work itself, and that something is power. Power is one of the shortest formulas in the Class 11 syllabus — a work done divided by a time, or a force dotted with a velocity — but it is also one of the most quietly powerful. It is the number stamped on every motor, light bulb, and car engine you will ever buy: a $100\ \text{W}$ bulb, a $1500\ \text{W}$ water heater, a $74.6\ \text{kW}$ (hundred-horsepower) car. It is the quantity your electricity bill is secretly built on, measured not in joules but in the everyday unit of kilowatt-hours. And it is the quiet reason a cyclist geared for a hill climb cannot also be geared for a flat-road sprint — the same engine, whether legs or motor, trades force against speed while its power output stays capped. This article builds power from the work-energy ideas you already have — work, kinetic energy, the work-energy theorem — and turns it into a working tool: how to compute average power over a stopwatch-timed interval, how to compute the instantaneous power of a force acting on a moving object using $P = Fv\cos\theta$ , why a force acting perpendicular to motion contributes nothing to power at all, and how the two worked problems from the original Class 11 course — a motor lifting cement, and a car engine cruising at constant speed — extend cleanly into the sharper, exam-style numbers you will meet in JEE and NEET.

Key takeaways

Power is the rate of doing work. Average power over a time interval is $P_{avg} = W/\Delta t$ ; instantaneous power at a moment is the derivative $P = dW/dt$ .
The working formula is $P = \vec F\cdot\vec v = Fv\cos\theta$ , obtained by dividing the small work $dW = \vec F\cdot d\vec s$ by $dt$ — force and velocity, not force alone, decide power.
The SI unit is the watt, $1\ \text{W} = 1\ \text{J/s}$ ; the practical non-SI unit is horsepower, $1\ \text{hp}\approx 746\ \text{W}$ .
Power is a scalar, even though it is built from the dot product of two vectors — it carries no direction of its own.
A force perpendicular to velocity delivers zero power — it can change direction but never speed, so it never transfers energy at that instant (uniform circular motion is the classic case).
The billed unit of energy, the kilowatt-hour, is power multiplied by time: $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$ , a direct rearrangement of $W = P\,t$ .
Key idea: power is work per unit time, computed as $P = dW/dt$ or equivalently $P = Fv\cos\theta$ , so the same work delivered faster means a larger power, and only the component of force along the velocity ever contributes.

The core insight

Power measures how fast work is done, $P = dW/dt = \vec F\cdot\vec v = Fv\cos\theta$ , so the same energy transfer can correspond to wildly different power depending on the time taken, and any force component perpendicular to the velocity contributes nothing at all.

Where it fits in the syllabus

Power closes the CBSE Class 11 unit Work, Energy and Power, coming right after work, kinetic energy, the work-energy theorem, and potential energy. It leans directly on the idea of work you have just built: power simply asks how quickly that work is delivered. The prescribed scope is exactly the definitions and formula given below, applied to translating bodies — motors lifting loads, engines driving cars, pumps lifting water — using SI units throughout. Power delivered in rotational systems, such as a torque turning a wheel ( $P = \tau\omega$ ), is explicitly kept for the later rotational-motion chapter and is out of scope here; anything phrased in terms of torque and angular velocity is a topic for another node.

The physical picture

Work done by a constant force over a displacement is $W = \vec F\cdot\vec d = Fd\cos\theta$ , a number with no clock attached — it does not care whether the displacement took a second or a year. Power repairs that omission by dividing work by the time it took. Over a finite stretch of time $\Delta t$ , the average power delivered is simply the total work divided by the total time:

P_{avg} = \frac{W}{\Delta t}

where: $W$ is the total work done (joules) and $\Delta t$ is the time interval over which it was done (seconds), so $P_{avg}$ comes out in joules per second, the watt. This is the number you compute whenever a problem gives you a total distance, a total force, and a total time and simply asks "how fast, on average, was work being delivered." It says nothing about whether the rate was steady throughout — a sprint followed by a rest, averaged over the whole interval, gives the same $P_{avg}$ as a smooth jog covering the same work in the same time. When the force or speed varies within the interval, you need the rate at a single instant rather than an average over the whole trip. Instantaneous power is defined as the derivative of work with respect to time, the limit of $P_{avg}$ as the time slice shrinks to zero:

P = \frac{dW}{dt}

where: $dW$ is the tiny work done in the tiny time $dt$ surrounding the instant of interest. Because the small work done by a force $\vec F$ over a small displacement $d\vec s$ is $dW = \vec F\cdot d\vec s$ , dividing straight through by $dt$ turns the displacement into a velocity and hands us the single most useful formula in this chapter:

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

where: $\vec v = d\vec s/dt$ is the instantaneous velocity of the point where the force acts, and $\theta$ is the angle between the force vector and the velocity vector at that instant. This form is the workhorse of every power problem you will see, because a problem almost always hands you a force and a speed directly, sparing you from ever computing $W$ and $\Delta t$ separately. Two special cases are worth burning into memory: when the force is entirely along the direction of motion, $\theta = 0°$ and $P = Fv$ ; when the force is entirely perpendicular to the motion, $\theta = 90°$ and $P = 0$ , however large the force. Tension in a string swinging a stone in uniform circular motion is exactly this second case — it changes the direction of velocity continuously, but does zero work and delivers zero power at every instant, because it never has a component along $\vec v$ . Note that power itself is a scalar quantity — the dot product $\vec F\cdot\vec v$ always collapses two vectors into a single signed number, with units of watts, and it carries no direction of its own. A negative value of $P$ (from $\theta$ between $90°$ and $180°$ ) simply means the force is doing negative work at that instant — draining kinetic energy from the object rather than supplying it, as with a retarding force such as friction or air drag.

Figure

Two carts of equal mass reach the same height on a ramp, one pulled by a small force over a long slow time and the other pushed by a large force over a short fast time, both bars showing equal work but very different power bars

A related everyday quantity built from power is energy consumed, billed by electricity companies in kilowatt-hours rather than joules: $1\ \text{kWh} = 1000\ \text{W}\times 3600\ \text{s} = 3.6\times10^{6}\ \text{J}$ , obtained by rearranging $W = P\,t$ into an energy from a power and a time. A $2\ \text{kW}$ heater run for three hours consumes $2\times3 = 6\ \text{kWh}$ of billed energy, regardless of how that consumption fluctuated minute to minute — the meter only cares about the average power times the total time.

Derivation APower delivered by a motor pulling a load up an incline at constant speed

The goal is to find the power a motor must supply to drag a block of mass $m$ up a rough incline of angle $\phi$ at constant speed $v$ , starting only from Newton's second law and the definition $P = Fv\cos\theta$ — no calculus beyond the derivative already used to define instantaneous power. This is the general version of the vertical-lift problem: it recovers the simple $mgh/\Delta t$ result when the incline is vertical and frictionless, and shows exactly which extra terms friction and slope angle add.

Step 1. Draw the forces on the block moving up the incline at constant velocity: the motor's pulling force $F$ along the incline (parallel to $\vec v$ , so $\theta=0°$ between $F$ and $v$ ), the component of gravity along the incline $mg\sin\phi$ acting down-slope, and kinetic friction $f = \mu mg\cos\phi$ also acting down-slope, opposing the motion. Because the incline is rigid, gravity's component perpendicular to the surface is balanced by the normal reaction and never appears in the along-incline equation.

Step 2. Apply Newton's second law along the incline. Since the speed is constant, the acceleration is zero, so the net force along the incline is zero: $F - mg\sin\phi - \mu mg\cos\phi = 0$ , which gives the pulling force the motor must maintain, $F = mg(\sin\phi + \mu\cos\phi)$ .

Step 3. Convert this force into power using $P = Fv\cos\theta$ with $\theta = 0°$ , since the motor's force acts along the direction of motion: $P = Fv = mgv(\sin\phi + \mu\cos\phi)$ .

Step 4. Check the two limits. If the incline is frictionless ( $\mu = 0$ ), this reduces to $P = mgv\sin\phi$ ; and if the incline is vertical ( $\phi = 90°$ , so $\sin\phi = 1$ ), it becomes $P = mgv$ , exactly the power needed to lift a mass straight up at constant speed $v$ — consistent with $P_{avg} = W/\Delta t = mgh/\Delta t = mg(v\,\Delta t)/\Delta t = mgv$ . The general formula therefore correctly contains the simple vertical-lift case as a special case, with friction and slope angle adding extra power demand on top of the pure lifting term.

Worked examples

Board level - a motor lifting a load

A motor lifts a $50\ \text{kg}$ load through a height of $12\ \text{m}$ in $8\ \text{s}$ , moving at constant velocity. Find the average power delivered by the motor. Use $g = 9.8\ \text{m/s}^2$ .

Step 1. Compute the work done against gravity. Since the load moves at constant velocity, the motor's upward force exactly balances gravity at every instant, so the total work done equals the gain in gravitational potential energy: $W = mgh = 50\times9.8\times12 = 5880\ \text{J}$ .

Step 2. Apply $P_{avg} = W/\Delta t$ , since only the total work and total time are given, not an instant-by-instant force or speed: $P_{avg} = \dfrac{5880}{8} = 735\ \text{W}$ . The motor delivers an average power of $735\ \text{W}$ , about $0.98\ \text{hp}$ — almost exactly one horsepower, a useful benchmark to remember for quick sanity checks.

Competitive level - power output of a car engine on a hill

A car of mass $1200\ \text{kg}$ climbs a straight incline of angle $\phi = 10°$ at a constant speed of $v = 20\ \text{m/s}$ . The coefficient of kinetic friction between the tyres' rolling resistance and the road is effectively $\mu = 0.04$ . Find the instantaneous power the engine must deliver. Use $g = 10\ \text{m/s}^2$ , $\sin10° \approx 0.174$ , $\cos10°\approx0.985$ .

Step 1. Use the derived incline formula $P = mgv(\sin\phi + \mu\cos\phi)$ , since the car moves at constant speed up a rough incline, exactly the situation just derived above.

Step 2. Compute the bracket: $\sin\phi + \mu\cos\phi = 0.174 + 0.04\times0.985 = 0.174 + 0.0394 = 0.2134$ .

Step 3. Substitute the numbers: $P = 1200\times10\times20\times0.2134 = 240{,}000\times0.2134$ .

Step 4. Evaluate: $P = 51{,}216\ \text{W} \approx 51.2\ \text{kW}$ , or about $68.7\ \text{hp}$ , dividing by $746\ \text{W/hp}$ . The engine delivers roughly $51.2\ \text{kW}$ of instantaneous power at that speed on that slope — close to, but a little above, the flat-road figure of $50\ \text{kW}$ computed for the same engine's driving force earlier in this course, because climbing costs extra power on top of overcoming friction alone. Timing note: under exam pressure, the single costliest slip in this problem is forgetting that $\theta = 0°$ between the net along-incline pulling force and the velocity — students sometimes try to multiply the full weight $mg$ by $v$ directly, ignoring that only the along-incline component of gravity and friction resist the motor. Write the free-body equation along the incline first, solve for $F$ , and only then multiply by $v$ ; do not shortcut straight to $P=mgv$ unless the incline actually is vertical.

Common traps

Confusing power with work or energy — power is the rate at which work is done, not the amount of work itself; two agents can do identical work but deliver very different power if the time taken differs, exactly as in the two labourers on the stairs.
Forgetting the $\cos\theta$ factor in $P = Fv\cos\theta$ — if force and velocity are not parallel, only the component of force along the velocity contributes to power; a force entirely perpendicular to velocity, however large, contributes nothing.
Assuming a bigger force always means bigger power — power depends on both force and speed together; a small force at very high speed can easily out-power a large force at low speed, which is exactly why gear ratios trade one for the other at fixed engine power.
Mixing up average power (over a finite interval, from $W/\Delta t$ ) with instantaneous power (at one moment, from $Fv\cos\theta$ ) — for a varying force or speed these can differ substantially, and a problem must be read carefully to see which one is actually being asked for.
On incline or friction problems, plugging $mg$ straight into $P = Fv$ without first finding the correct along-incline pulling force from Newton's second law — see the timing note above.

You should be able to…

State the definitions of average power ( $P_{avg}=W/\Delta t$ ) and instantaneous power ( $P=dW/dt$ ), with the SI unit watt.
Apply $P = \vec F\cdot\vec v = Fv\cos\theta$ to find the instantaneous power of a known force acting on a moving object.
Explain, with a concrete example, why a force perpendicular to velocity delivers exactly zero power.
Derive the power needed to move a load up a rough incline at constant speed, and recover the simple vertical-lift formula $P=mgv$ as its special case.
Convert cleanly between joules, watts, kilowatts, horsepower, and kilowatt-hours in everyday energy-billing and engine-rating contexts.
Spot and avoid the common trap of substituting the full weight into $P=mgv$ on an incline or friction problem without first solving the along-incline force equation.

Ask the tutor: Wondering how a fixed engine power caps a vehicle's maximum speed once drag is included — ask the AI tutor to work through $P = Fv$ against a speed-dependent drag force.

Is power the same thing as energy?

No, and mixing the two up is one of the most common exam slips. Energy (or work) is an amount — how much was transferred, measured in joules, with no clock attached. Power is a rate — how quickly that amount was transferred, measured in watts (joules per second). A small torch battery and a giant power-station generator can both deliver the same total energy over their lifetimes, but the generator does it at a vastly higher power, meaning far faster. Always ask "how much" for energy and "how fast" for power.

Why does a force perpendicular to velocity do zero work and deliver zero power?

Work and power both depend on the component of force along the direction of motion, captured by the $\cos\theta$ factor. When $\theta = 90°$ , that component is zero, so the force changes only the direction of the velocity, never its magnitude, and therefore never changes the object's kinetic energy at that instant. The tension in a string whirling a stone in a horizontal circle is the classic example: it constantly redirects the stone, but the stone's speed never changes because tension does zero work throughout the motion.

Why is the kilowatt-hour used for electricity bills instead of the joule?

Because the joule is an awkwardly small unit for household energy use — a typical home consumes tens of millions of joules a day, an unwieldy number to print on a bill. The kilowatt-hour is simply a bigger, more convenient energy unit built directly from power and time: run a one-kilowatt appliance for one hour and you have used one kilowatt-hour, which works out to $3.6\times10^{6}\ \text{J}$ . It is still an energy, not a power, despite the word "power" hiding inside "kilowatt" — always convert to joules before mixing it into an equation that expects SI energy units.

If two cars have engines of equal power but very different maximum forces at low speed, which one is faster at top speed?

At a fixed engine power $P$ , the relation $P = Fv$ means force and speed are locked in an inverse trade at any given moment: a gearbox can deliver a large $F$ at low $v$ (good for climbing or accelerating from rest) or a small $F$ at high $v$ (good for a high top speed on the flat), but the product $Fv$ stays pinned near the engine's rated power once friction and drag are accounted for. Two engines of equal power reach similar theoretical top speeds when drag is the limiting force; the car with the larger force at low speed instead wins the acceleration race off the line, not the top-speed contest. This trade-off between force and speed at fixed power is exactly what a vehicle's gears are built to exploit.