Power (rate of doing work)

Two workers can lift identical loads to the same height and do exactly the same work — yet if one finishes in half the time, everyone calls that worker "more powerful." That everyday word hides a precise physics quantity: power, the rate at which work is done or energy is transferred. It is the last idea in the CBSE Class 9 "Work and Energy" chapter, and it ties together everything learned earlier about work ( $W = F \times s$ ) and energy by adding the missing ingredient — time. This article defines power as $P = W/t$ , introduces its SI unit the watt, shows how to handle situations where the rate of doing work is not constant using average power, and derives the useful link between power, force, and velocity for steady motion. A single worked example — the same crane you may already know from Class 9 — is extended to show all these ideas working together.

The core insight

Power is the rate of doing work, $P = W/t$ , measured in watts (joules per second), and when the force stays constant while an object moves at steady velocity this becomes $P = F v$ .

The core idea

Power is defined as the amount of work done per unit time:

P = \frac{W}{t}

where: $P$ is power, $W$ is the work done (in joules), and $t$ is the time taken (in seconds) to do that work. Because work and energy share the same unit, power is equally the rate at which energy is transformed or transferred from one form or place to another.

Units. The SI unit of power is the watt (W), named after James Watt. One watt is defined directly from the formula: a rate of doing work of one joule every second.

1\ \text{W} = 1\ \text{J/s}

Because a watt is a small amount of power for many everyday machines, power is often expressed in kilowatts (kW), where $1\ \text{kW} = 1000\ \text{W}$ . (The kilowatt-hour, a unit of energy used on electricity bills, is not part of this topic and is not covered here.)

When the rate of doing work changes. Real machines rarely do work at a perfectly steady rate — a crane might strain harder against a heavy load, then ease off. To describe overall performance in such cases we use average power: the total work done divided by the total time taken, never a simple average of two separate power values.

P_{avg} = \frac{\text{total work done}}{\text{total time taken}}

This single figure summarises an entire task — however uneven the pace — without requiring a moment-by-moment record of how fast work was being done at each instant.

Figure

Two bar charts side by side comparing a slow worker and a fast worker who both do the same forty joules of work but the fast worker finishes in half the time and so has twice the power

Derivation APower in terms of force and velocity

This derivation starts from the basic definitions of power and work already used earlier in the chapter and shows that when a constant force acts on an object moving at a constant velocity in the direction of that force, the power delivered equals simply the product of the force and the velocity, a result that is extremely useful whenever a speed is known instead of a time.

Step 1. Begin with the definition of power as the rate of doing work: $P = \dfrac{W}{t}$ .

Step 2. Recall that when a constant force $F$ moves an object through a displacement $s$ in the direction of the force, the work done is $W = F \times s$ . Substituting this into the power formula gives $P = \dfrac{F \times s}{t}$ .

Step 3. Regroup the right-hand side as $P = F \times \dfrac{s}{t}$ . The term $\dfrac{s}{t}$ is simply the distance covered per unit time, which, for motion at constant velocity, is exactly the speed $v$ of the object.

Step 4. Replacing $\dfrac{s}{t}$ with $v$ gives the compact result:

P = F v

where: $F$ is the constant force (in newtons) causing the motion, and $v$ is the constant velocity (in m/s) of the object in the direction of that force. This form is only valid while both the force and the velocity stay steady — if either changes, $P = Fv$ gives only the instantaneous power at that moment, not an average over a longer, uneven process.

Worked example

A crane does 45,000 J of work in lifting a car engine straight up in 15 seconds. Find (a) the crane's power, (b) the weight of the engine if it is the only force the crane works against, and (c) verify the answer using $P = Fv$ .

Step 1. Find the power using the basic definition. Given $W = 45{,}000\ \text{J}$ and $t = 15\ \text{s}$ : $P = \dfrac{W}{t} = \dfrac{45{,}000}{15} = 3000\ \text{W} = 3\ \text{kW}$ .

Step 2. Find the height risen, using $W = F \times s$ . Since the crane works only against the engine's weight, the weight equals the force $F$ . We are told separately that the crane lifts the engine, whose weight is $F = 1500\ \text{N}$ , so the height risen is $s = \dfrac{W}{F} = \dfrac{45{,}000}{1500} = 30\ \text{m}$ .

Step 3. Find the (constant) speed of lifting: $v = \dfrac{s}{t} = \dfrac{30}{15} = 2\ \text{m/s}$ .

Step 4. Verify using the derived formula $P = Fv = 1500 \times 2 = 3000\ \text{W}$ , which exactly matches Step 1.

The crane's power is 3000 W (3 kW), confirmed two independent ways: from total work and time, and from force and velocity.

Competitive stretch

If the same crane later lifts a lighter, 900 N load through the same 30 m in only 9 s instead, its power becomes $P = Fv = 900 \times \dfrac{30}{9} = 3000\ \text{W}$ again — the same power can come from a smaller force moving faster, showing that power depends on the combination of force and speed, not on either alone.

Common traps

Confusing work and power: work is how much is done (joules), power is how fast it is done (watts). Doing the same work in less time means more power, not more work.
Writing the unit of power as the joule instead of the watt, or forgetting that a watt is joules per second, not joules alone.
Averaging two power values directly, as $\dfrac{P_1 + P_2}{2}$ , when the two stages take different times — average power must always be total work divided by total time.
Using $P = Fv$ when the force or velocity is not actually constant during the interval — in that case only $P = W/t$ (or $P_{avg}$ ) correctly describes the whole process.

Ask the tutor: If you are stuck combining several stages of a task — each with its own force, time, or work done — into one overall average power, ask the AI tutor to walk through the total work and total time with you step by step.

Is power the same thing as speed

No. Speed is distance covered per unit time and has nothing to do with force or work by itself. Power is work done per unit time. They coincide only in the special case $P = Fv$ , where power for a constant force equals that force multiplied by speed — but power always carries the idea of work (and hence force) built into it, while speed does not.

Why can I not just average the power values from two different stages of a task

Because power is not evenly weighted by work in a simple average — a stage that lasts longer contributes more to the total time (and often more work) than a short stage does. Averaging $P_1$ and $P_2$ directly ignores how long each stage lasted. The only way to get a true average power is $P_{avg} = \dfrac{\text{total work}}{\text{total time}}$ , which automatically weights each stage correctly.

Can power ever be zero even if a large force is acting

Yes. If a large force acts but produces no displacement in its direction — for example, someone pushing hard against a wall that does not move — then the work done is zero ( $W = F \times s$ with $s = 0$ ), so the power is also zero, however great the effort or force involved. Power requires actual work being done over time, not just a force being present.

Power (rate of doing work)

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads