Power

Two cranes can lift the same 500 kg beam to the same height and do exactly the same work — yet one finishes in half a minute and the other takes five minutes. Something clearly separates them, and it is not the work done, which is identical. It is the rate at which that work gets done. That rate is called power, and it is one of the shortest, most numerical ideas in the ICSE Class 10 "Force, Work, Power and Energy" unit. A single formula, $P = W/t$ , drives almost every question: convert time carefully, combine it with $W = mgh$ or $W = Fs\cos\theta$ , and watch the unwary horsepower conversion. This article builds that one formula up from the definition, works two full numericals (a lifting motor and a horsepower engine), and flags the traps examiners like to set.

The core insight

Power is the rate of doing work, $P = W/t$ , measured in watts, where one watt equals one joule of work done every second.

The core idea

Power is defined as the rate at which work is done, or equivalently the rate at which energy is transferred or converted from one form to another:

P = \frac{W}{t}

where: $P$ is power, $W$ is the work done (in joules), and $t$ is the time taken (in seconds). The SI unit of power is the watt (W), defined so that

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

where: one watt is exactly one joule of work per second — a small unit, so everyday devices are usually rated in kilowatts, $1\ \text{kW} = 1000\ \text{W}$ . A second, older unit still shows up in ICSE numericals: the horsepower, historically used to rate steam and petrol engines against the pulling power of a horse.

1\ \text{hp} = 746\ \text{W}

where: the conversion factor 746 must be applied before any horsepower value can be substituted into $P = W/t$ , since horsepower is not an SI unit. Because work along a straight line is $W = Fs\cos\theta$ (or simply $Fs$ when force and displacement point the same way), power can be rewritten in terms of force and speed for a body moving at constant velocity in the direction of the force:

P = \frac{Fs}{t} = F \times \frac{s}{t} = Fv

where: $F$ is the applied force, $v = s/t$ is the constant speed, and this form is handy whenever a problem states a force and a speed directly instead of a distance and a time — though most Class 10 numericals still lean on the basic $P = W/t$ combined with $W = mgh$ for lifting problems. The key conceptual point is that power depends on both how much work is done and how quickly it is done — never on either alone. A small job finished in a flash can demand more power than a huge job spread over a whole day. Two machines that do the identical total work over 24 hours can carry very different power ratings if one finishes in an hour and the other grinds on until nightfall: power is what tells apart the punchy device from the slow, steady one, even when their total energy output matches exactly. Power ratings surround us: light bulbs are rated in watts, vehicle engines in kilowatts or horsepower, and household electricity bills use the commercial energy unit kilowatt-hour — simply power in kilowatts multiplied by time in hours, the same relation $P = W/t$ rearranged as $W = P \times t$ .

Figure

A bar chart comparing two workers lifting the same load with the same work done but different times and different power, the faster worker showing a taller power bar

Derivation APower developed by a motor lifting a load against gravity

The goal is to show, step by step and with unit checks at every stage, how the definition of power $P = W/t$ combines with the work done against gravity $W = mgh$ to produce a single, exam-ready expression for the power a lifting motor must develop, using nothing beyond the definitions already stated above.

Step 1. Start from the definition of power, $P = W/t$ , and identify what "work" means for a motor lifting a load straight up: the motor must supply energy equal to the gravitational potential energy gained by the load, so $W = mgh$ , where $m$ is the mass lifted, $g$ is the acceleration due to gravity, and $h$ is the height risen.

Step 2. Substitute $W = mgh$ into $P = W/t$ to get the combined working formula for a vertical lift:

P = \frac{mgh}{t}

Step 3. Check the units. Mass in kilograms times $g$ in metres per second squared times height in metres gives joules for the numerator (since $\text{kg} \times \text{m/s}^2 \times \text{m} = \text{N} \times \text{m} = \text{J}$ ); dividing by time in seconds gives joules per second, which is exactly the watt. The formula is dimensionally consistent and ready to use directly with $m$ in kg, $g = 9.8\ \text{m/s}^2$ , $h$ in metres and $t$ in seconds.

Step 4. Note the two most common exam variants of this same idea: if the problem instead gives a horizontal force and distance, use $P = Fs/t$ (or $P = Fv$ for constant speed); if it gives horsepower, convert to watts first using $1\ \text{hp} = 746\ \text{W}$ before comparing with anything computed from $mgh/t$ .

Worked example

A motor lifts a load of 500 kg through a height of 20 m in 25 seconds. Calculate the power developed by the motor. (Take $g = 9.8\ \text{m/s}^2$ .)

Step 1. Calculate the work done against gravity using $W = mgh$ .

W = mgh = 500 \times 9.8 \times 20

W = 500 \times 9.8 = 4900; \quad 4900 \times 20 = 98000\ \text{J}

Step 2. Apply the power formula $P = W/t$ with $t = 25\ \text{s}$ .

P = \frac{W}{t} = \frac{98000}{25}

Step 3. Calculate the result.

P = 3920\ \text{W} = 3.92\ \text{kW}

The motor develops 3920 W, or 3.92 kW, of power.

Competitive stretch

An engine is rated at 15 hp. Express this power in watts, and calculate how much work the engine can do in 2 minutes, assuming it works at this constant rate.

Step 1. Convert horsepower to watts using $1\ \text{hp} = 746\ \text{W}$ .

P = 15 \times 746 = 11190\ \text{W}

Step 2. Convert the time to seconds, since the watt is defined per second.

t = 2\ \text{min} = 2 \times 60 = 120\ \text{s}

Step 3. Rearrange $P = W/t$ to $W = P \times t$ and substitute.

W = 11190 \times 120 = 1342800\ \text{J} \approx 1343\ \text{kJ}

The engine, working at a constant 11,190 W, does about 1343 kJ of work in 2 minutes.

Common traps

Confusing work and power — work ( $W = Fs\cos\theta$ , joules) tells you how much is done; power ( $P = W/t$ , watts) tells you how fast. A question asking for power but giving only force and distance is unsolvable without the time.
Forgetting to convert time into seconds before substituting into $P = W/t$ ; minutes and hours must become seconds first.
Plugging horsepower straight into $P = W/t$ without converting to watts using $1\ \text{hp} = 746\ \text{W}$ .
Assuming a heavier load or greater distance always means more power — power depends on the combination of work and time, so a light load moved quickly can need more power than a heavy load moved slowly.

Ask the tutor: Want more practice converting between horsepower and watts, or combining $mgh$ with $P = W/t$ in mixed numericals? Ask the AI tutor to work through a few with you.

Why does power need both work and time, and not just one of them?

Work alone only tells you the total energy transferred, with no sense of urgency; time alone tells you nothing about how much was actually done. Power ties the two together as a rate, $P = W/t$ , so it distinguishes a job done in a rush from an identical job done leisurely. That is exactly why two workers lifting the same load to the same height can have very different power ratings even though their work done is the same.

Why can I not just substitute horsepower directly into P = W/t?

Because the watt is an SI unit defined as one joule per second, while horsepower is an older, non-SI unit historically based on a horse's pulling rate. The two units are related but not numerically equal: $1\ \text{hp} = 746\ \text{W}$ . Any formula built from SI quantities like $mgh$ and seconds expects power in watts, so a horsepower value must always be multiplied by 746 first.

When should I use P = Fv instead of P = W/t?

Use $P = Fv$ when a problem directly gives you a constant force and a constant speed, rather than a distance and a time — it comes from the same definition, since $P = Fs/t = F \times (s/t) = Fv$ . Most Class 10 lifting numericals instead give a mass, height and time, in which case the simpler route is $W = mgh$ followed by $P = W/t$ ; both paths agree because they come from the same underlying definition of power.

Power

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads