Power as the rate of doing work or transferring energy, and the watt

Reading power from an energy–time graph and energy from a power–time graph

The power needed to lift a load with a crane, motor or lift

Converting between kilowatt-hours and joules

Deriving P = Fv and using it for vehicles at constant speed, on the level and on slopes

Efficiency from energies or from powers, as a ratio or a percentage

Sankey diagrams drawn to scale

Estimating the power of a person running upstairs, with uncertainties

Work and energy tell you how much has been transferred. They say nothing about how quickly. A small electric motor and a large one can both lift the same load to the same height, and both do exactly the same work on it, but one finishes in a fraction of the time. The quantity that captures "how fast" is power.

If E or W is the total for a whole interval, these equations give the average power over that interval.

A car accelerating hard, then cruising, then braking has a power output that changes from moment to moment, and E/t smooths all of that into one number.

The watt. 1 W = 1 J s⁻¹. In SI base units a joule is kg m² s⁻², so 1 W = 1 kg m² s⁻³.

Power is a scalar, like the work and energy it is built from. It has no direction.

Large powers are given in kW (10³ W), MW (10⁶ W) or GW (10⁹ W). A kettle is about 2–3 kW, a family car engine tens of kW, a large power station about 1–2 GW.

Graphs make the definition visual. If you plot the energy transferred against time, the gradient of the line is ΔE/Δt, which is the power. A steeper line means a more powerful machine.

The reverse is just as useful. On a graph of power against time, a thin strip of width Δt and height P has area P Δt = ΔE. Adding up all the strips, the area under a power–time graph is the energy transferred. This works even when the power changes, which is where the simple formula E = Pt would fail.

Lifting is the most common power calculation, because the useful energy is easy to find. Raising a mass m through a height Δh increases its gravitational potential energy by mgΔh. Divide by the time taken and you have the useful power:

When the load rises at constant velocity there is no resultant force on it (Newton's first law), so the tension in the cable equals the weight, T = mg. Also Δh/t is then simply the speed v. Putting these together gives a short cut, P = mgv, which you will see derived properly in the P = Fv section below.

The motor must supply more than this useful power, because some energy is always wasted, mostly by friction in gears and bearings and as thermal energy in the motor windings. The link between the two is efficiency, which comes later on this page.

WORKED EXAMPLE · QUESTION

A crane lifts a pallet of bricks of mass 850 kg through a height of 18.0 m at constant speed in 30.0 s. Take g = 9.81 N kg⁻¹. Calculate (a) the work done on the pallet by the crane cable and (b) the useful power output of the crane.

SOLUTION

At constant speed the tension equals the weight: T = mg = 850 ×9.81 = 8338.5 N.

(a) Work done = force × distance moved in the direction of the force: W = 8338.5 ×18.0 = 150 093 J, so W = 1.50 × 10⁵ J.

(b) P = W/t = 150 093 ÷30.0 = 5003 W, so P = 5.00 kW.

Check with P = Fv: v = 18.0 ÷30.0 = 0.600 m s⁻¹ and 8338.5 ×0.600 = 5003 W, the same answer.

The joule is a very small amount of energy on a household scale. A kettle boiling for three minutes transfers about half a million joules, and a home uses tens of millions of joules a day. Energy suppliers therefore use a bigger unit.

kWh → J: multiply by 3.6 × 10⁶. J → kWh: divide by 3.6 × 10⁶.

Energy in kWh = power in kW × time in hours. If you keep the units matched like this, no conversion is needed at all.

The name contains "watt", but kW × h is power × time, which is energy.

WORKED EXAMPLE · QUESTION

A small wind turbine on a farm has an average useful power output of 3.2 kW over one day. (a) Calculate the energy it supplies in one day, in kWh and in J. (b) A household uses 9.5 kWh per day. How many such households could the turbine supply on this day?

SOLUTION

(a) Power in kW × time in hours gives kWh directly: E = 3.2 ×24 = 76.8 kWh, so E ≈ 77 kWh.

Convert with 1 kWh = 3.6 × 10⁶ J: E = 76.8 ×3.6 ×10⁶ = 2.765 ×10⁸ J, so E = 2.8 × 10⁸ J.

Check in SI units: 3200 W ×(24 ×3600 s) = 2.765 ×10⁸ J.

(b) 76.8 ÷9.5 = 8.08. Round down, because a household cannot be part-supplied: 8 households.

Many power questions give you a force and a speed rather than an energy and a time. You can link them in two lines. Suppose a constant force F moves its point of application a distance Δs in its own direction, at constant velocity v, in a time Δt. Then

Here F is the force doing the work (N) and v is the velocity in the direction of that force (m s⁻¹). This equation is not given to you, so be ready to derive it from W = FΔs and P = W/t. If the velocity is changing, Fv gives the power at the instant when the velocity is v.

Vehicles at constant speed

A car cruising on a level motorway is not accelerating, so by Newton's first law the resultant force on it is zero. The forward driving force from the road on the tyres must exactly balance the total resistive force (air resistance plus rolling resistance and friction in moving parts).

The engine still has to work hard, because the resistive forces are always there, taking energy away to the surroundings.

So the useful power from the engine is (resistive force) × (speed). Resistive forces grow with speed, mainly because air resistance rises steeply, so the power needed grows faster than the speed.

A vehicle reaches its top speed when the power needed to overcome resistance equals the maximum useful power the engine can deliver.

Beyond that speed the driving force available, P_(max)/v, is smaller than the resistive force, so the car would decelerate back down.

WORKED EXAMPLE · QUESTION

A car travels along a level road at a constant 27.0 m s⁻¹. The total resistive force on it is 680 N. When it travels at a constant 31.0 m s⁻¹ the resistive force is 900 N. (a) Calculate the useful power output of the engine at each speed. (b) Compare the percentage increases in speed and in power. (c) Calculate the work done against resistive forces per kilometre at each speed.

SOLUTION

(a) Constant speed, so driving force = resistive force; then P = Fv.

P = 680 ×27.0 = 18 360 W → 18.4 kW; P = 900 ×31.0 = 27 900 W → 27.9 kW.

(b) Speed: (31.0 - 27.0) ÷27.0 = 0.148 → 15%. Power: (27 900 - 18 360) ÷18 360 = 0.520 → 52%. The power rises more than three times as fast as the speed.

(c) Work = force × distance, with 1 km = 1000 m: 680 ×1000 → 6.8 × 10⁵ J; 900 ×1000 → 9.0 × 10⁵ J.

Driving faster needs more power and more energy for every kilometre, because the resistive force is larger.

Vehicles on a slope

Going uphill at constant speed, the driving force must balance the resistive force and the component of the weight down the slope, mgsin θ. The extra power mgsin θ×v is the rate at which the vehicle gains gravitational potential energy, since vsin θ is its vertical speed.

WORKED EXAMPLE · QUESTION

A cyclist and bike of total mass 80 kg ride up a straight slope inclined at 10° to the horizontal at a constant 2.50 m s⁻¹. The total resistive force is 15 N. (a) Calculate the useful power the cyclist must supply. (b) What fraction of this power goes into gravitational potential energy?

SOLUTION

(a) Weight component down the slope: mgsin θ= 80 ×9.81 ×sin 10° = 136.3 N.

Constant speed, so driving force = 136.3 + 15 = 151.3 N.

P = Fv = 151.3 ×2.50 = 378 W → P = 378 W.

(b) Rate of gain of GPE = mgsin θ×v = 136.3 ×2.50 = 341 W. Fraction = 341 ÷378 = 0.90 → about 90%. On a steep hill at low speed, nearly all the effort goes into climbing.

No real machine transfers all of its input energy into the form you want. A lift motor raises the car, but it also warms its own windings, heats the gearbox through friction and makes some noise. Efficiency measures what fraction of the input actually does the job.

Useful output is the energy transferred in the form you want. For a lift it is the gain in gravitational potential energy; for a car at steady speed it is the work done against resistive forces.

The rest is wasted, usually dissipated as thermal energy in the surroundings (and some sound). Energy is conserved, so input = useful + wasted.

Use the power form when you are given rates, and the energy form when you are given amounts. Never divide a power by an energy.

Sankey diagrams

A Sankey diagram shows energy (or power) flows as arrows whose widths are proportional to the amounts. The input enters on the left, the useful output carries straight on, and wasted output bends away, conventionally downwards. Because energy is conserved, the widths of the output arrows add up to the width of the input arrow. A diagram drawn to scale lets you read the efficiency straight off the page by comparing widths.

WORKED EXAMPLE · QUESTION

A lift car and its passengers have a total mass of 1150 kg. The lift rises at a constant 0.95 m s⁻¹. The input power to the lift motor is 14.0 kW. Ignore any counterweight. (a) Calculate the efficiency of the lift system. (b) Calculate the energy wasted while the lift rises 24 m.

SOLUTION

Useful output power: constant speed, so F = mg and P = mgv = 1150 ×9.81 ×0.95 = 10 717 W.

(a) Efficiency = 10 717 ÷14 000 = 0.766 → 77%.

(b) Wasted power = 14 000 - 10 717 = 3283 W. Time to rise 24 m: t = 24 ÷0.95 = 25.3 s.

Energy = power × time: E_wasted = 3283 ×25.3 → 8.3 × 10⁴ J.

This is a classic practical. Running up a flight of stairs raises your body through a vertical height h, so the useful power you develop is the rate of gain of gravitational potential energy, P = mgh/t. You need three measurements.

Mass m: bathroom scales, with a resolution of 0.5 kg or 0.1 kg.

Height h: measure the rise of one step with a metre rule (mm scale) and multiply by the number of steps. Check that several steps are equal, or measure the total height directly with a tape measure.

Time t: a stopwatch started as the runner sets off from the bottom step and stopped as they reach the top. Repeat several times and take the mean.

Only the vertical height matters. The horizontal distance along the treads does not change the gravitational potential energy. For safety, use dry stairs, sensible footwear and a staircase with a handrail, with nobody else on the stairs.

WORKED EXAMPLE · QUESTION

A student of mass (58.0 ± 0.5) kg runs up a staircase of 16 steps. The rise of one step is (0.172 ± 0.001) m. Five timings are 3.71 s, 3.52 s, 3.64 s, 3.58 s and 3.65 s. Determine the student's useful power output, and identify the measurement with the largest percentage uncertainty.

SOLUTION

Height: h = 16 ×0.172 = 2.752 m. Multiplying by an exact 16 multiplies the uncertainty by 16: ±0.016 m.

Mean time: t = 18.10 ÷5 = 3.62 s; uncertainty = half the range = (3.71 - 3.52) ÷2 = ±0.095 s ≈±0.10 s.

Power: P = (58.0 ×9.81 ×2.752)/3.62 = 433 W → P ≈ 430 W.

Percentage uncertainties: mass 0.5/58.0 = 0.86%; height 0.016/2.752 = 0.58%; time 0.095/3.62 = 2.6%. The time has by far the largest percentage uncertainty (2.6%), so more repeats, or light gates at the bottom and top of the stairs, would improve the result most.

Combining the separate percentage uncertainties into one for P is a Unit 6 (A2) skill.

5 Work and energy

Power is built on work done, W = FΔs, and on the energy stores mgΔh and ½mv². Most power questions start by finding one of these.

2 Forces and Newton's laws

"Constant velocity, so the forces balance" is the step that turns a power question into P = (resistive force) ×v. On a slope the weight component along the slope joins the resistive force.

Electric circuits (Unit 2) and thermal physics (Unit 5)

Electrical power P = VI is often the input power to a motor; the efficiency equation links it to the mechanical output power on this page. Wasted power heats the device and its surroundings, and with specific heat capacity you can calculate the temperature rise it causes.

1. Show that the watt is equivalent to kg m² s⁻³.

2. A 1.2 kW pump runs for 15 minutes. Calculate the energy it transfers in J and in kWh.

3. A barge is towed at a constant 5.0 m s⁻¹; the water exerts a resistive force of 2.4 kN. Calculate the power of the towing force.

4. A motor is 40% efficient and gives a useful output power of 300 W. Calculate the input power and the wasted power.

5. A car's engine can deliver a maximum useful power of 45 kW. At its top speed on level ground the resistive force is 1.25 kN. Calculate the top speed.

6. Explain why mgh/t underestimates the power output of a person running upstairs.

1. 1 W = 1 J s⁻¹. 1 J = 1 N m = 1 kg m s⁻² × m = 1 kg m² s⁻². Dividing by s gives kg m² s⁻³.

2. 1200 ×15 ×60 = 1.08 ×10⁶ J. In kWh: 1.2 ×0.25 = 0.30 kWh. Check: 0.30 ×3.6×10⁶ = 1.08×10⁶ J.

3. Constant velocity, so towing force = 2.4 kN. P = Fv = 2400 ×5.0 = 12 000 W = 12 kW.

4. Input = 300 ÷0.40 = 750 W; wasted = 750 - 300 = 450 W.

5. At top speed the driving force equals the resistive force, so v = P/F = 45 000 ÷1250 = 36 m s⁻¹.

6. It counts only the gain in gravitational potential energy. The person also gains kinetic energy, and their muscles dissipate much of the chemical energy they use as thermal energy, so their total power output is larger.