1-6 Power and Efficiency

Chapter 1 · Mechanics

❓ Lesson question

Two lifts raise the same load through the same vertical height, but one completes the lift in half the time. Do both lifts perform the same useful work? Which lift has the greater power?

You will learn to

1Calculate the power of a machine from the work done and the time taken, or from force and constant velocity.
2Determine the efficiency of a device from its useful output energy and total input energy, expressed as a percentage.
3Interpret energy transfer in a Sankey diagram and identify where the wasted energy is dissipated.

Two lifts rise side by side in one building. Each carries the same load to the same floor — the same height. Lift $A$ takes 20 s; lift $B$ takes 10 s, half the time.

People sometimes say that the faster lift “does more work”. Is that true?

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) Did the faster lift perform more useful work on the load, or did both lifts perform the same useful work?

(b) A salesperson claims, "The more powerful motor uses less electrical energy." Could this be true?

Write a line for each before this opens.

🔬 Explore — climb the school stairs

What you’ll do: use a scale and a stopwatch to measure the power you produce while climbing stairs.

⏱️ 12 min🧰 a digital weighing scale · a stopwatch or phone timer · a measuring tape

1Measure your mass m in kilograms.
2Measure the vertical height h of one flight of stairs in metres — the vertical rise, not the length of the staircase.
3Calculate the work done in lifting yourself through one flight: W = mgh, taking g = 9.8 m/s².
4Climb the flight slowly while a partner times you. Record the result.
5Repeat the climb as rapidly as is safely possible, and record the shorter time. Then compare W ÷ t for the two climbs.

🎛️ Simulation — power and efficiency

Load mass500 kg

Height lifted6 m

Time taken12 s

Useful work W = mgh

29.40 kJ

Drag the time slider — this does not move. Work has no time in it.

Power P = W / t

2.45 kW

…but this does. Halve the time and it doubles.

Steady speed v = h / t

0.50 m/s

Check: P = Fv

2.45 kW

4900 N × 0.50 m/s — the same answer, the other route

Press Lift A then Lift B. The work is identical — 52.92 kJ both times, because the same load rose the same height. Only the power changes, from 2.21 kW to 4.41 kW. The faster lift did not do more work; it did the same work sooner. That is the whole distinction the lesson question is asking about.

Analyse what you collected

(a) Did the work you did change between the two climbs, or did it remain the same? Explain your answer.

(b) For which climb was the value of $W / t$ greater? What does that quantity represent?

✍️ Write your explanation in two lines before you read on.

Measurement note

Time each ascent twice and take the average. Human reaction time when starting and stopping a stopwatch introduces an uncertainty of a few tenths of a second — and that uncertainty is a larger percentage of the fast climb, because its time is shorter. The faster measurement is the less reliable one, which is worth remembering whenever you time something brief.

In the Explore you did the same amount of work in both climbs — your mass and the vertical height were unchanged, so $W = m g h$ was unchanged. But the faster climb completed that work in less time, and therefore gave a greater value of $W / t$ . That quantity is the average power.

Power — the rate of doing work

Two machines can do the same amount of work, or transfer the same amount of energy, but take different times over it. Power measures that difference:

P = \frac{Δ W}{Δ t}

The unit is the watt: $1 W = 1 J/s$ . Larger powers are given in $kW = 1 0^{3} W$ and $MW = 1 0^{6} W$ .

Power can also be found without measuring any time at all. If a force acts in the direction of motion at speed $v$ :

P = F v cos θ

Efficiency — the useful fraction

No device converts all of its input energy into useful output; some is always dissipated, usually as thermal energy. Efficiency measures the fraction that does become useful:

η = \frac{useful output energy}{total input energy}

It is dimensionless — a decimal between 0 and 1, or a percentage. And it is never 100%: some input energy always leaves as waste.

A Sankey diagram shows this visually: one input arrow splits into useful output and waste, with each arrow’s width proportional to the energy it carries. That proportionality is the whole point — read a Sankey by comparing widths, and draw one by measuring them.

Point

The work done by a constant force $F$ moving a body through a displacement $s$ at an angle $θ$ to the force is $W = F s cos θ$ . Two machines can do the same work in different times; the rate of doing work is the average power, $P = Δ W /Δ t$ , measured in watts.
When a force acts on a body moving at speed $v$ , the power can be written without measuring time: $P = F v cos θ$ .
The fraction of input energy leaving a device as useful output is its efficiency — dimensionless, written as a decimal or a percentage. No real device is 100% efficient.
In a Sankey diagram the total input equals the sum of the useful output and all wasted outputs. A missing output can therefore be found by subtracting the known outputs from the total input.

📐 Equation box

Work

W = F s cos θ

Power, from work and time

P = \frac{Δ W}{Δ t}

Power, from force and speed

P = F v cos θ

Efficiency

η = \frac{useful output}{total input}

Key terms

Work: — Energy transferred when a force moves its point of application: W = Fs cos θ.
Average power: — The rate of doing work or transferring energy, in watts (1 W = 1 J/s).
Efficiency: — The useful fraction of the input energy — dimensionless, never 100%.
Sankey diagram: — Energy flow drawn with arrow widths proportional to the energies carried.

🛠️ Worked example

Situation

A building-site hoist lifts a $5.0 \times 1 0^{2} kg$ pallet vertically through 6.0 m at constant speed in 12 s. Neglect air resistance. Determine (a) the work done, (b) the average power, and (c) check the power using $P = F v$ .

Step 1 of 6 — try it yourself before you open the next one.

A lift cabin of mass 400 kg rises vertically through 9.0 m at a constant speed in 15 s. Calculate the power output of its motor in kW, then check your answer using

P = F v

a-b

(a) An electric pump on a Nile Delta farm draws 2000 W of electrical power and delivers 1500 W of useful output lifting irrigation water. Determine its efficiency.
(b) A small electric water heater takes in 2400 W and delivers 2040 W of useful thermal power. Determine its efficiency.

c-d

(c) A lamp is supplied with 100 J of electrical energy. It emits 25 J of light, and the remainder is dissipated as heat. Determine the energy dissipated and the efficiency of the lamp.
(d) Sketch a Sankey diagram for that lamp, with arrow widths proportional to the energies.

In a New Context

For every 100 J of electrical energy supplied, a tungsten filament lamp may transfer only about 10 J as useful light — the remaining 90 J is dissipated to the surroundings, mainly as thermal energy. An LED lamp may transfer about 40 J of the same 100 J as useful light, dissipating about 60 J.

So the LED is four times as efficient: 40% against 10%. On a Sankey diagram the LED has a far wider useful-light arrow and a narrower wasted-energy arrow, even though both received the same 100 J input.

➡️ Explain, from the two Sankey diagrams, why an LED lamp needs less electrical energy than a filament lamp to produce the same amount of useful light.

The answer, if you want to check yourself

To produce the same 40 J of light, the LED needs 100 J of input while the filament lamp needs 400 J — because only a tenth of its input becomes light. The extra 300 J does not vanish; it heats the room. This is also why a filament bulb is hot to touch and an LED is not: the waste arrow is something you can feel.

🧪 Nature of Science — the origin of horsepower

When James Watt marketed his improved steam engines in the late eighteenth century, customers needed a familiar way to compare them with the working horses they already had. Watt used an estimate of the sustained power of a working horse as a practical basis for comparison, defining one horsepower as a rate of doing work of 33,000 foot-pounds-force per minute — equivalent to about 746 W.

This let engine power be expressed in terms customers could readily understand. Horsepower is therefore a conventional unit of power, not the exact power produced by every horse — a reminder that units are chosen for human convenience as much as for physics.

Your turn: Suggest one advantage of expressing engine power in horsepower rather than introducing a completely new unit.

✅ Lesson question, answered

Two lifts raise the same load through the same vertical height, but one completes the lift in half the time. Do both lifts perform the same useful work? Which lift has the greater power?

Both lifts perform the same useful work, because the same load is raised through the same vertical height: $W = m g h$ , and time does not appear in it.

The lift that finishes sooner has the greater power, since $P = W / t$ with the same $W$ and a smaller $t$ .

If their input energies differ, the more efficient lift is the one converting a larger fraction of input into useful energy — a wider useful arrow and a narrower waste arrow on a Sankey diagram. Power and efficiency are different questions about the same machine.

A goods lift raises a load of mass

1.5 \times 1 0^{3} kg

vertically through 18 m at constant speed, taking 30 s. (a) Calculate the work done on the load. (b) Determine the power output in kW. (c) Show that the speed is 0.60 m/s. (d) Using

P = F v

, confirm (b) and explain briefly why the two methods must agree.

Two lifts, A and B, each raise a load of mass 600 kg through the same vertical height of 9.0 m at constant speed. Lift A takes 24 s; lift B takes 12 s. (a) Explain why both do the same useful work, and state its value. (b) Calculate the power output of each in kW. (c) State the ratio

P_{B} : P_{A}

and describe how it relates to the times.

An electric winch is supplied with

5.0 \times 1 0^{4} J

of electrical energy. It does

3.5 \times 1 0^{4} J

of useful work raising a load. (a) Calculate the energy dissipated. (b) Determine the efficiency as a percentage. (c) Draw a Sankey diagram with widths proportional to the energies. (d) State the main form into which the dissipated energy is transferred.

An electric hoist raises a load of mass 250 kg through a vertical height of 6.0 m at constant speed in 8.0 s. While it does so, the hoist is supplied with electrical energy at a constant rate of 2.5 kW. (a) Calculate the useful power output in kW. (b) Determine the efficiency as a percentage. (c) Calculate the energy dissipated during the 8.0 s lift, and state the main form to which it is transferred.

📝 Exam-style question

A small crane lifts a load of mass 600 kg vertically through a height of 8.0 m at a constant speed in 16 s. Take $g = 9.8 m/s^{2}$ . Calculate the work done by the crane on the load, and the mechanical power output of the crane in kW.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

Part 1 — multiple choice (50% of the paper)

Most of the traps here are the work/power/efficiency confusion in different disguises.

Two identical loads are raised through the same height, one in 10 s and one in 20 s. Compared with the slower lift, the faster lift does:

A motor does 4500 J of work in 15 s. Its average power output is:

A device is supplied with 800 J and delivers 200 J of useful output. Its efficiency is:

A train travels at a constant 25 m/s against a total resistive force of 6000 N. The engine’s power output is:

On a Sankey diagram for a lamp, the input arrow is 100 J wide and the useful-light arrow is one fifth as wide. The energy wasted is:

Which statement about a real machine is correct?

Part 2 — written answer (the other 50%)

Ask three questions of every problem here: how much energy, how fast, and what fraction was useful. Each question has its own notebook page.

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

A motor does 6000 J of work in 20 s. Determine its average power.

1.2

A device takes in 500 J of energy and delivers 350 J of useful output. Determine its efficiency as a percentage.

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

A pump raises 200 kg of water through 12 m in 40 s. Determine (i) the useful work done and (ii) the useful power output.

2.2

A train moves at a constant 30 m/s against a total resistive force of 8000 N. Determine the power output of the engine.

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

An escalator carries 60 people of average mass 70 kg through a vertical height of 5.0 m in 60 s. Its motor draws 8.0 kW. Determine (i) the useful power output, (ii) the efficiency, (iii) the energy wasted in that 60 s.

3.2

A crane’s motor is 60% efficient and is supplied with 15 kW. It raises a load at a constant 0.40 m/s. Determine (i) the useful power, (ii) the mass of the load, (iii) the energy wasted in 30 s.

Keep reading — this lesson marks itself done automatically.