Power — energy per second — Physics

Before this

This leads to

Power and energy in continuous processes · coming soon
Intensity — power per unit area · coming soon

Two cranes, the same job, sixty times apart

Two cranes on a building site each lift a 500 kg load through 20 m. Take $g = 9.8$ N/kg. Each crane gives the load the same gravitational potential energy:

Δ E_{p} = m g Δ h = 500 \times 9.8 \times 20 = 98000 J

Crane A does the job in 10 seconds. Crane B takes 10 minutes, which is 600 seconds. The energy accounts of the last four lectures cannot tell them apart. Both transfer 98 000 J. Both do the same work. Yet nobody would call them equal machines. A site manager would pay far more for crane A.

This is the disruption. Everything so far has asked how much energy is transferred. It has never asked how fast. The last lecture ended with a promise: the next lecture adds time to the accounts. Here it is.

Power: energy per second

The quantity that separates the two cranes is power: the rate at which energy is transferred. A force that does work transfers energy, so power is also the rate of doing work.

Power

$P = \frac{Δ E}{t} = \frac{W}{t}$

$P$ is the power in watts (W), $Δ E$ the energy transferred in joules, $W$ the work done in joules, and $t$ the time in seconds. One watt is one joule per second: 1 W = 1 J/s.

Notice that the letter W does two jobs. In italics, $W$ is the work done, a quantity. Upright, W is the watt, a unit. The italics rule from the Notation sheet keeps them apart: " $W$ = 98 000 J" is a quantity, and "9800 W" is a number of watts.

Now compare the cranes. Crane A: $P = 98000 \div 10 = 9800$ W, or 9.8 kW. Crane B: $P = 98000 \div 600 \approx 163$ W. Crane A is exactly 60 times as powerful, because it does the same work in one sixtieth of the time.

Energy transferred against time for the two cranes. Both lines end at 98 kJ, but crane A gets there in 10 s and crane B needs 600 s. The power is the gradient of each line: steep means powerful.

The graph shows a second way to read power. Power is energy divided by time, so it is the gradient of a graph of energy against time. This is the same idea as speed, which is the gradient of a distance–time graph. A steady power gives a straight line. If the power changes, the gradient changes.

A sense of scale

Large powers use prefixes. A kilowatt (kW) is 1000 W, a megawatt (MW) is a million watts, and a gigawatt (GW) is a billion watts. A filament lamp might use 100 W, a toaster about 1 kW, and a large power station delivers of the order of 1 GW.

Older engines were rated in horsepower. James Watt introduced the unit in the 1700s to sell his steam engines as replacements for horses. He estimated how fast a horse could raise a load when turning a mill wheel, and rounded it to a fixed amount. One horsepower is about 746 W. It is a defined unit, a marketing comparison, not a measurement of every horse.

Kilowatt-hours: the energy on an electricity bill

Turn the power equation round and you get the energy:

Δ E = P \times t

This is how electricity is sold. A 2.0 kW kettle running for 3.0 minutes transfers $2000 \times 180 = 360000$ J. Joules are small, so power companies use a larger unit, the kilowatt-hour (kWh): the energy transferred by 1 kW in one hour.

1 kWh = 1000 W \times 3600 s = 3600000 J = 3.6 MJ

So the kettle uses $2.0 kW \times 0.050 h = 0.10$ kWh. The kilowatt-hour sounds like a power, because it contains "watt". It is not. A kilowatt-hour is an amount of energy, just as a kilometre is a distance and not a speed.

Stop and think

A 9 W LED and a 75 W filament lamp give about the same light (US Department of Energy, quoted in the last lecture). Both are switched on for 5 hours a day for a year. Does the LED save energy, power, or both?

Show the answer

Both, but they are different things. The LED's power is lower at every moment: 9 W instead of 75 W, so 66 W less. Over a year, the energy saved is that power times the time: $66 W \times 5 h \times 365 = 120450$ Wh, about 120 kWh. Power is the rate; energy is the rate added up over time. A bill charges for energy.

P = Fv: power from force and speed

For anything that moves, there is a second form of the power equation. Suppose a constant force $F$ pushes an object through a displacement $s$ in the direction of the force, in a time $t$ . The work done is $W = F s$ , so

P = \frac{W}{t} = \frac{F s}{t} = F \times \frac{s}{t} = F v

The last step uses $v = \frac{s}{t}$ , the speed. If the speed changes, this $v$ is the average speed, and $F v$ is the average power. At any single instant, the power delivered by the force is $F$ times the speed at that instant. If the force is at an angle $θ$ to the motion, only the part along the motion does work, as in the work lecture, and $P = F v cos θ$ .

Now a puzzle. A car drives along a level road at a steady speed. Its speed does not change, so the resultant force on it is zero. Does its engine need any power at all? Make a prediction.

Predict — does a car at steady speed need power?

A car moves at a steady 30 m/s on a level road. The driving force from its wheels exactly balances air resistance and rolling resistance, so the resultant force is zero. Choose the statement you believe, then read the explanation.

At a steady speed the resultant force is zero, so the car's kinetic energy does not change. But the resistive forces dissipate energy as thermal energy every second. The driving force must replace it, at a rate P = Fv, where F is the driving force and v = 30 m/s.

That is it. That is the whole idea of power at steady speed. Balanced forces mean the kinetic energy is not changing. They do not mean nothing is happening. Energy is flowing through the car, in from the fuel and out into the surroundings, at the rate $F v$ .

Why going faster costs so much

Air resistance grows quickly with speed. For a car, it is roughly proportional to the square of the speed: double the speed, and the air pushes back about four times as hard. The power needed to beat that force is force times speed. So the power for air resistance grows with the cube of the speed:

F_{air} \propto v^{2} so P_{air} = F_{air} v \propto v^{3}

Doubling the speed needs about $2^{3} = 8$ times the power for air resistance. Rolling resistance in the tyres behaves differently: it stays roughly the same at every speed, so its power grows only in proportion to $v$ . Explore an invented car that has both.

Explore — the power needed at each speed

An invented car on a level road has rolling resistance of 200 N at every speed, and air resistance of 0.4 × v² newtons. At a steady speed, the driving force equals the total. Change the speed and watch the power. Then compare a speed with twice that speed, such as 15 and 30 m/s.

Presets:

Speed v =

The invented car: rolling resistance 200 N, air resistance 0.4 × v² N. At a steady speed, P = (total resistive force) × v.

v (m/s)	Air (N)	Total (N)	Power (kW)
10	40	240	2.4
15	90	290	≈ 4.4
20	160	360	7.2
30	360	560	16.8
40	640	840	33.6

From 20 to 40 m/s the total power rises about 4.7 times. The air-resistance part alone rises from 3.2 kW to 25.6 kW, exactly 8 times.

What should you have seen? At low speeds the curve is almost a straight line, because rolling resistance dominates. At high speeds it bends upwards steeply, because air resistance takes over. Doubling the speed always multiplies the air-resistance power by exactly 8. The total rises by less, between 2 and 8 times, because the rolling part only doubles. This is why vehicles use so much more fuel per kilometre at high speed, and why designers work so hard to reduce drag.

Efficiency, measured in watts

The last lecture defined efficiency with energies. A machine that runs continuously is easier to describe with powers. Divide both energies by the same time, and the ratio does not change:

efficiency = (useful power output \div total power input) \times 100%

A Sankey diagram works in watts just as well as in joules. Each band then shows energy per second instead of total energy.

Worked examples

Take $g = 9.8$ N/kg throughout.

Worked Example 1 — A crane on a real schedule

A crane lifts a 500 kg load through 20 m in 40 s at a steady speed. What power does it develop?

Energy. $Δ E_{p} = 500 \times 9.8 \times 20 = 98000$ J.

Power. $P = \frac{98 000}{40} = 2450$ W, or 2.45 kW.

Check with P = Fv. At steady speed, the lifting force equals the weight: $F = 500 \times 9.8 = 4900$ N. The speed is $20 \div 40 = 0.50$ m/s. So $P = 4900 \times 0.50 = 2450$ W ✓. The two methods agree.

Worked Example 2 — The car at motorway speed

The invented car from the explorer travels at a steady 30 m/s. Find the power delivered to the wheels.

Forces. Air resistance $= 0.4 \times 3 0^{2} = 360$ N. Rolling resistance $= 200$ N. Steady speed, so the driving force equals the total: $F = 560$ N.

Power. $P = F v = 560 \times 30 = 16800$ W, about 16.8 kW.

Check. In one second, the car travels 30 m, and the driving force does $560 \times 30 = 16800$ J of work. All of it is dissipated as thermal energy by the resistive forces ✓. The kinetic energy does not change.

Worked Example 3 — How efficient is the motor?

An electric motor takes an input power of 800 W. It lifts a 50 kg load at a steady 0.30 m/s. Find its efficiency.

Force. At steady speed, the upward force equals the weight: $F = 50 \times 9.8 = 490$ N.

Useful power. $P = F v = 490 \times 0.30 = 147$ W.

Efficiency. $\frac{147}{800} \times 100% \approx 18.4%$ . The other 653 W is dissipated, mostly as thermal energy in the motor's wires and bearings.

Check. $147 + 653 = 800$ W ✓. The efficiency is below 100%, as it must be.

The motor's Sankey diagram, in watts. Every second, 800 J enters as electrical energy, 147 J raises the load and 653 J warms the motor.

Worked Example 4 — Finding a time

Lin, mass 60 kg, climbs stairs that rise 15 m in total. Her useful power output is 200 W. How long does the climb take?

Energy. $Δ E_{p} = 60 \times 9.8 \times 15 = 8820$ J.

Time. Rearrange $P = \frac{Δ E}{t}$ to $t = \frac{Δ E}{P} = \frac{8820}{200} = 44.1$ s.

Check. If each floor of a building is about 3 m high, 15 m is five floors, so this is about 9 s per floor: a brisk but realistic climb ✓. Her body uses far more than 200 W, because muscles also warm her up. The 200 W is only the part that becomes potential energy.

Three traps

Trap 1 — Treating the kilowatt-hour as a power

A kWh is an amount of energy, 3.6 MJ. "My heater uses 2 kWh" means energy; "my heater is 2 kW" means power. To convert power to energy, multiply by the time.

Trap 2 — Forgetting to convert minutes to seconds

$P = \frac{W}{t}$ gives watts only when the energy is in joules and the time in seconds. Crane B's 10 minutes is 600 s. Dividing by 10 instead would make it look as powerful as crane A.

Trap 3 — "Balanced forces, so no power"

At a steady speed the resultant force is zero, but the driving force still does work every second against the resistive forces. Use $P = F v$ with the driving force, not the resultant force.

Try it yourself

Take $g = 9.8$ N/kg. Work each one on paper first. Open a hint only when you are stuck.

Kofi, mass 55 kg, runs up a staircase 4.5 m high in 5.0 s. Find his useful power output. This is a real experiment you can do with a stopwatch.

Hint 1
The useful energy is his gain in gravitational potential energy.

Hint 2
$Δ E_{p} = 55 \times 9.8 \times 4.5$ . Then divide by the time.

Full answer
$Δ E_{p} = 55 \times 9.8 \times 4.5 = 2425.5$ J. $P = 2425.5 \div 5.0 \approx 485$ W. Check: about half a kilowatt, a little more than Lin's steady 200 W, which makes sense for a short sprint ✓. The measurement uses the vertical height of the stairs, not the length along them.
A 2.2 kW kettle runs for 2.5 minutes. How much energy does it transfer, in joules and in kWh? At an invented price of $0.20 per kWh, what does it cost?

Hint 1
$Δ E = P \times t$ , with $t$ in seconds for joules, or in hours for kWh.

Hint 2
2.5 minutes is 150 s, or $2.5 \div 60$ hours.

Full answer
In joules: $2200 \times 150 = 330000$ J. In kWh: $330000 \div 3600000 \approx 0.0917$ kWh (or $2.2 \times \frac{2.5}{60} \approx 0.0917$ ✓). Cost: $0.0917 \times 0.20 \approx $0.018$ , about two cents.
A lift motor raises an 800 kg lift at a steady 1.5 m/s. The motor's electrical input power is 15 kW. Find the useful output power and the efficiency.

Hint 1
At steady speed, the lifting force equals the weight.

Hint 2
$P = F v = (800 \times 9.8) \times 1.5$ .

Full answer
$F = 800 \times 9.8 = 7840$ N. Useful power $= 7840 \times 1.5 = 11760$ W. Efficiency $= \frac{11 760}{15 000} \times 100% = 78.4%$ . The other 3240 W is dissipated in the motor, the cables and the pulleys. Check: $11760 + 3240 = 15000$ ✓.
A cyclist on a flat road needs 200 W to ride at a steady 10 m/s. Assume that almost all the resistance is air resistance, proportional to $v^{2}$ . Estimate the power needed at 12 m/s. What steady speed could 400 W give?

Hint 1
If $F \propto v^{2}$ , then $P = F v \propto v^{3}$ .

Hint 2
At 12 m/s, $P = 200 \times (\frac{12}{10})^{3}$ . For 400 W, solve $(\frac{v}{10})^{3} = 2$ .

Full answer
$P = 200 \times 1. 2^{3} = 200 \times 1.728 \approx 346$ W. So a 20% rise in speed needs about 73% more power. For 400 W: $(\frac{v}{10})^{3} = 2$ , so $v = 10 \times 32 \approx 12.6$ m/s. Doubling the power gives only about 26% more speed. Check: $12. 6^{3} \div 1 0^{3} \approx 2.0$ ✓. This is why racing cyclists crouch low: reducing air resistance is cheaper than finding more power.

The bigger idea

Power turns energy into a flow. In the last lecture, energy sat in stores and moved between them. Now we can say how fast it moves: watts are joules per second, in the same way that a river's flow is litres per second. A power station, a kettle, a cyclist and a crane can all be compared on one scale.

The idea reaches far. Sunlight arriving on a solar panel is described as power per square metre, which a later lecture calls intensity. Electrical circuits deliver power as current times potential difference. In every case the question is the same one that separated the two cranes: not how much energy, but how much each second. This lecture closes the topic of work, energy and power. The next topic turns to materials, starting with springs and the energy they store.

Key concepts in this lesson

Power is the rate of transferring energy, or of doing work: $P = \frac{Δ E}{t} = \frac{W}{t}$ , measured in watts, where 1 W = 1 J/s.
Power is the gradient of a graph of energy transferred against time.
The kilowatt-hour is a unit of energy, equal to 3.6 MJ, found from power × time.
For a force acting along the motion, $P = F v$ , so a vehicle at steady speed needs power to replace the energy dissipated by resistive forces.
Efficiency can be written with powers: useful power output ÷ total power input × 100%, never more than 100%.