Physics · 1 Motion, forces and energy

1.7.4 Power

Before you start

Work done = force × distance moved in the direction of the force, $W = F d$ , and that work done is equal to the energy transferred (1.7.2).
Energy and work are measured in joules (J), and time in seconds (s).
Weight = mass × $g$ , and the change in gravitational potential energy $Δ E_{p} = m g Δ h$ .
Rearranging an equation of the form $a = b / c$ to make $b$ or $c$ the subject.
Converting kJ to J, and minutes or hours to seconds.

Syllabus objectives covered (1)

Define power as work done per unit time and also as energy transferred per unit time; recall and use the equations (a) P = W / t (b) P = ΔE / t

Part 1 of 5

Power measures how fast, not how much

Two cranes lift identical crates from the ground to the top of the same building. Crane A takes $20$ s and crane B takes $40$ s. Each crane uses the same force over the same height, so each does exactly the same work on its crate. Crane A is still the more powerful crane, because it does that work in less time.
That is the whole idea of power. It does not tell you how much work is done. It tells you how quickly the work is done.
The first definition: power is work done per unit time.
$P = \frac{W}{t}$
The second definition: power is energy transferred per unit time. Here $Δ E$ is the energy transferred, and the $Δ$ means 'change in'.
$P = \frac{Δ E}{t}$
The words 'per unit time' are the part that makes these definitions of power. 'Power is energy' or 'power is work' leaves the time out, and so describes a different quantity. When a question says Define, learn and write one of the two sentences above.
Time is measured in seconds, so 'per unit time' works out as 'in each second'. Power tells you how many joules are transferred every second.

Part 2 of 5

The watt, and the two meanings of W

The unit comes straight from the equation. Work or energy in joules divided by time in seconds gives joules per second. One joule per second has its own name: one watt, symbol W.
$1 W = 1 J/s$
Read every power rating as 'joules each second'. A $60$ W lamp transfers $60$ J of energy every second. A $2.0$ kW kettle transfers $2000$ J every second.
Large powers use prefixes: $1$ kW $= 1000$ W and $1$ MW $= 1 000 000$ W. Change kW to W before you put a power into an equation, so that it matches joules and seconds.
Here is the trap. In $P = W / t$ the letter W stands for work, measured in joules. After a number, as in ' $60$ W', the letter W is the unit watt. They are different things that share a letter.
To keep them apart, write 'work done' in words in your working, or substitute numbers with their units straight away. Then ' $W = 1200$ J' (work) and ' $P = 40$ W' (power) cannot be confused.

Part 3 of 5

Why there are two equations

From the work topic: work done is equal to the energy transferred. When a force does $500$ J of work on a box, $500$ J of energy is transferred.
So $W / t$ and $Δ E / t$ always give the same number when both apply. They are one idea written in two ways, and you must be able to state both.
Choose the form that matches what the question gives you. If a force moves an object through a distance, find the work done with $W = F d$ , then use $P = W / t$ .
If the question gives you an energy, use $P = Δ E / t$ . That includes energy supplied by an electrical appliance, and a change in a kinetic or gravitational potential energy store.
For an object lifted at a steady speed, both routes lead to the same work done. The lifting force equals the weight (if you ignore friction), so $W =$ weight × height, and that is the same as the gain in gravitational potential energy $m g Δ h$ .

Part 4 of 5

Rearranging, and getting the units right

To find the energy, multiply both sides of $P = Δ E / t$ by $t$ . You multiply because $t$ is dividing $Δ E$ , and doing the same to both sides keeps the equation balanced.
$Δ E = P t$
To find the time, start from $Δ E = P t$ and divide both sides by $P$ , because $P$ is multiplying $t$ .
$t = \frac{Δ E}{P}$
The same rearrangements work with $W$ in place of $Δ E$ : $W = P t$ and $t = W / P$ .
Use the units as a check. Watts are joules per second, so watts × seconds gives joules. If you multiply energy by time when you should divide, the units come out as J s, which is not a power.
Before you divide, put the time in seconds and the energy in joules. $1$ min $= 60$ s, $1$ h $= 3600$ s and $1$ kJ $= 1000$ J. If you divide joules by minutes, your answer is $60$ times too large to be in watts.

Part 5 of 5

Two-step questions: work or energy first, then power

A power question does not always give you the work done or the energy. When it does not, the calculation has two steps: find $W$ or $Δ E$ first, then divide by $t$ .
When something is lifted, the distance in $W = F d$ is the vertical height. The force acts upwards against the weight, and $d$ must be in the direction of the force. For stairs, the length along the slope is the wrong distance.
forces diagram — not drawn yet
Side view of a straight flight of stairs with a person at the bottom; a vertical double-headed arrow from the bottom step to the top step labelled h (vertical height), a dashed line along the slope of the stairs labelled 'distance along stairs', and a downward arrow from the person labelled 'weight'.
The work done against gravity uses the vertical height h, not the distance along the stairs.
If you are given a mass and not a weight, find the weight with $g$ . You do not have to memorise the value: the front of Paper 4 tells you to take the weight of $1.0$ kg as $9.8$ N. You do have to know that you need it.
$P = \frac{m g Δ h}{t}$
Keep the full calculator value of the work done, and round only the final power. Give the answer to the same number of significant figures as the data in the question.
Write the equation in symbols, then substitute. Under this marking system a C mark may be scored for working that is seen, even when the final answer is wrong. An answer with no working has nothing to give credit to if the number is wrong.

Worked examples

Example 1. A student of weight $540$ N runs up a flight of stairs. The vertical height of the stairs is $3.2$ m and the climb takes $4.0$ s. Calculate the power the student develops in lifting her weight up the stairs.

Show the working

Neither the power nor the work done is given, so this is a two-step question: work done first, then power.
The force is her weight, $540$ N. The distance in the direction of that force is the vertical height, $3.2$ m, not the length along the stairs.
$W = F d = 540 \times 3.2 = 1728 J$
Now divide by the time. Use the unrounded $1728$ J.
$P = \frac{W}{t} = \frac{1728}{4.0} = 432 W$
The data are given to 2 significant figures, so round to $430$ W. Writing $432$ W first shows where the rounded value came from.

Answer: $432$ W, which is $430$ W to 2 significant figures.

Example 2. An electric kettle transfers $252$ kJ of energy to the water inside it in $2.0$ minutes. Calculate the rate at which the kettle transfers energy to the water, in watts.

Show the working

'Rate at which energy is transferred' is power, so use $P = Δ E / t$ .
Convert the units first, because a watt is a joule per second. $252$ kJ $= 252 000$ J and $2.0$ min $= 2.0 \times 60 = 120$ s.
Substitute and divide.
$P = \frac{Δ E}{t} = \frac{252 000}{120} = 2100 W$
Check it the other way: $2100$ J every second for $120$ s is $2100 \times 120 = 252 000$ J, which is the energy in the question.
Notice what the unconverted numbers give: $252 \div 2.0 = 126$ . That is kilojoules per minute, not watts, and it is not the answer the question asks for.

Answer: $2100$ W (which is $2.1$ kW).

Example 3. A motor lifts a crate of mass $45$ kg at a steady speed through a vertical height of $12$ m. The useful power output of the motor is $600$ W. Calculate the time taken. Take the weight of $1.0$ kg to be $9.8$ N.

Show the working

The unknown is the time, so rearrange $P = Δ E / t$ . Multiply both sides by $t$ , then divide both sides by $P$ .
$t = \frac{Δ E}{P}$
The useful energy transferred is the gain in gravitational potential energy of the crate.
$Δ E_{p} = m g Δ h = 45 \times 9.8 \times 12 = 5292 J$
You get the same number from work done: the weight is $45 \times 9.8 = 441$ N, and $441 \times 12 = 5292$ J. That is because work done equals energy transferred.
Now divide by the power.
$t = \frac{5292}{600} = 8.82 s$
Sense check: the motor supplies $600$ J each second, so about $5300$ J should take a little under $9$ s. It does.

Answer: $8.82$ s, which is $8.8$ s to 2 significant figures.

Where people go wrong

Mistake: Thinking that a more powerful machine does more work on the same job.

Instead: A more powerful machine does the same work in less time, or more work in the same time.

Power is a rate: $P = W / t$ has time in it. Two cranes lifting identical crates to the same height do the same work, whatever their power.

Mistake: Dividing by the time in minutes, e.g. $252 000$ J $\div 2.0$ .

Instead: Convert the time to seconds first: $2.0$ min $= 120$ s.

A watt is a joule per second. Dividing by minutes gives joules per minute, a number $60$ times larger than the power in watts.

Mistake: Mixing up W for work and W for watt, and writing an answer such as 'work done $= 1728$ W'.

Instead: Work and energy are in joules (J); power is in watts (W).

The syllabus uses the letter W both as the symbol for work in $P = W / t$ and as the unit of power. A wrong unit on the final answer loses the A mark.

Mistake: Using the distance along the stairs, or along a slope, as $d$ when something is lifted.

Instead: Use the vertical height.

In $W = F d$ , $d$ is the distance moved in the direction of the force. The force that lifts an object acts vertically, against its weight.

Check yourself

Give the two definitions of power.
Answer
Power is work done per unit time. Power is also energy transferred per unit time.
A lamp transfers $1800$ J of energy in $30$ s. Calculate its power.
Answer
$P = Δ E / t = 1800 / 30 = 60$ W.
What does it mean to say that a heater has a power of $1500$ W?
Answer
The heater transfers $1500$ J of energy every second.
A force of $250$ N pushes a box $8.0$ m across a floor in $5.0$ s. Calculate the power.
Answer
Work done $W = F d = 250 \times 8.0 = 2000$ J. Power $P = W / t = 2000 / 5.0 = 400$ W.
A $3.0$ kW heater is switched on for $5.0$ minutes. Calculate the energy it transfers.
Answer
$P = 3000$ W and $t = 300$ s, so $Δ E = P t = 3000 \times 300 = 900 000$ J ( $9.0 \times 10^{5}$ J, or $900$ kJ).
A motor has a power of $2.0$ kW. How long does it take to do $50$ kJ of work?
Answer
$t = W / P = 50 000 / 2000 = 25$ s.
Crane A lifts a crate to the top of a building in $20$ s. Crane B lifts an identical crate to the same height in $40$ s. Compare the work done by the two cranes and their powers.
Answer
The work done is the same, because the force and the height are the same. Crane A has twice the power of crane B, because it does the same work in half the time ( $P = W / t$ ).

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A lift and its passengers have a total weight of $9600$ N. The lift rises at a steady speed through a vertical height of $24$ m in $32$ s.

(a) Define power. 1 mark No calculator

Mark scheme

B1 work done per unit time or energy transferred per unit time; 'energy' or 'work' alone with no reference to time is insufficient

(b) Calculate the work done by the motor in raising the lift. 2 marks

Mark scheme

C1 W = Fd or 9600 × 24 seen
A1 230 000 J or 2.3 × 10⁵ J (accept 230 400 J); unit required

(c) Calculate the useful power output of the motor. 2 marks

Mark scheme

C1 P = W / t (or P = ΔE / t) or 230 400 / 32 (or 230 000 / 32) seen; ecf from (b)
A1 7200 W (accept 7.2 kW); unit required

(d) On a later journey the lift and its passengers have a total weight of $8100$ N. The motor gives the same useful power output as in (c). Calculate the time taken for the lift to rise through the same $24$ m. 3 marks

Mark scheme

C1 work done = 8100 × 24 = 194 400 J; or, by an alternative method, t = 32 × 8100 / 9600 seen (same power, so time is proportional to work done)
C1 t = W / P or 194 400 / 7200 seen; ecf from (c)
A1 27 s; unit required

Question 2. A student compares two kitchen appliances. A toaster transfers $99$ kJ of energy in $1.5$ minutes. A microwave oven transfers $72$ kJ of energy in $60$ s.

(a) Calculate the power of the toaster, in W. 3 marks

Mark scheme

C1 time converted to 90 s and energy converted to 99 000 J
C1 P = ΔE / t or 99 000 / 90 seen
A1 1100 W; unit required

(b) Determine which appliance transfers energy at the greater rate. Show how you decide. 2 marks

Mark scheme

B1 power of the microwave oven = 72 000 / 60 = 1200 W
B1 microwave oven, because 1200 W is greater than the toaster's power; ecf from (a)

(c) State the energy transferred by the microwave oven in each second. 1 mark No calculator

Mark scheme

B1 1200 J; ecf from (b)

Question 3. Two students, A and B, have the same weight. They each climb the same flight of stairs. Student A takes $6.0$ s and student B takes $9.0$ s.

(a) Compare the work done against gravity by student A and by student B. Explain your answer. 2 marks No calculator

Mark scheme

B1 the work done is the same / equal
B1 same force (weight) moved through the same vertical height, so W = Fd is the same, owtte

(b) Explain which student develops the greater power. 2 marks No calculator

Mark scheme

B1 student A
B1 same work done in less time, and power is work done per unit time / P = W / t, ORA

(c) The power developed by student B is $P$ . Deduce the power developed by student A in terms of $P$ . 1 mark No calculator

Mark scheme

B1 1.5P (accept 3P / 2)

How this is examined

Finding the work done or energy first, then dividing by the time: a three- or four-mark Calculate in which you are given a force (or a mass) and a height, but not the work done $W = F d$ (or $m g Δ h$ ) with the vertical height as $d$ ; then $P = W / t$ using that work done; the final power with the unit W. A wrong work done, used correctly in $P = W / t$ , can still earn the mark for that second step.
Putting the time in seconds and the energy in joules before you divide: a two- or three-mark Calculate where the time is in minutes or the energy is in kJ The conversion written down (e.g. $1.5$ min $= 90$ s, $99$ kJ $= 99 000$ J); $P = Δ E / t$ ; the answer in W, or in kW if you label it kW.
Rearranging to find the energy or the time: a two- or three-mark Calculate that gives you the power and asks for the energy transferred or the time taken $Δ E = P t$ or $t = Δ E / P$ written; the power in W and the time in s; the answer in J or s.
Same work in less time means more power: a two-mark Compare or Explain about two people or machines doing the same job in different times 'The work done is the same, because the force and the height are the same' and 'the one that takes less time has the greater power, because power is work done per unit time'.
The definition of power in words: a one-mark Define 'Work done per unit time' or 'energy transferred per unit time'. The words 'per unit time' are what make it a definition of power and not of work or energy.

Words to know

Power: Work done per unit time, or energy transferred per unit time: $P = W / t = Δ E / t$ . Measured in watts.
Watt (W): The unit of power. One watt is one joule per second.
Kilowatt (kW): $1000$ watts.
Work done: Force × distance moved in the direction of the force, $W = F d$ . Equal to the energy transferred. Measured in joules.
Per unit time: For each unit of time; with time in seconds, for each second. It marks a quantity as a rate.

All Physics lessons Next: 1.8 Pressure