Work and power
Key idea: Work is energy transferred by a force: for a constant force, multiply it by distance moved in its direction. Power is the energy transferred per second.
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Guided lesson
Guided steps
- Choose the distance moved in the force’s direction.
- Explain the crate’s work and where its energy goes.
- Link steady lifting force with gained gravitational potential energy.
- Find work, then use kinetic energy to find speed.
- Explain when force and displacement give zero work.
- Compare support force direction with the bag’s movement.
- Explain watts as joules transferred each second.
- Find average power and check its unit.
- Convert power and time before using the energy equation.
- Compare the same height gain at different rates.
- Find the lifting work and power.
- Find both powers, then compare the total work.
- Find energy, elapsed time and average power between readings.
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ContinuePushing a crate across a floor or lifting a box onto a shelf transfers energy mechanically, and the energy a force transfers is the work it does. Running up the stairs and walking up them take the same work, but running does it faster: that rate is power. The calculations use Eₖ = (1/2)mv², Eₚ = mgh and weight = mg, with g = 10 N/kg.
Work: energy transferred by a force
A force transfers energy mechanically when an object moves with a component of displacement in the force’s direction. This transfer is the work done by the force. For a constant force acting in the direction of motion:
| Symbol | Quantity | SI unit |
|---|---|---|
| W | work done (energy transferred) | joule (J) |
| F | force | newton (N) |
| s | distance moved in the direction of the force | metre (m) |
1 joule is the work done when a force of 1 N moves an object 1 m in the direction of the force, so 1 J = 1 N m. Work is one way of transferring energy; both are measured in joules.
(In W = mg, W stands for weight. In W = Fs, it stands for work. Read which one a question means.)
Worked example 1
Pushing against friction
Problem
A worker pushes a crate 12 m across a level floor at a steady speed. She pushes horizontally with a force of 150 N. Find the work she does, and say where the energy goes.
Show full solution
Find the distance along the force
Method
Take the distance moved in the direction of the push.Reason
Work uses only the distance moved along the force.Working
Answer
s = 12 m
Find the work
Method
Multiply force by distance.Reason
Work done is force × distance moved in the direction of the force.Working
Answer
W = Fs = 150 × 12 = 1800 J
Say where the energy goes
Working
Answer
The speed is steady, so the crate’s kinetic store does not change. Her push balances friction, and friction transfers the 1800 J to the internal store of the crate and floor, which warm slightly.
Worked example 2
Lifting links work to mgh
Problem
A 12 kg box is lifted steadily from the floor onto a shelf 1.5 m higher. Find the work done by the lifting force.
Show full solution
Find the lifting force
Method
Set the upward force equal to the weight.Reason
A steady lift means the forces on the box balance.Working
Answer
F = mg = 12 × 10 = 120 N
Find the work
Method
Multiply the force by the height.Reason
The box moves 1.5 m upwards, in the direction of the force.Working
Answer
W = Fs = 120 × 1.5 = 180 J
Link it to mgh
Working
Answer
This equals the gain in gravitational potential energy: mgh = 12 × 10 × 1.5 = 180 J. Lifting force × height is mg × h, which is where Eₚ = mgh comes from.
Check your understanding 1
A constant resultant force of 8.0 N moves a 4.0 kg block 9.0 m along a level track from rest. Ignore resistance. Find the work done and the block’s final speed.
Show the steps Optional working
Show answer
The force transfers energy mechanically to the block’s kinetic store. W = Fs = 8.0 × 9.0 = 72 J. With no other transfer, all 72 J goes into the kinetic store: 1/2 × 4.0 × v² = 72, so v² = 36 and v = 6.0 m/s.
When a force does no work
Only the distance moved in the direction of the force counts.
- A student carries a bag along a level corridor at a steady speed. Her upward force on the bag is at right angles to its horizontal movement, so the bag moves no distance in the direction of that force. The force does no work on the bag: the bag’s speed and height, and so its energy, do not change. She still gets tired, because her muscles transfer energy inside her body.
- A car drives along a level road. Its weight acts straight down while it moves horizontally, so its weight does no work and its gravitational potential store does not change.
- A person pushes hard on a wall that does not move. The distance moved is zero, so no work is done on the wall.
Compare the solid force arrows with the dashed displacement arrows. In the last case, the load moves horizontally at constant height; it does not move upwards along the support force.
Check your understanding 2 A tiring carry can involve zero work on the load
A student holds a 40 N bag at constant height while walking 6.0 m along a level corridor. Find the work done on the bag by the upward support force. Which displacement belongs in the work equation?
Show answer
The upward force does 0 J of work on the bag: its displacement is horizontal, so it moves no distance in the force’s direction. The 6.0 m path is not an upward displacement and cannot be used for this force in W = Fs. The student’s muscles still transfer energy within her body and to the surroundings; zero work by this particular force on the bag is not zero energy use by the person.
Power: how quickly energy is transferred
Power is the energy transferred per unit time. It is also the rate of doing work. Over a time interval, the equation gives the average power:
Power is measured in watts (W): 1 W = 1 J/s. A 60 W lamp transfers 60 J every second. Larger powers are given in kilowatts: 1 kW = 1000 W. Rearranged, E = Pt and t = E ÷ P.
Worked example 3
Power of a motor
Problem
A motor transfers 3600 J in 30 s. Find its power.
Show full solution
Divide energy by time
Method
Use P = E/t.Reason
Power is the rate of energy transfer.Working
Answer
P = 3600 ÷ 30 = 120 W
Check the unit
Working
Answer
Joules divided by seconds gives watts.
Worked example 4
Kilowatts and minutes
Problem
A 2.0 kW kettle is switched on for 3.0 minutes. How much energy does it transfer?
Show full solution
Convert to SI units
Method
Change kilowatts to watts and minutes to seconds.Reason
E = Pt gives joules only from watts and seconds.Working
Answer
P = 2000 W, t = 3.0 × 60 = 180 s
Find the energy
Method
Multiply power by time.Reason
Power is energy transferred per second.Working
Answer
E = Pt = 2000 × 180 = 3.6 × 10⁵ J (360 kJ)
Check
Working
Answer
Using 2.0 and 3.0 without converting would give 6.0 J, far too small for boiling water.
Worked example 5
Finding a time
Problem
How long does a 1.5 kW motor take to transfer 90 kJ?
Show full solution
Convert and rearrange
Method
Write both quantities in base units and use t = E/P.Reason
Time is energy divided by the rate of transfer.Working
Answer
E = 90 000 J, P = 1500 W
Find the time
Working
Answer
t = 90 000 ÷ 1500 = 60 s
The same work in different times
A 50 kg student climbs a flight of stairs that rises 4.0 m. Running, she takes 5.0 s; walking, she takes 16 s.
- Either way, the work done against gravity is mgh = 50 × 10 × 4.0 = 2000 J.
- Running: P = 2000 ÷ 5.0 = 400 W.
- Walking: P = 2000 ÷ 16 = 125 W.
Her gravitational potential store increases by the same amount both times. Running produces that change more quickly, so her average power against gravity is greater. A higher power means a greater rate of energy transfer. The total energy also depends on how long the transfer lasts.
Check your understanding 3
A motor raises a 4.0 kg load steadily through 2.5 m in 5.0 s. No energy is dissipated. Calculate the work done and the motor’s power. Take g = 10 N/kg.
Show the steps Optional working
Show the steps Optional working
Show answer
Energy is transferred mechanically into the load’s gravitational potential store. Eₚ = mgh = 4.0 × 10 × 2.5 = 100 J. With no energy dissipated, the motor does 100 J of work. Its power is 100 ÷ 5.0 = 20 W.
Check your understanding 4
Motor A does 900 J of work in 30 s. Motor B does 600 J in 15 s. Calculate both powers. Which motor does more work?
Show answer
Motor A does more work: 900 J compared with 600 J. Motor B is more powerful: A’s power is 900 ÷ 30 = 30 W and B’s is 600 ÷ 15 = 40 W. Doing more work does not mean having more power; the time matters too.
Check your understanding 5 Use the change between meter readings
An energy meter records 400 J at 2.0 s and 1300 J at 5.0 s. Calculate the energy transferred and average power between those readings. Which time interval should you use?
Show the steps Optional working
Show answer
Between the readings, energy transferred is 1300 − 400 = 900 J and elapsed time is 5.0 − 2.0 = 3.0 s, giving average power 900/3.0 = 300 W. Dividing 1300 J by 5.0 s gives 260 W, but uses the final readings instead of their changes. The meter reading at 0 s is not given, so the average power over the first 5.0 s cannot be found from these two readings. The 2.0–5.0 s readings give an average; power may vary within that interval.
Challenge yourself
A lift raises the same load through the same height in 12 s and then in 8 s. Compare the work done and power in the two journeys.
Check your thinking
The work done against gravity is the same because the load and height are unchanged. The 8 s journey has greater power because the same energy is transferred in less time: P = W/t.
Common mistakes
- Using the whole path length in W = Fs. Only the distance moved in the direction of the force counts; for lifting, use the vertical height.
- Saying that carrying a bag along a level floor does work on the bag. The upward force is at right angles to the movement, so it does no work on the bag.
- Using the mass as the lifting force. The steady lifting force is the weight, mg, in newtons.
- Saying a more powerful machine transfers more energy. Power is a rate: the same energy can be transferred at different powers.
- Multiplying energy by time to find power. P = E ÷ t; it is energy that equals Pt.
- Leaving kJ, kW or minutes unconverted. Use J, W and s: 1 kJ = 1000 J, 1 kW = 1000 W, 1 min = 60 s.
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Syllabus and review details
- SEC G2 Science Physics component 2027 · 2027
Physics section, PDF pages 8–23
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