Work, Energy & Power
How much effort a machine transfers, how much energy is stored or moving, and how quickly that energy is used — the numbers behind every motor, hoist and hydraulic ram.
2Topics
8Examples
4Calculators
12Practice Qs
🔨Work & EnergyJoules, GPE and KE
⚡Power & EfficiencyWatts and useful output
⚙️Applications4 workshop calculators
🎮Game12 randomised questions
📝Practice12 graded questions
Why this matters in engineeringReal world
A hoist motor is sized by the power it must deliver: the work needed to lift the load, divided by the time allowed. Efficiency tells maintenance engineers how much of the electricity paid for actually moves the load — and how much becomes waste heat. Kinetic energy explains why a heavy vehicle needs so much braking distance.
🔨
Work & Energy
Work done when a force moves something, and the two energy stores engineers calculate most: gravitational potential and kinetic.
🧮Work & Energy CalculatorChoose a mode, enter values, press Calculate
Enter your values above and press Calculate to see every working step.
📚 Key vocabulary
- Work done
- energy transferred when a force moves an object, measured in joules (J).
- Joule (J)
- the energy transferred when a force of 1 N moves an object 1 m.
- Gravitational potential energy
- energy stored by lifting a mass higher.
- Kinetic energy
- energy of a moving object — it grows with the square of the speed.
- Energy transfer
- energy moving from one store to another; the total is conserved.
- Kilojoule (kJ)
- 1000 joules — used for larger energy values.
👁️ Visual aid — work done pushing a crate: the force arrow travels with the crate over the distance
✍️Worked example 1 — work done pushing a crate
QuestionA technician pushes a crate with a steady force of 150 N over a distance of 4 m. Calculate the work done.
Formula
W = F × dSubstitute
W = 150 × 4Multiply
150 × 4 = 600AnswerW = 600 J
✍️Worked example 2 — potential energy of a lifted component
QuestionA 12 kg gearbox is lifted 5 m onto a mezzanine. Calculate the gravitational potential energy gained. Take g = 9.81 N/kg.
Formula
GPE = m × g × hSubstitute
GPE = 12 × 9.81 × 5Step 1
12 × 9.81 = 117.72 (this is the weight in newtons)Step 2
117.72 × 5 = 588.6AnswerGPE = 588.6 J
✍️Worked example 3 — kinetic energy of a vehicle
QuestionA 900 kg workshop vehicle travels at 10 m/s. Calculate its kinetic energy in kJ.
Formula
KE = ½ × m × v²Square first
v² = 10 × 10 = 100Substitute
KE = 0.5 × 900 × 100Step 1
0.5 × 900 = 450Step 2
450 × 100 = 45000Convert
45000 ÷ 1000 = 45AnswerKE = 45000 J = 45 kJ
✍️Worked example 4 — rearranging for distance
QuestionA hydraulic ram does 2400 J of work with a force of 300 N. How far does the ram extend?
Formula
W = F × d rearranged to d = W ÷ FSubstitute
d = 2400 ÷ 300Divide
2400 ÷ 300 = 8Answerd = 8 m
⚡
Power & Efficiency
Power is how quickly energy is transferred. Efficiency is how much of the input energy does something useful.
⏱️Power & Efficiency CalculatorChoose a mode, enter values, press Calculate
Enter your values above and press Calculate to see every working step.
📚 Key vocabulary
- Power
- the rate of energy transfer — joules used every second.
- Watt (W)
- 1 joule per second. A 1 kW motor transfers 1000 J each second.
- Efficiency
- the fraction of input energy that becomes useful output, given as a percentage.
- Useful energy
- the energy that does the intended job (e.g. lifting the load).
- Wasted energy
- energy lost to the surroundings, usually as heat and sound.
- Kilowatt (kW)
- 1000 watts — the usual unit for motors and heaters.
👁️ Visual aid — energy flow through a motor: useful output grows, waste heat pulses away
✍️Worked example 5 — power of a bench drill
QuestionA bench drill transfers 3000 J of energy in 60 s. Calculate its power.
Formula
P = E ÷ tSubstitute
P = 3000 ÷ 60Divide
3000 ÷ 60 = 50AnswerP = 50 W
✍️Worked example 6 — energy used by a heater
QuestionA 2000 W workshop heater runs for 30 s. How much energy does it transfer, in kJ?
Formula
E = P × tSubstitute
E = 2000 × 30Multiply
2000 × 30 = 60000Convert
60000 ÷ 1000 = 60AnswerE = 60000 J = 60 kJ
✍️Worked example 7 — motor efficiency
QuestionA motor takes in 600 J of electrical energy and delivers 450 J of useful mechanical energy. Calculate its efficiency.
Formula
efficiency = (useful output ÷ total input) × 100%Substitute
efficiency = (450 ÷ 600) × 100Divide
450 ÷ 600 = 0.75Multiply
0.75 × 100 = 75AnswerEfficiency = 75% (150 J is wasted, mostly as heat)
✍️Worked example 8 — hoist power (combined)
QuestionA hoist lifts a 200 kg load through 3 m in 12 s. Calculate the useful power. Take g = 9.81 N/kg.
Step 1Work done lifting =
m × g × h = 200 × 9.81 × 3Multiply
200 × 9.81 = 1962, then 1962 × 3 = 5886 → W = 5886 JStep 2Power =
W ÷ t = 5886 ÷ 12Divide
5886 ÷ 12 = 490.5AnswerP = 490.5 W (≈ 0.49 kW)
⚙️
Engineering Applications
Four workshop calculators that put work, energy and power to use.
🔧 Hydraulic ram — work done
Fitters check the work a ram must do to push a press tool through its stroke.
W = F × d
🧮 Calculator
Enter the force and stroke, then press Calculate.
🏗️ Hoist motor — power to lift
Size a hoist motor from the load, lift height and time allowed for the lift.
P = (m × 9.81 × h) ÷ t
🧮 Calculator
Enter mass, height and time, then press Calculate.
⚡ Motor test — efficiency
Compare the useful energy delivered at the shaft with the electrical energy supplied.
efficiency = (useful ÷ input) × 100%
🧮 Calculator
Enter both energies and press Calculate.
🚗 Vehicle safety — kinetic energy
Brake designers work out the energy the brakes must absorb to stop a moving vehicle.
KE = ½ × m × v²
🧮 Calculator
Enter mass and speed, then press Calculate.
🎮
Energy Challenge
12 randomised questions per round. New numbers every time you play.
⚡ Quick-fire round
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Press Start to begin!
📝
Practice Questions
12 graded questions. Attempt each one on paper first, then reveal the full worked solution.
Take g = 9.81 N/kg throughout. Show the formula, the substitution and every calculation step.
Question 1 · Work1 mark
A force of 50 N pushes a box 2 m. Calculate the work done.
W = F × d = 50 × 2 = 100 J
Question 2 · Work2 marks
A winch pulls a trolley with a force of 250 N over 6 m. Calculate the work done.
W = F × d
W = 250 × 6
W = 1500 J (= 1.5 kJ)
W = 250 × 6
W = 1500 J (= 1.5 kJ)
Question 3 · Work2 marks
A machine does 1800 J of work with a force of 450 N. How far does the load move?
d = W ÷ F
d = 1800 ÷ 450
d = 4 m
d = 1800 ÷ 450
d = 4 m
Question 4 · Energy2 marks
An 8 kg toolbox is lifted 2.5 m onto a shelf. Calculate the gravitational potential energy gained.
GPE = m × g × h
GPE = 8 × 9.81 × 2.5
8 × 9.81 = 78.48, then 78.48 × 2.5 = 196.2 J
GPE = 8 × 9.81 × 2.5
8 × 9.81 = 78.48, then 78.48 × 2.5 = 196.2 J
Question 5 · Energy2 marks
A 70 kg cyclist rides at 4 m/s. Calculate their kinetic energy.
KE = ½ × m × v²
v² = 4 × 4 = 16
KE = 0.5 × 70 × 16 = 560 J
v² = 4 × 4 = 16
KE = 0.5 × 70 × 16 = 560 J
Question 6 · Power1 mark
A machine transfers 1200 J in 40 s. Calculate its power.
P = E ÷ t = 1200 ÷ 40 = 30 W
Question 7 · Power2 marks
A 1500 W compressor runs for 120 s. How much energy does it use, in kJ?
E = P × t
E = 1500 × 120 = 180000 J
180000 ÷ 1000 = 180 kJ
E = 1500 × 120 = 180000 J
180000 ÷ 1000 = 180 kJ
Question 8 · Power2 marks
How long does a 750 W motor take to transfer 9000 J?
t = E ÷ P
t = 9000 ÷ 750
t = 12 s
t = 9000 ÷ 750
t = 12 s
Question 9 · Efficiency2 marks
A pump receives 500 J of input energy and delivers 300 J of useful output. Calculate its efficiency.
efficiency = (useful ÷ input) × 100
= (300 ÷ 500) × 100 = 0.6 × 100 = 60%
= (300 ÷ 500) × 100 = 0.6 × 100 = 60%
Question 10 · Efficiency2 marks
A motor is 85% efficient and receives 2000 J of input. How much useful energy does it deliver?
useful = efficiency × input = 0.85 × 2000
useful = 1700 J (300 J wasted)
useful = 1700 J (300 J wasted)
Question 11 · Combined3 marks
A hoist lifts a 150 kg load through 4 m in 20 s. Calculate the work done, then the useful power.
W = m × g × h = 150 × 9.81 × 4
150 × 9.81 = 1471.5, then 1471.5 × 4 = 5886 J
P = W ÷ t = 5886 ÷ 20 = 294.3 W
150 × 9.81 = 1471.5, then 1471.5 × 4 = 5886 J
P = W ÷ t = 5886 ÷ 20 = 294.3 W
Question 12 · Combined3 marks
A 1200 kg van travels at 15 m/s. Calculate the kinetic energy the brakes must absorb to stop it, giving your answer in kJ.
KE = ½ × m × v²
v² = 15 × 15 = 225
KE = 0.5 × 1200 × 225 = 135000 J
135000 ÷ 1000 = 135 kJ
v² = 15 × 15 = 225
KE = 0.5 × 1200 × 225 = 135000 J
135000 ÷ 1000 = 135 kJ