Work and energy are measured in joules. Power tells you how quickly energy is transferred and is measured in watts. Efficiency compares useful output with total input. Before calculating, identify which of these quantities the question asks for and which energy transfer counts as useful.
The work, energy, power and efficiency examples below use constructed data for calculation practice. Unless otherwise stated, take gravitational field strength g = 10 N/kg. This is a supplied approximation for these questions, not a claim that g has exactly that value everywhere.
Follow the energy from input to useful output
| Situation | Quantity | Equation and condition |
|---|---|---|
| Constant force along a displacement | Work done | W = Fs |
| Raised through vertical height h | Increase in gravitational potential energy | ΔE = mgh, approximately constant g |
| Moving at speed v | Kinetic energy | Ek = mv²/2 |
| Energy E transferred in time t | Average power | P = E/t |
| Useful output compared with total input | Efficiency | Useful energy ÷ input energy × 100% |
The symbol W can mean work in an equation, while W after a number is the unit watt. Write units beside calculated answers to make the distinction clear. A result of 400 J is an energy; 400 W means 400 joules transferred per second.
In W = Fs, the displacement must be along the force. More generally, a constant force at angle θ to displacement does work Fs cos θ. A force perpendicular to displacement does zero work on that motion. “Force × distance” is therefore not a universal instruction to multiply any two numbers in a question.
Worked example 1: the same lift, two different quantities
A model lifting system raises a 40 kg load vertically through 2.4 m in 30 s at constant speed. Find the load’s gain in gravitational potential energy and the average useful lifting power.
Weight = mg = 40 × 10 = 400 N. Useful energy gained = mgh = 40 × 10 × 2.4 = 960 J. The same value follows from upward lifting force × vertical displacement, because constant speed and the stated ideal model make the lifting force equal to the weight.
Average useful power = 960/30 = 32 W. If the same lift takes 15 s instead, the useful energy remains 960 J but the average useful power becomes 64 W. Halving the time doubles the rate without doubling the energy for the same height.
We have not found the total input energy of a real machine. Friction and other losses could require additional input. The calculation also assumes no change in the load’s kinetic energy and excludes energy stored in moving parts of the mechanism.
Worked example 2: calculate the change in kinetic energy
A 5.0 kg model trolley speeds up from 2.0 m/s to 6.0 m/s on a level track. Find the net work done on it. Neglect changes in rotational energy and gravitational potential energy.
Initial kinetic energy = 5 × 2²/2 = 10 J. Final kinetic energy = 5 × 6²/2 = 90 J. Therefore the change is 90 − 10 = 80 J, and the net work done on the trolley is 80 J.
It is incorrect to use 5(6 − 2)²/2 = 40 J. The rule subtracts the two squared speeds: ΔEk = m(v² − u²)/2. Squaring a difference does not produce the difference of two squares.
As an independent check, a constant net forward force of 10 N acting through 8.0 m would do 10 × 8 = 80 J of work. The word “net” matters. If an applied force has to overcome resistance as well, its work can exceed the kinetic-energy increase because some energy is transferred to the surroundings.
If the trolley instead slows from 6.0 m/s to 2.0 m/s, its change in kinetic energy is −80 J. Negative net work represents a decrease in kinetic energy; it is not an instruction to discard the sign.
Worked example 3: complete an energy account
A fictional motor system receives 24 kJ of energy over 8.0 s. Its useful mechanical output over the same interval is 18 kJ. Assume no net increase in energy stored inside the system. Find efficiency, wasted energy, input power and useful output power.
| Transfer | Energy |
|---|---|
| Total input | 24,000 J |
| Useful mechanical output | 18,000 J |
| Energy transferred to unwanted outputs | 6,000 J |
Efficiency = 18,000/24,000 × 100% = 75%. Wasted energy = 24,000 − 18,000 = 6,000 J. “Wasted” means not serving the stated purpose; the energy has not vanished.
Average input power = 24,000/8 = 3,000 W. Average useful output power = 18,000/8 = 2,250 W. The difference, 750 W, is the average power going to unwanted outputs. Checking 2,250/3,000 × 100% gives the same 75% efficiency.
Energy efficiency and the corresponding average-power ratio agree here because both energy transfers cover the same time interval. Comparing an input measured over eight seconds with an output measured over a different interval would not be valid.
Worked example 4: work backwards from useful energy
A lifting device has efficiency 80% and must supply 1,600 J of useful energy. How much input energy is needed?
Use the fraction 0.80: useful energy = 0.80 × input energy. Rearranging gives input = 1,600/0.80 = 2,000 J. Unwanted output is 2,000 − 1,600 = 400 J, under the no-storage-change assumption.
Multiplying 1,600 by 0.80 would incorrectly make input smaller than useful output. The size check is decisive for an ordinary energy-conversion device: a device below 100% efficiency requires more input energy than the useful energy it delivers.
Keep units and system boundaries consistent
Convert kilojoules to joules and minutes to seconds when calculating watts: 1 kJ = 1,000 J and 1 min = 60 s. If a machine transfers 9 kJ in 3 min, its average power is 9,000/180 = 50 W, not 3 W.
Energy conservation applies to a properly defined system. In general, input energy equals useful output plus unwanted output plus any increase in stored energy. The simple input = useful + wasted account assumes that stored energy does not change. State that assumption rather than silently forcing every remainder into “waste”.
Dimensional checks help: N × m gives joules for work; kg × (m/s)² gives joules for kinetic energy; J/s gives watts. Efficiency is a ratio of like quantities and has no physical unit, although it is often expressed as a percentage.
Three solved practice tasks
Practice 1: work and average power
A constant 75 N force acts along a 12 m displacement over 45 s. Calculate its work and average power.
Solution: Work = 75 × 12 = 900 J. Average power = 900/45 = 20 W. These are the work and power associated with that force, not automatically the net values if other forces also do work.
Practice 2: a kinetic-energy increase
An 8 kg object increases speed from 2 m/s to 5 m/s. Find its kinetic-energy increase.
Solution: Initial energy = 8 × 4/2 = 16 J; final energy = 8 × 25/2 = 100 J. Increase = 84 J. Alternatively, 8(25 − 4)/2 = 84 J. Keeping the initial term avoids incorrectly calling the entire final energy an increase.
Practice 3: input, output and losses
A system delivers 3,600 J of useful energy in 120 s at 60% efficiency. Find input energy, unwanted output and average input power. Assume no change in stored energy.
Solution: Input = 3,600/0.60 = 6,000 J. Unwanted output = 6,000 − 3,600 = 2,400 J. Average input power = 6,000/120 = 50 W. Useful power is 30 W, and 30/50 = 0.60 verifies the efficiency.
Frequently asked questions
Can an object have energy while stationary?
Yes. It may have gravitational potential energy relative to a chosen level, chemical energy or other stored energy, even with zero kinetic energy.
Is a more powerful machine always more efficient?
No. Power describes rate; efficiency describes the useful fraction of input. Compare both quantities for the particular task.
What height belongs in mgh?
Use the vertical rise between the initial and final positions. The length of a sloping route is generally larger and is not the height change.
Why does doubling speed quadruple kinetic energy?
Speed is squared in mv²/2. For unchanged mass, replacing v by 2v multiplies kinetic energy by four.
Can ordinary energy efficiency exceed 100%?
Not when all energy inputs are included and useful output is defined consistently. A result above 100% signals a missing input, mismatched intervals or a calculation error.
Must wasted energy always be heat?
No. Depending on the device and its purpose, unwanted outputs may include sound, light, movement or thermal transfers. State what the question establishes.
When can I use power instead of energy for efficiency?
Use useful output power divided by input power under matching operating conditions, or matching average powers over the same interval.
Further reading and revision
The standard power relationship is explained in OpenStax’s power section. All worked data here are original. The existing Physics topical revision resource requires login; this lesson makes no claim about particular downloadable files.
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