8867 / 2027
Energy and Fields overview

Lesson 6 of 6

Power, losses and efficiency

  • I can give examples of stores and transfers and apply conservation of energy to problems.Syllabus

    Syllabus 8867, 4(b). Give examples of stores and transfers and apply conservation of energy to problems.

  • I can define power as the rate of energy transfer.Syllabus

    Syllabus 8867, 4(l). Define power as the rate of energy transfer.

  • I can use force times the velocity component in its direction to explain mechanical power.Syllabus

    Syllabus 8867, 4(m). Use force times the velocity component in its direction to explain mechanical power.

  • I can explain the implications of energy losses in practical devices and calculate useful-output to total-input efficiency.Syllabus

    Syllabus 8867, 4(n). Explain the implications of energy losses in practical devices and calculate useful-output to total-input efficiency.

Energy measures how much transfers. Power measures the rate of transfer. Efficiency compares the useful output with the total input for a stated task.

Average and instantaneous power

Average power = energy transferred / elapsed time
Paverage = ΔE/Δt

The unit is W = J/s. Instantaneous power is the rate at a particular time. Always identify the transfer being described: total input, useful output or the work of a named force.

Over a short interval, a force's work is Fs cos θ. Divide by the interval and use displacement per unit time in the limit of a short interval:

P = Fv cos θ

Here θ is the angle between force and velocity. Mechanical power is the force multiplied by the velocity component in its direction. P = Fv is the aligned special case. An opposing force has negative power; a perpendicular force has zero power.

For the crate in the angled-pull example, consider an instant when its speed is 2.0 m/s right. The 8.0 N pull at 60 degrees supplies 8.0 × 2.0 × cos 60° = 8.0 W. The 2.0 N resistance contributes -4.0 W, giving net mechanical power +4.0 W, the rate of kinetic-energy increase. The perpendicular support contributes zero.

Compare the useful output and total input

Efficiency = useful energy output / total energy input

Efficiency is unitless and is often expressed as a percentage. For powers averaged over the same interval, the same ratio can be calculated as useful output power / total input power.

Worked power and efficiency

The steady lifting motor

A motor raises a 40.0 N load through 5.00 m in 4.00 s at steady speed. Its measured electrical input energy is 320 J. The useful gravitational increase is 40.0 × 5.00 = 200 J, and the remaining 120 J increases internal energy of the device and surroundings.

Useful output power = 200/4.00 = 50.0 W
Input power = 320/4.00 = 80.0 W
Efficiency = 200/320 = 50.0/80.0 = 0.625 = 62.5%

The load's speed is 5.00/4.00 = 1.25 m/s. Lifting-force power is therefore 40.0 × 1.25 = 50.0 W. Use the lifting force, not the zero resultant force. The resultant's zero power accounts for unchanged kinetic energy; it does not mean the motor transfers no energy.

Try it step by step

Work it out in steps: Useful power = mgh / t in W, then Input power = useful power / efficiency in W, then Input energy in one minute in J.

  1. Problem 1. A motor lifts a 50 kg load through 12 m in 20 s. Its efficiency is 60%. Take g = 9.81 m/s2. Find the useful power, the input power and the input energy in one minute.
  2. Problem 2. A hoist lifts an 80 kg load through 10 m in 16 s at 75% efficiency. Take g = 9.81 m/s2. Find the useful power, the input power and the input energy in one minute.
  3. Problem 3. A winch lifts a 20 kg load through 15 m in 10 s at 70% efficiency. Take g = 9.81 m/s2. Find the useful power, the input power and the input energy in one minute.
Show the answer

Problem 1.

Useful power = mgh / t in W: Useful power = 50 x 9.81 x 12 / 20 = 294 W

Input power = useful power / efficiency in W: Input power = 294 / 0.60 = 491 W

Input energy in one minute in J: Input energy = 491 x 60 = 29400 J (3 sf)

Problem 2.

Useful power = mgh / t in W: Useful power = 80 x 9.81 x 10 / 16 = 491 W

Input power = useful power / efficiency in W: Input power = 491 / 0.75 = 654 W

Input energy in one minute in J: Input energy = 654 x 60 = 39200 J (3 sf)

Problem 3.

Useful power = mgh / t in W: Useful power = 20 x 9.81 x 15 / 10 = 294 W

Input power = useful power / efficiency in W: Input power = 294 / 0.70 = 420 W

Input energy in one minute in J: Input energy = 420 x 60 = 25200 J (3 sf)

Useful lifting power and input power describe different transfers

A steady lift can transfer energy with zero resultant force

A forty-newton lifting force at 1.25 metres per second supplies fifty wattsA motor pulls a load steadily upward. On the load, the upward lifting force and downward weight are each forty newtons, shown by equal-length purple arrows. A separate blue arrow shows upward velocity 1.25 metres per second. The steady interval covers five metres in four seconds. Although the resultant force is zero, the lifting force supplies fifty watts and gravity's mechanical power on the load is negative fifty watts. The load's kinetic energy does not change, while load-Earth gravitational potential energy increases. Arrangement and velocity-arrow lengths are schematic.MotorLift: 40.0 NWeight: 40.0 N1.25 m/supwardSteady interval: 5.00 m in 4.00 sUseful P = 40.0 × 1.25 = 50.0 W

Use the lifting force when finding its power, not the zero resultant force. Gravity's negative work balances the positive work on the load, so its kinetic energy stays constant while the load-Earth gravitational store increases.

Same energies in half the time

Two supplied runs have equal energy and efficiency but different powersBoth runs receive 320 joules and provide 200 joules usefully. Their input bars are both 288 drawing units wide and useful bars both 180, using one energy scale. Run A takes four seconds and Run B two; their separate time bars are 240 and 120 units wide. Average input and useful powers are eighty and fifty watts for A, and 160 and one hundred watts for B. Both efficiencies are 200 divided by 320, or 62.5 percent. The equal energies are supplied conditions; a real device's efficiency need not remain constant when its speed changes.Run A: 4.00 sInput: 320 JUseful: 200 JTime: 4.00 sInput 80.0 W; useful 50.0 WRun B: 2.00 sInput: 320 JUseful: 200 JTime: 2.00 sInput 160 W; useful 100 WBoth efficiencies: 62.5%

Energy bars use one scale; time bars use another. Halving the time doubles both powers here. Equal input and useful energies keep the efficiency at 62.5% under the supplied conditions.

The lifting-force calculation identifies the useful transfer. In the two supplied runs, equal input and useful energies are transferred over different times, so both powers change while their ratio stays the same.

If a second supplied run transfers the same 320 J input and 200 J useful output in 2.00 s, its input power is 160 W and useful output power 100 W. Both powers double; efficiency remains 62.5%. The equal energies are supplied conditions, not a general claim that changing a real motor's speed leaves efficiency unchanged.

The 120 J internal increase does not lift the load, so the motor needs more than 200 J input. Reducing unwanted mechanical friction can reduce that loss in an otherwise comparable lifting task. All energy is still accounted for; the loss is a reduction in useful output for this task.

A quoted efficiency alone does not give an energy amount: the input is also needed. For a complete account with the stated useful output and total input, efficiency cannot exceed 100%.

Check your understanding

Two supplied motor runs each transfer 320 J in and raise the load-Earth gravitational store by 200 J. One run takes 4.00 s and the other 2.00 s. Which comparison follows from these data?

  1. The faster run transfers twice the useful energy and has twice the efficiency.
  2. The useful output powers are equal because the useful energies are equal.
  3. The faster run has twice the input and output powers; efficiencies are both 62.5%.
Show the answer

The faster run has twice the input and output powers; efficiencies are both 62.5%.

Input powers are 80 and 160 W; useful output powers are 50 and 100 W. Each efficiency is 200/320 = 0.625. This comparison relies on the supplied equal energies; real efficiency need not stay constant at another operating speed.

Kilowatt-hours measure energy

A kilowatt-hour is the energy transferred at 1 kW for 1 hour. Its symbol is kWh; multiplying power by time gives energy.

1 kWh = 1000 J/s × 3600 s = 3.6 × 106 J

For a supplied constant power of 2.0 kW over 3.0 minutes:

E = Pt = 2.0 × (3.0/60) = 0.10 kWh
= 3.6 × 105 J

The 2.0 kW value is a rate. The 0.10 kWh value is the amount transferred during the stated interval.

Investigate a lift with matching measurements

To determine a lifting device's work, power and efficiency, measure quantities for the same defined interval:

  1. Load force: use a suitable force reading, or measure mass and multiply by the stated local g. Check zero and choose a range that covers the load.
  2. Vertical rise: measure the difference between start and finish heights with a suitable rule or tape. A sloping string length is not the vertical rise. View a pointer perpendicular to the scale to avoid parallax, and retain the same reference.
  3. Elapsed time: identify the same start and finish events used for the displacement and energy readings. Use a steady interval, or account for kinetic-energy changes if the speed changes.
  4. Input energy: use the change in a suitable calibrated energy reading over that interval. A motor's rated power is not a measurement of the energy taken in during a particular run.
  5. Analysis: calculate useful work from force and vertical rise, then divide each measured energy by the same time. Use useful energy / input energy for efficiency.

Choose ranges and resolutions that suit the force, length and time scales. Record raw values with units, repeat comparable runs to assess scatter, and retain guard digits in calculations. An underestimated rise makes the calculated useful work and efficiency too low if the other readings are correct. Repeating the same incorrect height reference will not remove that error.

Load force × speed gives useful mechanical output power here, not electrical input power. An input-energy reading supplies the latter account without assuming that every input joule lifts the load. Friction and warming may explain a real energy difference; they are not automatically faults in the instruments.

Estimate before calculating: a roughly 1 kg load has weight about 10 N near Earth. Raising it about 1 m in about 1 s at steady speed takes roughly 10 J and needs useful power of order 10 W. An inefficient device needs greater input power. The mass, rise, time and local field are the stated assumptions behind this estimate.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

A car travels at a constant speed of 30 m s-1 on a level road against a total resistive force of 900 N. Its engine is 30% efficient. Calculate the useful power output and the rate at which the engine uses energy from the fuel. [3]

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Exam-style question

Data-based question: The range of an electric car [17]

The range of an electric car depends on how fast it is driven. Engineers measured the total resistive force F on a car of mass 1500 kg driven at constant speed v on a level road with no wind. Part of F is rolling resistance from the tyres, which is about 180 N at every speed; the rest is air resistance. The car's battery stores 60 kW h of usable energy, and 85% of the energy taken from the battery does work at the wheels.

Table 1. Resistive force at different speeds
v / m s-11015202530
F / N2804055808051080
  1. (a) The data fit F = 180 + kv2. Use the data at 10 m s-1 to find k, and check it at 20 m s-1. [2]
  2. (b) Calculate the power needed at the wheels at a constant 25 m s-1. [2]
  3. (c) Calculate the power taken from the battery at this speed. [1]
  4. (d) Calculate the range of the car at a constant 25 m s-1. [3]
  5. (e) Calculate the range at a constant 15 m s-1. [2]
  6. (f) A driver says: "If I drive twice as fast, my range halves." Use the data to evaluate this statement. [3]
  7. (g) The car climbs a hill that rises 1 m for every 20 m along the road, at a constant 25 m s-1. Calculate the power needed at the wheels. [3]
  8. (h) State one assumption you made in (d). [1]

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