Skip to main content
Northern IrelandTechnology and DesignSyllabus dot point

How are work, power and efficiency calculated for mechanical systems, and where is energy lost?

Work, power, torque and efficiency in mechanical systems, and the relationship between mechanical advantage, velocity ratio and efficiency.

A CCEA A-Level Technology and Design answer on work, power, torque and efficiency in mechanical systems, and how efficiency relates mechanical advantage to velocity ratio, with the sources of energy loss such as friction.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

CCEA expects you to calculate work, power, torque and efficiency for mechanical systems, and to relate mechanical advantage and velocity ratio through efficiency. You should explain where energy is lost (chiefly friction). These calculations are common in the A2 mechanical paper.

The answer

Work, power and torque

Efficiency

Where energy is lost

Worked example: efficiency of a lifting machine

Examples in context

Example 1. Electric motor rating
A motor is rated by its torque and speed; its output power TωT\omega and its efficiency (output over electrical input) decide how much heat it wastes, the calculation used to size a drive.
Example 2. Gearbox losses
Each gear mesh and bearing adds friction, so a multi-stage gearbox is less than 100 per cent efficient; lubrication and precision reduce the heat lost, the practical side of efficiency. A compound gearbox with several stages multiplies these small per-stage losses, so overall efficiency falls as more stages are added, which is why designers use the fewest stages that achieve the required ratio.
Example 3. Pulley hoist on a building site
A block and tackle has a high velocity ratio but real friction in its sheaves, so its mechanical advantage is well below the velocity ratio and the worker must pull through a long length of rope. Comparing the work done by the effort (force times the long rope length) with the useful work on the load (load times the short lift) directly measures the efficiency, and the difference is the energy lost to friction as heat.

Try this

Q1. State the equation for work done. [1 mark]

  • Cue. W=F×dW = F \times d (force times distance in the direction of the force).

Q2. A machine has MA = 3 and VR = 4. Find its efficiency. [2 marks]

  • Cue. η=(3/4)×100%=75%\eta = (3/4) \times 100\% = 75\%.

Q3. Why is the efficiency of a real machine always less than 100 per cent? [2 marks]

  • Cue. Some input energy is lost, mainly to friction (turned into heat), so the useful output is less than the input.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20196 marksA machine has a velocity ratio of 5. An effort of 250 N raises a load of 1000 N. Calculate the mechanical advantage and the efficiency, and explain why the efficiency is less than 100 per cent.
Show worked answer →

Mechanical advantage (MA) is load over effort:

MA=loadeffort=1000250=4.\text{MA} = \frac{\text{load}}{\text{effort}} = \frac{1000}{250} = 4.

Efficiency is the mechanical advantage divided by the velocity ratio (or useful output over total input), expressed as a percentage:

η=MAVR×100%=45×100%=80%.\eta = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{4}{5} \times 100\% = 80\%.

So the machine is 80 per cent efficient.

It is less than 100 per cent because some input energy is lost, mainly to friction between moving parts (and a little to moving the machine's own weight and to deformation/noise/heat). The lost energy is turned into heat, so the useful work out is always less than the work put in. This is why MA is less than VR in a real machine.

Markers reward MA = load/effort = 4, efficiency = MA/VR x 100 = 80%, and friction (energy lost as heat) as the reason efficiency is below 100%.

CCEA 20214 marksA motor produces a torque of 4.0 N m at a rotational speed of 1500 rev/min. Calculate the output power. (Use power = torque x angular velocity in rad/s.)
Show worked answer →

Output power for a rotating shaft is torque times angular velocity (with angular velocity in rad/s):

P=T×ω.P = T \times \omega.

First convert the speed to rad/s:

ω=2πN60=2π×150060=157 rad/s.\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1500}{60} = 157\ \text{rad/s}.

Then:

P=T×ω=4.0×157=628 W≈0.63 kW.P = T \times \omega = 4.0 \times 157 = 628\ \text{W} \approx 0.63\ \text{kW}.

So the output power is about 628 W.

Markers reward converting rev/min to rad/s (2 pi N over 60), using P = T omega, and the answer of about 628 W.

Related dot points

Sources & how we know this