Skip to main content
ScotlandEngineering ScienceSyllabus dot point

How do we calculate the work done by a force and the power of a machine?

Work done by a force, mechanical power as work done per second, and the relationships work equals force times distance and power equals work over time.

An SQA National 5 Engineering Science answer on work, energy and power, covering work done as force times distance, energy transferred equal to work done, mechanical power as work done per second, and the relationships needed to calculate each in an engineering context.

Generated by Claude Opus 4.89 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 key area is asking
  2. Work done
  3. Energy transfer
  4. Power
  5. Putting work and power together
  6. Why this matters
  7. Try this

What this key area is asking

The SQA wants you to calculate the work done by a force using Ew=F×dE_w = F \times d, understand that work transfers energy, and calculate mechanical power as work done per second.

Work done

If a force does not move (no movement, or the force is at right angles to the motion), then no work is done by that force. Lifting a crate a height hh does work F×hF \times h against gravity, where FF is the weight being lifted.

Energy transfer

Because work done equals energy transferred, the joule is the unit of both. When a forklift lifts a load it does work against gravity, transferring energy to the load as gravitational potential energy. When a motor drives a mechanism, it does work that becomes the useful output plus the wasted heat. This links straight back to the conservation of energy: the energy put in equals the useful work out plus the energy wasted.

This connection also lets you find efficiency from work and power. Since efficiency is the useful output energy divided by the total input energy (times 100%), and work done is the energy transferred, you can compare the useful work a machine delivers with the total work put into it. A hoist that takes in 8000 J of input work but delivers only 6000 J of useful lifting work is 75% efficient, with the missing 2000 J wasted as heat through friction - the same idea as comparing mechanical advantage with velocity ratio.

Power

Putting work and power together

Many questions give a force, a distance and a time, and ask for the power. The method is always: first find the work done with Ew=F×dE_w = F \times d, then divide by the time with P=Ew/tP = E_w / t.

Why this matters

Work and power tie the whole mechanical area together. A mechanism's efficiency compares the useful work out with the energy in; a motor's power rating tells you how quickly it can do work; and lifting, pushing and driving are all work calculations. These are among the most frequently examined relationships in the course.

Try this

Q1. A force of 80 N80 \text{ N} moves a box 3.0 m3.0 \text{ m} in the direction of the force. Calculate the work done. [2 marks]

  • Cue. Ew=F×d=80×3.0=240 JE_w = F \times d = 80 \times 3.0 = 240 \text{ J}.

Q2. State the unit of power and what it is equivalent to. [1 mark]

  • Cue. The watt (W), equal to one joule per second.

Q3. A machine does 9000 J9000 \text{ J} of work in 45 s45 \text{ s}. Calculate its power. [2 marks]

  • Cue. P=Ew/t=9000/45=200 WP = E_w/t = 9000/45 = 200 \text{ W}.

Exam-style practice questions

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

SQA N5 style3 marksA forklift lifts a crate with a force of 250 N through a height of 1.8 m. Calculate the work done.
Show worked answer →

Use the work done relationship.

Relationship: Ew=F×dE_w = F \times d (work done equals force times distance moved in the direction of the force).

Substitution: Ew=250×1.8=450 JE_w = 250 \times 1.8 = 450 \text{ J}.

Markers reward selecting work equals force times distance, correct substitution, and a final answer in joules (J). The distance must be in the direction of the force - here, the height lifted.

SQA N5 style4 marksA motor does 6000 J of work in 30 s. Calculate the power developed by the motor.
Show worked answer →

Power is the rate of doing work.

Relationship: P=EwtP = \dfrac{E_w}{t} (power equals work done divided by time).

Substitution: P=600030=200 WP = \dfrac{6000}{30} = 200 \text{ W}.

Markers reward selecting power equals work over time, correct substitution, and a final answer in watts (W). Time must be in seconds. A more powerful motor would do the same work in less time.

Related dot points

Sources & how we know this