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Work

Key idea: Learn what work done means, when no work is done, and how to calculate work using W = Fd with correct units (O Level Physics).

  • SEC G3 Physics 2027
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Learning objectives

  • Recognise kinetic, potential, nuclear and internal energy stores
  • Describe mechanical energy transfer by a force acting over a distance
  • Describe electrical energy transfer by an electric current
  • Describe energy transfer by heating due to a temperature difference
  • Describe energy transfer by electromagnetic and mechanical waves
  • Recall and apply Ek = ½mv² in new situations
  • Recall and apply Ep = mgh near the Earth's surface in new situations
  • State and apply the principle of conservation of energy
  • Recall and apply work done = force × distance moved in the force direction
  • Recall and apply power = energy transfer / time taken
  • Calculate efficiency as useful energy output / total energy input
  • Evaluate prescribed electricity-generation resources by efficiency, cost, reliability and environmental impact

1. Definition

A. Work done

Work done is the energy transferred when a force causes an object to move in the direction of the force.

For a constant force parallel to the motion:

  • W = Fd
  • W = work done (J), F = force (N), d = distance moved in the direction of the force (m)
  • 1 J = 1 N m
Three force and displacement cases for work doneThree rows compare a box pushed in the force direction, a wall pushed without moving, and a carried load moving horizontally while the support force is vertical. The first force does positive work; the other two forces do zero work.Does this force do work?Compare the direction of the named force with the object's displacement.A Force and displacement in the same directionforce, Fdisplacement, dwork done = FdB Force applied, but the wall does not movepushd = 0work done by the push = 0C Support force perpendicular to horizontal displacementsupport forcedisplacementwork done by the support force = 0
Scroll diagram horizontally to read all labels.
Ask about one force at a time. Work is done by that force only when the object has a displacement component in the force's direction.

2. Key Ideas

  • Use the distance moved in the direction of the force (not the total path length).
  • If there is no displacement in the force direction, work done by that force is zero.
  • Work done is a way of transferring energy mechanically (a force acting over a distance).
  • On a rough surface, some work is done against friction, increasing the internal energy stores of the surfaces and surroundings.
  • Convert units first (cm → m, kN → N). The final unit for work is J.

3. Detailed Explanations

A. When does a force do work?

Work is about energy transfer. A force does work on an object only if the object has a displacement in the direction of that force.

This is why the work formula includes a direction condition:

W = Fd (distance moved in the direction of the force)

B. When is no work done?

No work is done by the force being considered when:

  1. the object does not move (e.g. pushing against a wall), or
  2. the displacement is perpendicular to the force (e.g. carrying a load horizontally at constant height).

C. Work done and energy stores (O Level)

Work done is energy transferred mechanically (a force acting over a distance). The energy transferred can:

  • increase the object’s kinetic energy if it speeds up, or
  • increase the object’s gravitational potential energy if it is lifted, or
  • increase the internal energy stores of the object and surroundings when resistive forces such as friction act.
Mechanical work transfers energy into different storesTwo panels show a force acting through a distance. In the first, a horizontal push speeds up a trolley and increases its kinetic energy. In the second, an upward force lifts a box and increases its gravitational potential energy.Work done = energy transferred mechanicallySpeeding up on a smooth surfaceforcemotionwork done → kinetic energy increasesif resistive forces are negligibleLifting at constant speedlifting forceheight, hwork done → GPE increasesW = Fh = mgh when F = mg
Scroll diagram horizontally to read all labels.
Mechanical work is an energy transfer. A horizontal push can increase a kinetic energy store, while an upward force can increase a gravitational potential energy store.

D. Lifting and Eₚ = mgh

If you lift an object at constant speed, the lifting force equals the weight:

F = mg

The displacement in the direction of the lifting force is the vertical height h, so:

W = Fh = mgh

This work done equals the gain in gravitational potential energy.

4. Common Mistakes

  • Using the wrong distance: use the distance moved in the direction of the force.
  • Mixing up work (J) and power (W).
  • Forgetting unit conversions (cm → m, g → kg, kN → N).
  • Saying “work is done because a force is applied” (if there is no displacement in that direction, work done is zero).
  • Confusing work (N m = J) with moment (N m): they share units but represent different quantities.

5. Exam Tips

  • Write the formula first: W = Fd.
  • Check that the force is constant and the motion is in the same direction as the force.
  • If lifting at constant speed with no losses, work done = gain in GPE = mgh.
  • For “describe the energy transfer” questions, use the syllabus wording: energy transferred mechanically by a force acting over a distance.
  • Always end with the correct unit: J.

6. Worked Examples

Modelled example 1

Work done on a trolley

Core

Problem

A librarian applies a constant horizontal force of 8.0 N to push a trolley through 5.0 m in the direction of the force. Calculate the work done and describe the main energy transfer.

Study the worked solution

  1. Use the displacement in the force direction

    Method

    Apply W = Fd with the 5.0 m horizontal displacement.

    Reason

    The force is constant and parallel to the motion.

    Working

    W = (8.0)(5.0) = 40 J

  2. Describe the transfer

    Method

    State that energy is transferred mechanically.

    Reason

    A force acting over a distance is a mechanical transfer pathway.

    Working

    If resistance is negligible and the trolley speeds up, its kinetic energy store increases.

Guided practice 2

Work done when pushing a box

About 3 min

Problem

A boy pushes a box with a force of 5.0 N through 2.0 m in the direction of the force. Find the work done by the boy.

Apply the force-direction condition

Unit: J

Hints

Hint 1: select the relationship
For a constant parallel force, use W = Fd.
Hint 2: substitute
Multiply 5.0 N by 2.0 m.
View solution step by step

  1. Calculate the work

    Reason

    The displacement is entirely in the direction of the applied force.

    Working

    W = Fd = (5.0)(2.0) = 10 J

Common misconception 3

Pulling a wagon (linking to F = ma)

Find and correct the mistake

Learner response

A 1.0 kg toy wagon is pulled along a smooth horizontal floor through 5.0 m with acceleration 2.0 m s⁻². A student writes W = md = (1.0)(5.0) = 5.0 J. Locate the first error and find the work done by the pulling force.

Identify the missing physical quantity

What must be found before using W = Fd?
Unit: J

View solution step by step

  1. Find the pulling force

    Method

    Use Newton’s second law before the work equation.

    Reason

    On the smooth floor, friction is negligible and the pulling force is the resultant force.

    Working

    F = ma = (1.0)(2.0) = 2.0 N

  2. Calculate the work

    Reason

    Use force, not mass, in W = Fd.

    Working

    W = (2.0)(5.0) = 10 J

Examiner practice 4

Lifting and GPE

3 marks

Examination question

A 5.0 kg box is raised vertically through 50 m at constant speed. Take g = 10 N kg⁻¹ and ignore air resistance. Find the gain in GPE and the work done by the lifting force. [3 marks]

Show the calculation and energy link

View solution step by step

  1. Select the GPE relationship

    1 mark

    Method

    Use the vertical height in Δ Eₚ = mgh.

    Reason

    The box changes vertical position in the gravitational field.

    Working

    Δ Eₚ = (5.0)(10)(50)

  2. Calculate the GPE gain

    1 mark

    Reason

    All quantities are in SI units.

    Working

    Δ Eₚ = 2.5 × 10³ J

  3. Relate lifting work to GPE

    1 mark

    Method

    State the work done by the lifting force.

    Reason

    At constant speed with losses neglected, the lifting work equals the GPE gain.

    Working

    W = 2.5 × 10³ J

Challenge 5

Dropping a ball with air resistance

Minimal support

Resistive transfer added

A 0.500 kg ball is dropped from rest through 10 m. During the fall, 10 J is dissipated by air resistance. Take g = 10 N kg⁻¹. Find its speed just before impact.

Track the energy transferred to the surroundings

Unit: m s^-1

Hints

Hint 1: find initial GPE
Use mgh = (0.500)(10)(10).
Hint 2: include air resistance
Subtract 10 J before using Eₖ = (1/2)mv².
View solution step by step

  1. Find the initial GPE

    Method

    Calculate the energy available at release.

    Reason

    The ball starts from rest 10 m above the chosen zero level.

    Working

    Eₚ = (0.500)(10)(10) = 50 J

  2. Account for dissipation

    Method

    Subtract the transfer caused by air resistance.

    Reason

    Total energy is conserved, but only 40 J remains in the ball’s kinetic store.

    Working

    Eₖ = 50-10 = 40 J

  3. Calculate impact speed

    Reason

    Use the remaining kinetic energy in Eₖ = (1/2)mv².

    Working

    40 = (1/2)(0.500)v² ⇒ v = 12.6 m s⁻¹

7. Mind Stretchers

Mind stretcher 1: Energy dissipated in a reboundExtension

A 0.500 kg ball is thrown vertically downwards from a height of 10 m with speed 5.0 m s⁻¹. It rebounds and rises to a maximum height of 8.0 m. Take g = 10 N kg⁻¹. Find the energy dissipated during the bounce (ignore air resistance).

Show Answer

Initial energy: Eᵢₙᵢₜᵢₐₗ = (1/2)(0.500)(5.0²) + (0.500)(10)(10) = 56.25 J

Final energy at the top after rebound (KE = 0): E_final = (0.500)(10)(8.0) = 40.0 J

Energy dissipated: E_dissipated = 56.25 - 40.0 = 16.25 ≈ 16.3 J

Mind stretcher 2: Work done by a constant driving forceExtension

A constant resultant force of 10 kN acts on a car. The car accelerates from rest to 30 m s⁻¹ in 10 s with constant acceleration. Find:

  1. the distance travelled in 10 s,
  2. the work done by the force, and
  3. the kinetic energy of the car at 30 m s⁻¹.
Show Answer

Acceleration: a = (v-u)/t = (30-0)/10 = 3.0 m s⁻²

Distance travelled: d = (1/2)(u + v)t = (1/2)(0 + 30)(10) = 150 m

Work done: W = Fd = (10 000 N)(150 m) = 1.5 × 10⁶ J

Mass of car (since F = ma): m = F/a = (10 000)/3.0 = 3.33 × 10³ kg

Kinetic energy at 30 m s⁻¹: Eₖ = (1/2)mv² = (1/2)(3.33 × 10³)(30²) ≈ 1.5 × 10⁶ J

8. Practice

Practice Time!

Use the Work, Energy & Efficiency Explorer to practise choosing W = Fd, then try the focused Work, Energy & Power Quiz.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027