Mini PhysicsPart of Mini Education

Work

Key idea: Learn what work done means, when no work is done, and how to calculate work using W = Fd with correct units (G3 Physics and O-Level Physics).

  • G3 Physics / O-Level Physics
  • Reviewed Jul 19, 2026

By the end, you can

  • Calculate work done when a constant force moves an object through a distance in the force's direction.
  • Explain work done as a mechanical energy transfer and identify when a force does no work.
Topic lessons
  1. Work (W = Fd)
  2. Energy (Stores, Transfers & Conservation)
  3. Energy Calculations: KE, GPE and Conservation
  4. Power and Efficiency (P = E/t)
  5. Energy Resources (Renewable & Non-Renewable)

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
  • be transferred to internal energy (heat) of the object/surroundings when resistive forces like 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.

A Level extension (optional)

At A Level, you will learn the work–energy theorem (linking net work done to change in kinetic energy). See Kinetic Energy From Work Done.

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

Example 1: Work done on a trolleyCore

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.

  1. Calculate the work done on the trolley.
  2. Describe the main energy transfer.
Show Answer

Work done: W = Fd = (8.0 N)(5.0 m) = 40 J

Energy transfer: energy is transferred mechanically by the force. If resistive forces are negligible and the trolley speeds up, its kinetic energy store increases.

Example 2: Work done when pushing a boxCore

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.

Show Answer

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

Example 3: Pulling a wagon (linking to F = ma)Core

A 1.0 kg toy wagon is pulled along a smooth horizontal floor through 5.0 m. Its acceleration is 2.0 m s⁻². Find the work done by the pulling force.

Show Answer

Smooth floor ⇒ negligible friction, so the pulling force is the resultant force. F = ma = (1.0)(2.0) = 2.0 N

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

Example 4: Lifting and GPECore

A 5.0 kg box is raised vertically through 50 m at constant speed. Take g = 10 N kg⁻¹. Find:

  1. the gain in gravitational potential energy, and
  2. the work done by the lifting force (ignore air resistance).
Show Answer

Gain in GPE: Δ Eₚ = mgh = (5.0)(10)(50) = 2.5 × 10³ J

Since the lift is at constant speed and losses are neglected, the work done by the lifting force equals the gain in GPE: W = 2.5 × 10³ J

Example 5: Dropping a ball (no air resistance)Core

A 0.500 kg ball is dropped from rest from a height of 10 m above the ground. Take g = 10 N kg⁻¹.

  1. Find the ball’s initial gravitational potential energy (relative to the ground).
  2. Find its speed just before it hits the ground (ignore air resistance).
Show Answer

  1. Initial GPE: Eₚ = mgh = (0.500)(10)(10) = 50 J

  2. With resistive transfers neglected, decrease in GPE = increase in KE: 50 = 1/2(0.500)v² Rightarrow v = 14.1 m s⁻¹

Example 6: Dropping a ball with air resistanceCore

The same 0.500 kg ball is dropped from 10 m, but 10 J is dissipated due to air resistance. Find its speed just before it hits the ground.

Show Answer

Initial GPE = 50 J. With 10 J dissipated, final KE is: Eₖ = 50 - 10 = 40 J

40 = 1/2(0.500)v² Rightarrow 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): Efᵢₙₐₗ = (0.500)(10)(8.0) = 40.0 J

Energy dissipated: Edᵢₛₛᵢₚₐₜₑd = 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 = 1/2mv² = 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.

Recommended next step

Why this will help: Use one focused question set to check that you can apply the lesson without prompts.

About 10 minutes

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