Work done = force × distance moved in the direction of the force, or W = Fd. It gives the energy transferred by a force, in joules, and appears in almost every mechanics and energy question.
It follows on from separating energy transfer from force. It also reuses weight from mass, weight and density, because lifting questions need W = mg for the force.
How does W = Fd work, step by step?
The formula has three parts, and each one must match the others.
- Identify the force that does the work. It is the force doing the pushing, pulling or lifting, not every force on the object.
- Identify the distance moved in the direction of that force. For lifting, this is the vertical height.
- Convert to standard units. Force in N, distance in m, and then work is in J.
- Multiply and write the unit with the answer.
Rearranging gives F = W ÷ d and d = W ÷ F. The units help you check. A value in N × m is a value in J.
Worked example
A student lifts a 2.4 kg bag from the floor onto a shelf 1.5 m above the floor, at a steady speed. Take g = 10 N/kg. (Invented example data.)
Step 1, find the force. Steady speed means the upward force equals the weight. Weight = mg = 2.4 × 10 = 24 N.
Step 2, find the distance. The vertical height gained is 1.5 m.
Step 3, apply the relationship. W = Fd = 24 × 1.5 = 36 J.
Step 4, interpret. The student transfers 36 J to the bag’s gravitational potential store.
Check by estimation: about 25 N over about 1.5 m is about 37 J, so 36 J is sensible.
What about a horizontal push with a unit change?
A shopper pushes a trolley with a steady force of 30 N along an aisle 250 cm long.
Convert first: 250 cm = 2.5 m. Then W = 30 × 2.5 = 75 J.
The same idea works for friction. If friction of 20 N acts over a 6.0 m slide, the work done against friction is 20 × 6.0 = 120 J. That energy goes to the thermal store of the surfaces.
What mistake should I watch for?
The most common slip is to use the distance without converting it.
Mistaken answer: W = 30 × 250 = 7500 J.
The student used centimetres with newtons.
The correction is to change 250 cm to 2.5 m before multiplying. A second slip is to use the object’s mass as the force, as in W = 2.4 × 1.5 for the bag, which gives 3.6 and is not in joules. Always turn mass into weight with W = mg first.
A sense check helps: 7500 J is about the energy needed to lift a 75 kg person 10 m, which is far more than a shopper’s push along an aisle.
Check yourself
Try these without a calculator, then open each answer. Use g = 10 N/kg.
1. A person lifts a 2.0 kg bag steadily by 1.5 m. How much work is done on the bag?
Show answer
Weight = 2.0 × 10 = 20 N. Work = 20 × 1.5 = 30 J.
2. A force does 240 J of work on a cart that moves 4.0 m in the direction of the force. What is the force?
Show answer
F = W ÷ d = 240 ÷ 4.0 = 60 N.
3. A steady force of 30 N pushes a box, and 900 J of work is done. How far does the box move in the direction of the force?
Show answer
d = W ÷ F = 900 ÷ 30 = 30 m.
Where this leads next
Now that you can calculate work, move to tracking energy stores in a simple system. You can also revisit the work, energy and efficiency topic guide to see the full route.
Questions that combine weight, unit conversion and rearranging are where slips multiply. A teacher in online one-to-one Physics tuition can watch each step as you work and steady the order you use.