Mechanical work:
W = force × distance

What is the mechanical work formula?

Mechanical work is force multiplied by distance: W = F × d. If you push with 25 newtons and the object moves 12 metres in the direction you push, you have done 25 × 12 = 300 joules of work.

Physicists use work to describe energy transferred by a force. One joule is one newton-metre, so the unit comes straight from the formula. The Calcylator calculator multiplies the force by the distance, so it assumes the force acts along the direction of travel.

The formula with worked examples

Lifting a weight: find the force first

To lift a 20 kg bag you must supply a force equal to its weight, which is mass × 9.81 m/s². That is 196.2 N. Raising it 1.5 m onto a shelf is 196.2 × 1.5 = 294.3 J of work against gravity.

A builder hoisting bricks, a porter carrying a suitcase up a staircase and a person loading a boot all do work against gravity. Work depends on the height gained, so ten bricks lifted one metre need the same work as one brick lifted ten metres.

Zero work and negative work

Carrying a bag horizontally at constant speed does no work on the bag in the physics sense, because the force you apply upward is at right angles to the motion. You still get tired, because your muscles use energy, but the formula gives zero.

Work can also be negative. If a force acts opposite to the movement, such as friction slowing a sliding box, the work is negative. For example, a 5 N force against 4 m of motion gives −20 J. The calculator accepts a negative force for this case.

Joules in everyday energy units

Joules are small. The 300 J above is only 0.0717 kilocalories, using 1 kcal = 4,184 J. A kilowatt-hour is 3,600,000 J, so 10,000 J of work is about 0.0028 kWh.

That is why the work done in everyday lifting looks tiny next to an electricity bill or a food label. A motor or a body also loses energy as heat, so the energy used is always more than the useful work done.

Where this formula is used

A gym coach estimating the work in a deadlift, a crane operator comparing loads and a student solving a physics problem all use it. The result is the useful work only, so a real machine such as a motor or a pulley system needs more input energy than the answer shows.

To find the average power, divide the work by the time. Doing 294.3 J of lifting in 3 seconds is about 98 watts, and the same lift done in 6 seconds is half that. Work stays the same, but the power changes with speed.

Common mistakes with mechanical work

• Using mass in kilograms as if it were force in newtons.
• Using the total path length when the force acts only along part of the path.
• Ignoring the angle when the force is not in line with the motion.
• Mixing centimetres or kilometres with metres.
• Confusing work with power, which is work divided by time.