Work & Power Calculator
Calculate work done, power output, force, distance, or time using physics formulas. W = F × d and P = W / t.
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How to use this calculator
W = work (J), F = force (N), d = distance (m), θ = angle between force and displacement, P = power (W), t = time (s).
- 1
Enter the applied force in Newtons and the displacement in metres.
- 2
If the force is not parallel to the motion (e.g., pulling a sled at an angle), enter the angle between force and direction of motion.
- 3
Enter the time for which the force is applied to get power output.
Frequently asked questions
What does this tool actually calculate?
Work W = F × d × cos θ and, if a time is entered, power P = W ÷ t, with work also shown in kJ, power in kW and power in horsepower. It does not solve backwards for force, distance or time; to get those, rearrange the equations yourself, for example d = W ÷ (F cos θ).
What angle should I enter?
The angle between the force direction and the direction of travel, in degrees. Pulling straight along the motion is 0°. A 50 N pull over 20 m gives 1000 J at 0°, 866.0254 J at 30°, 707.1068 J at 45° and 500 J at 60°. Use the angle between force and displacement, not the angle of a slope.
Why is the work zero at 90° and negative above it?
At 90° the force is perpendicular to the motion, so it does no work; the display shows 0.0000 J because cos 90° is about 6e-17 in floating point. Between 90° and 180° the force opposes the motion and the work is negative: 50 N over 20 m at 120° gives -500.0000 J, and at 180° gives -1000.0000 J. Negative work removes energy, as friction does.
What happens if time is left at 0?
Power needs a positive time. With a time of 0 or less the Power and Horsepower rows and the time line are simply not displayed, and you only see work. A blank field counts as 0, so enter the real duration to see power.
Which horsepower is used?
Power in watts is divided by 745.7, the mechanical horsepower. 1000 W is therefore 1.3410 hp. Electric, metric and boiler horsepower use slightly different watt values, so a catalogue rating may not match exactly.
Why does a negative force or distance give an error?
Both must be zero or above, and the sign of the result is controlled by the angle instead. To model a force against motion, enter a positive force and an angle of 180°, not a negative force.
Is the power average or instantaneous?
Average. P = W ÷ t divides total work by total time, so a force that varies during the motion still produces a single number. Peak power during the motion can be higher, which matters for motors sized by their maximum rather than mean load.
Work and power calculator — joules, watts and horsepower from force and distance
Only the parallel part of the force counts
Work is force multiplied by the distance moved along the force’s own direction, which is why the formula carries cos θ. Hold a suitcase still and the displacement is zero; carry it level and the lifting force is at 90° to your path. Neither does work on the suitcase, whatever the effort feels like. The calculator reflects this: with the angle set to 90° it reports zero work however large the force or distance, which is the physically correct answer rather than a bug.
Worked example: dragging a crate
A crate is dragged 20 m by a 50 N pull, taking 8 s. With the pull along the motion the calculator gives 1000 J of work and 125 W of power. Raise the rope to 30° above the ground and the figures become 866.0254 J and 108.2532 W, because only 50 × cos 30° = 43.30 N acts along the floor. The other component, 50 × sin 30° = 25 N, pulls upward and shares the load with the floor without doing any work along the direction of travel.
Effect of the angle
The output for 50 N over 20 m in 8 s at several angles shows how quickly work falls away.
| Angle | Work | Power |
|---|---|---|
| 0° | 1000 J | 125 W |
| 30° | 866.0254 J | 108.2532 W |
| 45° | 707.1068 J | 88.3883 W |
| 60° | 500 J | 62.5 W |
| 90° | 0 J | 0 W |
| 120° | -500 J | -62.5 W |
Worked example: lifting a load
Lifting something that needs 735 N through 3 m in 2 s requires 2205 J of work and 1102.5 W of power, which the tool also gives as 1.1025 kW or 1.4785 hp. A force of 1500 N applied over 100 m in 10 s shows 150 kJ and 15 kW. Dividing the same work by a longer time lowers the power while the work stays put, so lifting the 735 N load in 4 s instead of 2 s would need half the watts for the same 2205 J.
Joules, watts and kilowatt-hours
A joule is a newton-metre and a watt is one joule per second. Multiply kilowatts by hours to get kilowatt-hours: a steady 1.1025 kW sustained for one hour amounts to 1.1025 kWh, the unit electricity is billed in. A joule is far smaller, since one kilowatt-hour equals 3.6 million joules, which is why household energy is quoted in kWh and not in joules.
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Results are estimates for informational purposes only and do not constitute professional financial, medical, legal, or technical advice. Read full disclaimer →