Work and power calculator
Enter a force, distance and time, get the work and power.
How do I calculate work?
Work = force × distance × cos(angle between them). Pushing with 100 N over 10 m in the direction of motion does 1,000 joules of work.
What if the force is at an angle?
Only the part along the motion counts: at 60°, cos 60° = 0.5, so the same push does 500 J.
How do I calculate power?
Power = work ÷ time: 1,000 J in 5 seconds is 200 watts — about 0.27 horsepower.
What is a joule?
The work of 1 newton over 1 meter. 1 foot-pound is 1.356 J.
Is carrying a box across a room work?
In physics, not by the force holding it up: that force is vertical and the motion horizontal, so cos 90° = 0.
Can work be negative?
Yes. Work is negative when the force points against the displacement, so the angle is more than 90° and cos θ is negative (OpenStax University Physics, Volume 1, section 7.1). Lifting a 10 kg box 2 m, you do 196.1 J of work and gravity does −196.1 J, since it pulls down while the box goes up (W = −mg·Δy, Eq. 7.4).
How does work change an object’s speed?
By the work-energy theorem, the net work on an object equals the change in its kinetic energy (OpenStax, section 7.3, Eq. 7.9). If 1,000 J of net work is done on a 10 kg object starting from rest, it ends with ½ × 10 × v² = 1,000 J, so v = 14.1 m/s. The work you enter here is the work of one force; use the net force to apply the theorem.
How do I convert between watts, horsepower and kilowatt-hours?
One horsepower is 745.7 W (NIST SP 811; OpenStax rounds it to 746 W), so 1 kW is 1.341 hp. A kilowatt-hour is power times time, 1 kW × 3,600 s = 3.6 × 10⁶ J, an amount of work, not a rate. The sled in the example below needs only 0.197 hp over 40 s.
The formula and what it assumes
For a constant force, the work is W = F·d·cos θ, where d is the distance moved and θ is the angle between the force and the displacement (OpenStax University Physics, Volume 1, section 7.1). Only the component of the force along the motion does work. The SI unit is the joule, 1 J = 1 N·m, and the US unit is the foot-pound. Average power is the work divided by the time, P = W ÷ t (section 7.4, Eq. 7.10), in watts (1 W = 1 J/s). The calculator assumes a constant force on a straight path.
Worked example: pulling a sled with a slanted rope
A rope pulls a sled with 50 lbf, 222.4 N, at 30° above the horizontal over 100 ft, 30.48 m, in 40 s. The work is W = 222.4 × 30.48 × cos 30° = 5,871 J, which is 4,330 ft·lbf. A pull along the ground would do 5,000 ft·lbf (50 × 100). The power is 5,871 ÷ 40 = 146.8 W, or 0.197 hp. OpenStax Example 7.1 does the same for a lawn mower: 75.0 N at 35.0° over 25.0 m is 1.54 kJ.
Common mistakes
The angle is the one between the force and the direction of motion, not the angle with the vertical. Carrying a bag horizontally, the lifting force is at 90° to the motion and does no work (OpenStax, section 7.1), and holding it still, with no displacement, does none either. Work and power are also confused: a joule is an amount of work and a watt a rate, so doubling the time halves the power but not the work. Finally, the force unit matters: in this calculator pounds are pounds-force, and a mass in pounds or kilograms must first be turned into a weight.
Sources: OpenStax University Physics, Volume 1, section 7.1 (work, Eq. 7.4; Example 7.1), section 7.3 (work-energy theorem, Eq. 7.9) and section 7.4 (power, Eq. 7.10); NIST SP 811, Appendix B.8, for the pound-force (4.448222 N), the foot-pound force (1.355818 J) and the horsepower (745.6999 W). Related: kinetic energy calculator, horsepower calculator, torque calculator and potential energy calculator.