Work Calculator (W = Fd)

Find work, force, displacement or the angle from W = Fd cos θ, with positive, negative and zero work explained.

Solve for
Magnitude of the applied force.
Distance the point of application moves.
0° when you push straight along the direction of motion.
In kJ
0.433013 kJ
In cal
103.493 cal
In ft·lbf
319.374 ft·lbf
In Wh
0.120281 Wh
Force along the motion
43.3013 NF cos θ — only this part does work
Work (W)433.013 J
  • Positive work: the force adds energy to the object.

Show the work

  1. Start from the formula W = Fd cos θ
  2. Substitute the known values: W = 50 N × 10 m × cos 30° = 433.013 J

In physics, work is the energy a force transfers to an object by moving it. Only the part of the force that points along the motion counts, which is why the angle matters as much as the force and the distance. This calculator solves W = Fd cos θ for work, force, displacement or the angle, and tells you whether the work is positive, negative or zero.

How to use the work calculator

  1. Under Solve for, choose Work, Force, Displacement or Angle.
  2. Enter Force (F) as a positive magnitude in N, kN, lbf, kgf, dyn or ozf.
  3. Enter Displacement (d), the distance the point where the force is applied moves, in m, cm, mm, km, in, ft, yd or mi.
  4. Enter Angle between force and motion (θ) in degrees or radians: 0° when pushing straight along the motion, 180° when pushing directly against it.
  5. When solving for force, displacement or angle, enter Work (W) in J, kJ, MJ, cal, kcal, Wh, kWh, BTU, ft·lbf, eV or erg. Negative work is allowed.

The result shows conversions, the force component along the motion (F cos θ) and a note saying whether the force adds energy, removes it or does nothing.

Work formula

W = Fd cos θ

The calculator rearranges it for each unknown:

F = W ÷ (d cos θ)  ·  d = W ÷ (F cos θ)  ·  θ = arccos(W ÷ Fd)

Force and displacement must be positive and the angle must lie between 0° and 180°. At exactly 90° the force does no work, so the force or distance cannot be recovered from the work. When solving for force or distance, work and cos θ must share a sign. When solving for the angle, the size of the work cannot exceed F × d, the most a force can do when aimed straight along the motion.

Worked example: pulling a sled

You pull a sled 10 m across level snow with a rope held at 30° above the ground, keeping a steady 50 N of tension.

Component along the motion: 50 N × cos 30° = 50 × 0.8660 = 43.30 N.

Work: W = 50 N × 10 m × 0.8660 = 433.0 J, about 319.4 ft·lbf, 103.5 cal or 0.120 Wh.

The upward part of the pull, 50 × sin 30° = 25 N, does no work on the sled; it only lifts some of the sled’s weight off the snow. Run the same force and distance backward and an answer of 250 J corresponds to an angle of 60°, while −250 J corresponds to 120°.

Positive, negative and zero work

Angle θ cos θ Work Everyday case
0° 1 +Fd, the maximum Pushing a cart straight ahead
0° to 90° positive positive Towing a sled with a rising rope
90° 0 zero Carrying a bag level; a car’s sideways grip on a curve
90° to 180° negative negative Lowering a box slowly by hand
180° −1 −Fd Sliding friction; brakes

Positive work hands energy to the object, negative work takes it away, and a force at right angles to the motion only changes direction. A 150 N friction force on a crate pushed 10 m does −1,500 J.

Work and energy

The work–energy theorem ties this calculator to motion: the net work done by all forces equals the change in kinetic energy, Wnet = ½mv² − ½mu². Push a crate with +1,800 J of work while friction does −1,500 J, and the crate leaves with 300 J more kinetic energy than it started with. When the works cancel, the speed stays the same. Lift a box at a steady pace and your +mgh is matched by gravity’s −mgh; the energy you spend is stored as potential energy instead.

Units of work

Unit Joules
joule (N·m) 1
foot-pound-force (ft·lbf) 1.3558179483314
calorie (thermochemical) 4.184
kilocalorie (food Calorie) 4,184
watt-hour 3,600
kilowatt-hour 3,600,000
British thermal unit (IT) 1,055.05585262

A US example: lifting a 20 lbf load 5 ft straight up takes 100 ft·lbf, or about 135.6 J.

To see how fast work gets done, use the power calculator. The kinetic energy calculator and potential energy calculator show where the energy ends up, and the friction calculator gives the force for the negative-work case.

Frequently asked questions

What is the formula for work?

Work equals force times displacement times the cosine of the angle between them: W = Fd cos θ. When the force points straight along the motion, cos 0° = 1 and the formula reduces to W = Fd. A 50 N pull at 30° over 10 m does about 433 J of work.

Can work be negative?

Yes. When a force has a component opposite to the motion (an angle above 90°), it takes energy away from the object. Friction on a sliding box and brakes on a moving car both do negative work. A 150 N friction force acting over 10 m does −1,500 J.

Why is no work done when I carry a bag across a level room?

Your arms push up on the bag while it moves sideways, so the angle is 90° and cos 90° = 0. In the physics sense you do no work on the bag, even though your muscles use energy to hold it up.

How does a joule compare with a foot-pound and a kilowatt-hour?

One joule is one newton acting through one meter. One foot-pound-force is about 1.356 J, one calorie is 4.184 J, and one kilowatt-hour is 3.6 million joules, enough to push with a steady 1,000 N for 3.6 km.

Work and torque both use N·m. Are they the same thing?

No. Work is force times distance moved along the force and is an energy, so it is written in joules. Torque is force times a lever arm at right angles to it, a turning effect that stays in N·m by convention and is never expressed in joules.

Last reviewed October 2026 by the CalcFluent editorial team. How we check our calculators.