Work (W = Fd cos θ)

Also known as W = Fd · work done by a force

W=Fdcos⁡θW = F d \cos\theta
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Worked example: 50 N over 10 m at 60 deg → 250 J. Press Try an example to run it live, then adjust anything.

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See also

Power (P = W/t): Both use work done.Displacement (Uniform Acceleration): Both use kinematic displacement.Displacement from Average Velocity: Both use kinematic displacement.Displacement from Final Velocity (d = vt − ½at²): Both use kinematic displacement.Velocity-Displacement Relation (v² = v₀² + 2ad): Both use kinematic displacement.Impulse (J = FΔt): Both use applied force.
4 more relatedPower from Force and Velocity (P = Fv): Both use applied force.Pressure (P = F/A): Both use applied force.Torque: Both use applied force.Torque with a Lever Arm (τ = rF sin θ): Both use applied force.
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Work (W = Fd cos θ) explained

Fθd

In physics, work is force applied through a distance, and only the component of force along the motion counts, which is where the cos θ comes from. Pull a sled with 100 N on a rope angled 30° above the snow for 20 m, and you do W = 100 × 20 × cos 30° ≈ 1732 J, not the full 2000 J. The term itself was coined by the French engineer Gaspard-Gustave de Coriolis in 1826, precisely to compare what steam engines and horses could deliver.

Two useful edge cases: a force perpendicular to the motion (θ = 90°) does no work at all, the Moon's orbit costs gravity nothing, and a force opposing the motion, like friction, does negative work, showing up as cos θ below zero. Solving for θ uses the arccos principal branch, 0° to 180°, which conveniently covers the entire physical range of angles between two directions.

Work (W = Fd cos θ) formula

W=Fdcos⁡θW = F d \cos\theta
Where
  • WW= Work (J)
  • FF= Force (N)
  • dd= Displacement (m)
  • θ\theta= Angle between force and motion (°)
Revised 16 October 2026
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formula.expert. (2026, October 16). Work (W = Fd cos θ) Formula & Calculator. https://formula.expert/solvers/work-force-distance