Free Work Calculator (W = F·d·cos θ)
This work calculator computes the mechanical work done by a constant force using the equation W = F·d·cos θ, where force F is in newtons (N), displacement d is in metres (m), and θ is the angle between the force and the direction of motion. Enter the three values and the calculator returns the work done in joules (J), along with the result in kilojoules for larger numbers. In physics, work is the energy transferred to or from an object when a force moves it through a distance — and only the component of the force that lies along the displacement actually does work, which is exactly why the cosine of the angle appears in the formula.
Enter force, distance, and the angle θ between them to find the work done.
W = F · d · cos θ. Work is in joules (J) when force is in newtons and distance is in metres. A perpendicular force (θ = 90°) does zero work; a force opposing the motion (θ > 90°) does negative work.
Quick answer
In physics, work is calculated with the equation W = F·d·cos θ, where W is work in joules (J), F is the applied force in newtons (N), d is the displacement in metres (m), and θ is the angle between the force and the direction of motion. For a force acting in the same direction as the motion, θ = 0° and cos θ = 1, so the formula simplifies to W = F·d. For example, pushing with 100 N over 5 m in line with the motion does 100 × 5 × cos 0° = 500 joules of work.
Formula & method
W = F · d · cos θ (when θ = 0°, W = F · d)
- W — Work done, in joules (J)
- F — Applied force, in newtons (N)
- d — Displacement, in metres (m)
- θ — Angle between force and motion, in degrees
W = work done (joules, J), F = magnitude of the applied force (newtons, N), d = displacement of the object (metres, m), and θ = the angle between the force vector and the displacement vector (degrees). One joule is the work done by a force of one newton acting through one metre in its own direction (1 J = 1 N·m). Only the force component along the motion, F·cos θ, does work; a force perpendicular to the motion (θ = 90°) does zero work because cos 90° = 0.
Examples
- Input
- F = 150 N, d = 4 m, θ = 0°
- Result
- W = 600 J
- Why
- The push is in the same direction as the motion, so θ = 0° and cos 0° = 1. Apply W = F·d·cos θ = 150 × 4 × 1 = 600 joules. All 150 N of force acts along the displacement, so the full force does work.
- Input
- F = 200 N, d = 3 m, θ = 60°
- Result
- W = 300 J
- Why
- Only the horizontal component of the pull moves the sled forward. With cos 60° = 0.5, W = F·d·cos θ = 200 × 3 × 0.5 = 300 joules. Even though you pull with 200 N, only half of it (100 N) acts along the motion.
- Input
- F = 100 N, d = 8 m, θ = 90°
- Result
- W = 0 J
- Why
- Holding a load while walking, the support force points straight up while the motion is horizontal, so θ = 90° and cos 90° = 0. W = 100 × 8 × 0 = 0 joules. No work is done on the load because the force has no component along the direction of travel.
- Input
- F = 80 N, d = 15 m, θ = 30°
- Result
- W ≈ 1039.23 J
- Why
- With cos 30° ≈ 0.8660254, W = F·d·cos θ = 80 × 15 × 0.8660254 ≈ 1039.23 joules. The 80 N force is partly along the path, so its effective component is 80 × 0.866 ≈ 69.3 N over the 15 m distance.
When to use this tool
- Calculating the work done in joules when you push, pull, or lift an object with a known constant force over a known distance.
- Solving introductory physics or engineering problems that involve a force applied at an angle to the direction of motion (W = F·d·cos θ).
- Checking how much energy a force transfers to an object, since work done equals the energy added (or removed, when the work is negative).
- Demonstrating why a force perpendicular to motion does no work, or why pulling at an angle is less efficient than pulling straight ahead.
- Verifying a homework answer and seeing the cosine factor and arithmetic, not just the final number of joules.
Common mistakes
- Ignoring the angle and using W = F·d when the force is not aligned with the motion. The factor cos θ is essential: a force pulling at 60° does only half the work (cos 60° = 0.5) compared with the same force pulling straight along the path.
- Confusing distance travelled with displacement along the force. The d in this formula is the displacement of the object; if the object moves back and forth, only the net displacement in the relevant direction counts, not the total path length.
- Forgetting that perpendicular forces do zero work. The normal force from a floor, the tension in a string for circular motion, and the force of carrying a bag while walking horizontally all act at 90° to the motion, so each does no work even though a real force is present.
- Using inconsistent units. Work comes out in joules only when force is in newtons and distance is in metres. Pounds, grams-force, centimetres, or feet must be converted to SI units first, or the answer will be wrong.
- Misreading θ as the angle of the slope rather than the angle between the force and the displacement. These can differ — always measure θ from the direction the object actually moves to the line of the applied force.
Frequently asked questions
What is the formula for work in physics?
Work is calculated with W = F·d·cos θ, where W is work in joules (J), F is the applied force in newtons (N), d is the displacement in metres (m), and θ is the angle between the force and the direction of motion. When the force acts along the motion, θ = 0° and cos θ = 1, so the formula reduces to the familiar W = F·d. Work measures the energy transferred to an object by the force.
What is one joule of work?
One joule (J) is the work done when a force of one newton moves an object one metre in the direction of that force (1 J = 1 N·m). It is the SI unit of both work and energy, so doing one joule of work on an object adds one joule of energy to it. For comparison, lifting a small apple (about 1 N of weight) by one metre takes roughly one joule of work.
Why does the work formula include cos θ?
Only the part of the force that points along the direction of motion does work. The cos θ factor extracts that component: F·cos θ is the portion of the force aligned with the displacement. When the force is parallel to the motion (θ = 0°), cos θ = 1 and the full force counts; when it is perpendicular (θ = 90°), cos θ = 0 and no work is done.
Can work be zero even when a force is applied?
Yes. If the force is perpendicular to the motion, θ = 90° and cos 90° = 0, so the work is zero — for example, carrying a heavy bag while walking on level ground, or the centripetal force on an object in uniform circular motion. Work is also zero if there is no displacement at all (d = 0), such as pushing hard against an immovable wall.
Can work be negative?
Yes. When the angle θ is between 90° and 180°, cos θ is negative, so the work is negative. This happens when a force opposes the motion, such as friction or air resistance acting backward, or when you lower an object slowly against gravity. Negative work means the force removes energy from the object rather than adding it.
What units should I use in the work calculator?
Use SI units: force in newtons (N), distance in metres (m), and the angle in degrees. The result then comes out in joules (J). If your data is in other units — pounds-force, centimetres, feet, or grams — convert to newtons and metres first so the joule result is correct.
What is the difference between work and power?
Work (in joules) is the total energy transferred by a force over a distance, while power (in watts) is the rate at which that work is done, equal to work divided by time: P = W / t. Two people who push the same crate the same distance do the same amount of work, but the one who finishes faster delivers more power.
Sources & references
- Work (physics) — Wikipedia
- Work — HyperPhysics (Georgia State University)
- Joule — NIST SI unit definitions
External references open in a new tab. We are independent and not affiliated with these organizations.
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