Work Done by a Force
Calculate mechanical work from force and angle
Result
How it is calculated
The tool calculates W = F × d × cos(θ). F is the force in newtons, d the displacement in metres, and θ the angle in degrees between the force and the direction of movement. The angle is converted to radians internally, and the work W is shown in joules (J) with two decimals.
The result can be negative: an angle between 90° and 270° means the force points partly against the motion (friction or braking), so it removes energy. At exactly 90° the work is 0.00 J. The tool does not reject any numbers, so a negative force or distance simply flips the sign. Use consistent SI units; if you type pounds-force or feet, the result will not be in joules.
Worked example
You pull a sled with a 50 N force over 10 m, and the rope makes a 30° angle with the ground. The component of the force along the motion is 50 × cos 30° = 50 × 0.8660 = 43.30 N. Multiply by the distance: 43.30 × 10 = 433.0. The work done is 433.01 J. A perfectly horizontal pull (0°) would give 500 J, so the angled rope wastes some effort.
Common mistakes
- Taking the angle from the wrong reference: θ is measured between the force and the direction of motion, not between the force and the vertical.
- Typing the angle in radians; the tool expects degrees, so 0.52 is read as about half a degree and the work comes out almost equal to F × d.
- Entering the distance travelled by the object when the force acted for only part of it; only the displacement during which the force was applied counts.
- Ignoring the sign: a friction force opposite to the motion has θ = 180° and gives negative work, not a positive amount.
- Confusing work with power; work is energy in joules, and the time taken does not appear in this formula.
About this calculator
This calculator uses the physics definition of mechanical work, W = F × d × cos(θ), where F is the applied force, d is the displacement, and θ is the angle between the force and the direction of motion. The cosine term matters because only the component of force actually aligned with the movement does work — a force applied straight down on a horizontally-sliding box (θ = 90°) does zero work no matter how hard you push.
Physics students use it to solve classic problems like pulling a sled at an angle or pushing a lawnmower, and engineers use the same formula to size motors and actuators by working out exactly how much energy a task requires when the applied force isn't perfectly aligned with the direction of travel.
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