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Work Done Calculator

Find the work done by a force.

How it is calculated

Mechanical work is the part of a force that acts along the direction of motion, multiplied by the distance moved: W = F × d × cos(θ). F is the force in newtons (N), d the displacement in metres (m) and θ the angle in degrees between the force and the direction of displacement; the result is in joules (J). If the angle is left empty it is taken as 0, so all the force counts. At 90° the cosine is 0 and no work is done; above 90° the cosine is negative, which means the force opposes the motion and removes energy. The result is rounded to two decimals.

Worked example

A 50 N force pulls a crate 10 m, with the rope at 30° above the ground. The component along the motion is cos 30° = 0.8660. Work: 50 × 10 × 0.8660 = 433.01 J. If the pull were exactly horizontal (0°), the work would be the full 500 J; the tilt wastes about 67 J because part of the pull lifts the crate instead of moving it forward.

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Common mistakes

  • Typing the angle in radians. The field is in degrees, so 0.52 is read as roughly half a degree.
  • Measuring the angle from the vertical or from the surface normal instead of from the direction of displacement.
  • Using the total path length when the object returns. Work here uses the displacement along the force direction you entered.
  • Entering mass instead of force. Weight in newtons is mass in kg × 9.81, not the kg figure.

About this calculator

This calculator applies the physics definition of mechanical work, W = F·d·cosθ, where F is the applied force, d is the displacement, and θ is the angle between the force and displacement directions — work is zero when the force is perpendicular to motion (like carrying a bag horizontally against gravity) and maximum when force and displacement are aligned. Work is measured in joules and represents energy transferred to or from an object by the force.

Physics students use this constantly for problems involving pushing, pulling, and lifting objects, and it's the foundation for the work-energy theorem linking work done to change in kinetic energy.

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Comments (1)

  • Editorial team

    Only the part of the force that points along the motion does work. Carrying a bag horizontally at constant height means the force is vertical, 90 degrees to the movement, and the work is essentially zero. Pulling a sled with a 50 N rope at 30 degrees over 10 m gives about 433 J. Enter the angle in degrees; if you leave it empty, 0 is used. Describe your situation and the angle you used.

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