Work Calculator

Free Online Work Calculator (W = F × d × cos θ)

This work calculator instantly solves for work done, force, or distance using the fundamental formula W = F × d × cos θ — Work equals Force times Distance times the cosine of the angle between them. Enter any set of known values — force, distance, and angle; work, distance, and angle; or work, force, and angle — in any common unit, and the calculator converts everything automatically, works through the full step-by-step solution, and draws a labeled diagram showing the force vector, its working component, and the displacement, so the entire calculation is visible at a glance.

Whether you're a physics student solving a work-energy word problem, an engineer estimating the energy needed to move a load, or just curious how much work is done pushing a box across a floor or hauling a sled at an angle, this tool handles newtons, kilonewtons, pounds-force, dynes, and kilogram-force for force, meters, centimeters, kilometers, feet, inches, and miles for distance, and joules, kilojoules, calories, kilocalories, foot-pounds, and ergs for work — and always shows its complete working, not just the final number.

What Is Work? (Definition and Formula)

In physics, work is done whenever a force causes an object to move through a distance in the direction of that force. Simply pushing on a wall that doesn't move does zero work, no matter how hard you push, because there is no displacement. Work is a scalar quantity — it has magnitude but no direction — and it represents the amount of energy transferred to or from an object by a force.

The formula for work is one of the cornerstone equations of classical mechanics and the work-energy theorem:

• Work (W) = Force (F) × Distance (d) × cos(θ)
• θ is the angle between the direction of the applied force and the direction of motion (displacement).
• The SI unit of work is the joule (J), where 1 joule = 1 newton-meter (N·m).
• Example: pushing a box with 50 N of force straight along the floor (θ = 0°) for 10 m does work = 50 × 10 × cos(0°) = 500 J

Why the Angle Matters: The Role of cos(θ)

Only the component of force that acts along the direction of motion actually does work — this is exactly what the cos(θ) term captures. When the force is applied in the same direction as the motion (θ = 0°), cos(0°) = 1 and the full force contributes to the work done. When the force is applied at an angle, only part of it — F cos θ — pushes the object forward, while the rest acts to lift or press the object without moving it in that direction. When the force is applied completely perpendicular to the motion (θ = 90°), cos(90°) = 0 and no work is done at all, no matter how large the force is. This is why carrying a heavy bag while walking horizontally does no work on the bag in the physics sense — the force (holding it up) is perpendicular to the motion (walking forward).

How to Find Force from Work and Distance

If you already know the work done and the distance over which it acted, you can rearrange the formula to solve for the force involved.

• Force (F) = Work (W) ÷ (Distance (d) × cos(θ))
• Example: 500 J of work is done moving an object 25 m in a straight line (θ = 0°) → Force = 500 ÷ (25 × 1) = 20 N
• This rearrangement comes directly from dividing both sides of W = F × d × cos(θ) by d × cos(θ).

How to Find Distance from Work and Force

If you know the work done and the force applied, you can solve for the distance the object moved by rearranging the formula once more.

• Distance (d) = Work (W) ÷ (Force (F) × cos(θ))
• Example: a 400 N force does 1000 J of work moving an object in a straight line (θ = 0°) → Distance = 1000 ÷ (400 × 1) = 2.5 m
• This rearrangement comes from dividing both sides of W = F × d × cos(θ) by F × cos(θ).

The Work–Force–Distance Relationship

For the common case where the force acts entirely along the direction of motion (θ = 0°, so cos θ = 1), the work formula simplifies to the familiar W = F × d, and behaves just like the Speed-Distance-Time or Force-Mass-Acceleration triangles: cover the quantity you're solving for and what remains gives the formula — covering W leaves F next to d (F × d); covering F leaves W over d (W ÷ d); covering d leaves W over F (W ÷ F). The diagram on this page extends that idea by also drawing the angled force vector and highlighting its horizontal (working) component, F cos θ, so you can see exactly how much of the applied force is actually doing work.

Common Work, Force, and Distance Units

Work, force, and distance can each be expressed in several different unit systems depending on whether you're working in SI (metric), CGS, or imperial units. This calculator supports the most common ones used in physics classes and engineering.

• Work: joules (J), kilojoules (kJ), calories (cal), kilocalories (kcal), foot-pounds (ft·lb), ergs
• Force: newtons (N), kilonewtons (kN), pounds-force (lbf), dynes (dyn), kilogram-force (kgf)
• Distance: meters (m), centimeters (cm), kilometers (km), feet (ft), inches (in), miles (mi)
• 1 J = 0.239 cal = 0.7376 ft·lb = 10,000,000 erg

How to Use This Work Calculator

Choose what you want to find — Work, Force, or Distance — from the dropdown. Then fill in the values you already know, picking whichever unit matches your problem (newtons, kilonewtons, pounds-force, dynes, or kilogram-force for force; meters, centimeters, kilometers, feet, inches, or miles for distance; joules, kilojoules, calories, kilocalories, foot-pounds, or ergs for work), and enter the angle between the force and the direction of motion — leave it at 0° if the force acts straight along the path of motion. The calculator instantly converts everything to a common base unit, applies W = F × d × cos θ, and shows the result in your chosen unit — plus automatically converted into joules, kilojoules, and foot-pounds for quick reference.

Scroll down for the complete step-by-step solution, which mirrors exactly how you'd solve it by hand: the formula first, then your real values substituted in with unit conversion and the cosine of the angle shown explicitly, then the arithmetic worked through to the final answer.

Worked Examples

Pushing a box straight ahead: 50 N of force over 10 m at 0° → Work = 50 × 10 × cos(0°) = 500 J.

Pulling a sled at an angle: 200 N of force at 30° above the ground over 15 m → Work = 200 × 15 × cos(30°) ≈ 200 × 15 × 0.866 ≈ 2598 J.

Lifting a box straight up: 100 N of force over 2 m at 0° (force and motion aligned) → Work = 100 × 2 × cos(0°) = 200 J.

Finding force: 500 J of work done over 25 m in a straight line → Force = 500 ÷ (25 × cos 0°) = 20 N.

Finding distance: 1000 J of work done by a 400 N force in a straight line → Distance = 1000 ÷ (400 × cos 0°) = 2.5 m.

Work, Energy, and the Work-Energy Theorem

Work is intimately tied to energy: doing positive work on an object transfers energy into it, typically increasing its kinetic energy, while negative work (such as friction acting opposite to motion) removes energy from it. The work-energy theorem formalizes this relationship, stating that the net work done on an object equals its change in kinetic energy: W_net = ΔKE = ½mv₂² − ½mv₁². This is why work and energy share the same unit, the joule, and why work calculations show up constantly alongside kinetic and potential energy problems in physics coursework.

Real-World Applications of the Work Formula

The work formula and the concept of a force's working component appear throughout engineering, sports, and everyday mechanics:

• Mechanical engineering — calculating the energy input required by motors, cranes, and conveyor systems to move loads over a distance.
• Automotive engineering — estimating the work an engine must do to accelerate a vehicle or overcome friction and air resistance.
• Construction and lifting equipment — sizing hoists and pulleys based on the work needed to raise materials to a given height.
• Sports biomechanics — analyzing how much work an athlete's muscles do when pushing, pulling, or lifting equipment at an angle.
• Renewable energy — work done by wind or water on a turbine blade depends on the angle between the flow and the blade's motion.

Tips for Solving Work Word Problems

Most textbook and exam word problems involving work follow a predictable pattern, and a few habits make them far easier to solve correctly:

• Identify which quantities are given and which one is being asked for — this tells you whether to use W = F × d × cos θ, F = W ÷ (d cos θ), or d = W ÷ (F cos θ).
• Always check whether an angle is mentioned — if the force acts directly along the motion, θ = 0° and cos θ = 1, simplifying the formula to W = F × d.
• Remember that a force perpendicular to motion (θ = 90°) does zero work, even if the force itself is very large — this trips up many students working with circular motion or carrying loads.
• Watch the sign of work: if the force has a component opposite to the motion (θ between 90° and 180°), the work done is negative, meaning energy is being removed from the object rather than added.

Frequently Asked Questions

What is the formula for work in physics? Work = Force × Distance × cos(θ), where θ is the angle between the force and the direction of motion. When the force acts entirely along the motion, this simplifies to W = F × d.

How do you calculate work done in joules? Multiply the force in newtons by the distance in meters and by the cosine of the angle between them. For example, a 20 N force moving an object 5 m in the same direction does 20 × 5 × cos(0°) = 100 J of work.

What is the difference between work and energy? Work is the process of transferring energy to or from an object via a force acting through a distance; energy is the capacity to do work. They share the same unit — the joule.

Can this calculator solve for force or distance, not just work? Yes — switch the 'What do you want to find?' dropdown to Force or Distance, and the calculator rearranges the formula automatically and solves for whichever value you need.

What is the SI unit of work? The SI unit of work is the joule (J), equal to one newton-meter (N·m), and equivalent to the SI unit of energy.

Does work depend on time? No — the amount of work done depends only on force, distance, and angle, not on how quickly it happens. However, power (the rate of doing work) does depend on time: Power = Work ÷ Time.