Work Calculator
Compute mechanical work, force, distance or angle using the dot-product relation W = F × d × cos(θ). Use the tabs below to toggle whether the angle between force and displacement is considered. Enter any two known values (plus angle if used) and leave the target blank.
Work: definition, formula, units, and the role of angle
Work is a central concept in mechanics and energy analysis. Informally, work measures the transfer of energy that occurs when a force causes a displacement. In its most general vector form, mechanical work done by a force \(\mathbf{F}\) acting on a particle undergoing an infinitesimal displacement \(\mathrm{d}\mathbf{s}\) is defined as the line integral \( W = \int \mathbf{F}\cdot \mathrm{d}\mathbf{s} \). For the common practical case of a constant force acting over a straight-line displacement, this reduces to the scalar expression:
W = F × d × cos(θ)
Here:
- W is the work (SI unit: joule, J).
- F is the magnitude of the applied force (SI unit: newton, N).
- d is the magnitude of the displacement (SI unit: metre, m).
- θ is the angle between the force vector and the displacement vector.
Why the cosine appears
The dot product \(\mathbf{F}\cdot\mathbf{d}=|\mathbf{F}||\mathbf{d}|\cos(\theta)\) projects the force onto the displacement direction. Only the component of the force that acts along the displacement performs work — perpendicular components do no work on the object in that displacement. For example, when you carry a suitcase horizontally at constant height, the vertical contact force from your hand balances weight but does little work on the horizontal displacement because the angle between the vertical force and the horizontal displacement is 90°, and cos(90°)=0.
Work is a scalar
Despite being computed from vectors, work is a scalar quantity (it has magnitude and sign but no direction). The sign of work indicates the direction of energy transfer: if the force component along displacement is in the same direction (0° ≤ θ < 90°), work is positive (energy transferred into the object); if the force component opposes the displacement (90° < θ ≤ 180°), work is negative (force removes energy from the object). Zero work can occur when force is perpendicular to displacement or when magnitude or displacement is zero.
Units and conversions
Work is measured in joules (J) where 1 J = 1 N·m. Practically, alternate units are often used:
- kilojoule (kJ) = 1000 J.
- calorie (chemistry small calorie) where 1 cal = 4.184 J; dietary calories (kcal) are 1000 cal.
Force units: newton (N) is standard. 1 lbf (pound-force) = 4.4482216152605 N. 1 dyne (cgs) = 10^-5 N. Distance units include metres, centimetres (1 cm = 0.01 m), millimetres, kilometres, feet (1 ft = 0.3048 m) and inches (1 in = 0.0254 m). Accurate unit handling is essential for correct numerical results; this calculator converts all inputs to SI internally before computing.
Special cases and intuition
Force parallel to displacement (θ = 0°): cos θ = 1, so W = F × d. This is the simplest and often implicitly assumed case when angle is not specified.
Force perpendicular to displacement (θ = 90°): cos θ = 0, so W = 0 — the force does no work for that displacement.
Force opposite displacement (θ = 180°): cos θ = −1, work is negative: W = −F × d.
Solving for different variables
The algebraic rearrangements are straightforward when cos(θ) ≠ 0:
- W known, find F: \( F = \dfrac{W}{d \cos\theta} \)
- W known, find d: \( d = \dfrac{W}{F \cos\theta} \)
- W known, find θ: \( \theta = \arccos\!\left(\dfrac{W}{F d}\right) \) provided \( |W| \le F d \)
In practice check domain constraints — the argument of arccos must be within [−1, 1]. If it is outside due to inconsistent inputs or measurement error, the calculator will flag the issue. Also, when cos θ ≈ 0 small measurement errors can cause very large uncertainties in computed F or d because division by a small number amplifies uncertainty.
Negative work and energy bookkeeping
Negative work is not an error — it means energy is being removed from the object. For instance, friction does negative work on moving objects (it converts kinetic energy to heat). In conservative force fields (e.g., gravity), work around closed paths is zero; potential energy formalism provides a convenient bookkeeping alternative.
Angle units and numeric stability
This calculator accepts angle in degrees or radians; degrees are used by default. Internally trigonometric calculations use radians, so degree inputs are converted. For angles very close to 90°, cos θ approaches zero and small changes in θ lead to large relative changes in cos θ — be careful when solving for F or d near that region.
Practical examples (worked)
Example 1 — Work from force parallel to displacement: A worker applies 150 N horizontally to push a crate 4.0 m. θ = 0°, so W = 150 × 4.0 × cos(0°) = 600 J. The energy transferred to the crate is 600 J.
Example 2 — Work with angle: A rope pulls a sled with tension 200 N at an angle 30° above horizontal for 10 m. The horizontal component does the work on the sled: W = F d cos θ = 200 × 10 × cos(30°) ≈ 2000 × 0.8660 = 1732 J.
Example 3 — Solve for angle: Suppose a force of 50 N acting over 2.0 m does 50 J of work. cos θ = W/(F d) = 50 / (50 × 2) = 0.5 so θ = arccos(0.5) = 60°.
Measurement and errors
When performing experiments measure forces with calibrated load cells and displacements with appropriate rulers or sensors. Estimate uncertainties (± values) for each input and propagate uncertainties when reporting work — for multiplication/division this involves fractional uncertainties. Be careful converting units and ensure consistent significant figures.
When this formula doesn't apply directly
If force varies along the path (non-constant F) or motion follows a curved path, the work requires integration: \( W = \int_{s_1}^{s_2} \mathbf{F}(s)\cdot \mathrm{d}\mathbf{s} \). For example, lifting an object against gravity at varying height with non-uniform force distribution or doing work against nonlinear springs needs calculus-based treatment. This calculator handles the common uniform-force straight-line case and provides useful results for most introductory and applied problems.
Summary
Work connects force and displacement through the projection of force onto the displacement direction. Correct unit handling and careful attention to angle and sign produce meaningful energy bookkeeping for mechanical and engineering problems. Use this interactive tool to compute W, F, d or θ, export results, and include the step-by-step derivations when documenting lab work or design calculations.
Frequently Asked Questions
Because only the component of force along the displacement does work — cos projects the force onto the displacement direction.
Yes — if the force is perpendicular to displacement (θ = 90°), or if either force or displacement is zero.
Use θ = arccos(W/(F d)) provided |W| ≤ F d. If outside domain, inputs are inconsistent or rounding error introduced issues.
Yes — the form handles common units for work, force and distance and converts internally to SI before computing.
No — work is a scalar (but it's computed from vector quantities via dot product).
Yes — negative work indicates the force component is opposite the displacement direction (θ > 90°).
Then integrate: W = ∫ F·ds. This calculator assumes constant force along a straight path.
No — friction is a separate force. Include friction force in F or compute work due to friction explicitly.
Either is fine — the calculator supports both and converts to radians internally for cos/arccos operations.
Yes — enable step-by-step and use Copy or Download CSV to save the math and values.