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Work Done by a Force

Calculates the mechanical work done by a constant force acting on an object as it moves, accounting for the angle between force and displacement.

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Work Done by a Force CalculatorPhysics · Mechanics

W = F · d · cos(θ)
W = work  ·  F = force  ·  d = displacement  ·  θ = angle between F and d
⟹ SolveW, F, d, θ
J
N
m
°
Please fix the errors above.
Solve for:
Presets:
Work (W)
W: F: d: θ:
✓ Copied!
Angle (θ) Gauge
Positive Work (0°–90°) Zero Work (90°) Negative Work (90°–180°)
W = F · d · cos(θ)  ·  Work in joules (J), force in newtons (N), displacement in metres (m), angle in degrees

Interpretation

Work done by a force: W = F·d·cosθ, where F is force, d is displacement, θ is angle between them. Work is energy transferred. Example: 10 N force at 60° to displacement of 5 m → W = 10×5×0.5 = 25 J.

W = F * d * cos(theta)
Work Done by a Force

Variables

SymbolQuantityUnit
WWork doneJ
FApplied force magnitudeN
dDisplacement of the objectm
thetaAngle between the force and displacement directiondegrees

What it means

Work is defined as the energy transferred to or from an object by a force acting over a displacement. The formula W = F d cosθ accounts for the component of force in the direction of motion. If the force is parallel to displacement, W = Fd; if perpendicular, work is zero. Work is a scalar quantity measured in joules. The work‑energy theorem states that the net work done on an object equals its change in kinetic energy. Work is a central concept in mechanics and thermodynamics, used in analysing machines, engines, and energy systems. In everyday life, work is done when lifting, pushing, or pulling. Understanding work is essential for energy analysis and efficiency calculations.

Worked example

Work Done – Two Examples

Real‑World
Scenario: A 100 N force pushes a box 5 m in the direction of motion. Find work done.
ParameterValue
F100 N
d5 m
θ
1W = F·d·cosθ = 100 × 5 × cos(0°) = 500 J
Result 500 J ✓ Positive work
Scenario: A 50 N force acts at 60° to the displacement of 10 m. Find work.
ParameterValue
F50 N
d10 m
θ60°
1W = 50 × 10 × cos(60°) = 500 × 0.5 = 250 J
Result 250 J ✓ Partial work
Key insight: Work = F·d·cosθ – only the component of force in the direction of motion does work.

Common mistakes

  • Force F and displacement d: Both are vectors – the angle θ is between their directions.
  • Work W: Scalar – can be positive, negative, or zero.
  • Units: F in N, d in m → J (joules).
  • Constant force: This formula assumes constant force; for variable force, use integral ∫F·dr.
  • Work done by a force: Only the component of force along displacement does work – perpendicular component does zero work.

Applications

Work done by a force, W = F·d·cosθ, is the energy transferred by a force acting over a distance. It is fundamental to all energy analyses in engineering and physics. Engineers use this formula to calculate the work required to move loads, to design lifting equipment, and to evaluate motor and engine output. In thermodynamics, work is a key term in energy balances. In civil engineering, it is used in the design of dams and excavation. The formula also appears in biomechanics to assess muscular work. By understanding work, professionals can quantify energy consumption, size actuators and motors, and optimise mechanical systems for efficiency and performance.

  • Design of lifting and conveying equipment
  • Motor and engine power calculations
  • Energy balance in thermodynamic cycles
  • Excavation and earthmoving work estimation
  • Biomechanical analysis of human movement

Frequently Asked Questions

Q01What is the formula for work done by a constant force?
A01

The work done by a constant force is W = F·d·cosθ, where F is the magnitude of the force, d is the magnitude of the displacement, and θ is the angle between the force and displacement vectors.

Q02What is the common mistake when using this formula?
A02

Forgetting the cosine term. Work is only done by the component of the force in the direction of displacement. If the force is perpendicular to displacement (θ=90°), work is zero.

Q03What are the units of work?
A03

The SI unit is the joule (J) = N·m = kg·m²/s².

Q04When is work positive, negative, or zero?
A04

  • Positive: force and displacement are in the same direction (θ < 90°), increasing kinetic energy.
  • Negative: force and displacement are opposite (θ > 90°), decreasing kinetic energy (e.g., friction).
  • Zero: θ = 90° (e.g., centripetal force in uniform circular motion).

Q05How does work relate to kinetic energy?
A05

The work‑energy theorem states that the net work done on an object equals its change in kinetic energy: W_net = ΔKE. This is a powerful tool for solving problems without knowing acceleration.

Q06How do you calculate work done by a variable force?
A06

If the force varies with position, use integration: W = ∫F(x)·dx (or the appropriate vector dot product). For example, the work done by a spring is ½kx².

Q07What is the difference between work and torque?
A07

Work is force times displacement (linear); torque is force times lever arm (rotational). Work changes translational kinetic energy; torque changes rotational kinetic energy.

Q08How is work used in energy conservation?
A08

Work is the mechanism by which energy is transferred. In a closed system, work done by conservative forces (like gravity) is equal to the negative change in potential energy. This leads to conservation of mechanical energy.