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UY1: Concept of Work

Define mechanical work as energy transfer by a force: dot product, 1D integrals for variable forces, and line integrals along curved paths.

  • University Physics Year 1
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Learning objectives

  • Apply work–energy and momentum methods, then interpret the physical result.

This lesson introduces mechanical work: a scalar that measures energy transferred by a force as a particle moves. Work is the bridge from force-based dynamics to energy methods.

At a glance

  • Prerequisites: vectors + dot product (Mathematics for Undergraduate Physics); basic kinematics (Basics & Kinematics (UY1)); basic integration (Integration Techniques)
  • You will learn: compute work for constant forces; compute work from Fₓ(x) via an integral; write work along a curved path as a line integral
  • Key result: W_(1 → 2) = ∫₁² vector F · d vector r (and for constant vector F, W = vector F · Δ vector r = Fs cos φ)
  • Common trap: using W = Fs when the force is not parallel to the displacement, or forgetting the sign from the dot product

Setup (what “work” means)

For a small displacement d vector r, the small amount of work done by a force vector F is: dW = vector F · d vector r.

Integrate along the actual path from point 1 to point 2: W_(1 → 2) = ∫₁² vector F · d vector r.

Units: N · m = J.

Dot product = “only the parallel component counts”

Write the force component along the displacement as F_∥. Then: dW = F_∥ ds.

  • If F_∥ points along the displacement, work is positive.
  • If F_∥ points opposite the displacement (e.g. friction), work is negative.
  • If the force is always perpendicular to the motion, work is zero (e.g. ideal centripetal force in uniform circular motion).
Modelling workflow (work problems)
  1. Choose the object/system you care about (what gains/loses mechanical energy?).
  2. Draw a quick force diagram to identify which forces can do work along the motion.
  3. Choose a path coordinate (x in 1D, or arc length s along a curve).
  4. Compute W = ∫ vector F · d vector r and keep the sign from the dot product.
Sign convention: the integral already knows which way you moved

In 1D, if you move in the negative direction then dx < 0. For example, a constant force Fₓ = +5 N acting while an object moves from x = 2 to x = 0 gives W = ∫₂⁰ 5 dx = 5(0-2) = -10 J. Don’t “force” s to be positive and then guess the sign afterwards.

1) Work done by a constant force

If vector F is constant and the displacement is Δ vector r: W = vector F · Δ vector r = F Δ r cos φ, where φ is the angle between the force and the displacement.

Special cases:

  • φ = 0°: W = FΔ r (force along motion).
  • φ = 90°: W = 0 (force perpendicular to motion).
  • φ = 180°: W = -FΔ r (force opposite motion).

2) Work done by a varying force (1D)

If motion is along the x-axis and the force component depends on position, Fₓ(x), approximate the path by small steps:

Approximating work with narrow displacement stepsA force against displacement curve is covered by narrow rectangles. Each rectangle has area force times delta x and their sum approaches the integral as the widths shrink.xF
Σ F(x_i)Δx_i approaches ∫F(x) dx as the displacement intervals become narrower.

For a small step Δ xᵢ: Δ Wᵢ ≈ Fₓ(xᵢ) Δ xᵢ.

Sum and take the limit Δ x → 0: W = ∫_x₁^x₂Fₓ(x) dx.

Geometric interpretation: the work is the signed area under the Fₓ–x graph.

Signed area under a force-displacement graphA force curve lies above the displacement axis first and below it later. The area above contributes positive work and the area below contributes negative work.xFpositive worknegative work
Work is signed area: regions below the displacement axis subtract from regions above it.

3) Work along a curved path (line integral)

For curved motion, the displacement element is tangent to the path, d vector l.

Work along a curved pathAn object follows a curved path. At one point the displacement element d l is tangent to the path, while a force vector makes angle theta to it. The tangential force component is highlighted.dℓFθ
Only the component of force parallel to the tangent contributes: dW = F cos θ dl.

The small work element is: dW = vector F · d vector l = F cos θ dl = F_∥ dl, so the total work from P₁ to P₂ is: W = ∫_P₁^P₂ vector F · d vector l.

In general, this depends on the path. (A key exception: conservative forces, covered in the Potential Energy lesson.)

Worked example

Example: pulling a crate at an angle

Given: a constant pull F = 120 N at φ = 25° above the horizontal; the crate moves s = 8.0 m horizontally.

Find: the work done by the pull.

Working: only the horizontal component does work: W = Fs cos φ = (120)(8.0) cos 25° ≈ 8.7 × 10² J.

Answer + check: W ≈ 8.7 × 10² J (positive, since the force has a forward component).

Practice set (with hints + answers)

  1. A 10 N horizontal force pushes a box 3.0 m. Find the work done.
  2. A 50 N force acts at 60° to the displacement over 2.0 m. Find the work done.
  3. A 1D force is Fₓ(x) = kx with k = 5.0 N m⁻¹. Find the work done from x = 0 to x = 0.40 m.
  4. In ideal uniform circular motion, the centripetal force is always perpendicular to the instantaneous displacement. What is the work done by the centripetal force over one full circle?
  5. A force increases linearly from 0 to 8.0 N over 4.0 m. Find the work done over that interval.

Hints

  1. Use the dot product: W = Fs cos φ.
  2. Integrate: W = ∫₀^(0.40)kx dx.
  3. Work is the area under the F–x graph (a triangle).

Answers

  1. W = (10)(3.0) = 30 J.
  2. W = (50)(2.0) cos 60° = 50 J.
  3. W = 1/2 kx²|₀^(0.40) = 0.40 J.
  4. W = 0.
  5. W = (1/2)(4.0)(8.0) = 16 J.

Summary + next steps

  • Work is defined by W = ∫ vector F · d vector r (a scalar).
  • The dot product means only the component of force along the motion transfers energy.
  • In 1D, work is the signed area under the Fₓ(x) curve.
  • Along curved paths, use the line integral ∫ vector F · d vector l.

Next: Work-Energy Theorem Previous: Uniform Circular Motion & Non-uniform Circular Motion Back To Mechanics (UY1)