Physics I: Mechanics › Work & energy › full formula sheet

W = Fd cos θ

Say it: “the work done by a constant force equals force times distance times cosine theta”

Work done by a constant force

How a push over a distance becomes energy — and why only the part of the push along the motion counts.

F is the (constant) force in newtons, d the displacement magnitude in meters, θ the angle between the force and the displacement vectors. Work W is in joules: 1 J = 1 N·m.

Where it comes from

You push a heavy crate across the floor. Your arms ache, the crate speeds up — clearly something is being transferred from you to the crate. But how much? And what if you pull with a rope angled upward instead of pushing straight?

Before reading on: a rope pulls a cart with tension T at 30° above the horizontal. Does the full tension T do work on the cart, or only part of it? Guess, then read the verdict.

The verdict: only the horizontal part counts. The upward part of the pull lifts against nothing that moves — the cart moves horizontally, so the vertical component acts over zero vertical displacement and contributes zero work. Work is selective: a force earns credit only for the part pointing along the motion.

The clean way to say it uses the dot product of vectors:

W = F · dwork is the dot product of force and displacement — parallel parts multiply, perpendicular parts dieSay it: “work equals the dot product of force and displacement”

A dot product multiplies magnitudes and the cosine of the angle between the vectors — which is exactly where cos θ comes from. The full formula W = Fd cos θ is just this dot product written out.

Derivation

Derivation needs no new physics — just the definition of the dot product. Line the displacement up with the x-axis, so d = (d, 0), and split the force into components:

F
=
(F cos θ, F sin θ)
Step 1 — split F. The angle θ is measured tail to tail between F and d: adjacent component F cos θ, opposite F sin θ.
W
=
F · d = (F cos θ)(d) + (F sin θ)(0)
Step 2 — dot product. Multiply matching components and add. The second term dies: there is no displacement in the y-direction, so the perpendicular component has nothing to act over.
=
Fd cos θ
Step 3 — the formula. Only the parallel component survives. ∎

The three landmark angles tell the whole story:

θ = 0°
=
W = +Fd
Force along the motion: maximum positive work — energy flows into the object.
θ = 90°
=
W = 0
Force perpendicular to motion: zero work. This is why gravity does no work on a level-moving object.
θ = 180°
=
W = −Fd
Force opposing the motion: negative work — friction drains energy out of the object.

How to use it

The procedure, every time:

  1. Confirm the force is constant. Same magnitude and same direction for the whole motion. If it varies, you need the variable-force page (Work (variable force)).
  2. Identify the displacement d. Final position minus initial position — a straight-line magnitude in meters. Not the path length: a round trip has zero displacement.
  3. Measure θ tail to tail. Slide the force arrow (without rotating it) so its tail sits on the displacement arrow's tail, then read the angle between them. This is the #1 error on this formula.
  4. Compute W = Fd cos θ. Units: N·m = J. A missing joule is a missing mark.
  5. Read the sign. Positive: the force feeds energy in. Negative: it drains energy out. Zero: no energy transfer at all.
Common mistake: measuring θ from the wrong reference — e.g. using the angle above the vertical in cos θ. θ is always between the force and the displacement, tail to tail. When in doubt, draw both arrows from the same point.

Worked examples

Four problems, easiest first. Watch the angle and the sign in each one.

Example 1 — straight push: F = 40 N, d = 5 m

  1. Check the setup. Force is constant and parallel to the displacement: θ = 0°, cos 0° = 1.
  2. Compute. W = 40 × 5 × 1 = 200 J.
  3. Sanity check. Positive — the push feeds energy into the crate. Units: N·m = J. ✓
Common mistake: writing W = 40 × 5 = 200 “N” — forgetting the unit is joules. A number with the wrong unit is a wrong answer in physics.
Your turn — F = 60 N parallel to a 8 m displacement. W = ?

Answer: 480 J. θ = 0°, so W = 60 × 8 × 1 = 480 J.

Example 2 — angled pull: F = 100 N at θ = 30°, d = 10 m

  1. Identify θ. The rope makes 30° with the horizontal displacement: θ = 30°, cos 30° ≈ 0.866.
  2. Compute. W = 100 × 10 × 0.866 = 866 J.
  3. Compare. A straight 100 N pull would give 1000 J — the angle costs about 134 J. Only the 86.6 N horizontal component works.
Common mistake: answering 1000 J — using the full force and dropping cos θ. Ask: “is any of this force pointing off-axis?” If yes, cosine it.
Your turn — F = 80 N at 60° to the displacement, d = 12 m. W = ?

Answer: 480 J. cos 60° = 0.5, so W = 80 × 12 × 0.5 = 480 J.

Example 3 — friction: f = 15 N opposing a 4 m slide

  1. Identify θ. Friction points opposite the displacement: θ = 180°, cos 180° = −1.
  2. Compute. W = 15 × 4 × (−1) = −60 J.
  3. Read the sign. Negative — friction drains 60 J of the crate's energy (into heat).
Common mistake: reporting +60 J (“work is work”). The minus is the physics: it tells you energy left the object. Dropping it reverses the meaning.
Your turn — kinetic friction 25 N opposes a 6 m slide. W = ?

Answer: −150 J. 25 × 6 × (−1) = −150 J — energy drained.

Example 4 — the normal force: N = 98 N while the block slides 3 m

  1. Identify θ. The normal force points straight up; the block slides horizontally: θ = 90°, cos 90° = 0.
  2. Compute. W = 98 × 3 × 0 = 0 J.
  3. Why. The normal force has no component along the motion — it changes the floor's compression, not the block's speed.
Common mistake: writing W = 98 × 3 = 294 J. A perpendicular force does no work, ever — no matter how big it is. cos 90° is the off switch.
Your turn — a 50 N force acts perpendicular to a 2 m displacement. W = ?

Answer: 0 J. cos 90° = 0 kills the whole product.

Memorization tips

  • Say it aloud: “work equals force times distance times cosine theta.” The cosine is the part students forget — saying it keeps it in the formula.
  • The cosine landmarks: cos 0° = 1 (full credit), cos 90° = 0 (off switch), cos 180° = −1 (drain). Memorize these three and most problems fall out.
  • Tail-to-tail ritual: before touching the calculator, sketch both arrows from one point. Ten seconds of drawing prevents the most common error on this page.
  • The units check: N·m = J. If your answer's units aren't joules, something went wrong upstream.
  • The sign is the story: positive = energy in, negative = energy out, zero = no transfer. Read it like a bank statement.
  • Gravity's rule of thumb: on level ground gravity is perpendicular to motion, so Wg = 0. Gravity only works when height changes.

Final challenge

Five mixed questions — angles, signs, and the zero-work trap. Score 5/5 and constant-force work is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Why isn't work just Fd?

Because only the part of the force along the displacement moves the object. The dot product F · d = Fd cos θ keeps the parallel part and discards the perpendicular part. Fd alone overcharges every angled push.

Can work really be negative?

Yes — negative work means the force drains energy from the object instead of feeding it in. Friction opposing motion does W = −Fd, turning the object's kinetic energy into heat.

What exactly is a joule?

One joule is the work of one newton acting over one meter: 1 J = 1 N·m. It is also the SI unit of energy — work and energy are measured in the same unit because work is energy in transit.

How do I measure the angle θ?

Slide the force arrow (without rotating it) until its tail sits on the displacement arrow's tail, then read the angle between them. It is always the angle between the two vectors, never the angle to some axis you invented.

Does a perpendicular force ever do work?

Only if the displacement gains a component along the force. A force at exactly 90° to the displacement gives cos 90° = 0 — zero work, no exceptions. Carrying a bag horizontally: your upward force does no work on the bag.

More from the codex

Live

Physics I formula sheet

All the mechanics formulas — kinematics, forces, energy, momentum — printable and quiz-ready.

Open sheet → Live

Formula Sheet Builder

Mix and match any sections into your own printable sheet.

Open tool → Live

Prompt Simulator

Practice prompt engineering with deterministic scoring.

Open tool →