Work — definition, dot product F·d, sign convention: Worked examples

Before anything, one reminder of the only three symbols we use:

$F$ = how strong the push is, in newtons (N).
$d$ = how far the object actually moved, in metres (m) — its displacement, a straight arrow from start to finish.
$θ$ = the angle between the force arrow and the displacement arrow, drawn tail-to-tail.

Everything below is just those three quantities combined as $F d cos θ$ , or the component form $W = F_{x} d_{x} + F_{y} d_{y}$ when we are handed vectors instead of an angle. See Vectors & Components if resolving arrows into $x$ and $y$ parts is not yet automatic.

The scenario matrix

Every work problem you will ever meet is one of these cells. The examples that follow each carry a tag like (C1) so you can see the whole space getting covered.

Cell	What makes it different	Sign / result you expect
C1	Force along motion, $0° \leq θ < 90°$	$W > 0$ (energy added)
C2	Force at $θ = 90°$ (perpendicular)	$W = 0$ (only steers)
C3	Force against motion, $90° < θ \leq 180°$	$W < 0$ (energy removed)
C4	Given as vectors — use $F_{x} d_{x} + F_{y} d_{y}$	sign pops out automatically
C5	Degenerate: $d = 0$ (no displacement)	$W = 0$ no matter how big $F$
C6	Degenerate: $F = 0$ (no force)	$W = 0$
C7	Multiple forces on one body — sum the works	net $W = \sum W_{i}$
C8	Limiting behaviour: sweep $θ$ from $0° \to 180°$	$W$ traces a cosine curve
C9	Real-world word problem (ramp / lifting)	mix of C1–C3, pick angles carefully
C10	Exam twist: closed path with a constant force	net $W = 0$ (displacement returns to start)

Figure — Work — definition, dot product F·d, sign convention

The single red curve above is your compass for the whole page: work is proportional to $cos θ$ . Read off where each cell lives — C1 on the positive hump, C2 at the zero-crossing, C3 on the negative dip.

The worked examples

Worked example Example 2 — perpendicular force does nothing

(C2) A satellite of mass moves in a circle. Earth's gravitational pull on it is $F = 500 N$ , always pointing to the centre; the satellite sweeps a displacement $d = 2000 m$ along its circular arc. Work done by gravity over that arc?

Forecast: gravity points to the centre; the motion is along the circle. What is the angle between "toward centre" and "along the rim"?

Find the angle. The radius (force direction) meets the tangent (motion direction) at exactly $90°$ . Why this step? For circular motion the centre-pointing force is always $⊥$ to the velocity.
Plug in. $cos 90° = 0$ , so $W = 500 \times 2000 \times cos 90° = 500 \times 2000 \times 0 = 0 J .$ Why this step? A perpendicular force produces zero displacement in its own direction, so it transfers no energy — it only bends the path.

Verify: $cos 90° = 0$ kills the whole product regardless of how big $F$ or $d$ are ✓. This is why orbital speed stays constant — see Power ( $P = F \cdot v$ is also zero here).

Worked example Example 7 — many forces, sum the works

(C7) A $2 kg$ block is dragged $d = 5 m$ right along the floor. Three forces act: applied pull $F_{1} = 30 N$ at $0°$ (rightward), friction $F_{2} = 8 N$ at $180°$ , and gravity $F_{3} = m g = 19.6 N$ straight down ( $90°$ to the horizontal motion). Find the work done by each force and the net work.

Forecast: which of the three does zero work? Which is negative?

Work by the pull (C1). $θ = 0°$ : $W_{1} = 30 \times 5 \times cos 0° = 150 J$ . Why this step? It is fully along the motion — maximum positive contribution.
Work by friction (C3). $θ = 180°$ : $W_{2} = 8 \times 5 \times cos 180° = - 40 J$ . Why this step? Opposes motion ⇒ removes energy (heat).
Work by gravity (C2). $θ = 90°$ : $W_{3} = 19.6 \times 5 \times cos 90° = 0 J$ . Why this step? Vertical force, horizontal motion ⇒ perpendicular ⇒ zero.
Add them. $W_{net} = 150 + (- 40) + 0 = 110 J$ . Why this step? Work is a scalar, so total work is a plain sum — no vector adding needed.

Verify: $W_{net} = 110 J > 0$ , so the block speeds up, exactly what Kinetic Energy & Work-Energy Theorem predicts. Units all joules ✓.

Worked example Example 9 — real-world word problem: pushing up a ramp

(C9) A worker pushes a $50 kg$ crate a distance $d = 4 m$ up a frictionless ramp inclined at $30°$ . Gravity is $g = 9.8 m/s^{2}$ . Find the work done by gravity on the crate.

Forecast: the crate rises as it moves up the slope. Gravity points straight down; the motion is up-slope. Positive or negative?

Find the angle between gravity and the up-slope displacement. Displacement is $30°$ above horizontal (up the ramp); gravity points straight down. The angle between "straight down" and " $30°$ above horizontal" is $90° + 30° = 120°$ . Why this step? We need the true angle between the two arrows, not the ramp angle itself — a classic trap.
Plug in. $F = m g = 50 \times 9.8 = 490 N$ , and $W_{grav} = 490 \times 4 \times cos 120° = 490 \times 4 \times (- 0.5) = - 980 J .$ Why this step? $θ > 90°$ ⇒ negative work — gravity resists the upward climb, draining kinetic energy into Potential Energy & Conservative Forces.

Verify (independent route): the crate's height gain is $h = d sin 30° = 4 \times 0.5 = 2 m$ , so gravity's work is $- m g h = - 490 \times 2 = - 980 J$ ✓ — the two methods agree, confirming the $120°$ .

Worked example Example 10 — exam twist: constant force around a closed loop

(C10) A constant horizontal force $F = (6 \overset{}{^}) N$ pushes a bead that travels $d_{1} = (3 \overset{}{^} + 2 \overset{}{^}) m$ , then $d_{2} = (- 3 \overset{}{^} - 2 \overset{}{^}) m$ , returning to the start. Find the total work by $F$ .

Forecast: the bead ends where it began. Does a constant force do net work over a closed loop?

Work on leg 1. $W_{1} = F_{x} d_{1 x} + F_{y} d_{1 y} = (6) (3) + (0) (2) = 18 J$ . Why this step? Component form — only the $x$ -part of $F$ is nonzero.
Work on leg 2. $W_{2} = (6) (- 3) + (0) (- 2) = - 18 J$ . Why this step? Same force, reversed displacement ⇒ equal-and-opposite work.
Total. $W_{net} = 18 + (- 18) = 0 J$ . Why this step? A constant force is conservative; over any closed path the net displacement is zero, so net work is zero.

Verify: total displacement $d_{1} + d_{2} = 0$ , and $W_{net} = F \cdot (d_{1} + d_{2}) = F \cdot 0 = 0$ ✓. This is exactly the "closed loop with gravity" fact from the parent's mistakes section.

Active Recall

Recall Did you cover every cell?

Which example is the $d = 0$ degenerate case? ::: Example 5 (C5) Which example shows net work as a plain scalar sum? ::: Example 7 (C7) In Example 9, why is the angle $120°$ and not $30°$ ? ::: it is the angle between straight-down gravity and the $30°$ up-slope displacement, $90° + 30° = 120°$ What are the maximum and minimum possible values of $W$ for fixed $F, d$ ? ::: $+ F d$ (at $θ = 0°$ ) and $- F d$ (at $θ = 180°$ ) Why is the net work around the closed loop in Example 10 zero? ::: a constant force is conservative; net displacement is zero so $F \cdot 0 = 0$

Connections

Parent: Work — definition & sign convention
Dot Product (Scalar Product) — the component form used in Examples 4 and 10
Kinetic Energy & Work-Energy Theorem — net work changes $\frac{1}{2} m v^{2}$ (Examples 1, 7)
Potential Energy & Conservative Forces — negative gravity work stores energy (Example 9, 10)
Power — perpendicular force gives $P = 0$ too (Example 2)
Friction — the negative-work force in Examples 3 and 7
Vectors & Components — resolving arrows into $x, y$ parts