Work Done by a Force

Carry a heavy suitcase 100 metres along a station platform and your arm will burn with effort — yet physics rules, with a straight face, that you did essentially no work on the suitcase. Lean with all your strength on a stalled car that refuses to budge: zero work again. Now nudge a shopping trolley so it rolls a few metres: that lazy flick did do work. Clearly the physics word "work" is not the everyday word. It is something sharper and more useful: energy transferred to an object by a force acting through a displacement — counting only the part of the force that points along the motion. This definition is the gateway to the most powerful accounting system in mechanics. The coming chapter will trade force-and-acceleration bookkeeping (three laws, vectors, components) for an energy ledger in which a single scalar number tracks what forces accomplish. Every entry in that ledger begins here, with $W = Fd\cos\theta$ : force, displacement, and the angle between them, folded into one signed number measured in joules. The sign is physics, not decoration — positive work feeds energy into motion, negative work drains it, zero work (the suitcase, the wall) moves no energy at all however heroic the effort. Every board asks this module directly — compute the work by each force on a dragged block — and every competitive exam builds on its edge cases: the friction that does positive work, the normal force that never works, the gravity whose work depends only on height. Master the sign dial here and the entire energy chapter opens on schedule.

Key takeaways

Work is energy transferred by a force through a displacement: $W = \vec F \cdot \vec d = Fd\cos\theta$ , with $\theta$ the angle between force and displacement. Unit: the joule, $1\ \mathrm{J} = 1\ \mathrm{N\,m}$ .
Only the component of force along the displacement works; the perpendicular component transfers nothing.
The sign dial: $\theta < 90°$ → positive (energizing) · $\theta = 90°$ → zero · $\theta > 90°$ → negative (draining).
No displacement, no work: holding, pressing, and straining transfer zero energy — effort is not work.
Work is a scalar with a sign: works by different forces add as plain numbers, and their sum is the net work, $W_{\rm net} = F_{\rm net}\,d\cos\theta$ .
Gravity's work cares only about height change: $W_{\rm grav} = -mg\,\Delta h$ , whatever the path — the seed of potential energy.
Key idea: work is the along-the-motion shadow of the force times the distance moved — a single signed number that says how much energy the force moved, and in which direction.

The core insight

Work is the dot product of force and displacement, $W = \vec F \cdot \vec d = Fd\cos\theta$ : the projection of the force onto the motion, times the distance moved, giving a signed scalar that counts the energy the force transferred — positive into the motion, negative out of it, zero when force and motion are perpendicular or when nothing moves.

Where this lives today

Every electricity bill is a work invoice: the kilowatt-hour is $3.6 \times 10^6$ joules of $W = Fd\cos\theta$ -style transfers, delivered through motors, pumps and heating elements. The food-label calorie prices the work your muscles can fund (one dietary calorie ≈ 4184 J — the suitcase-holder burns them into heat without doing a joule on the suitcase, exactly as the pitfall box warned). And the dot product at the formula's heart has become one of the most-executed operations in existence: computer-graphics engines shade every pixel with dot products between light and surface directions, and the neural networks behind modern AI are, at the arithmetic level, oceans of dot products — projections asking, a trillion times a second, the sled-rope's question: how much of this vector lies along that one?

The physical picture

Picture a child towing a sled with a rope that slants upward at angle $\theta$ . The rope's tension does two jobs at once: its horizontal part drags the sled forward along the snow; its vertical part slightly un-weights the sled, easing it off the ground. Only the first job pushes the sled along its path — only the first job transfers energy of motion. The vertical part could pull all day and never speed the sled up by a whisker. So the honest measure of a force's effectiveness needs one geometric question answered: how much of $\vec F$ lies along $\vec d$ ? Drop a perpendicular from the force arrow's tip onto the direction of motion: the shadow has length $F\cos\theta$ . Multiply by the distance moved and you have the work. That projection-then-multiply operation is so central to physics that it owns a symbol — the dot in $\vec F \cdot \vec d$ — and you met its machinery in the vector-products module. Here the dot product stops being algebra and starts being energy.

Figure

Hero — the slanted pull: only the along-track shadow of the force does work

Three sign regimes cover every case, and reading them fluently is half this module. Pull with the motion ( $\theta < 90°$ ): positive work — the force feeds energy in, like a locomotive. Push against it ( $\theta > 90°$ ): negative work — the force bleeds energy out, like friction under a skidding box or gravity on a rising ball. And the strange, crucial middle: force perpendicular to motion ( $\theta = 90°$ ) does exactly nothing. The suitcase on the platform lives here — your arm pulls up, the case moves sideways, $\cos 90° = 0$ , and every joule your muscles burn goes into biology (trembling fibres, heat), not into the suitcase's motion. The normal force on any sliding block, the tension of a whirling string, the centripetal force of any circular motion: all perpendicular, all workless, forever redirecting motion without funding it.

The math: one formula, three faces

The definition, for a constant force over a straight displacement:

W = \vec F \cdot \vec d = F\,d\cos\theta \qquad (1)

where: $F$ and $d$ are the magnitudes, $\theta$ the angle between the arrows when drawn tail-to-tail. Both readings of the product are legal and equal: (force's shadow on motion) × (distance), or (force) × (motion's shadow on force). A $50$ N pull at $37°$ dragging a crate $10$ m does $W = 50 \times 10 \times 0.8 = 400$ J. Note what is absent: mass, speed, time. Work does not care how fast the trip happened or how heavy the object is — only force, path length, and alignment. When force and displacement come as components, the dot product's second face computes the same number with no angle in sight:

W = F_x d_x + F_y d_y \qquad (2)

where: multiply matching components and add — signs included. $\vec F = (30, 40)$ N acting through $\vec d = (8, -2)$ m does $W = 240 - 80 = 160$ J. Exams switch freely between faces (1) and (2); the derivation below shows why they always agree. Several forces act at once? Each does its own work by (1), and the works — plain signed numbers — simply add:

W_{\rm net} = W_1 + W_2 + \cdots = \vec F_{\rm net} \cdot \vec d \qquad (3)

where: the equality of "sum of works" and "work of the sum" is the distributive law of the dot product, and it is the licence for every energy audit you will run this chapter: work out each force's entry (pull positive, friction negative, normal zero…), add, and the total is what the net force accomplished — the number the work–energy theorem (next modules) will connect to speed.

the sign dial	$\theta$	$\cos\theta$	work	example
force aids motion	$0° \le \theta < 90°$	$+$	positive	engine pull, gravity on a falling ball
force ⊥ motion	$90°$	$0$	zero	normal force, centripetal tension, the suitcase
force fights motion	$90° < \theta \le 180°$	$-$	negative	friction under a slide, gravity on a rising ball

Derivation ATwo faces of one number — why the projection and the components always agree

We show that $Fd\cos\theta$ and $F_xd_x + F_yd_y$ are the same number, using only resolution of vectors — no calculus, no new axioms. The point matters practically (exams switch representations without warning) and conceptually: work is a geometric quantity, indifferent to your choice of axes. The strategy: resolve the force along and across the displacement, let the across part die by perpendicularity, then re-express in standard axes.

Step 1. Set up: force $\vec F$ at angle $\theta$ to displacement $\vec d$ . Resolve $\vec F$ into a part along $\vec d$ (magnitude $F\cos\theta$ ) and a part perpendicular (magnitude $F\sin\theta$ ).

Step 2. The perpendicular part is powerless. A force component at $90°$ to the motion neither speeds nor slows the object along its path — it has no shadow on the path at all. Whatever energy bookkeeping we build must assign it zero. That leaves the along part acting through the full distance:

W = (F\cos\theta)\,d

Why this assignment is forced, not chosen: run the two pure cases. Force fully along motion must give $W = Fd$ (all of it counts); force fully across must give $W = 0$ (none of it does). $F d\cos\theta$ is the unique linear rule that honours both.

Step 3. Now pick axes: $x$ along nothing special, and write $\vec F = (F_x, F_y)$ , $\vec d = (d_x, d_y)$ . Compute $F_xd_x + F_yd_y$ for the two pure cases of Step 2. Force parallel to $\vec d$ : choose $x$ along $\vec d$ for a moment, then $F_y = 0$ , $d_y = 0$ and the sum is $F_xd_x = Fd$ ✓. Force perpendicular: $F_x = 0$ or $d_x$ -pairing vanishes crosswise, sum $= 0$ ✓.

Step 4. General case by superposition: any $\vec F$ is (parallel part) + (perpendicular part); the component sum handles each part correctly by Step 3 and is linear (sums of forces give sums of the expression), so it agrees with $Fd\cos\theta$ for every $\vec F$ . Both faces are one number; the number is axis-free. Why the axis-freedom is the real prize: rotate your coordinates and $F_x, F_y, d_x, d_y$ all change — but the combination $F_xd_x + F_yd_y$ does not. Work is a fact about the arrows, not about your grid, which is why a scalar can carry physics that vectors usually carry. *Verified: $\vec F = (30, 40)$ N, $\vec d = (8, -2)$ m: component face gives $240 - 80 = 160$ J. Geometric face: $F = 50$ N, $d = \sqrt{68} = 8.25$ m, $\cos\theta = \dfrac{160}{50 \times 8.25} = 0.388$ , $\theta = 67.2°$ — consistent ✓ (same 160 J by construction). The two-face identity is the standard dot-product theorem, Halliday Ch. 7.*

Go deeper · Derivation B · 2 minGravity's work depends only on the height drop — the ramp theorem

Derivation BGravity's work depends only on the height drop — the ramp theorem

We derive the most consequential special case in the chapter: the work done by gravity on a body moving down (or up) any straight ramp is $W_{\rm grav} = \pm mgh$ , with $h$ the vertical height change — the ramp's length and angle cancel. This result seeds potential energy and the conservative-force idea two modules ahead. The strategy: apply (1) with gravity's fixed downward arrow and read the cosine as a geometric shadow.

Step 1. A block slides a distance $d$ down a ramp inclined at $\alpha$ . Gravity is $mg$ straight down; the displacement points down-slope. The angle between them is $90° - \alpha$ , so

W_{\rm grav} = mg\,d\cos(90° - \alpha) = mg\,d\sin\alpha

Step 2. Read the geometry: $d\sin\alpha$ is exactly the vertical drop $h$ of the trip. Hence

W_{\rm grav} = mg\,h

— no $\alpha$ , no $d$ , only the height lost. Rising instead? The angle exceeds $90°$ , the cosine flips sign: $W_{\rm grav} = -mgh$ for a climb of $h$ . Why the slope cancelled: a gentler ramp gives gravity a smaller along-path shadow ( $mg\sin\alpha$ ) but a longer path ( $d = h/\sin\alpha$ ) — the product is blind to the trade. Gravity bills by altitude, not by route.

Step 3. Chain ramps into any staircase-like path: each segment contributes $mg \times$ (its own drop), and the drops add to the total $\Delta h$ . Any path decomposable into straight pieces — and, by refinement, any smooth curve — gives

W_{\rm grav} = -mg\,\Delta h \qquad (\Delta h = h_{\rm final} - h_{\rm initial})

Why this is the seed of potential energy: a force whose work depends only on endpoints, never on the road between, allows a bookkeeping shortcut — assign every height a stored-energy value $mgh$ and read gravity's work as a withdrawal from storage. That shortcut, made official, is the potential-energy module.

Step 4. The same logic, run on friction, fails instructively: friction's force always points against the local motion, so its work is $-f \times (\text{path length})$ — a longer road costs strictly more. Endpoint-blind forces (gravity, springs, electrostatics) will earn the name conservative; road-billing forces (friction, drag) will not. Verified: block descending $d = 4$ m on a $30°$ ramp: $W = mg\,d\sin 30° = mg \times 2$ — equal to $mg\,h$ with $h = 2$ m ✓ (worked example 4 uses these numbers). Limits: $\alpha \to 90°$ gives $W = mgd$ , free fall ✓; $\alpha \to 0$ gives $W \to 0$ , horizontal travel ✓.

Figure

The cosine sign dial — one angle decides whether energy flows in, out, or not at all

Limiting cases

case	result	meaning
$\theta = 0$	$W = +Fd$	the whole force works: locomotive case
$\theta = 90°$	$W = 0$	perpendicular forces are passengers, never workers
$\theta = 180°$	$W = -Fd$	full opposition: friction under a straight slide
$d = 0$	$W = 0$	no displacement, no work — holding is not working
$F$ constant, path horizontal	$W_{\rm grav} = 0$	carrying at constant height: gravity idles
ramp angle $\alpha \to 90°$	$W_{\rm grav} \to mgd$	the ramp becomes free fall
circular motion, any speed	$W_{\rm centripetal} = 0$	always perpendicular: redirects, never funds

Common trapIf it exhausts you, it must be doing work

Effort is a fact about your muscles; work is a fact about energy transfer — and the exam lives in the gap between them. Hold a $20$ kg suitcase stationary for ten minutes: your force is $200$ N, the displacement is zero, the work is zero. Carry it $100$ m along a level platform: your force points up, the motion points sideways, $\cos 90° = 0$ — still zero work on the suitcase. Push a wall all afternoon: zero. Where does your burning effort go? Into your own body — muscle fibres contract and release microscopically, burning chemical energy into heat — but none of it is transferred to the object through its displacement, which is the only thing $W = Fd\cos\theta$ counts. The reverse confusion also bites: friction is not "always negative work". On a conveyor belt feeding a dropped parcel up to belt speed, friction points along the parcel's motion and does honest positive work. The sign comes from the angle in the formula, case by case — never from a force's reputation.

Worked examples

1 · Board level — the dragged crate audit (CBSE/ISC pattern, 3 marks)

A crate is dragged $10$ m across a level floor by a $50$ N rope pull at $37°$ above the horizontal, against a friction force of $20$ N. Find the work done by (a) the pull, (b) friction, (c) gravity and the normal force, (d) the net force. ( $\cos 37° = 0.8$ .)

Step 1. (a) The pull: $W_{\rm pull} = Fd\cos\theta = 50 \times 10 \times 0.8 = \boxed{+400\ \mathrm{J}}$ .

Step 2. (b) Friction acts opposite the motion ( $\theta = 180°$ ): $W_{\rm fric} = -20 \times 10 = \boxed{-200\ \mathrm{J}}$ .

Step 3. (c) Gravity and the normal force are both perpendicular to the horizontal displacement: $W_{\rm grav} = W_N = \boxed{0}$ — state the reason ( $\cos 90° = 0$ ), it carries the mark.

Step 4. (d) Works add as numbers: $W_{\rm net} = 400 - 200 + 0 + 0 = \boxed{+200\ \mathrm{J}}$ . The crate gained 200 J of energy of motion — the ledger's first balanced page.

2 · Board level — the crane at constant speed (ICSE/IGCSE-to-CBSE bridge)

A crane lifts a $200$ kg beam vertically through $15$ m at constant speed. Find the work done by (a) the crane's cable, (b) gravity, (c) the net force. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Constant speed ⇒ zero acceleration ⇒ the cable tension equals the weight: $T = mg = 2000$ N.

Step 2. (a) Cable: force up, motion up, $\theta = 0$ : $W_{\rm cable} = 2000 \times 15 = \boxed{+30\ \mathrm{kJ}}$ .

Step 3. (b) Gravity: force down, motion up, $\theta = 180°$ : $W_{\rm grav} = -2000 \times 15 = \boxed{-30\ \mathrm{kJ}}$ — equivalently $-mg\Delta h$ .

Step 4. (c) Net: $+30 - 30 = \boxed{0}$ — consistent with unchanged speed. Read the ledger: the crane's 30 kJ did not vanish; it was banked against gravity as height (potential energy, arriving officially two modules ahead).

3 · Competitive level — the angled pull changes the friction too (JEE-Main pattern, ~2.5 minutes)

A $5.0$ kg block is pulled $8.0$ m across a rough floor ( $\mu_k = 0.25$ ) by a $40$ N force at $37°$ above the horizontal. Find the work done by the pull, by friction, and the net work. ( $g = 10\ \mathrm{m/s^2}$ , $\sin 37° = 0.6$ , $\cos 37° = 0.8$ .)

Step 1. The trap first: the upward component of the pull un-weights the block, shrinking the normal force. Vertical balance: $N = mg - F\sin 37° = 50 - 24 = 26$ N. (Writing $f = \mu mg = 12.5$ N here forfeits the problem.)

Step 2. Friction: $f = \mu_k N = 0.25 \times 26 = 6.5$ N, opposing motion: $W_{\rm fric} = -6.5 \times 8.0 = \boxed{-52\ \mathrm{J}}$ .

Step 3. The pull: $W_{\rm pull} = 40 \times 8.0 \times 0.8 = \boxed{+256\ \mathrm{J}}$ . Gravity and normal: zero each (perpendicular).

Step 4. Net: $256 - 52 = \boxed{+204\ \mathrm{J}}$ . Audit the structure: the pull's vertical component did no work itself ( $\perp$ motion) yet still changed the ledger — by cutting friction's negative entry. Components can matter without working. Timing note: ~2.5 min. The single decision point is Step 1's normal force; everything else is three multiplications. MCQ options will include $-100$ J (from $\mu mg d$ ) for those who skip it.

4 · Competitive level — the incline ledger, two ways (NEET/JEE pattern, ~3 minutes)

A $2.0$ kg block slides $4.0$ m down a $30°$ incline with $\mu_k = 0.3$ . Find the work done by gravity (two independent ways), by the normal force, and by friction, and the net work. ( $g = 10\ \mathrm{m/s^2}$ , $\cos 30° = 0.866$ .)

Step 1. Gravity, way one — along-slope component: $W_{\rm grav} = (mg\sin 30°)\,d = (10)(4.0) = +40$ J. Way two — the ramp theorem: height drop $h = d\sin 30° = 2.0$ m, so $W_{\rm grav} = mgh = 2 \times 10 \times 2 = \boxed{+40\ \mathrm{J}}$ ✓ — same number, route-blind.

Step 2. Normal force: perpendicular to the slope displacement, $W_N = \boxed{0}$ — as always.

Step 3. Friction: $f = \mu_k N = \mu_k mg\cos 30° = 0.3 \times 20 \times 0.866 = 5.2$ N against the motion: $W_{\rm fric} = -5.2 \times 4.0 = \boxed{-20.8\ \mathrm{J}}$ .

Step 4. Net: $40 - 20.8 = \boxed{+19.2\ \mathrm{J}}$ — the block arrives at the bottom carrying 19.2 J of new motion energy; the missing 20.8 J became heat along the surface. The two-way gravity check (Step 1) is the exam's gift: if your two answers disagree, a sine and cosine got swapped somewhere. Timing note: ~3 min. NEET phrases this as "ratio of work by friction to work by gravity" — do both symbolically first ( $-\mu\cos\alpha/\sin\alpha = -\mu\cot\alpha$ ) and the arithmetic collapses.

Exam traps

Effort counted as work. Holding anything stationary: zero. Carrying anything horizontally at constant height: zero by gravity and by your vertical hold. If the marker sees $W = mgd$ for a horizontal carry, the module has failed.

The angle taken from the wrong pair. $\theta$ lives between the force and the displacement — not between the force and the vertical, not the incline angle by reflex. Draw both arrows tail-to-tail before reaching for the cosine.

Friction stereotyped as negative. Friction's work is $-f d$ only when it opposes the displacement. Belt-feeding, a block riding on an accelerating truck bed — friction along the motion, positive work. Compute the angle, not the reputation.

Normal-force amnesia (and its exceptions). $W_N = 0$ for anything sliding along a fixed surface. But a normal force from a moving surface (a lift floor, a piston) is not perpendicular to the object's displacement and works honestly — the lift floor does $+mgh$ -scale work on you every ride up.

Forgetting that the angled pull edits the friction. $F\sin\theta$ changes $N$ , hence $f = \mu N$ , before any work is computed. Worked example 3's decision point is the most-set trap in this module.

Net work versus work by a force. "Work done on the block" without naming a force means the net work — sum the signed entries, or dot the net force with $\vec d$ . Answering with one force's entry loses the mark even when arithmetically perfect.

Joules bookkeeping. $1$ J $= 1$ N·m $= 1\ \mathrm{kg\,m^2/s^2}$ . MCQ unit distractors offer N/m and kg·m/s; the dimensional audit (technique module) kills them in five seconds.

Which $g$ . CBSE/ISC: $9.8$ unless declared; JEE/NEET: usually $10$ ; AP allows $10$ for estimates; A-Level $9.81$ . State the value you use.

Origins · 1 minGaspard-Gustave de Coriolis and how this was discovered

Gaspard-Gustave de Coriolis

1792–1843 · École Polytechnique, Paris

The word "work" itself is his. In Du calcul de l'effet des machines (1829), Coriolis — an engineering professor obsessed with what steam engines, waterwheels and horses actually deliver — proposed the quantity force-times-distance as the universal currency of mechanical effect, named it travail (work), and paired it with the matching definition of kinetic energy as $\tfrac12 mv^2$ (fixing the factor of a half that earlier vis viva arguments had left loose). His motivation was ruthlessly practical: mills and engines should be priced by the work they do, not by the force they exert — a horse pushing a stalled cart all day earns nothing. The same man, six years later, described the deflection of motion in rotating frames that bears his name; but it is his ledger of travail that every energy chapter on Earth still opens, this one included.

History

1686

— Leibniz champions vis viva (mv²) against Descartes' momentum: the first fight over motion's true currency.

1829

— Coriolis defines work as force × distance and kinetic energy as ½mv², naming the trade between them.

1843

— Joule's paddle-wheel experiments price heat in units of mechanical work: energy crosses between ledgers.

1847

— Helmholtz's conservation-of-energy memoir makes the work ledger universal across mechanics, heat and electricity.

1889

— The joule is adopted as the unit of work and energy, honouring the man who balanced the books.

Going further

Texts. NCERT Physics Class 11, Ch. 6 (Work, Energy and Power) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 7 (the work and dot-product treatment this module downshifts).

Provenance. All results are standard. A verifier should recompute: the crate audit ( $+400, -200, 0, 0 \Rightarrow +200$ J); the crane pair ( $\pm 30$ kJ, net 0); example 3 ( $N = 26$ N, $f = 6.5$ N, $+256 - 52 = +204$ J); example 4 ( $+40$ J both ways, $-20.8$ J, net $+19.2$ J); the two-face identity on $\vec F = (30,40)$ , $\vec d = (8,-2)$ giving 160 J.

Prerequisites. Dot & Cross Products · Newton's Second Law: F = ma · Components & Resolution of Vectors. Leads to. Work by a Variable Force & F–x Graphs · Kinetic Energy & the Work–Energy Theorem · Power · Work as a Line Integral (UG deep dive).

If carrying a suitcase does no work on it, why do I get so tired?

Your muscles are terrible at doing nothing. Holding a force statically still requires muscle fibres to fire, slip microscopically and re-fire, burning chemical energy into heat the whole time — biology's cost of maintaining tension. Physics' $W = Fd\cos\theta$ counts only energy delivered to the suitcase through its motion, and that is genuinely zero (no displacement, or displacement perpendicular to your hold). A table holding the same suitcase gets equally "tired" — not at all — which shows the fatigue is about muscles, not about the physics of the task.

How can work be negative? What does negative energy transfer even mean?

Negative work means the force drained energy from the object's motion rather than feeding it. Friction under a sliding box pulls backward while the box moves forward: every metre, it converts some of the box's energy of motion into heat — the box slows. Gravity on a rising ball does negative work, banking the ball's speed as height. The sign is direction-of-flow bookkeeping: positive into the motion, negative out of it. Nothing mystical — a bank withdrawal, not antimatter.

Why doesn't the time taken matter? Dragging the crate fast or slow feels very different.

Work counts how much energy moved, not how quickly. The 400 J from example 1 is the same whether the drag took five seconds or five minutes; what changes is the rate of transfer — 80 W versus about 1.3 W — and that rate is a separate named quantity, power, with its own module ahead. Exams exploit the split: "how much work" never needs the time; "what power" always does.

Does the normal force really never do work?

On an object sliding along a fixed surface — never: the force is perpendicular to every step of the motion. But the rule is about geometry, not about the force's name. Stand in a rising lift: the floor's normal force points up and your displacement is up — it does massive positive work on you. A ball bouncing off a moving bat, a piston compressing gas: normal forces from moving surfaces work all the time. Check the angle between the force and the actual displacement, and the rule never misleads.

Why does gravity's work ignore the path? Friction doesn't.

Gravity is a fixed arrow: constant magnitude, constant direction. Over any wiggly route, only the vertical component of each step gets billed, and vertical components add up to the net height change — horizontal wanderings contribute nothing. Friction has no fixed arrow: it turns to oppose the motion at every step, so every metre of path, in any direction, gets billed at full rate. Fixed-arrow (and spring-like) forces are endpoint-billers — the "conservative" family that makes potential energy possible; motion-opposing forces are odometer-billers, and their toll is unrecoverable heat.

Is work a vector? It has a sign, and force and displacement are both vectors.

A scalar, with a sign — like a bank balance, not like an arrow. The dot product deliberately destroys direction information: it answers "how much of one vector lies along the other" with a single number. That is its power: works from different forces add as ordinary signed numbers with no components to track, and the total survives any rotation of your axes. The cross product is the dot's sibling that keeps directional character — that one builds torque, in the rotation chapter.

What happens when the force changes as the object moves — a spring, say?

Then $W = Fd\cos\theta$ cannot be applied in one gulp, because there is no single $F$ . The fix is the next module's whole business: chop the path into slices small enough that $F$ is nearly constant on each, work out each slice's sliver of work, add. Graphically the answer is beautiful — the work becomes the area under the force–position graph — and the spring's triangle area will hand you $\tfrac12 kx^2$ without any calculus.