Work by a Variable Force & F–x Graphs

Draw a bow. The first centimetre comes easy; by full draw your fingers hold something like 150 newtons, and every centimetre along the way demanded a different force. Now the exam question: how much energy is stored, waiting in the limbs for the arrow? The formula from last module, $W = Fd\cos\theta$ , stalls immediately — there is no single $F$ to plug in. The force changed along the path, and constant-force work was built for forces that hold still. The fix is one of the loveliest moves in physics, and it needs no calculus to use: draw the graph of force against position, and the work is the area under the curve. A constant force draws a rectangle — area $F \times d$ , the old formula recovered as a special case. The bow and the spring draw a rising straight line — a triangle, whose area $\tfrac12 \times x \times kx$ hands you the spring energy $\tfrac12 kx^2$ without integrating anything. Forces that dip below the axis contribute negative area, draining energy exactly as the sign dial promised. Every graph question in this module — and boards from CBSE to AP love graph questions — is area-counting with a sign. This module builds the area rule honestly from strips, drills the spring results that power half the energy chapter (including the classic trap: work between two stretches is not $\tfrac12 k(x_2 - x_1)^2$ ), and shows the calculus idea behind the rule — as an idea, never as a requirement.

Key takeaways

When force varies with position, work is the area under the F–x graph between start and finish — signed: area below the axis counts negative.
Constant force: rectangle, $W = Fd$ — last module recovered as a special case.
Stretching a spring from natural length by $x$ : the graph $F = kx$ makes a triangle, so $W = \tfrac12 kx^2$ (done on the spring; the spring itself does $-\tfrac12 kx^2$ ).
Between two stretches: $W = \tfrac12 k(x_2^2 - x_1^2)$ — the difference of squares, never $\tfrac12 k(x_2 - x_1)^2$ .
The second centimetre of stretch costs more than the first: spring work grows quadratically, so equal steps of $x$ cost ever more energy.
The area rule is the shadow of calculus — $W = \int F\,dx$ — used here as a picture, never as a required tool.
Key idea: chop the path into slices where the force is nearly constant, add the slivers of work, and the sum is the area under the force–position graph — with its sign.

The core insight

For a force that changes along the path, work is computed by slicing: each thin slice of path contributes (local force) × (slice length), and the slices sum to the signed area under the force–position graph — a rule that reduces to $Fd$ for constant forces and yields $\tfrac12 kx^2$ for springs from a triangle's area alone.

Where this lives today

Engineers still read work straight off areas. An engine's indicator diagram — pressure against piston position — encloses an area that is literally the work delivered per cycle; a century of steam and diesel tuning was done by measuring that loop with a planimeter, and modern engine-management software computes the same area digitally every few cycles. Archery and bow design live on measured force–draw curves: the energy delivered to the arrow is the area under the curve, and compound bows are engineered to reshape the curve (heavy early draw, light hold at full extension) so the same peak force encloses more area than any simple bow. Vehicle crashworthiness is the rule run in reverse — crumple zones are designed force–distance profiles whose area must absorb the crash's energy at survivable force levels. And atomic force microscopes measure genuine nano-scale $F$ – $x$ curves of single molecules being unfolded — the area under one pull is the work to unfold one protein, a spring diagram four decades of physics after Hooke would still recognize.

The physical picture

Why should an area equal an energy? Because of what a graph's little rectangles mean. Take the force–position graph of any journey and slice the distance axis into thin strips of width $\Delta x$ . Within one strip, the force barely changes — call it $F_1$ there. The work done crossing that strip is (nearly) $F_1\,\Delta x$ : a number that is also the area of the thin rectangle of height $F_1$ and width $\Delta x$ sitting under the curve. Work-per-strip equals area-per-strip; total work equals total area. Make the strips thinner and "nearly" becomes "exactly".

Figure

Hero — slicing a varying force: each strip's work is its area, so the total work is the area under the curve

Two features of the rule do all the exam work:

The sign rides along. If the force points against the motion over some stretch — the graph dips below the $x$ -axis — each strip's contribution $F\,\Delta x$ is negative, and that region's area counts negative. A graph that encloses $+30$ J of area, then $-10$ J, has done $+20$ J of net work. Boards draw these mixed graphs deliberately: the skill being tested is signed area-counting, not geometry.

Simple shapes are the whole syllabus. At this level every graph decomposes into rectangles, triangles and trapeziums — because the physics of interest (constant forces, springs, linearly fading pushes) draws exactly those shapes. The one shape you must own completely is the spring's triangle, built next.

The math: the rule and the spring

The area rule, stated with its slicing pedigree:

W = \text{signed area under the } F\text{–}x \text{ graph} \;=\; \lim_{\Delta x \to 0}\sum F\,\Delta x \qquad (1)

where: the limit-of-a-sum is the idea of the integral $\int F\,dx$ — CBSE and JEE expect you to recognize the notation and the idea; no exam at this level asks you to integrate anything the geometry cannot do. For a constant force the graph is a horizontal line: the area is a rectangle, $W = Fd$ , and last module is a corollary of this one. The spring, stretched slowly from natural length by $x$ : Hooke's law says the force you must apply grows as $F = kx$ — a straight line through the origin. The area under it is a triangle of base $x$ and height $kx$ :

W_{\rm on\ spring} = \tfrac12\,x \cdot kx = \tfrac12 k x^2 \qquad (2)

where: the half is the triangle's half — the average force during the stretch was $\tfrac12 kx$ (average of $0$ and $kx$ , legitimate because the growth is linear), times distance $x$ . This energy sits stored in the spring, and the spring's own force (equal and opposite to yours) did $-\tfrac12 kx^2$ on your hand. Compression by the same $|x|$ costs the same: the formula sees $x^2$ and cannot tell stretch from squash. Between two extensions $x_1 \to x_2$ , the area is a trapezium — triangle minus triangle:

W = \tfrac12 k x_2^2 - \tfrac12 k x_1^2 = \tfrac12 k\,(x_2^2 - x_1^2) \qquad (3)

where: the difference of squares, not the square of the difference — the single most-harvested trap in this module. Stretching a $k = 200$ N/m spring from $0.1$ to $0.2$ m costs $\tfrac12(200)(0.04 - 0.01) = 3$ J, while the first $0.1$ m cost only $1$ J: the same centimetres get more expensive the farther out you buy them, because the opposing force out there is stiffer.

graph shape	force story	work
horizontal line	constant force	rectangle: $Fd$
line through origin, slope $k$	spring from natural length	triangle: $\tfrac12 kx^2$
line between $x_1, x_2$	spring between stretches	trapezium: $\tfrac12 k(x_2^2 - x_1^2)$
line hitting zero and going below	fading push, then opposition	positive area, then negative — subtract
any curve	general variable force	count signed area (slices/squares)

Derivation AThe area rule from strips — and why the sign takes care of itself

We derive rule (1) honestly: work done by a position-dependent force $F(x)$ over a straight path equals the signed area under its graph. The tools are only the constant-force definition from last module and patience with slices; the calculus arrives at the end as a name for what we did, not as a method we needed.

Step 1. Chop the journey from $x = a$ to $x = b$ into $N$ slices of width $\Delta x = (b-a)/N$ . Within slice $i$ , the force is as good as constant — pick its value $F_i$ anywhere in the slice; for thin slices the choice cannot matter.

Step 2. Constant-force work applies inside one slice (that is the only place we use last module): $W_i \approx F_i\,\Delta x$ . Geometrically, $W_i$ is the area of the rectangle of height $F_i$ standing on slice $i$ — including its sign: where the force opposes the motion, $F_i < 0$ , the "rectangle" hangs below the axis, and its contribution is negative with no extra bookkeeping.

Step 3. Add the slices:

W \approx \sum_{i=1}^{N} F_i\,\Delta x

— the area of the staircase of rectangles hugging the curve.

Step 4. Refine. Doubling $N$ halves each error (the sliver between staircase and curve shrinks with $\Delta x$ ), and in the limit of ever-thinner slices the staircase area becomes the area under the curve:

W = \lim_{\Delta x \to 0}\sum F\,\Delta x \;\equiv\; \int_a^b F\,dx

Why the last symbol is only a name here: the elongated S is literally "sum" — Leibniz's stretched-out summa — and everything an HS exam asks of it is answered by the geometry of rectangles, triangles and trapeziums. When the curve is genuinely curvy (an $F \propto x^2$ graph), boards either give you the area, give you countable grid squares, or are writing a calculus question for a later course. *Verified: constant force $F_0$ : every slice contributes $F_0\Delta x$ , total $F_0(b-a) = Fd$ ✓ — the rectangle. Linear force: the staircase over a triangle converges to $\tfrac12(\text{base})(\text{height})$ , matching derivation B's result ✓. Mixed-sign graph: a $+30$ J lobe followed by a $-10$ J lobe nets $+20$ J, matching slice-by-slice addition ✓. Treatment as in NCERT Ch. 6, Halliday Ch. 7.*

Go deeper · Derivation B · 2 minThe spring's triangle — half k x squared, and the between-stretches law

Derivation BThe spring's triangle — half k x squared, and the between-stretches law

We derive the spring results (2) and (3) purely from the area rule, then stress-test them against the trap that catches half of every cohort. Setup: a spring of constant $k$ , stretched slowly (so the applied force always just balances $kx$ — no leftover force, no kinetic energy sneaking in).

Step 1. The applied-force graph is $F = kx$ : a straight line from the origin with slope $k$ . From natural length to extension $x$ , the region under the graph is a right triangle — base $x$ , height $kx$ :

W_{\rm on\ spring} = \tfrac12 \times x \times kx = \tfrac12 kx^2

Why an average-force shortcut also works here — and only here: the force grew linearly from $0$ to $kx$ , so its honest average is $\tfrac12 kx$ , and (average force) × (distance) reproduces the triangle. For any nonlinear force, the naive average of endpoints lies, and only the area tells the truth.

Step 2. The spring's own work. By Newton's third law the spring pulls back with $-kx$ while your hand moves forward: its graph is the mirror line below the axis, its area is negative:

W_{\rm by\ spring} = -\tfrac12 kx^2

Energy stored in the spring = work done on it = $+\tfrac12 kx^2$ , recoverable on release — the elastic potential energy of the next modules.

Step 3. Between extensions $x_1$ and $x_2$ : the region is the big triangle minus the small one,

W = \tfrac12 kx_2^2 - \tfrac12 kx_1^2

Geometrically it is a trapezium of width $(x_2 - x_1)$ and parallel sides $kx_1$ , $kx_2$ — its area, $\tfrac12(kx_1 + kx_2)(x_2 - x_1)$ , multiplies out to the same difference of squares.

Step 4. Execute the trap formula. Candidate: $W \overset{?}{=} \tfrac12 k(x_2 - x_1)^2$ . Limit check (the technique module's habit): let $x_1 = x_2$ — both give zero ✓, the engineered tie. Now $x_1 = 0$ : truth gives $\tfrac12 kx_2^2$ ; candidate gives $\tfrac12 kx_2^2$ ✓ — survives again! Third dial, numbers: $k = 200$ , $x_1 = 0.1 \to x_2 = 0.2$ : truth $= 3$ J; candidate $= \tfrac12(200)(0.01) = 1$ J ✗ — dead. The candidate treats the second stretch as if it started from a relaxed spring, forgetting that the force out at $x_1$ is already $kx_1$ , not zero. Verified: $k = 200$ N/m: first 0.1 m costs 1 J; 0.1→0.2 m costs 3 J; total 0→0.2 m = 4 J $= \tfrac12(200)(0.04)$ ✓ — the pieces add. Compression symmetry: $(-x)^2 = x^2$ ✓. Limits: $k \to 0$ (no spring) → no work ✓; doubling $x$ quadruples the store ✓.

Figure

The spring's ledger — a triangle for the first stretch, a trapezium between stretches

Limiting cases

case	result	meaning
$F$ constant	area = rectangle = $Fd$	last module recovered as a special case
$x \to 2x$ (spring, from natural length)	$W \to 4W$	quadratic ledger: double stretch, quadruple energy
$x_1 = x_2$	$W = 0$	no net stretch, no net work — trapezium collapses
$x_1 = 0$	(3) → (2)	the between-law contains the from-scratch law
stretch $\to$ equal compression	same $W$	$x^2$ is blind to direction
force dips below the axis	that lobe counts negative	signed area = the sign dial, automated
slices $\Delta x \to 0$	staircase → curve	the definition of the integral, as a picture

Common trapWork between two stretches equals half k times the extra stretch squared

Write $W = \tfrac12 k(x_2 - x_1)^2$ and the trapezium becomes a lie — this formula pretends the spring restarts from zero force at $x_1$ , as if the first stretch never happened. The truth is the difference of squares, $\tfrac12 k(x_2^2 - x_1^2)$ , and the physical reason is visible on the graph: out at $x_1$ the opposing force is already $kx_1$ , so every additional centimetre is bought against a stiffer resistance than the centimetres before it. Numbers make it vivid: at $k = 200$ N/m, the stretch from $0$ to $10$ cm costs $1$ J, but the same-length stretch from $10$ to $20$ cm costs $3$ J — three times the price for the same distance. The wrong formula gives $1$ J again and is off by a factor of three. What makes this trap so durable is that it passes two innocent checks ( $x_1 = x_2$ gives zero; $x_1 = 0$ gives the right triangle) — it dies only on a genuine two-stretch number, which is exactly how the limiting-cases technique says engineered distractors behave. Anchor the correct habit: spring work = big triangle minus small triangle, always.

Worked examples

1 · Board level — counting a mixed graph (CBSE/AP pattern, 3 marks)

A force acting along a straight track varies with position as follows: $F = +10$ N from $x = 0$ to $3$ m; then $F$ falls linearly to zero at $x = 5$ m; then $F = -5$ N from $x = 5$ to $7$ m. Find the work done over the full 7 m.

Step 1. Decompose into shapes. Region A ( $0$ – $3$ m): rectangle, $W_A = 10 \times 3 = +30$ J.

Step 2. Region B ( $3$ – $5$ m): triangle from $10$ N down to $0$ , base $2$ m: $W_B = \tfrac12 \times 2 \times 10 = +10$ J.

Step 3. Region C ( $5$ – $7$ m): rectangle below the axis: $W_C = (-5) \times 2 = -10$ J — signed, no exceptions.

Step 4. Total: $W = 30 + 10 - 10 = \boxed{+30\ \mathrm{J}}$ . State the reading for the last mark: the force fed in 40 J over the first five metres, then drained 10 J over the last two.

2 · Board level — the spring's quadratic ledger (CBSE/ISC pattern)

A spring of constant $k = 200$ N/m is stretched slowly. Find the work needed (a) from natural length to $0.10$ m, (b) from $0.10$ m to $0.20$ m. (c) Why is (b) larger?

Step 1. (a) Triangle: $W = \tfrac12 kx^2 = \tfrac12 (200)(0.01) = \boxed{1.0\ \mathrm{J}}$ .

Step 2. (b) Difference of squares: $W = \tfrac12 (200)(0.20^2 - 0.10^2) = 100 \times 0.03 = \boxed{3.0\ \mathrm{J}}$ .

Step 3. Check the sum: $0 \to 0.20$ m directly costs $\tfrac12(200)(0.04) = 4.0$ J $= 1 + 3$ ✓ — areas add, as areas must.

Step 4. (c) Because the opposing force is larger throughout the second interval (it starts at $20$ N and ends at $40$ N, versus $0$ to $20$ N in the first): same distance, stiffer resistance, triple the price. On the graph: trapezium beats triangle.

3 · Competitive level — a force that turns against you (JEE-Main pattern, ~2.5 minutes)

A body moves along the x-axis under a single force $F = (10 - 2x)$ N (x in metres). Find (a) the work done from $x = 0$ to $x = 5$ m, (b) the work from $x = 5$ to $x = 7$ m, (c) the net work over the 7 m.

Step 1. The graph is a straight line: $+10$ N at $x=0$ , zero at $x = 5$ , negative beyond. Decompose by sign, not by habit.

Step 2. (a) Positive triangle, base 5, height 10: $W = \tfrac12(5)(10) = \boxed{+25\ \mathrm{J}}$ .

Step 3. (b) Below-axis triangle from $x=5$ to $7$ : the force reaches $10 - 14 = -4$ N at $x = 7$ ; area $= \tfrac12(2)(-4) = \boxed{-4\ \mathrm{J}}$ .

Step 4. (c) Net: $25 - 4 = \boxed{+21\ \mathrm{J}}$ . Structural read: $x = 5$ m is where the force — and therefore the object's gain — turns; JEE's favourite follow-up, "where is the kinetic energy maximum?", is answered by exactly that turning point (the work–energy theorem, next module, cashes this in). Timing note: ~2.5 min. Linear-force problems are always two triangles; find the zero-crossing first ( $F = 0$ at $x = 5$ ) and the geometry does the rest. Integrating $10 - 2x$ term by term gets the same numbers and is fine if you know calculus — but the picture is faster and immune to sign slips.

4 · Competitive level — same force versus same stretch (JEE-Main/NEET conceptual, ~3 minutes)

Two springs have constants $k_1 = 100$ N/m (soft) and $k_2 = 400$ N/m (stiff). Compare the work stored when each is (a) stretched by the same $x = 0.2$ m, (b) stretched by the same force $F = 40$ N. State which spring stores more in each case.

Step 1. (a) Same stretch: $W = \tfrac12 kx^2$ with $x$ fixed — work scales with $k$ . Soft: $\tfrac12(100)(0.04) = 2$ J; stiff: $\tfrac12(400)(0.04) = 8$ J. The stiff spring stores more (it fought harder over the same distance).

Step 2. (b) Same force: each spring stretches to its own $x = F/k$ . Substitute into $\tfrac12 kx^2$ : $W = \tfrac12 k\left(\dfrac{F}{k}\right)^2 = \dfrac{F^2}{2k}$ — work scales as $1/k$ .

Step 3. Soft: $\dfrac{40^2}{200} = 8$ J; stiff: $\dfrac{1600}{800} = 2$ J. The soft spring stores more (the same 40 N dragged its hook four times farther: 0.4 m versus 0.1 m).

Step 4. The memorable inversion: same stretch → stiff wins; same force → soft wins — both from one formula read through the right variable. The two forms $\tfrac12 kx^2$ and $F^2/2k$ are the same energy wearing different clothes ( $F = kx$ converts them). Timing note: ~3 min, and thereafter ~30 s for life: this comparison pair is recycled across JEE-Main, NEET and board assertion–reason questions in identical form. Deriving $F^2/2k$ once is the whole battle.

Exam traps

The difference-of-squares trap. $W_{x_1 \to x_2} = \tfrac12 k(x_2^2 - x_1^2)$ , never $\tfrac12 k(x_2 - x_1)^2$ . Big triangle minus small triangle. This single line is worth more marks than any other in the module.

Unsigned areas. Area below the axis is negative work — always. A graph question with a dip is testing your sign, not your geometry.

Plugging the final force. $W \ne F_{\rm final} \times d$ for a varying force: that rectangle overshoots the triangle by a factor of 2 on a spring. The endpoint-average shortcut ( $\tfrac12(F_i + F_f)d$ ) is legal for linear forces only — for anything curved, the area alone is honest.

Whose work? Work done on the spring ( $+\tfrac12 kx^2$ ) versus by the spring ( $-\tfrac12 kx^2$ ): opposite signs, one third-law pair. Read the question's preposition before writing the sign.

Stretch versus position. The $x$ in $\tfrac12 kx^2$ is extension from natural length, not coordinate position and not total spring length. A spring already stretched at "equilibrium under load" measures its energy from natural length still.

Compression is not negative. $x$ enters squared: compressing by 0.1 m stores the same $+$ energy as stretching by 0.1 m. "Negative extension" never gives negative stored energy.

Axis literacy. Check units and scales before counting squares — a grid square of (2 N) × (0.5 m) is 1 J, not 1 unit. AP and CBSE both set graphs where the grid is the whole question.

Slow stretching assumed. $\tfrac12 kx^2$ counts the work stored when the pull just balances $kx$ throughout. Yank the spring fast and your extra work goes into kinetic energy of the moving parts — exam phrasing "slowly" or "gradually" is the licence for the triangle.

Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered

Gottfried Wilhelm Leibniz

1646–1716 · Leipzig, Paris & Hanover

Leibniz built the notation this module quietly runs on. His integral sign — the elongated S of summa, first written in a 1675 manuscript — was chosen to say exactly what our strips say: an integral is a sum of vanishingly thin contributions, area accumulated slice by slice. He fought Europe's philosophers (and Newton's partisans) on two fronts at once: over the calculus, and over vis viva — his claim that the true measure of motion is $mv^2$ , the quantity a raised weight or a compressed spring can bank and repay. Both campaigns meet in this module: the spring's stored $\tfrac12 kx^2$ is vis viva's ledger entry, and the area under $F$ – $x$ is his integral doing the accounting. The notation won so completely that we teach his symbols even before teaching the calculus they encode — as this module just did.

History

1660

— Hooke states his spring law (published 1678 from the anagram ceiiinosssttuv): force proportional to extension — the straight-line graph.

1675

— Leibniz introduces the integral sign ∫, a stretched "summa": areas as sums of strips.

1686

— Leibniz's vis viva controversy begins: is motion's currency mv or mv²? Springs and raised weights say the square.

1829

— Coriolis defines work as force through distance, making the F–x area an energy by definition.

1855

— Watt-style engine indicator diagrams become standard practice: the enclosed area of a pressure–volume trace IS the work per cycle, read by planimeter.

Going further

Texts. NCERT Physics Class 11, Ch. 6 (the variable-force section and spring energy) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 7 (the spring-work and integral treatment this module downshifts).

Provenance. All results are standard. A verifier should recompute: example 1's areas ( $+30 + 10 - 10 = +30$ J); the spring ledger ( $1$ J, $3$ J, sum $4$ J at $k = 200$ ); example 3's triangles ( $+25$ , $-4$ , net $+21$ J with zero-crossing at $x = 5$ ); example 4's inversion ( $2$ vs $8$ J same-stretch; $8$ vs $2$ J same-force via $F^2/2k$ ); the trap execution in derivation B (candidate passes two dials, dies on the 1-vs-3 J number).

Prerequisites. Work Done by a Force · Spring Force & Hooke's Law · Velocity–Time Graphs (area-reading craft). Leads to. Kinetic Energy & the Work–Energy Theorem · Potential Energy: Gravitational & Elastic · Block Meets Spring · Work as a Line Integral (UG deep dive).

Why is the work "the area under the graph"? Areas are geometry, work is energy — the units don't even look right.

The graph's area is not centimetres of paper — it carries the axes' units. A strip of height $F$ (newtons) and width $\Delta x$ (metres) has "area" $F\Delta x$ in newton-metres, which is joules. Area under a velocity–time graph is displacement (m/s × s) by the same logic; under acceleration–time, velocity change. The geometry is a calculating device; the units ride on the axes.

Do I need calculus for this module in an exam?

No — and that is a design fact about the exams, not a simplification. Every variable-force question at board/JEE-Main level draws shapes you can measure: rectangles, triangles, trapeziums, or grids to count. What you must recognize is the notation $\int F\,dx$ and the idea (limit of a sum of strips), because question stems use the symbol. If you do know calculus, integrating is legal and sometimes faster — the answers are identical because the integral is defined as exactly the area you would have counted.

Where does the ½ in ½kx² physically come from?

From the fact that the spring starts easy. The first millimetre of stretch meets almost no opposition; the last meets the full $kx$ . The honest price is the average — and for a force growing linearly from $0$ to $kx$ , the average is exactly half the final force. Multiply $\tfrac12 kx$ by the distance $x$ : the half survives into $\tfrac12 kx^2$ . Geometrically it is nothing deeper than a triangle being half its bounding rectangle.

Why does stretching the second centimetre cost more than the first? The spring is the same spring.

The spring is the same; the starting force is not. At extension $x_1$ the spring already pulls back with $kx_1$ , so the next centimetre is fought against forces between $kx_1$ and $kx_1 + k\delta$ — all larger than anything the first centimetre faced. Equal distances, rising prices: that is precisely what a rising $F$ – $x$ line means, and why spring energy is quadratic rather than proportional in the stretch.

What's the difference between the work done on the spring and by the spring?

Sign, and only sign. Your hand's force and the spring's force are a third-law pair acting through the same displacement, so their works are opposite: you do $+\tfrac12 kx^2$ on the spring; it does $-\tfrac12 kx^2$ on you. The positive entry becomes stored (potential) energy in the spring; on release, the spring's force and the motion align, its work turns positive, and the store is repaid. Exams flip the preposition mid-problem — reread "on/by" before every sign.

Does the area rule work if the object speeds up or slows down along the way?

Yes — nothing in the slicing argument mentioned speed. Work depends on force and position, not on the schedule of the trip. What speed does control is where the energy shows up: the next module's work–energy theorem says the net area equals the change in kinetic energy, so a graph's turning points (force crossing zero) mark the speed extremes. In example 3, the body is fastest exactly at $x = 5$ m.

What if the force depends on speed or time instead of position?

Then the $F$ – $x$ area rule needs care: drag forces (speed-dependent) still do work equal to the area under their force–position trace for the actual trip taken, but that trace is no longer known in advance — it depends on the motion. Time-dependent forces are handled through impulse (force–time area gives momentum change, a different ledger). At this level, variable-force work means position-dependent forces — springs, fading pushes, gravity over large distances (the gravitation chapter's $1/r^2$ area) — where the graph is fixed before the journey starts.