Work done by variable force — integration: Worked examples

Before anything else, one reminder about the symbols we use, so nobody is lost from line one:

The scenario matrix

Every variable-force work problem is one (or a blend) of these cells. The right column names the example that covers it.

#	Case class	What's tricky about it	Covered by
C1	Force from a formula, positive throughout	routine integrate-and-plug	Ex 1
C2	Force negative throughout (opposes motion)	sign of the answer	Ex 2
C3	Force changes sign mid-path	area above and below axis cancel	Ex 3
C4	Force given only as a graph (piecewise)	compute geometric area, no formula	Ex 4
C5	Spring / Hooke's law — "by spring" vs "against spring"	opposite signs, $\frac{1}{2} k x^{2}$	Ex 5
C6	Reversed limits (moving backwards)	swapping limits flips the sign	Ex 6
C7	Degenerate / limiting inputs ( $F = 0$ , zero displacement, $k \to \infty$ )	edge behaviour, sanity checks	Ex 7
C8	Force at an angle — the $cos θ$ dot product	only the along-motion component works	Ex 8
C9	Real-world word problem (stickier floor)	translating words → integral	Ex 9
C10	Exam twist: inverse-square force to infinity	improper integral, limit at $\infty$	Ex 10

Now we hit every cell.

C1 — Force from a formula, positive throughout

Worked example Ex 1 — a growing force

$F (x) = (4 x + 3) N$ pushes a block from $x = 0$ to $x = 5 m$ , force always along motion. Find $W$ .

Forecast: Guess first — the force starts at $3$ N and ends at $23$ N. The average is somewhere near $13$ N over $5$ m, so expect roughly $65$ J. Hold that number.

Write $W = \int_{0}^{5} (4 x + 3) d x$ . Why this step? $F$ depends on $x$ , so a single " $F d$ " is a lie; we slice-push-sum, which is the integral.
Antiderivative term-by-term: $\int 4 x d x = 2 x^{2}$ , $\int 3 d x = 3 x$ . So $W = [2 x^{2} + 3 x]_{0}^{5}$ . Why this step? The antiderivative is the running-total-of-area function; evaluating it at the ends gives the net area — the Fundamental Theorem of Calculus.
Plug: $(2 \cdot 25 + 3 \cdot 5) - (0) = 50 + 15 = 65 J$ .

Verify: Average force $= \frac{3 + 23}{2} = 13$ N (valid because $F$ is linear), times $5$ m $= 65$ J. Matches step 3 and our forecast. Units: N·m = J. ✓

C2 — Force negative throughout

C3 — Force changes sign mid-path

This is the one students miss: part of the journey the force helps, part it hinders. See the figure — the green area (helping) and red area (hindering) partly cancel.

Figure — Work done by variable force — integration

Worked example Ex 3 — help then hinder

$F (x) = (6 - 3 x) N$ , body moves $x = 0 \to x = 4 m$ . Find $W$ . Where does the force flip sign?

Forecast: $F = 0$ when $6 - 3 x = 0 \Rightarrow x = 2$ . Before $x = 2$ , $F > 0$ (helps); after, $F < 0$ (hinders). The two areas may nearly cancel — expect something small.

$W = \int_{0}^{4} (6 - 3 x) d x = [6 x - \frac{3}{2} x^{2}]_{0}^{4}$ . Why this step? One integral automatically handles both signs — positive area above the axis, negative below, added with their signs. No need to split.
$= (24 - \frac{3}{2} \cdot 16) - 0 = 24 - 24 = 0 J$ . Why this step? The helping triangle and hindering triangle happen to be equal here, so the net work is exactly zero.

Verify — split and check:

$0 \to 2$ : triangle $\frac{1}{2} (2) (6) = + 6$ J (above axis, green).
$2 \to 4$ : triangle $\frac{1}{2} (2) (- 6) = - 6$ J (below axis, red).
Sum $= 0$ J. ✓ Matches step 2, confirming the "areas cancel" reading.

C4 — Force given only as a graph

C5 — Spring: "by spring" vs "against spring"

Worked example Ex 5 — the two-sign spring

A spring has $k = 200 N/m$ . It is stretched from $x = 0$ to $x_{0} = 0.1 m$ . (a) Work done by you pulling it. (b) Work done by the spring.

Forecast: These must be equal and opposite — you pour energy in, the spring resists, storing it. See Hooke's law and spring potential energy.

The spring force is $F_{spring} = - k x$ (pulls back toward $x = 0$ ). You must apply $F_{you} = + k x$ to balance it. Why this step? Defining which force we integrate is the whole battle here.
(a) $W_{you} = \int_{0}^{0.1} k x d x = \frac{1}{2} k x_{0}^{2} = \frac{1}{2} (200) (0.1)^{2} = 1 J$ . Why this step? Positive force along positive displacement → positive work, stored as spring energy.
(b) $W_{spring} = \int_{0}^{0.1} (- k x) d x = - \frac{1}{2} k x_{0}^{2} = - 1 J$ . Why this step? Spring force opposes the stretch, so it does negative work.

Verify: $W_{you} = + 1$ J, $W_{spring} = - 1$ J — equal and opposite as forecast, sum zero (no net kinetic gain if pulled slowly). Geometrically each is the triangle $\frac{1}{2} x_{0} \cdot (k x_{0}) = \frac{1}{2} (0.1) (20) = 1$ J. ✓

C6 — Reversed limits (moving backwards)

C7 — Degenerate & limiting inputs

Worked example Ex 7 — the edge cases (never trip on these)

Check three degenerate scenarios.

Forecast: Each should collapse to an "obvious" answer; if the machine is right, it agrees.

Zero force, $F (x) = 0$ over any path: $W = \int_{a}^{b} 0 d x = 0$ J. Why this step? No push ⇒ no work, regardless of distance. Area under the $x$ -axis line is zero.
Zero displacement, any $F$ , $a = b$ : $W = \int_{a}^{a} F d x = 0$ J. Why this step? Width of every slice is zero ⇒ no area. You can strain forever holding a wall: zero mechanical work.
Stiff-spring limit: for $W = \frac{1}{2} k x_{0}^{2}$ , hold the energy fixed at $E$ and let $k \to \infty$ . Then $x_{0} = 2 E / k \to 0$ . Why this step? An infinitely stiff spring barely stretches; to store the same energy it needs almost no displacement. Consistent, not paradoxical.

Verify: (1) $0$ J, (2) $0$ J, (3) $x_{0} = 2 E / k \to 0$ as $k \to \infty$ . All three behave as physical intuition demands. ✓

C8 — Force at an angle (the $cos θ$ )

Only the part of the force along the motion does work. The figure resolves $F$ into a useful (along-path) piece and a wasted (perpendicular) piece — see Dot product and components of vectors.

Worked example Ex 8 — pushing a sled at an angle

You pull a sled a distance $d = 10$ m with a rope at $θ = 6 0^{\circ}$ above horizontal. The tension is not constant: $T (x) = (2 x + 5) N$ where $x$ is distance travelled. Motion is horizontal. Find $W$ .

Forecast: Only $cos 6 0^{\circ} = \frac{1}{2}$ of the tension pulls forward; the vertical part does zero work. So expect half of what a straight pull would give.

General work: $d W = F \cdot d r = T (x) cos θ d x$ . Why this step? The dot product keeps only the component of force along the displacement; $cos 6 0^{\circ} = \frac{1}{2}$ .
$W = \int_{0}^{10} (2 x + 5) (\frac{1}{2}) d x = \frac{1}{2} [x^{2} + 5 x]_{0}^{10}$ . Why this step? $cos θ$ is a constant here, so it pulls outside the integral.
$= \frac{1}{2} (100 + 50) = \frac{1}{2} (150) = 75 J$ .

Verify: A straight ( $θ = 0$ ) pull would give $\int_{0}^{10} (2 x + 5) d x = 150$ J; at $6 0^{\circ}$ we get half, $75$ J, as forecast. The vertical component does $0$ J (perpendicular to horizontal motion). ✓

C9 — Real-world word problem

C10 — Exam twist: force to infinity (improper integral)

Worked example Ex 10 — escaping to infinity against

$1/ x^{2}$ A repulsive force $F (x) = \frac{c}{x ^{2}}$ with $c = 8 N\cdotm^{2}$ pushes a particle outward along $x$ . Find the work done by this force as the particle moves from $x = 2$ m all the way to $x = \infty$ .

Forecast: The force weakens fast ( $\propto 1/ x^{2}$ ), so even though the distance is infinite, the area under the curve may be finite. Expect a modest number.

$W = \int_{2}^{\infty} \frac{c}{x ^{2}} d x$ . Why this step? Still slice-push-sum, but the upper limit is infinite — an improper integral, handled by taking a limit.
Antiderivative of $x^{- 2}$ is $- x^{- 1}$ : $W = c [- \frac{1}{x}]_{2}^{\infty} = c (lim_{b \to \infty} (- \frac{1}{b}) - (- \frac{1}{2}))$ . Why this step? $- 1/ x$ is the running-area function; we evaluate the top end as a limit.
As $b \to \infty$ , $- 1/ b \to 0$ . So $W = c (0 + \frac{1}{2}) = \frac{c}{2} = \frac{8}{2} = 4 J$ . Why this step? The tail contributes nothing in the limit; the whole infinite journey costs a finite $4$ J.

Verify: $\int_{2}^{\infty} 8 x^{- 2} d x = 8 \cdot \frac{1}{2} = 4$ J — finite despite infinite range, exactly as forecast. This is the mechanism behind escape energy in Conservative forces and potential energy. ✓

Recall Quick self-test (cover the answers)

Ex 3 net work, and why? ::: $0$ J — equal helping and hindering triangles cancel. Ex 6 vs Ex 1 relationship? ::: Reversed limits flip the sign: $- 65$ J vs $+ 65$ J. Ex 8 why factor of $\frac{1}{2}$ ? ::: $cos 6 0^{\circ} = \frac{1}{2}$ ; only the along-motion component works. Ex 10 why is the answer finite? ::: Force $\propto 1/ x^{2}$ decays fast enough that the infinite-range area converges. Work over zero displacement? ::: Always $0$ J — every slice has zero width.

Connections

Parent: Work done by a variable force — the machine these examples exercise.
Work done by a constant force — the $F =$ const cell (Ex 4's rectangle piece).
Hooke's law and spring potential energy — Ex 5.
Work–Energy Theorem — turn any $W$ above into a change in kinetic energy.
Area under curves and the definite integral — the graph/area reading (Ex 3, 4).
Conservative forces and potential energy — Ex 10's improper integral is an escape-energy calculation.
Dot product and components of vectors — the $cos θ$ in Ex 8.

Worked examples — Work done by variable force — integration

The scenario matrix

C1 — Force from a formula, positive throughout

C2 — Force negative throughout

C3 — Force changes sign mid-path

C4 — Force given only as a graph

C5 — Spring: "by spring" vs "against spring"

C6 — Reversed limits (moving backwards)

C7 — Degenerate & limiting inputs

C8 — Force at an angle (the $cos θ$ )

C9 — Real-world word problem

C10 — Exam twist: force to infinity (improper integral)

Connections

Case Map