Work and the Line Integral of Force

Carry a heavy suitcase 100 m along a station platform and your arm will ache — yet physics insists you did essentially no work on the suitcase. Lean on a stalled car with all your strength while it refuses to budge: still zero work. Now give a shopping trolley the gentlest nudge and let it roll a few metres — that tiny push did do work. Clearly the physics meaning of "work" is not effort. It is something sharper: energy transferred to an object by a force, through a displacement, counting only the part of the force that points along the motion. For a steady force and a straight path, one line of trigonometry settles the account: $W = \vec F \cdot \vec d = Fd\cos\theta$ . But real forces vary and real paths bend — a spring pulls harder as it stretches, a mountain road climbs and dips. The fix is one of the loveliest moves in physics: chop the path into pieces so small that each is straight and the force on it constant, apply the dot product to every piece, and add. In the limit of infinitely fine chopping the sum becomes the line integral $W = \int_C \vec F\cdot\mathrm{d}\vec r$ — the general definition of work, valid for any force along any curve. That integral is the gateway to the entire energy formulation of mechanics. Integrated against Newton's second law it yields the work–energy theorem; asked whether its value depends on the route or only the endpoints, it sorts every force into conservative and non-conservative and hands you potential energy. This article builds the machine from the suitcase up, runs honest numbers through it, and shows you exactly when a path matters and when it cannot.

Key takeaways

Work is energy transferred by a force acting through a displacement. Constant force, straight path: $W = \vec F\cdot\vec d = Fd\cos\theta$ , measured in joules ( $1\ \mathrm{J} = 1\ \mathrm{N\,m}$ ).
Only the component of force along the motion does work; the perpendicular component transfers nothing. The sign of $W$ is the sign of $\cos\theta$ : helper, spectator, or brake.
For a varying force on a curved path, chop–dot–add gives the line integral $W = \int_C \vec F\cdot\mathrm{d}\vec r$ — every piece of it is still $Fd\cos\theta$ applied locally.
Works are signed scalars and simply add: $W_{\text{net}} = W_1 + W_2 + \cdots$ , and the work–energy theorem promises $W_{\text{net}} = \Delta\!\left(\tfrac12 mv^2\right)$ .
Some forces (gravity, springs) give a path-independent integral — conservative forces, which admit a potential energy. Others (friction) charge by the metre travelled: path-dependent, no potential energy possible.
Key idea: work is the line integral $W=\int_C\vec F\cdot\mathrm{d}\vec r$ — it accumulates the along-path component of force over the actual route, and by the work–energy theorem it equals the change in kinetic energy.

The core insight

Work is the line integral $W=\int_C\vec F\cdot\mathrm{d}\vec r$ : it accumulates the component of force along the actual path, and by the work–energy theorem the net work equals the change in kinetic energy.

Where this lives today

The dot product in equation (5) is arguably the most-executed operation in modern engineering. Every video-game physics engine evaluates millions of $\vec F\cdot\vec v$ products per frame to track power flow; a robot arm's controller continuously dots motor forces with joint velocities to keep the machine from fighting itself. Sign-of-work reasoning is the beating heart of regenerative braking: when an electric car decelerates, its motor is reconfigured so the electromagnetic force on the wheels points against the motion — deliberate negative work — but instead of shedding that energy as brake heat, the motor runs as a generator and pumps it back into the battery. The control software is doing the crate audit of Worked Example 1 thousands of times per second: which way is the force, which way is the motion, which direction is the energy flowing? The zero-work case runs the sky. Satellites in circular orbit need no fuel to maintain speed precisely because gravity sits at $90^\circ$ to their velocity — mission designers treat "gravity does no work" as an engineering resource. In elliptical orbits the sign flips rhythmically: gravity does positive work on the infalling half and negative work on the outbound half, a perfectly balanced account closed over every orbit, for free, forever. And where paths genuinely matter — atmospheric drag on a re-entering capsule, tyre friction in lap-time simulation — engineers integrate $\vec F\cdot\mathrm{d}\vec r$ numerically along the actual trajectory: the chop–dot–add of Derivation A, run on silicon.

The physical picture

Picture a child towing a sled with a rope that slants upward. The rope's tension does two jobs at once: its horizontal part drags the sled forward, while its vertical part slightly un-weights the sled, easing it off the snow. Only the first job moves the sled along its path — only the first job transfers energy of motion. The second part, perpendicular to the displacement, could act all day and never speed the sled up. So the honest measure of work must answer one geometric question: how much of $\vec F$ lies along $\vec d$ ? Drop a perpendicular from the tip of the force vector onto the displacement direction and you get the answer — a shadow of length $F\cos\theta$ . Multiply that effective force by the distance moved and you have the work.

Figure

Hero — only the along-track part of the force does work

The cosine is a sign dial as much as a size dial, and it is worth reading like a physicist before any algebra. Think of an object's kinetic energy as a bank account and every force acting on it as a client. A force leaning even slightly forward along the motion ( $\theta < 90^\circ$ ) is making a deposit. A force leaning backward ( $\theta > 90^\circ$ ) is making a withdrawal — friction and brakes do very real, very negative work. And a force at exactly $90^\circ$ is a client who walks in, rearranges the furniture, and leaves the balance untouched: the string whirling a conker, the floor under a sliding crate, gravity on the orbiting Moon — all enormous forces, all workless, forever. A perpendicular force can bend a path as violently as it likes, but it cannot change a speed.

Figure

Three forces, three signs — the angle decides the direction of energy flow

Now let the path curve and the force vary — a cyclist on a mountain road, gravity pulling straight down the whole way. Over any short stretch the road is straight and gravity constant, so $Fd\cos\theta$ applies locally: negative on a climb, positive on a descent, zero on the flat. The total work is the running sum of all these signed contributions, and the sum is exactly what the line integral computes. For gravity the deposits and withdrawals settle to a total that depends only on the net height change — gravity is conservative. For friction every metre of road charges the same toll regardless of direction, so the total grows with the length of the route — friction is not. That single distinction, visible already in the picture, is where potential energy will come from.

From one push to any path — the math

The physical picture demanded a measure that multiplies three things: how hard the force pushes, how far the object moves, and how well the two directions agree. The cosine is the agreement dial — reading $+1$ for perfect alignment, $0$ at right angles, $-1$ for dead opposition — and the constant-force definition wires the three together:

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

where: $F$ — magnitude of the (constant) force, in newtons; $d$ — magnitude of the straight-line displacement, in metres; $\theta$ — angle between $\vec F$ and $\vec d$ drawn tail-to-tail; $W$ — work in joules, $1\ \mathrm{J} = 1\ \mathrm{N\cdot m}$ . Magnitudes only ever enter positively, so the sign of the work is decided entirely by the angle. Note the two equivalent slicings of the same product: (effective force $F\cos\theta$ ) × (full distance $d$ ), or (full force $F$ ) × (displacement shadow $d\cos\theta$ ) — same rectangle of numbers, cut two ways. Often no one hands you the angle — a drone's force and displacement arrive as coordinate lists from its sensors, not as a tidy diagram. The dot product has a second, angle-free form, computable straight from components:

\vec A \cdot \vec B = A_x B_x + A_y B_y \;(+\,A_z B_z \text{ in 3D}) \qquad (2)

where: $A_x, A_y$ — components of $\vec A$ along the coordinate axes; likewise for $\vec B$ . Multiply matching components and add. The two forms always agree — for the drone force $\vec F = (12\,\hat x + 5\,\hat y)\ \mathrm{N}$ acting over $\vec d = (30\,\hat x - 8\,\hat y)\ \mathrm{m}$ , equation (2) gives $W = 360 - 40 = 320\ \mathrm{J}$ , and running (1) backwards with $F = 13\ \mathrm{N}$ , $d = 31.05\ \mathrm{m}$ returns $\cos\theta = 0.793$ , $\theta = 37.6^\circ$ — mostly helping, partly wasted. Equation (1) is the atom. The molecule — work for any force along any path — is built by the chop–dot–add construction, and it deserves a proper derivation.

Derivation AFrom force times distance to the line integral

We want the general definition of work when the force varies from point to point and the path curves — the situation of a spring, a planet, or a cyclist on a real road. The strategy is the standard one of calculus: reduce the hard problem to many easy ones. We chop the path into segments short enough that each is straight and the force on it effectively constant, apply the constant-force formula (1) to each, and pass to the limit. The thing to watch is that both approximations — straight segment, constant force — become exact together as the chopping gets finer.

Step 1. Break the curve $C$ into $N$ short directed segments $\Delta\vec r_1, \Delta\vec r_2, \ldots, \Delta\vec r_N$ , laid tip-to-tail along the path from start to finish. Why this helps: a smooth curve looks straight under a magnifying glass, and a continuous force barely changes across a small enough neighbourhood — each segment is now a legal constant-force problem.

Step 2. On segment $i$ , apply the constant-force formula with the local force value:

\Delta W_i \approx \vec F(\vec r_i)\cdot\Delta\vec r_i

Why the dot product survives: on each little piece the geometry is exactly the sled picture — only the component of the local force along the local direction of travel transfers energy.

Step 3. Add the pieces and refine the chopping:

W = \lim_{N\to\infty}\sum_{i=1}^{N} \vec F(\vec r_i)\cdot\Delta\vec r_i \;=\; \int_C \vec F\cdot\mathrm{d}\vec r

The limit of the Riemann sum is, by definition, the line integral of $\vec F$ along $C$ .

Step 4. In components the recipe becomes concrete: $\int_C \vec F\cdot\mathrm{d}\vec r = \int_C (F_x\,\mathrm{d}x + F_y\,\mathrm{d}y + F_z\,\mathrm{d}z)$ , evaluated by parameterizing the path — by arc length, by $x$ , or most physically by time $t$ , using $\mathrm{d}\vec r = \vec v\,\mathrm{d}t$ . Why this is self-consistent: for a constant force on a straight path the integrand is constant and the integral collapses back to $\vec F\cdot\vec d$ , equation (1) — the general construction reduces to the atom it was built from.

The result is the working definition for everything that follows:

W = \int_C \vec F\cdot\mathrm{d}\vec r \qquad (3)

where: $C$ — the specific path traversed through space; $\mathrm{d}\vec r$ — an infinitesimal displacement element tangent to $C$ ; $\vec F$ — the force, possibly varying with position along the path. Read it as an itinerary, not a formula: walk the actual route, and at every step credit the account with the along-step component of the local force. The value can, and in general does, depend on the route — that is why the path label $C$ sits on the integral sign.

Figure

Chop, dot, add — the line integral is the constant-force formula applied piece by piece

Two frequently used special faces of equation (3) are worth naming. Confine the motion to a line, and the vector integral collapses to an ordinary one — this is the version that turns a force–position graph into a work-reading machine, because an integral of $F(x)$ is the area between the curve and the axis:

W = \int_{x_1}^{x_2} F_x(x)\,\mathrm{d}x \qquad (4)

where: $F_x$ — the component of force along the line of motion; $x_1, x_2$ — start and end positions. For a constant force the area is a rectangle, $W = F\,(x_2 - x_1)$ , and you recover the locomotive's rule that energy is paid by the metre, like a toll — which is why a small force over a long distance can transfer the same energy as a huge force over a short one. Stopping a supertanker with a tug takes kilometres; with a harbour wall, centimetres. Parameterize by time instead, using $\mathrm{d}\vec r = \vec v\,\mathrm{d}t$ , and the line integral becomes a time integral whose integrand earns its own name:

W = \int_{t_1}^{t_2} \vec F\cdot\vec v\,\mathrm{d}t, \qquad P = \frac{\mathrm{d}W}{\mathrm{d}t} = \vec F\cdot\vec v \qquad (5)

where: $\vec v$ — the velocity along the trajectory; $P$ — the instantaneous power, in watts, the rate at which the force feeds or drains energy. A force perpendicular to $\vec v$ has $P = 0$ identically — the workless spectator, caught in the act at every instant.

you are given	reach for	equation
constant force, straight path, angle known	$W = Fd\cos\theta$	(1)
force and displacement as components	$W = F_xd_x + F_yd_y + F_zd_z$	(2)
force varying along a curved path	parameterize and integrate $\vec F\cdot\mathrm{d}\vec r$	(3)
1D motion or a force–position graph	area under $F_x(x)$	(4)
force and velocity as functions of time	integrate the power $\vec F\cdot\vec v$	(5)

The payoff: work moves kinetic energy

Why organize all of mechanics around this particular integral? Because Newton's second law, pushed through it, produces the single most useful scalar statement in dynamics.

Go deeper · Derivation B · 1 minThe work–energy theorem from Newton's second law

Derivation BThe work–energy theorem from Newton's second law

We want to show that the net work done on a particle — by all forces together, over the actual path — equals the change in its kinetic energy $T = \tfrac12 mv^2$ . The strategy is to trade the path parameterization for time using $\mathrm{d}\vec r = \vec v\,\mathrm{d}t$ , insert $\vec F_{\text{net}} = m\,\mathrm{d}\vec v/\mathrm{d}t$ , and watch the integrand become a perfect time derivative. The one clever move is a chain-rule identity for $\vec v\cdot\mathrm{d}\vec v/\mathrm{d}t$ ; everything else is bookkeeping.

Step 1. Express the net work as a time integral:

W_{\text{net}} = \int_C \vec F_{\text{net}}\cdot\mathrm{d}\vec r = \int_{t_1}^{t_2} \vec F_{\text{net}}\cdot\vec v\,\mathrm{d}t

Why this is legal: the particle's own trajectory is the path, so time parameterizes it perfectly — every point of $C$ is visited at a definite instant with velocity $\vec v$ .

Step 2. Insert Newton's second law $\vec F_{\text{net}} = m\,\mathrm{d}\vec v/\mathrm{d}t$ :

\vec F_{\text{net}}\cdot\vec v = m\,\frac{\mathrm{d}\vec v}{\mathrm{d}t}\cdot\vec v

Step 3. Use the chain-rule identity on $\vec v\cdot\vec v = v^2$ :

\frac{\mathrm{d}\vec v}{\mathrm{d}t}\cdot\vec v = \tfrac12\,\frac{\mathrm{d}}{\mathrm{d}t}\left(\vec v\cdot\vec v\right) = \tfrac12\,\frac{\mathrm{d}(v^2)}{\mathrm{d}t}

Why the half appears: differentiating $\vec v\cdot\vec v$ by the product rule produces the same cross term twice — the same bookkeeping that gives every quadratic its factor of 2.

Step 4. The integrand is now a total time derivative, so the integral collapses to endpoint values:

W_{\text{net}} = \int_{t_1}^{t_2} \tfrac12\,m\,\frac{\mathrm{d}(v^2)}{\mathrm{d}t}\,\mathrm{d}t = \tfrac12 m v_2^2 - \tfrac12 m v_1^2

Why this is reassuring: in 1D with constant force, $W = Fd = mad$ and the kinematic identity $v_2^2 = v_1^2 + 2ad$ give $W = \tfrac12 m(v_2^2 - v_1^2)$ — the same statement with no calculus at all.

W_{\text{net}} = \Delta T = \tfrac12 m v_2^2 - \tfrac12 m v_1^2 \qquad (6)

where: $W_{\text{net}}$ — the work done by the total force, equivalently the signed sum of the works of the individual forces; $v_1, v_2$ — initial and final speeds. This converts the vector, instant-by-instant content of $\vec F = m\vec a$ into a scalar relation between two endpoint states. Positive net work means the object leaves faster than it arrived; negative, slower; zero, same speed — even if the direction changed completely.

When the route matters — and when it cannot

Equation (3) carries a path label, and the deepest question you can ask it is whether the label matters. For some forces it does not. Near-surface gravity is the standard example: with $\vec F = -mg\,\hat y$ and $\mathrm{d}\vec r = \mathrm{d}x\,\hat x + \mathrm{d}y\,\hat y$ , the dot product keeps only $-mg\,\mathrm{d}y$ , and the integral telescopes to

W_{\text{grav}} = -mg\,(y_2 - y_1) \qquad (7)

where: $m$ — the mass; $g \approx 9.8\ \mathrm{m/s^2}$ ; $y$ — height above any convenient reference. Every horizontal wiggle of the route is workless and every vertical metre costs the same, so only the net height change survives — a ball thrown straight up 2 m and a ball carried up a winding 2 m ramp receive identical work from gravity. The same telescoping happens for a spring, by direct integration of Hooke's law $F = -kx$ along the line of motion:

W_{\text{spring}} = \int_{x_1}^{x_2}(-kx)\,\mathrm{d}x = -\tfrac12 k\left(x_2^2 - x_1^2\right) \qquad (8)

where: $k$ — the spring constant, in N/m; $x$ — displacement from natural length. Again the answer is a function of the endpoints only. Forces with this property are called conservative; each admits a potential energy $U$ with $\vec F = -\nabla U$ , and their closed-loop work vanishes:

\vec F = -\nabla U \iff \oint_C \vec F\cdot\mathrm{d}\vec r = 0 \qquad (9)

where: $U$ — the potential energy, a scalar field of position; $\oint$ — the integral around any closed loop. The equivalence holds on any region without holes; the closed-loop condition on the right is the criterion that is safe in every case. But path independence is a special property, not a general truth, and the cleanest way to respect that is to catch a force red-handed charging different prices for different routes.

Go deeper · Derivation C · 2 minA force whose work depends on the path

Derivation CA force whose work depends on the path

We want an explicit, fully computed example of path dependence — two different routes between the same two points, two different amounts of work. Take the force field $\vec F = c\,y\,\hat x$ with $c = 2\ \mathrm{N/m}$ : a horizontal push whose strength grows with height, like a wind that blows harder the higher you climb. We will carry a particle from the origin $O = (0,0)$ to the point $P = (2\ \mathrm{m},\,3\ \mathrm{m})$ along two elbow-shaped routes and evaluate equation (3) honestly on each. Watch which legs contribute: this force is purely horizontal, so only horizontal legs can ever do work — but how much they do depends on the height at which they are walked.

Step 1. Route A — first along the ground, then up. On the leg $(0,0)\to(2,0)$ the height is $y = 0$ , so $F_x = c\,y = 0$ everywhere on it: $\int F_x\,\mathrm{d}x = 0$ . On the vertical leg $(2,0)\to(2,3)$ the displacement is $\mathrm{d}\vec r = \mathrm{d}y\,\hat y$ , perpendicular to the purely-horizontal force: contribution $0$ .

W_A = 0 + 0 = 0\ \mathrm{J}

Step 2. Route B — first up, then across the top. The vertical leg $(0,0)\to(0,3)$ is again workless (force horizontal, motion vertical). On the top leg $(0,3)\to(2,3)$ the height is fixed at $y = 3\ \mathrm{m}$ , so $F_x = 2\times 3 = 6\ \mathrm{N}$ , constant:

W_B = \int_0^{2} 6\,\mathrm{d}x = 12\ \mathrm{J}

Step 3. Same endpoints, different answers: $W_A = 0$ but $W_B = 12\ \mathrm{J}$ . Now run route A out and route B backwards to close a loop from $O$ back to $O$ : the circuit work is $W_A - W_B = -12\ \mathrm{J} \neq 0$ . Why this kills any potential energy: if a $U(\vec r)$ existed with $W = U(\text{start}) - U(\text{end})$ , the work would be forced to agree on every route between the same endpoints — and every closed loop would return exactly zero.

Step 4. The physical archetype of this behaviour is kinetic friction, $\vec f = -\mu_k N\,\hat v$ , which always points against the motion. Along every element of any path, $\vec f\cdot\mathrm{d}\vec r = -\mu_k N\,|\mathrm{d}\vec r|$ , so the work is $-\int_C \mu_k N\,\mathrm{d}s$ , which for constant $N$ is $-\mu_k N L$ with $L$ the total path length — a longer detour between the same endpoints always costs more, and no potential energy can be defined for it. Why the curl test agrees: the loop integral of $\vec F = cy\,\hat x$ around any rectangle equals $-c\times(\text{area})$ by direct summation of the four legs — for the $2\ \mathrm{m}\times 3\ \mathrm{m}$ circuit just walked, $-2\times 6 = -12\ \mathrm{J}$ , exactly the Step 3 result, consistent with the curl test $\partial F_x/\partial y = c \neq 0$ — a non-zero curl is the algebraic fingerprint of path dependence.

Figure

Two routes, two answers — the fingerprint of a non-conservative force

Limiting cases

case	result	meaning
$\theta = 0^\circ$	$W = +Fd$	force fully aligned — maximum possible work
$\theta = 90^\circ$	$W = 0$	perpendicular force — steers or supports, never energizes
$\theta = 180^\circ$	$W = -Fd$	dead opposition — maximum drain (friction, brakes)
$d = 0$	$W = 0$	pushing a wall: effort without displacement is workless
constant $\vec F$ , any path	$W = \vec F\cdot\vec d_{\text{net}}$	integral collapses — only the net displacement survives
1D motion	$W = \int F_x\,\mathrm{d}x$	area under the force–position graph
conservative force	$W = -\Delta U$ , $\oint\vec F\cdot\mathrm{d}\vec r = 0$	endpoints only; potential energy exists
kinetic friction	$W = -\mu_k N L$	charges by total path length $L$ — path-dependent, always negative

Common trapI am exhausted, so I must have done work

Physiological effort and physical work are different currencies. Hold a 200 N barbell perfectly still overhead: $d = 0$ , so $W = 0$ — no work done on the barbell, however badly your arms shake. Carry a suitcase at steady speed across a level floor: your supporting force is vertical, the displacement horizontal, $\theta = 90^\circ$ , so again $W = 0$ . Your muscles burn chemical energy in both cases — fibres twitch and re-grip internally, biology is inefficient — but none of that energy is transferred to the object as work. The companion trap points the other way: never rank forces by size to guess energy flow. The Moon feels about $2\times10^{20}$ N of gravitational pull and receives zero work from it, eternally, because the pull stays perpendicular to the motion. Rank forces by their angle to the displacement, and remember that negative work is not "no work" — a braking force does very real work; the energy leaves the car and turns up as heat in the discs.

Worked examples

1 · Four forces, one crate (confidence builder)

A 5.0 kg crate is dragged 8.0 m across a level floor by a horizontal pull of 40 N, against a friction force of 12 N. Find the work done by each force and the crate's final speed if it starts from rest. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Work by the pull ( $\theta = 0^\circ$ ): $W_{\text{pull}} = 40 \times 8.0 = \mathbf{+320\ J}$ .

Step 2. Work by friction ( $\theta = 180^\circ$ ): $W_{\text{fric}} = -12 \times 8.0 = \mathbf{-96\ J}$ .

Step 3. Gravity and the normal force both act vertically while the crate moves horizontally — $\theta = 90^\circ$ for each, so $W_{\text{grav}} = W_N = \mathbf{0}$ .

Step 4. Works are signed scalars and simply add: $W_{\text{net}} = 320 - 96 + 0 + 0 = +224\ \mathrm{J}$ .

Step 5. The work–energy theorem (6) turns the ledger into speed: $\tfrac12(5.0)v_f^2 = 224$ , so $v_f^2 = 89.6$ and $v_f = \mathbf{9.5\ m/s}$ . The sign audit did all the physics; the algebra was one line.

2 · Up the ramp, force by force (exam level)

An 18 kg crate is pulled 6.0 m up a $30^\circ$ incline at constant speed by a force $P$ parallel to the slope. The kinetic friction coefficient is 0.25. Find $P$ , the work done by each force, and verify the net work is zero. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Weight: $mg = 18 \times 9.8 = 176.4\ \mathrm{N}$ . Resolve it along and across the slope: $mg\sin 30^\circ = 88.2\ \mathrm{N}$ down-slope, $mg\cos 30^\circ = 152.8\ \mathrm{N}$ into the surface.

Step 2. The normal force balances the across component, $N = 152.8\ \mathrm{N}$ , so kinetic friction opposing the climb is $f = 0.25 \times 152.8 = 38.2\ \mathrm{N}$ .

Step 3. Constant speed means the along-slope forces balance: $P = 88.2 + 38.2 = \mathbf{126.4\ N}$ .

Step 4. Work by $P$ ( $\theta = 0^\circ$ to the 6.0 m displacement): $W_P = 126.4 \times 6.0 = \mathbf{+758\ J}$ .

Step 5. Work by gravity two ways. Via the along-slope component: $W_{\text{grav}} = -88.2 \times 6.0 = -529\ \mathrm{J}$ . Via the conservative shortcut (7): height gained $h = 6.0\sin 30^\circ = 3.0\ \mathrm{m}$ , so $W_{\text{grav}} = -176.4 \times 3.0 = \mathbf{-529\ J}$ — the slope cancels out of gravity's bill, exactly as path independence promises.

Step 6. Work by friction ( $\theta = 180^\circ$ ): $W_f = -38.2 \times 6.0 = \mathbf{-229\ J}$ . The normal force: $\mathbf{0}$ .

Step 7. Audit: $758 - 529 - 229 = 0$ ✓ — zero net work at constant speed, as equation (6) demands. Of the 758 J invested, 529 J went into height (recoverable — it is stored against gravity) and 229 J went to heat in the surfaces (gone: friction's work is path-dependent and unrecoverable).

3 · A spring launches, friction collects (exam level)

A 0.50 kg block is pressed against a spring of constant $k = 250\ \mathrm{N/m}$ , compressing it by 0.12 m, and released on a horizontal surface. The first stretch of the surface is frictionless; beyond it lies a rough patch with $\mu_k = 0.30$ . How fast is the block moving when it leaves the spring, and how far into the rough patch does it slide? (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Work done by the spring on the block as it relaxes from $x_1 = -0.12\ \mathrm{m}$ to $x_2 = 0$ , from equation (8): $W_{\text{spring}} = -\tfrac12 k(x_2^2 - x_1^2) = +\tfrac12 \times 250 \times (0.12)^2 = \mathbf{+1.8\ J}$ . Note the variable force made this an honest integral — the average force $\tfrac12 kx_{\max} = 15\ \mathrm{N}$ times 0.12 m gives the same 1.8 J.

Step 2. Work–energy theorem across the launch: $\tfrac12(0.50)v^2 = 1.8\ \mathrm{J}$ , so $v^2 = 7.2$ and $v = \mathbf{2.7\ m/s}$ .

Step 3. In the rough patch, friction is the only horizontal force: $f = \mu_k m g = 0.30 \times 0.50 \times 9.8 = 1.47\ \mathrm{N}$ , doing work $-fL$ over a slide of length $L$ .

Step 4. The block stops when friction has withdrawn the whole deposit: $-1.47\,L = 0 - 1.8\ \mathrm{J}$ , so $L = 1.8/1.47 = \mathbf{1.2\ m}$ .

Step 5. Sanity-check the chain: spring's stored energy (endpoint quantity, path-free) became kinetic energy, which friction — charging $1.47\ \mathrm{J}$ per metre, by the metre — converted to heat in 1.2 m. Three different work formulas, one unbroken energy ledger.

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

Gaspard-Gustave de Coriolis

1792–1843 · École Polytechnique, Paris

Coriolis is famous for the deflection force that bears his name, but his quieter revolution was giving work its modern physics meaning. In Du calcul de l'effet des machines (1829), analysing waterwheels and steam engines, he defined "travail" — work — as force times the distance through which it acts, and paired it with $\tfrac12 mv^2$ (fixing the crucial factor of one-half) so that work done and energy of motion would balance exactly. Engineers had been arguing about how to rate machines for a century; Coriolis handed them the correct bookkeeping. Every line integral of force you compute is his ledger, written in vectors that came half a century later.

History

1686

— Leibniz champions vis viva,

mv^2

, igniting a century-long debate over what a moving body carries.

1807

— Thomas Young coins the word "energy" for

mv^2

-type quantities, reviving Leibniz's idea under the name it keeps today.

1829

— Coriolis's Du calcul de l'effet des machines defines work as force × distance and fixes kinetic energy as

\tfrac12 mv^2

, so that the two balance; Poncelet popularizes the term across engineering.

1845

— Joule's paddle-wheel experiments fix the exchange rate between mechanical work and heat — the energy that negative work removes is found, not lost.

1881

— Gibbs's Elements of Vector Analysis frees the dot product from Hamilton's quaternions; by the Gibbs–Wilson textbook of 1901,

W = \int\vec F\cdot\mathrm{d}\vec r

is written the way you write it now.

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 5 (Conservation of Energy and Momentum) · Taylor, Classical Mechanics, ch. 4 (Energy) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.2 · Halliday, Resnick & Walker, Fundamentals of Physics, ch. 7 · Young & Freedman, University Physics, ch. 6. All equations follow from the work–energy theorem derived via the chain-rule identity $\vec v\cdot\mathrm{d}\vec v/\mathrm{d}t = \tfrac12\,\mathrm{d}(v^2)/\mathrm{d}t$ , with $W_{\text{spring}} = -\tfrac12 k(x_2^2 - x_1^2)$ and $W_{\text{grav}} = -mg\,\Delta y$ obtained by direct integration, and the friction path-dependence confirmed by the closed-loop criterion. Worked-example numbers use $g = 9.8\ \mathrm{m/s^2}$ throughout: the crate audit gives $+320 - 96 = +224\ \mathrm{J}$ and $v_f = \sqrt{2\times224/5.0} = 9.5\ \mathrm{m/s}$ ; the ramp audit gives $P = 126.4\ \mathrm{N}$ and $758 - 529 - 229 = 0$ ; the spring–friction chain gives $\tfrac12\times250\times0.12^2 = 1.8\ \mathrm{J}$ , $v = 2.7\ \mathrm{m/s}$ , and $L = 1.8/1.47 = 1.2\ \mathrm{m}$ ; and the two-route computation for $\vec F = 2y\,\hat x$ gives $W_A = 0$ and $W_B = 12\ \mathrm{J}$ .

Builds on: Vectors, Coordinate Systems and Kinematics · Position, Displacement and Reference Frames · Newton's Second Law · Velocity and Acceleration. Leads to: Normal Force · Gravitational Force Near Earth's Surface · Circular Motion and Centripetal Acceleration.

If I do zero work carrying a suitcase, why do my muscles burn energy doing it?

Because muscles are not rigid brackets. To hold a load, muscle fibres twitch — they contract, slip, and re-grip many times a second, doing microscopic internal work that ends up as heat in your tissue. That chemical energy is genuinely spent, but none of it is transferred to the suitcase: the supporting force is vertical, the displacement horizontal, and $Fd\cos 90^\circ = 0$ . Physics is auditing the suitcase's energy account, not your metabolism. Replace your arm with a table on wheels and the suitcase travels identically with no energy spent at all — proof the burn was a fact about biology, not about the work done on the load.

How can work be negative? What does minus 96 joules physically mean?

Negative work is energy flowing out of the object's kinetic energy account. When friction does $-96\ \mathrm{J}$ on the dragged crate, the crate ends the interval with 96 J less kinetic energy than the pull alone would have given it — energy that reappears as heat in the two surfaces. Nothing mysterious happens with the arithmetic: in $W = Fd\cos\theta$ the magnitudes are positive, and $\cos\theta$ goes negative the moment the force leans backward of $90^\circ$ from the motion. A backward-leaning force acting over a forward displacement is a withdrawal, and the sign records the direction of the flow.

What does the line integral actually ask me to do, concretely?

Three things: parameterize, dot, integrate. First write the path as a function of one parameter — time $t$ , arc length, or $x$ itself — so that $\mathrm{d}\vec r$ becomes an explicit vector of differentials. Second, take the dot product with the force evaluated on the path: substitute the path's $y(x)$ or $\vec r(t)$ into $\vec F$ before integrating (this is the step everyone forgets). Third, integrate the resulting ordinary function between the endpoint parameter values. Derivation C runs the full recipe twice on the field $\vec F = 2y\,\hat x$ ; notice how route A gives 0 J and route B gives 12 J purely because $\vec F$ takes different values on the legs actually walked.

Why is work a scalar when force and displacement are both vectors?

Because the dot product is defined to produce a scalar — that is its whole personality (its other name is the scalar product). Physically, work answers a magnitude-only question: how much energy crossed from the force's agent into the object's motion? Energy has no direction; 320 J from an eastward push and 320 J from a northward push are indistinguishable once deposited. This is also why works from different forces combine by ordinary signed addition, with no components or angle bookkeeping — a huge practical simplification over adding the forces as vectors first, and one reason energy methods often beat force methods.

Does the normal force ever do work on anything?

On anything sliding along the surface, never — perpendicular is perpendicular. But the rule is "perpendicular to the displacement", not "normal forces are magically workless". Stand in an accelerating elevator: the floor's normal force points up and your displacement is up, so it does positive work on you. A curling stone pushed by a broom held at an angle receives work from the normal component of the broom's push. The reliable habit is to check the angle between each force and the actual displacement in each problem, rather than memorizing a list of "forces that do no work" — the list has exceptions; the angle test has none.

Gravity's work came out the same for every path. Friction's did not. What is the structural difference?

Gravity's integrand is a perfect differential: $\vec F\cdot\mathrm{d}\vec r = -mg\,\mathrm{d}y$ , so the integral telescopes to endpoint values and every route between the same points yields $-mg\,\Delta y$ . Friction can never be written that way, because its direction is slaved to the velocity — reverse the motion and friction reverses too, always charging $\mu_k N$ per metre travelled. Its work is $-\mu_k N L$ , proportional to path length, not endpoint separation. The clean test is the closed loop, equation (9): gravity refunds on the way down what it charged on the way up ( $\oint = 0$ ), while friction charges in both directions ( $\oint < 0$ always). Path-independent forces earn a potential energy; friction structurally cannot.

Work done by the force or against the force — which sign convention should I trust?

Fix the object, then ask of each force: what is $\vec F\cdot\vec d$ for this force over this displacement? That signed number is "the work done by that force on the object", and it is the only quantity the work–energy theorem wants. "Work done against gravity" is shorthand for the positive work an external agent does while gravity does the equal-and-opposite negative work at constant speed — harmless as vocabulary, dangerous as bookkeeping, because it tempts you to flip signs twice. In an audit like Worked Example 2, list every force with its own $Fd\cos\theta$ and let the signs fall where the angles put them; never insert a minus "because it opposes".

If the net work on an object is zero, does that mean nothing happened to it?

No — it means only that the speed ends where it started. The crate lowered at constant velocity had $-588\ \mathrm{J}$ from the rope and $+588\ \mathrm{J}$ from gravity: net zero, speed constant, yet the crate is now 4 m lower and 588 J of gravitational potential energy has been handed through the rope to whatever lowered it. Uniform circular motion is the other classic: net work zero at every instant, speed constant, while the velocity's direction changes continuously. Zero net work is a statement about the kinetic-energy balance, not a claim that the system is static or that no energy moved through it.

Work and the Line Integral of Force

Where this lives today

The physical picture

From one push to any path — the math

The payoff: work moves kinetic energy

When the route matters — and when it cannot

Limiting cases

Worked examples

1 · Four forces, one crate (confidence builder)

2 · Up the ramp, force by force (exam level)

3 · A spring launches, friction collects (exam level)

Going further

Before this

Where this leads