Work-energy theorem — derivation from Newton's second law: Visual walkthrough

We start from absolutely nothing — not even the meaning of "work". Every symbol is earned before it is used.

Step 1 — What are we even measuring?

Figure — Work-energy theorem — derivation from Newton's second law

WHAT: we set up a bead on a horizontal wire so everything happens along one straight line. That way a single number (with a $+$ or $-$ sign) captures each quantity — no angles yet.

WHY: the whole theorem is easiest to see in 1-D. Once we've seen it once, the 3-D version (Step 8) is just the same story with arrows.

PICTURE: the bead sits at position $x$ ; the orange arrow is the force $F$ ; the teal arrow is the velocity $v$ . Right now they point the same way (speeding up).

Step 2 — Newton's law: the only thing we're allowed to assume

Let's read the new symbol. $\frac{d v}{d t}$ means "the change in velocity $d v$ over a tiny sliver of time $d t$ " — the rate at which speed is changing. This rate is exactly what we call acceleration.

WHY this tool? We need some law connecting force to motion, and Newton's Second Law is the deepest one we have. Everything below is squeezed out of this single equation — we invent nothing new.

PICTURE: two snapshots of the bead a tiny time $d t$ apart. The velocity arrow grew a little; that little growth is $d v$ . Newton says the size of the force sets how big $d v$ is per unit time.

Step 3 — Introduce distance by multiplying by $d x$

WHAT: we multiplied both sides by the same tiny distance $d x$ — a legal move, like multiplying both sides of $3 = 3$ by $5$ .

WHY: on the left we now have $F d x$ , which is force over distance — the raw material of work. We're deliberately steering toward $\int F d x$ .

PICTURE: the bead nudges forward by a sliver $d x$ (plum). During that sliver, the force $F$ paints out a thin rectangle of "work" — height $F$ , width $d x$ . Add up all such slivers and you get total work; that's the shaded strip.

Step 4 — The chain-rule swap: kill time, keep distance

Here is the clever heart of the whole thing. Look at the right side: $m \frac{d v}{d t} d x$ . The pieces $d t$ and $d x$ are just tiny numbers, so we may shuffle them:

WHY this and not something else? We wanted the $d t$ gone. By pairing $d x$ with $\frac{1}{d t}$ we form $\frac{d x}{d t} = v$ , and time vanishes. What's left, $v d v$ , speaks only the language of speed — no clocks.

PICTURE: think of it as a change of viewpoint. On the left, we asked "how does speed change each second?" On the right, we ask "how does speed change each metre?" The figure shows the same growing velocity arrow read against a distance ruler instead of a stopwatch — metres are exactly what work counts.

Step 5 — Add up all the slivers: define net work

WHAT: we integrate both sides of Step 4's equation over the whole journey.

$\int_{x_{i}}^{x_{f}} F_{n e t} d x = \int_{v_{i}}^{v_{f}} m v d v$

WHY the limits change — the change-of-variables rule. This is not sloppy relabelling; it is a genuine substitution. Read it slowly:

WHY (the piling-up part): a single sliver $F d x$ is too small to be useful. The real quantity — the work over the whole path — is the pile of all slivers, and $\int$ is the machine that stacks them.

PICTURE: the force graph. Each thin plum rectangle is one $F d x$ ; the total orange area is $W_{n e t}$ . This is exactly the picture of Work done by a variable force — the area, not "force times distance", is the honest definition.

Step 6 — Evaluate the right side: kinetic energy appears

We need the value of $\int_{v_{i}}^{v_{f}} m v d v$ . Because $m$ is a constant (our Step 1 assumption), it is not part of the summing — we may slide it out in front of the integral sign:

$\int_{v_{i}}^{v_{f}} m v d v = m \int_{v_{i}}^{v_{f}} v d v .$

Why is that allowed? Pulling a fixed number out of a sum is just the distributive law: $ma + mb + \dots = m (a + b + \dots)$ . If $m$ changed as the bead moved, it would be trapped inside and this step would be illegal — that is exactly why the constant-mass assumption earned its keep. Now use the pocket note from Step 4 ( $v d v$ builds up $\frac{1}{2} v^{2}$ ):

Putting Step 5 and Step 6 together:

$W_{n e t} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = Δ K ■$

PICTURE: the parabola $K = \frac{1}{2} m v^{2}$ . Mark $v_{i}$ and $v_{f}$ on the horizontal axis; the vertical gap between the two heights on the curve is $Δ K$ — and that gap equals the orange work-area from Step 5. The theorem is that these two pictures are the same number. See Kinetic Energy for why $K$ is always $\geq 0$ .

Step 7 — Every case: signs, zero, and moving backwards

The boxed result must survive all scenarios. Let's check each on one figure.

Case A — Force with motion ( $F > 0$ , moving $+ x$ ): slivers $F d x$ are positive ( $F > 0$ , $d x > 0$ ), so $W_{n e t} > 0$ , so $Δ K > 0$ : the bead speeds up. (Left panel.)

Case B — Force against motion ( $F < 0$ , moving $+ x$ ): each sliver $F d x$ is negative ( $F < 0$ , $d x > 0$ ), so $W_{n e t} < 0$ , so $Δ K < 0$ : the bead slows down. This is braking/friction. Work being negative is not a paradox — the force simply steals kinetic energy. (Middle panel.)

Case C — Constant velocity ( $Δ K = 0$ ): if speed never changes, the positive and negative slivers must exactly cancel, so $W_{n e t} = 0$ even if individual forces do plenty of work. (Right panel: a dragged block — your push and friction cancel.)

Case D — Bead moving in the $- x$ direction ( $v < 0$ ). Here is the case the panels above deliberately left for careful thought. When the bead slides left, each step is $d x < 0$ . The sign of the sliver $F d x$ is now the product of two signs:

Force also pointing left ( $F < 0$ ) with leftward motion ( $d x < 0$ ): $F d x = (-) (-) = +$ ⇒ positive work, the bead speeds up in the leftward direction. Correct — a leftward push on a leftward-moving bead should energise it.

Force pointing right ( $F > 0$ ) against leftward motion ( $d x < 0$ ): $F d x = (+) (-) = -$ ⇒ negative work, the bead slows down. Correct — it opposes the motion.

Notice $K = \frac{1}{2} m v^{2}$ uses $v^{2}$ , so a leftward-moving bead at speed $5$ m/s has the same $K$ as a rightward one. The theorem doesn't care about direction of travel — only $F d x$ getting its sign right, which it always does because both $F$ and $d x$ carry their own signs.

Degenerate case — zero displacement ( $d x = 0$ ): if the bead doesn't move, every sliver has width $0$ , so $W_{n e t} = 0$ no matter how hard you push. Pushing a wall changes no kinetic energy.

Degenerate case — starts and ends at rest ( $v_{i} = v_{f}$ ): then $Δ K = 0$ , so total net work over the trip is zero, even if the bead sped up and slowed down in between (positive then negative work).

Step 8 — The 3-D version: arrows and the dot product

In real life motion curves through space. Position becomes a vector $r$ , velocity $v$ , force $F$ . The tiny step is $d r$ . But work must stay a single number, so we need a way to multiply two arrows into one number — that tool is the dot product $F \cdot d r$ .

Now we run the same four moves as before, but each deserves its own WHY because arrows behave a little differently from plain numbers:

Integrating — and where the limits go. Summing every tiny piece along the path from the start point to the end point: $\int_{start}^{end} F \cdot d r = \int_{v_{i}}^{v_{f}} m v \cdot d v = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = Δ K .$ WHY the limits transfer: identical change-of-variables logic as Step 5 — the left integral is taken over position along the path, so its endpoints are the start and end points; the right integral is over velocity, so its endpoints are the velocities $v_{i}, v_{f}$ at those same two instants. Only the speeds $v_{i}, v_{f}$ survive in the final answer because $v \cdot v = v^{2}$ throws away direction.

PICTURE: a curved path. At one point the force arrow is split into a piece along the motion (this does the work) and a piece across it (does nothing). Only the along-piece survives the dot product — the geometry that the sign in 1-D was quietly handling all along.

The one-picture summary

Everything on one canvas: start at $F = ma$ (time-language) → multiply by $d x$ → chain-rule swap to $v d v$ (distance-language) → integrate → the force-area on the left equals the $K$ -gap on the right.

Recall Feynman retelling — say it to a 12-year-old

Imagine pushing a bead along a wire. Newton's rule tells you how your push changes the bead's speed each second — but you want to know the effect each metre, because "work" is about metres. So you play a tiny swap: instead of tracking speed against a stopwatch, you track it against a ruler. When you do that swap honestly, the messy "speed-change-per-second" turns into a tidy " $v$ times a nudge in $v$ ". Pile up all the tiny nudges (and because the bead keeps the same mass the whole time, you can factor that mass out cleanly) and you get exactly one-half-mass-times-speed-squared — the go-fast energy. So the total pushing you did over the whole distance (the shaded area under your force) is precisely the change in the bead's go-fast energy. Push forward, it gains; push backward, it loses; push sideways, nothing happens; even if it's sliding leftward, the plus-and-minus signs sort themselves out. That's the whole theorem — bookkeeping for motion.

Recall checkpoints

Recall Self-test

Which language does Newton's law speak, and which do we need? ::: Newton speaks time ( $d v / d t$ ); work needs distance ( $\int F d x$ ). What does multiplying by $d x$ set up? ::: The left side becomes $\int F d x$ , the definition of net work. What hidden assumption lets us pull $m$ out of the integral? ::: The mass is constant throughout the motion. What does the chain-rule swap eliminate? ::: Time — $\frac{d v}{d t} d x = v d v$ has no $d t$ left. When changing $\int F d x$ to $\int m v d v$ , what happens to the limits? ::: They become the speeds at the same two instants: $x_{i} \to v_{i}$ , $x_{f} \to v_{f}$ . A bead slides in the $- x$ direction under a leftward force — is the work positive? ::: Yes: $F < 0$ and $d x < 0$ give $F d x > 0$ , so it gains kinetic energy. Why does a sideways force do no work? ::: $F \cdot d r = ∣ F ∣∣ d r ∣ cos 9 0^{\circ} = 0$ .

Connections

Parent topic — the symbolic derivation this page illustrates.
Newton's Second Law — Step 2, our only assumed input.
Kinetic Energy — the $\frac{1}{2} m v^{2}$ that emerges in Step 6.
Work done by a variable force — Step 5, why work is an area, not a product.
Power — the rate $\frac{d W}{d t} = F \cdot v$ , if you keep the time-clock instead of dropping it.
Impulse–Momentum Theorem — the sibling derivation that multiplies by $d t$ instead of $d x$ .
Conservation of Mechanical Energy — what the theorem becomes when forces are conservative.