Visual walkthrough — Kinetic energy — derivation

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Step 1 — What "moving fast" costs: a picture of a push

WHAT: We draw the block, the arrow of the push $F$, the length of the slide $s$, and the final speed $v$.

WHY these four things: They are the only actors in the story. $m$ = how much stuff there is to shove. $F$ = how hard we shove. $s$ = how far we shove. $v$ = the payoff, the speed we bought. Everything that follows is just a relationship between these four.

PICTURE: the block starts at rest (speed $0$) on the left and ends moving at $v$ on the right — the pale-yellow arrow is our push, the pink number is the distance it acted over.

Recall What is each symbol?

$m$ ::: the mass — how much matter, in kilograms. $F$ ::: the steady force we apply, in newtons. $s$ ::: the distance the block moves while the force acts. $v$ ::: the final speed, starting from rest.

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Step 2 — Turning "push over distance" into a number: work

WHAT: We multiply the push $F$ by the slide length $s$ and call the result $W$, the work.

WHY multiply, and why this tool? A big push over a tiny distance and a tiny push over a huge distance can transfer the same energy. Neither $F$ alone nor $s$ alone captures "how much energy went in" — only their product does. That product is exactly the definition of work. (Because our push and the motion point the same way, the dot product $\vec{F}\cdot \vec{s}$ is just $Fs$.)

PICTURE: think of a rectangle. Its height is $F$, its width is $s$, and its area is the work $W$. Energy = area of the force-vs-distance rectangle.

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Step 3 — Trading force for motion: Newton's second law

WHAT: We replace the letter $F$ with $ma$ inside our work equation.

WHY do this swap? We want the final answer written in the language of motion ($v$), not of forces. Newton's second law is the bridge: it says a force $F$ is precisely what it takes to give mass $m$ an acceleration $a$. So $F$ and $ma$ are two names for the same thing — and $ma$ is the name that talks about movement.

PICTURE: the same rectangle from Step 2, but now the height labelled $F$ is re-labelled $ma$ — the acceleration $a$ is the block speeding up, drawn as a growing speed arrow underneath.

Recall New symbol

$a$ ::: the acceleration — how fast the speed itself is increasing, in ${\text{m/s}}^2$.

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Step 4 — Getting rid of $a$ and $s$: the kinematics link

WHAT: We use an equation of motion to rewrite the combination $as$ as $\frac{1}{2}{v}^2$.

WHY this exact equation? Our work formula still contains $a$ and $s$, but we don't care about those — we care about the final speed $v$. This particular kinematic equation is special: it contains no time. It links speed, acceleration, and distance directly, so it lets us swap the unwanted $as$ for the wanted $v^2$ in one clean move. Since the block starts from rest, the starting speed $u=0$, and the $u^2$ term simply vanishes.

PICTURE: a speed-vs-distance graph. The curve climbs from $0$ up to $v$; the algebra says the quantity $as$ equals exactly half of $v^2$. Notice the square already sneaking in — this is where the squaring comes from.

Recall New symbol

$u$ ::: the initial speed. Here $u=0$ because the block starts at rest.

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Step 5 — The substitution: watch the formula appear

WHAT: We drop the result of Step 4 into the equation of Step 3. The $as$ becomes $\frac{1}{2}{v}^2$, and out pops the famous formula.

WHY does $W$ become $K$? The block had no motion energy at the start (it was still). Every joule of work we did had nowhere to go but into motion. So the work we spent is the kinetic energy the block now carries. That is the whole idea: $K$ is bottled-up work.

PICTURE: the term-by-term anatomy of the final formula — the $\frac{1}{2}$ (born in the kinematics), the $m$ (the mass we shoved), and the $v^2$ (the speed, squared, because $as$ carried a square).

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Step 6 — Edge case: what if the force varies?

WHAT: We repeat the derivation without assuming $F$ is constant, by slicing the slide into infinitely many tiny pieces and adding up the work in each.

WHY bother? Step 2's rectangle only works if $F$ stays the same the whole way. Real pushes wobble. The integral $\int Fds$ just means "add up $F\times (\text{tiny }ds)$ over every tiny slice" — the area under a curvy force graph instead of a rectangle. The trick $\frac{ds}{dt}=v$ swaps the variable of integration from distance to speed, and the answer collapses to the same $\frac{1}{2}m{v}^2$. Constant or varying, the result is identical — this is the general Work–Energy Theorem.

PICTURE: a wiggly force curve chopped into thin strips; the shaded total area is still the work, and it still equals $\frac{1}{2}m{v}^2$.

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Step 7 — Edge case: signs, zero, and reversed motion

WHAT: We cover the cases our clean "start from rest" story skipped.

WHY: A reader must never hit a scenario we didn't show. Three cases:

• Force along motion → $W>0$ → speed rises → $K$ increases. (Our whole derivation.)
• Force against motion (like friction) → the dot product $\vec{F}\cdot \vec{s}$ is negative → $W<0$ → $K$ decreases. Energy is drained, not added.
• Zero speed → $v=0$ → $K=\frac{1}{2}m(0{)}^2=0$. A still object carries no kinetic energy, exactly as common sense demands.

Never negative $K$: because $v^2\ge 0$ for any $v$ — moving left ($v<0$) gives the same $v^2$ as moving right. Direction is squared away. Only the change $K_f-K_i$ can be negative.

PICTURE: three little blocks — one speeding up (blue, $W>0$), one slowing under friction (pink, $W<0$), one frozen ($K=0$) — with the speed→$v^2$ curve showing why both $\pm v$ land on the same non-negative energy.

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The one-picture summary

Here is the whole journey on one board: push → work (rectangle area) → swap force for $ma$ → swap $as$ for $\frac{1}{2}{v}^2$ → out falls $\frac{1}{2}m{v}^2$.

Recall Feynman: the walkthrough in plain words

We shoved a still block across the ice. "How much energy did we spend?" is just "how hard × how far" — a rectangle's area, and we named it work. But we wanted the answer in terms of speed, not force, so we used Newton's rule that a push is really mass-times-speeding-up. Then a no-time equation of motion told us that speeding-up-times-distance is exactly half the speed squared. Slot that in, and the work we spent turns out to be one-half, times the mass, times the speed times itself. Because it's the speed times itself, going twice as fast costs four times the energy. And since anything times itself can't be negative, a moving thing's go-energy is never below zero — only the change can be, like when friction quietly steals it back.

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Connections

• Work — definition and dot product
• Work–Energy Theorem
• Newton's Second Law
• Equations of Motion (kinematics)
• Power — rate of doing work
• Momentum vs Energy
• Potential Energy
• Conservation of Mechanical Energy