Conservation of mechanical energy — derivation: Visual walkthrough

Step 1 — What "energy" even is (two pockets)

WHAT. Before any algebra, meet the two quantities we track.

Kinetic energy $K$ — the "moving money." A thing that moves has it; a thing at rest has none. Its size is $K = \frac{1}{2} m v^{2}$ , where $m$ is how much stuff there is (mass) and $v$ is how fast it moves (speed).
Potential energy $U$ — the "stored money." A ball held high, or a squished spring, is loaded — release it and it will move. Its size depends on position, not speed.

WHY these two. They are the only two ways a mechanical system can hold energy: in its motion or in its configuration. Everything below is bookkeeping between these pockets.

PICTURE. Two jars. Right now the ball is high and still: the $U$ jar is full, the $K$ jar empty. The total liquid is what we will prove never changes.

Step 2 — What "work" measures (force times how far it pushes with you)

WHAT. A force $F$ acting while an object moves a tiny distance $d x$ does a tiny amount of work $d W = F d x$ . Work is the currency that moves money between the two jars.

WHY this quantity. We need a number that says "how much did this force change the motion?" A force that pushes along the motion speeds things up; one that pushes against it slows them. $F d x$ captures exactly that: positive when force and motion agree, negative when they fight.

PICTURE. A blue arrow (force) pushing a block rightward over a small gap $d x$ . The shaded strip is one sliver of work. Add up all the slivers along the path → total work $W$ .

Step 3 — Work–energy theorem (net work fills the moving jar)

WHAT. We show: the net work (from all forces added) equals the change in kinetic energy.

Start from Newton's second law — force causes acceleration: $F_{n e t} = ma = m \frac{d v}{d t} .$ Here $\frac{d v}{d t}$ is how fast the speed itself is changing (acceleration). Put this into the work sliver: $d W_{n e t} = F_{n e t} d x = m \frac{d v}{d t} d x .$

WHY the chain-rule swap. The right side still mentions time $t$ , but $K$ only cares about speed $v$ . We want time gone. Since a tiny step is speed times a tiny time, $d x = v d t$ : $m \frac{d v}{d t} d x = m \frac{d v}{d t} (v d t) = m v d v .$ The two $d t$ 's cancel — time has vanished, exactly what we wanted, leaving only $v$ .

PICTURE. A staircase of speed: as the object speeds from $v_{1}$ to $v_{2}$ , each thin vertical strip is $m v d v$ ; the whole triangle-ish area under the line $y = m v$ is the total, $\frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$ .

$W_{n e t} = \int_{v_{1}}^{v_{2}} m v d v = K_{2} \frac{1}{2} m v_{2}^{2} - K_{1} \frac{1}{2} m v_{1}^{2} = Δ K .$

Step 4 — What makes a force conservative (path doesn't matter)

WHAT. A conservative force does work that depends only on where you start and end, never on the route. Equivalently, a round trip back to the start does zero net work.

WHY this matters. If the work only depends on the two endpoints, we can bottle it as a number attached to position — a stored quantity. That bottled number is potential energy. Path-dependent forces (friction) can't be bottled this way: the longer the road, the more they steal.

PICTURE. Two paths from A to B under gravity — a straight drop and a lazy curve. The work by gravity is identical for both (pink labels equal). Beside it, a friction path: the long wiggly route (blue) loses more, so friction is not conservative.

Step 5 — Building potential energy (bottling the conservative work)

WHAT. For a conservative force we define its potential energy $U$ so that the work it does equals the drop in $U$ : $W_{co n s} = - Δ U = - (U_{2} - U_{1}) .$

WHY the minus sign. Lift a ball up: gravity pulls down while the ball goes up, so gravity does negative work — yet the ball is now more loaded, so $U$ went up. The two must have opposite signs, so we bake a minus into the definition. Sanity check: ball falls ⇒ gravity does $+$ work ⇒ $U$ drops. ✓ (See Potential energy.)

Now build gravity's $U$ concretely. Near Earth the force is $F = - m g$ (the minus = "points down"). Moving from height $h_{1}$ to $h_{2}$ : $W_{g r a v} = \int_{h_{1}}^{h_{2}} (- m g) d h = - m g (h_{2} - h_{1}) = - (U_{2} m g h_{2} - U_{1} m g h_{1}) .$ Match against $W_{co n s} = - Δ U$ and read off $U = m g h$ — derived, not assumed (Gravitational potential energy). The same recipe on a spring $F = - k x$ gives $U = \frac{1}{2} k x^{2}$ (Spring potential energy (Hooke's law)).

PICTURE. A ball climbing: down-arrow (gravity, negative work) alongside a rising green $U$ column labelled $m g h$ . Opposite signs, drawn as a see-saw.

Step 6 — Snapping the two halves together

WHAT. Suppose the only force doing work is conservative. Then the "net work" of Step 3 is the "conservative work" of Step 5: $Step 3 Δ K = W_{n e t} = W_{co n s} = Step 5 - Δ U .$

Move everything to one side: $Δ K + Δ U = 0 ⟹ Δ (K + U) = 0.$

WHY this is the whole game. $Δ (something) = 0$ means that something doesn't change. So $K + U$ is frozen — that is conservation of mechanical energy.

PICTURE. The two jars again, mid-fall: liquid pouring from the $U$ jar straight into the $K$ jar, level of the combined total marked by a flat dashed line that never moves.

Step 7 — The degenerate case: enter friction (the leak)

WHAT. Now let a non-conservative force also act — friction. Net work now has two parts: $W_{n e t} = W_{co n s} + W_{n c} .$ Feed this into the work–energy theorem: $Δ K = W_{co n s} + W_{n c} = - Δ U + W_{n c} \Rightarrow Δ (K + U) = W_{n c} .$

WHY it must not be ignored. Friction always opposes motion, so $W_{n c} < 0$ : the combined jar drops. The lost liquid isn't destroyed — it became heat, which lives outside the $K + U$ ledger (see Energy lost to friction). This is exactly the "no friction, no problem" caveat.

PICTURE. The same two jars, but now a small drip labelled "heat" leaks out the bottom — the flat total line tilts gently downward.

Step 8 — Degenerate case: zero-work forces (why tension is invisible)

WHAT. A force can be huge yet do no work, if it points perpendicular to the motion. Work is $d W = F d x$ measured along the motion; a sideways force contributes nothing.

WHY it matters. On a pendulum the string tension is large, but it always points along the string — perpendicular to the bob's swing. So $W_{t e n s i o n} = 0$ , gravity alone works, and energy is conserved. This is what licenses using conservation on a pendulum.

PICTURE. A swinging bob: velocity arrow (yellow) along the arc, tension arrow (blue) pointing to the pivot at a right angle — the little right-angle square shows their $9 0^{\circ}$ , hence zero work.

Recall Quick checks

Ball dropped from rest at height $h$ : speed at ground? ::: $m g h = \frac{1}{2} m v^{2} \Rightarrow v = 2 g h$ Spring compressed $x$ launches block ( $m$ ): speed? ::: $\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} \Rightarrow v = x k / m$ Ball rolls down $h$ , compresses spring by max $x$ : find $k$ ? ::: at max compression $K = 0$ , so $m g h = \frac{1}{2} k x^{2} \Rightarrow k = 2 m g h / x^{2}$

The one-picture summary

Everything above is one diagram: Newton's law → work–energy theorem gives $Δ K$ ; the definition of $U$ gives $- Δ U$ ; when only conservative forces work they are the same number, so $K + U$ is a flat line. Add friction and the line droops.

Recall Feynman: the whole walkthrough in plain words

You carry money in two pockets — a "moving" pocket and a "stored" pocket. Work is the only way to move money into your moving pocket, and Newton's law promises that every bit of net work lands there exactly (Step 3). For nice forces like gravity and springs, we noticed the work depends only on where you go, not how — so we bottled it as stored money, defining it with a minus sign so the arithmetic lines up (Step 5). Now if only those nice forces push you, the money leaving the stored pocket is precisely the money arriving in the moving pocket — your grand total never changes (Step 6). The only way to lose money is a thief called friction, who quietly siphons some off as heat that lives outside your two pockets (Step 7). And beware: some big-looking forces, like a string's pull, push sideways and move no money at all (Step 8). No thief, no sideways surprises — total frozen forever.

Visual walkthrough — Conservation of mechanical energy — derivation