Work-energy theorem — derivation from Newton's second law: Worked examples

First: what do all those integral symbols mean?

Before any example, we must nail down the one notation that every scenario below reuses. The parent note wrote work three ways — $\int F \cdot d r$ , $\int F d x$ , and $F d cos θ$ . A beginner deserves to see why these are the same thing.

Definition From the vector line integral down to

$\int F d x$ What is $d r$ ? As the particle moves along its path, $d r$ is one tiny step of displacement — a little arrow pointing in the direction the particle is currently heading, with length equal to the tiny distance covered in that step. Break the whole path into millions of such tiny arrows.

The work of one step is the dot product of the force there with that tiny step: $d W = F \cdot d r .$ Total work adds up every step along the path — that summation-over-a-path is what the symbol $\int F \cdot d r$ means (it is a line integral: an integral taken as you travel along the actual route).

Writing the dot product in components. If $F = (F_{x}, F_{y}, F_{z})$ and the tiny step is $d r = (d x, d y, d z)$ , then by the component rule for dot products: $F \cdot d r = F_{x} d x + F_{y} d y + F_{z} d z .$ This is the bridge: the abstract $F \cdot d r$ is literally "force-in- $x$ times step-in- $x$ , plus the same for $y$ and $z$ ."

The 1-D collapse. If the motion is only along the $x$ -axis, then $d y = d z = 0$ and only the first term survives: $F \cdot d r = F_{x} d x ⟶ W = \int F d x,$ where we now write $F$ for the (single) force component along the direction of motion. That is why the parent's $\int F d x$ is the same object as $\int F \cdot d r$ — it's just the 3-D form with two of its three pieces switched off.

The angle form. Using $F \cdot d r = ∣ F ∣ ∣ d r ∣ cos θ$ , and writing $d s = ∣ d r ∣$ for the tiny distance (a positive scalar) and $θ$ for the angle between force and motion: $d W = F cos θ d s ⟶ W = \int F cos θ d s .$ If additionally the force is constant and its angle $θ$ doesn't change, the integrand $F cos θ$ is a fixed number and pulls out of the sum, leaving $W = F cos θ \int d s = F d cos θ,$ where $d = \int d s$ is the total distance travelled (a single scalar magnitude, defined here so we never write a bare $d$ without meaning). This last form is the familiar $W = F d cos θ$ — but now you can see it is the most special case, not the definition.

Recall The four faces of "work"

Same quantity, four notations — pick by situation. $\int F \cdot d r$ ::: the general definition, any path, any geometry. $\int (F_{x} d x + F_{y} d y)$ ::: the component form, useful when you know $F$ and $d r$ coordinate-wise. $\int F d x$ ::: 1-D motion along $x$ ; $F$ is the force component along $x$ . $F d cos θ$ ::: force constant AND angle fixed — pulls out of the integral.

What does "net work" actually mean?

Every example below says $W_{n e t}$ . Let us pin down exactly what it is — because half the traps on this page come from misreading this one symbol.

Definition Net work = sum of the works of all forces

Suppose several forces $F_{1}, F_{2}, \dots, F_{n}$ act on a particle at once (your push, friction, gravity, the normal force...). Crucially, they all act on the same particle moving along the same single path, so every one of them is integrated over the same sequence of tiny steps $d r$ . Each does its own work $W_{i} = \int F_{i} \cdot d r$ . The net work is their sum: $W_{n e t} \equiv \sum_{i} W_{i} = \sum_{i} \int F_{i} \cdot d r = \int (F_{n e t} \sum_{i} F_{i}) \cdot d r = \int F_{n e t} \cdot d r .$ Read left to right: adding up each force's work is the same as first adding the forces into one net force $F_{n e t}$ and then finding its work. This works only because the path is common to all forces — the integral of a sum of things travelling the same route is the sum of the integrals.

(Caveat for later: for some forces — friction — the work depends on the actual path taken, not just start and end points; for others — gravity, springs — it depends only on endpoints. That path-dependence never breaks the sum above, but it matters when we meet Conservation of Mechanical Energy.)

So there are two equivalent recipes, and we use whichever is easier:

Recipe 1 (add the works): compute $W_{1}, W_{2}, \dots$ separately, then add — best when you know each force. (Used in Ex 6, 7, 9.)
Recipe 2 (add the forces first): find $F_{n e t}$ , then integrate once — best when forces combine simply. (Used in the kinematics checks.)

The theorem then reads, in full: $W_{n e t} = \int F_{n e t} \cdot d r = Δ K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} .$

The scenario matrix

Everything a work–energy problem can throw at you falls into one of these cells. We will hit every cell below.

#	Case class	What's tricky about it	Example
A	Constant force, parallel, speeding up	the "textbook" baseline	Ex 1
B	Constant force, opposing motion (negative work)	sign of $F \cdot d r$	Ex 2
C	Variable force (must integrate)	can't use $F d$	Ex 3
D	Force that changes sign mid-path	work can partly cancel	Ex 4
E	Force at an angle to motion	only the parallel part works	Ex 5
F	Multiple forces — need net work	one force cancels another	Ex 6
G	Degenerate: zero net work but motion happens	$Δ K = 0$ trap	Ex 7
H	Limiting / real-world word problem (stopping distance)	reason with the theorem symbolically	Ex 8
I	Exam twist: curved path, speed at the bottom	direction + variable geometry combined	Ex 9

Prerequisites we lean on: Newton's Second Law, Kinetic Energy, Work done by a variable force, and the Power relation at the very end.

Before the examples, one reminder that unlocks every sign question below.

Reading the figure (s01): three copies of the same rightward displacement arrow (grey, labelled "step $d r$ "). On the left, a coral force arrow lies flat along the motion ( $θ = 0^{\circ}$ ) — the label reads "positive W". In the middle, a butter-yellow force stands straight up, perpendicular to the motion ( $θ = 9 0^{\circ}$ ) — "zero W". On the right, a mint force points straight back against the motion ( $θ = 18 0^{\circ}$ ) — "negative W". The mint arc at each base shows the growing angle. As the force swings from flat, to upright, to reversed, $cos θ$ slides from $+ 1$ to $0$ to $- 1$ , and the work's sign follows it.

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

cos θ

sign

positive for

θ < 9 0^{\circ}

, zero at

9 0^{\circ}

, negative for

θ > 9 0^{\circ}

Example 1 — Cell A: constant force, speeding up

Worked example Baseline: sled on frictionless ice

A $4 kg$ sled starts at rest. A constant force $12 N$ pushes it (in the direction of motion) over $3 m$ of frictionless ice. Find the final speed.

Forecast: Guess before computing — will the answer be around $2$ , $3$ , or $6 m/s$ ? Jot it down.

Step 1. Motion is 1-D along $x$ , so the general $\int F \cdot d r$ collapses to $\int_{0}^{3} F d x$ . The force is constant ( $F = 12 N$ ) and parallel ( $θ = 0$ , $cos θ = + 1$ ), so it pulls out: $W_{n e t} = \int_{0}^{3} F cos θ d x = F cos 0^{\circ} \int_{0}^{3} d x = 12 (1) (3) = 36 J .$ Why this step? Only one force acts along the motion, so $W_{n e t} = W_{p u s h}$ . Because that force is constant and parallel, the integrand is a fixed number and the line integral becomes the plain product $F d$ over the distance $d = 3 m$ — this is the only situation where $W = F d$ replaces the full $\int F \cdot d r$ .

Step 2. $Δ K = W_{n e t} = 36 J$ . Since $K_{i} = \frac{1}{2} (4) (0)^{2} = 0$ , we get $K_{f} = 36 J$ . Why this step? The theorem: net work is the change in kinetic energy. Starting from rest means all of it becomes final $K$ .

Step 3. $\frac{1}{2} (4) v_{f}^{2} = 36 \Rightarrow v_{f}^{2} = 18 \Rightarrow v_{f} = 18 = 32 \approx 4.24 m/s$ . Why this step? Invert $K = \frac{1}{2} m v^{2}$ to recover speed.

Reading the figure (s02, single panel): a mint sled sits on a grey ice line at rest ( $v_{i} = 0$ ); a coral force arrow labelled $F = 12 N$ points along the motion, and a grey displacement bracket marks the $3 m$ path. Small grey tick-arrows along the axis are the tiny steps $d r$ — the force lies parallel to each, so every step's $cos θ = 1$ , which is the geometric reason $W = F d$ .

Verify: Cross-check with kinematics. $a = F / m = 12/4 = 3 m/s^{2}$ , and $v_{f}^{2} = v_{i}^{2} + 2 a d = 0 + 2 (3) (3) = 18$ ✓. Units: $[J] = [N\cdotm] = [kg\cdotm^{2} / s^{2}]$ , so $2 K / m$ has units $m/s$ ✓.

Example 2 — Cell B: constant force opposing motion

Worked example Friction only

A $1000 kg$ car moving at $20 m/s$ hits a rough patch where friction is a constant $5000 N$ directed backward. It skids $30 m$ . Find its speed at the end of the patch.

Forecast: Will it stop before $30 m$ , or still be rolling? Guess.

Step 1. $W_{f r i c} = \int_{0}^{30} F cos 18 0^{\circ} d x = 5000 \times 30 \times (- 1) = - 150, 000 J$ . Why this step? Friction points opposite to motion, so at every tiny step $θ = 18 0^{\circ}$ and $cos θ = - 1$ . Force is constant so it pulls out of the integral. It is the only horizontal force, so $W_{n e t} = W_{f r i c}$ , and it is negative — it removes kinetic energy.

Step 2. $K_{i} = \frac{1}{2} (1000) (20)^{2} = 200, 000 J$ . Why this step? We need the starting kinetic energy to see how much is left.

Step 3. $K_{f} = K_{i} + W_{n e t} = 200, 000 - 150, 000 = 50, 000 J$ . Why this step? Rearranged theorem: $K_{f} = K_{i} + W_{n e t}$ . Since $K_{f} > 0$ , the car is still moving.

Step 4. $\frac{1}{2} (1000) v_{f}^{2} = 50, 000 \Rightarrow v_{f}^{2} = 100 \Rightarrow v_{f} = 10 m/s$ .

Reading the figure (s03): the road line, with the car moving right (grey velocity arrow, $v_{i} = 20 m/s$ ) while a coral friction arrow points left, against the motion. This head-to-tail opposition is $θ = 18 0^{\circ}$ made visible — the arrow "fights" every tiny step $d r$ , which is exactly why its work carries a minus sign.

Verify: Kinetic energy fell to a quarter, so speed should fall to half ( $1/4 = 1/2$ ): $20 \to 10$ ✓. Sign sanity: work negative ⇒ slower ⇒ matches braking intuition ✓.

Example 3 — Cell C: variable force (must integrate)

Worked example A force that grows with position

A force $F (x) = 8 x N$ (with $x$ in metres), always pointing along the motion, acts on a $2 kg$ mass starting from rest, from $x = 0$ to $x = 3 m$ . Find the final speed.

Forecast: Would using $F (3) \times 3 = 24 \times 3 = 72 J$ be right or too big? Predict.

Step 1. Motion is 1-D and the force points along it, so $θ = 0$ and $cos θ = 1$ at every point — the general $\int F \cdot d r$ becomes $\int F (x) d x$ : $W = \int_{0}^{3} 8 x d x = [4 x^{2}]_{0}^{3} = 4 (9) - 0 = 36 J .$ Why this step? Two things happen here. First, because the force is always parallel to the motion we may set $cos θ = 1$ and drop the vector notation — this is what turns $F \cdot d r$ into $F d x$ . Second, the force is not constant, so $F cos θ$ can not be pulled out; the shortcut $F d$ is illegal and we must keep it inside the integral, per the variable-force rule. (Guessing $72 J$ overcounts because it uses the final, largest force everywhere.)

Step 2. $Δ K = 36 J$ , and $K_{i} = 0$ , so $K_{f} = 36 J$ . Why this step? Work–energy theorem applies to variable forces unchanged — the integral is the net work.

Step 3. $\frac{1}{2} (2) v_{f}^{2} = 36 \Rightarrow v_{f}^{2} = 36 \Rightarrow v_{f} = 6 m/s$ .

Verify: The work $36 J$ is the area of the triangle under the line $F = 8 x$ from $0$ to $3$ : base $3$ , height $F (3) = 24$ , area $\frac{1}{2} (3) (24) = 36$ ✓ — a picture-check with no calculus.

Example 4 — Cell D: force that changes sign mid-path

Worked example Push then pull

A $1 kg$ block, initially at rest at $x = 0$ , feels $F (x) = (4 - 2 x) N$ (always directed along the $x$ -axis) from $x = 0$ to $x = 4 m$ . Notice $F > 0$ for $x < 2$ and $F < 0$ for $x > 2$ . Find its speed at $x = 4$ .

Forecast: The force pushes forward for the first half and backward for the second half. Will the block end up moving, or back at rest?

Step 1. The force is along the $x$ -axis, so again $F \cdot d r = F (x) d x$ and we integrate the component directly: $W = \int_{0}^{4} (4 - 2 x) d x = [4 x - x^{2}]_{0}^{4} = (16 - 16) - 0 = 0 J .$ Why this step? Because the force stays parallel (or anti-parallel) to the $x$ -axis, its component is the signed value $4 - 2 x$ , so no separate $cos θ$ factor is needed — the sign of $F (x)$ itself carries the direction. Integrating through the sign change is automatic: the positive work from $0 \to 2$ exactly cancels the negative work from $2 \to 4$ .

Step 2. $Δ K = 0 \Rightarrow K_{f} = K_{i} = 0 \Rightarrow v_{f} = 0$ . Why this step? Zero net work means zero change in kinetic energy. It sped up then slowed down back to rest.

Reading the figure (s04): the lavender line is the force $F (x) = 4 - 2 x$ ; it starts at $+ 4 N$ and slopes down, crossing zero at $x = 2 m$ (marked by the arrow "force flips sign at x = 2 m"). The mint shaded triangle (from $0$ to $2$ ) is the positive work, $+ 4 J$ ; the coral triangle (from $2$ to $4$ ) hangs below the axis and is the negative work, $- 4 J$ . The two triangles are mirror images — identical size, opposite sign — so their areas cancel and the net work is exactly zero. That cancellation is the whole story of this example.

Verify: Split the integral. Forward part: $\int_{0}^{2} (4 - 2 x) d x = [4 x - x^{2}]_{0}^{2} = 8 - 4 = 4 J$ . Backward part: $\int_{2}^{4} (4 - 2 x) d x = [4 x - x^{2}]_{2}^{4} = (16 - 16) - (8 - 4) = - 4 J$ . Sum $= 0$ ✓.

Example 5 — Cell E: force at an angle

Worked example Pulling a suitcase by its handle

You pull a $10 kg$ suitcase across smooth floor with a $50 N$ force directed $6 0^{\circ}$ above horizontal, over $4 m$ . It starts at rest. (Floor is frictionless; the vertical part of your pull is taken up by a reduced normal force, doing no work since there's no vertical motion.) Find its final speed.

Forecast: Only part of your $50 N$ helps it move forward. More or less than half of $50$ ?

Step 1. Displacement is horizontal, force is tilted. Using the component form $F \cdot d r = F_{x} d x + F_{y} d y$ with $d y = 0$ (no vertical motion), only the horizontal component $F_{x} = F cos θ$ survives: $W = \int_{0}^{4} F cos θ d x = F d cos θ = 50 \times 4 \times cos 6 0^{\circ} = 200 \times 0.5 = 100 J .$ Why this step? The component form makes it explicit: the vertical part of the force ( $F_{y}$ ) multiplies $d y = 0$ and contributes nothing, so only $F cos θ$ does work. Force and angle are both constant, so $F cos θ$ pulls out to give $F d cos θ$ . This is the dot product doing its job — see the figure for the projection.

Step 2. $Δ K = 100 J$ , $K_{i} = 0 \Rightarrow K_{f} = 100 J$ .

Step 3. $\frac{1}{2} (10) v_{f}^{2} = 100 \Rightarrow v_{f}^{2} = 20 \Rightarrow v_{f} = 20 = 25 \approx 4.47 m/s$ .

Reading the figure (s05): the grey arrow along the floor is the $4 m$ displacement. The coral arrow tilted up at $6 0^{\circ}$ (mint arc marks the angle) is the full $50 N$ pull. Drop a dashed vertical line from its tip to the floor: the lavender arrow along the ground is the shadow (projection $F_{x} = F cos 60$ ) of the force onto the direction of motion, and its length is $25 N$ . Only that shadow does work — the taller the tilt, the shorter the shadow, the less work.

Verify: Horizontal force $= 50 cos 60 = 25 N$ , so $a = 25/10 = 2.5 m/s^{2}$ ; kinematics $v_{f}^{2} = 2 a d = 2 (2.5) (4) = 20$ ✓. Sanity: $100 J$ is exactly half of what a straight $50 N$ pull ( $200 J$ ) would give, matching $cos 60 = 0.5$ ✓.

Example 6 — Cell F: multiple forces, use NET work

Worked example Dragged against friction

A $6 kg$ crate at rest is pulled by a $40 N$ horizontal rope over $5 m$ . Kinetic friction opposes it with $16 N$ . Find the final speed.

Forecast: Applied work is $40 \times 5 = 200 J$ . Will the final $K$ be $200 J$ or less?

Step 1. $W_{p u l l} = \int_{0}^{5} 40 d x = + 40 \times 5 = + 200 J$ (parallel to motion, $cos 0 = 1$ ). Why this step? The rope pulls along the motion, constant and parallel, so its work is positive $F d$ .

Step 2. $W_{f r i c} = \int_{0}^{5} 16 cos 18 0^{\circ} d x = - 16 \times 5 = - 80 J$ (opposes motion). Why this step? Friction points backward, $cos 180 = - 1$ , negative work.

Step 3. $W_{n e t} = W_{p u l l} + W_{f r i c} = 200 + (- 80) = 120 J$ . Why this step? Here we use Recipe 1 from the net-work definition: add each force's work. The theorem uses net work — the sum over all forces, not just the one you apply. (Gravity and normal force are vertical, motion is horizontal, so their $F \cdot d r = F_{y} d y = 0$ and they drop out of the sum.)

Step 4. $Δ K = W_{n e t} = 120 J$ , and $K_{i} = 0$ , so $K_{f} = 120 J$ ; then $\frac{1}{2} (6) v_{f}^{2} = 120 \Rightarrow v_{f}^{2} = 40 \Rightarrow v_{f} = 40 = 210 \approx 6.32 m/s$ . Why this step? Apply the theorem: net work is the change in kinetic energy; from rest, $K_{f} = W_{n e t}$ , then invert $K = \frac{1}{2} m v^{2}$ for the speed.

Verify: Cross-check with Recipe 2 (add forces first): net force $= 40 - 16 = 24 N$ , $a = 24/6 = 4 m/s^{2}$ , $v_{f}^{2} = 2 a d = 2 (4) (5) = 40$ ✓ — both recipes agree, as the definition promised.

Example 7 — Cell G: zero net work, yet it moves (degenerate)

Worked example The classic trap: constant velocity

A $20 kg$ box is dragged $8 m$ along a floor at constant velocity by a horizontal force. Friction is $30 N$ . Find (a) the applied force, (b) the work you do, (c) the net work.

Forecast: If it's moving, surely net work is positive... isn't it? Watch out.

Step 1. Constant velocity ⇒ zero acceleration ⇒ net force $= 0$ , so applied force $=$ friction $= 30 N$ . Why this step? Newton's Second Law: $Δ v = 0 \Rightarrow a = 0 \Rightarrow F_{n e t} = 0$ .

Step 2. Your work: $W_{y o u} = + 30 \times 8 = + 240 J$ . Friction's work: $W_{f r i c} = - 30 \times 8 = - 240 J$ . Why this step? You push forward (positive), friction pushes back over the same distance (negative), equal magnitudes.

Step 3. $W_{n e t} = W_{y o u} + W_{f r i c} = 240 - 240 = 0 J$ . Why this step? Summing the works (Recipe 1) must give zero, because $Δ K = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = 0$ when speed doesn't change. The theorem predicted it before we added anything.

Verify: $Δ K = 0$ (no speed change) and $W_{n e t} = 0$ agree ✓. Lesson: "the object moved" tells you nothing about net work; only the change in speed does. This is the degenerate cell.

Example 8 — Cell H: limiting / real-world (stopping distance)

Worked example Why braking distance quadruples with speed

A car of mass $m$ brakes with a constant friction force $f$ . Show algebraically that stopping distance $d$ is proportional to $v^{2}$ , then compute: if $30 m/s$ needs $45 m$ to stop, what distance is needed from $60 m/s$ ?

Forecast: Double the speed — does stopping distance double, triple, or quadruple?

Step 1. Stopping means $K_{f} = 0$ . Net work is friction only: $W_{n e t} = - f d$ . Why this step? Only friction acts along the motion; it brings the car to rest.

Step 2. Theorem: $- f d = 0 - \frac{1}{2} m v^{2} \Rightarrow d = \frac{m v ^{2}}{2 f}$ . Why this step? Solve for $d$ . With $m$ and $f$ fixed, $d \propto v^{2}$ — that's the key structural result.

Step 3. Since $d \propto v^{2}$ , doubling $v$ multiplies $d$ by $2^{2} = 4$ : $d_{60} = 4 \times 45 = 180 m$ . Why this step? Ratio method — the unknown constants $m$ and $f$ cancel because they are the same in both cases: $\frac{d _{60}}{d _{30}} = \frac{m ( 60 ) ^{2} /2 f}{m ( 30 ) ^{2} /2 f} = (\frac{60}{30})^{2} = 4$ . We never need the numerical values of $m$ or $f$ at all.

Verify: Instead of guessing $m$ and $f$ , extract the single combination the data fixes. From $d = \frac{m v ^{2}}{2 f}$ , the deceleration is $a = \frac{f}{m} = \frac{v ^{2}}{2 d} = \frac{( 30 ) ^{2}}{2 ( 45 )} = \frac{900}{90} = 10 m/s^{2}$ — this one number is determined by the first case alone (no separate $m$ or $f$ needed). Reusing it at $60 m/s$ : $d = \frac{v ^{2}}{2 a} = \frac{( 60 ) ^{2}}{2 ( 10 )} = \frac{3600}{20} = 180 m$ ✓. This is the physics behind "twice as fast = four times the crash distance."

Example 9 — Cell I: exam twist (genuinely curved path)

Worked example Down a curved frictionless track (the full line-integral payoff)

A $0.5 kg$ bead slides from rest down a frictionless curved wire from height $h = 1.8 m$ to the bottom. The wire twists left and right as it descends. Gravity is $g = 10 m/s^{2}$ . Find its speed at the bottom — without knowing the wire's shape.

Forecast: The wire twists and turns; can we still get an answer? What does the normal force contribute?

Step 1. This is where the general $\int F \cdot d r$ earns its keep: the path is curved, so $d r$ changes direction at every point. Two forces act — gravity $F_{g}$ (straight down) and the wire's normal force $N$ . The normal force is always perpendicular to the motion, so at every tiny step $N \cdot d r = 0$ : it does zero work, no matter how the wire curves. Why this step? $θ = 9 0^{\circ}$ everywhere between $N$ and $d r$ ⇒ $cos 90 = 0$ . This is precisely why the shape doesn't matter — only gravity survives the line integral.

Step 2. For gravity, use the component form $F_{g} \cdot d r = F_{g, x} d x + F_{g, y} d y$ . Gravity has no horizontal component ( $F_{g, x} = 0$ ) and $F_{g, y} = - m g$ , so $F_{g} \cdot d r = - m g d y$ . Summing over the whole descent (total drop $h$ , with $d y$ negative going down): $W_{g r a v} = \int (- m g) d y = m g h = 0.5 \times 10 \times 1.8 = 9 J .$ Why this step? The component form isolates what matters: only the vertical travel $d y$ pairs with gravity. All the left–right wandering ( $d x$ ) multiplies $F_{g, x} = 0$ and contributes nothing — so the twisty path gives the same work as a straight vertical drop of $h$ .

Step 3. $W_{n e t} = W_{g r a v} + W_{N} = 9 + 0 = 9 J$ (Recipe 1). By the theorem $Δ K = W_{n e t} = 9 J$ ; with $K_{i} = 0$ , we get $K_{f} = 9 J$ . Why this step? Net work = sum of all forces' work; the normal force contributed zero, so gravity's work is the whole net work.

Step 4. $\frac{1}{2} (0.5) v_{f}^{2} = 9 \Rightarrow v_{f}^{2} = 36 \Rightarrow v_{f} = 6 m/s$ . Why this step? Invert $K = \frac{1}{2} m v^{2}$ for the final speed at the bottom.

Reading the figure (s06): a lavender curved wire snakes down from height $h = 1.8 m$ to the ground. At three points along it, grey tiny-step arrows $d r$ point along the wire (each in a different direction), a coral gravity arrow points straight down at each, and a mint normal-force arrow points perpendicular to the wire. Notice: the mint arrows are always at $9 0^{\circ}$ to the grey steps (zero work), while gravity's overlap with each step depends only on how much that step drops. Summed, only the total vertical drop $h$ counts.

Verify: The mass cancels in $\frac{1}{2} m v^{2} = m g h \Rightarrow v = 2 g h = 2 (10) (1.8) = 36 = 6 m/s$ ✓ — the same $2 g h$ as a free vertical drop, confirming the shape is irrelevant. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ ✓. This bridges to Conservation of Mechanical Energy.

Recall checkpoints

Recall The single sentence to remember

In every example above, the only equation was $\int F_{n e t} \cdot d r = Δ K$ , where net work means add up the work of every force. The dot product (via $F_{x} d x + F_{y} d y$ ) handled direction, the integral handled variation, and "net" meant sum over all forces. Master those three and no scenario is new.

Connections

Newton's Second Law — every "verify" step falls back to $F = ma$ .
Kinetic Energy — the $\frac{1}{2} m v^{2}$ we solve for each time.
Work done by a variable force — Examples 3, 4, 9.
Conservation of Mechanical Energy — Example 9 is its gateway.
Power — divide any of these works by the time taken to get average power.
Impulse–Momentum Theorem — the time-integrated sibling; contrast with these distance problems.