Classical-Mechanics · Unit 16 · Video 2 · Interactive Practice

Energy on Loan: The Closed-Loop Theorem for Conservative Forces

IKey Formulas

Formula	Name	What you need
$W = W_1 + W_2 = \displaystyle\int_A^B \vec{F}_c(1)\cdot d\vec{r}_1 + \int_B^A \vec{F}_c(2)\cdot d\vec{r}_2$	Loop work, leg by leg	An outbound path and a return path
$\displaystyle\int_B^A \vec{F}_c(2)\cdot d\vec{r}_2 = -\int_A^B \vec{F}_c(2)\cdot d\vec{r}_2$	Limit reversal	Traversing the return leg backwards flips every $d\vec{r}$
$\displaystyle\oint \vec{F}_c \cdot d\vec{r} = 0$	Closed-loop theorem	Path independence of $\displaystyle\int_A^B \vec{F}_c\cdot d\vec{r}$
$\Delta K_{\text{sys}} + \Delta U_{\text{sys}} = 0$	Closed system, internal conservative forces only	$\Delta U_{\text{sys}} \equiv -W_c$

Key Insight: Path independence and zero loop work are the same statement read two ways — reverse the return leg and the two identical integrals cancel. So a conservative force only ever lends kinetic energy; a non-conservative force such as drag does negative work on both legs and the deficit leaves as thermal energy.

IIVisualization 1 — Two Legs That Cancel

Does the work around a closed loop depend on which routes the outbound and return legs take?

IIIVisualization 2 — The Round Trip's Energy Books

A ball leaves your hand at $14$ m/s and returns to the same height — with how much of its kinetic energy?

Air drag b (kg/s) 0 Flight 0.00

IVVisualization 3 — Where You Draw the Boundary

The same flight, three boundaries: which forces count as internal, and how do the books change?

System

Gravity is external.

$W_{\text{ext}} = \displaystyle\oint \vec{F}^{\,g}\cdot d\vec{r} = 0$ over the flight, so $\Delta K_{\text{sys}} = 0$ : the ball is caught at its launch speed.

No $\Delta U_{\text{sys}}$ exists here — potential energy belongs to an interaction between two bodies inside the system.

Gravity is internal and conservative.

$\Delta U_{\text{sys}} = -W_c$ , and with no external work $\Delta K_{\text{sys}} + \Delta U_{\text{sys}} = 0$ .

The earth's own kinetic energy now sits inside the boundary, so its imperceptible share is counted — and returned.

Air resistance is internal and non-conservative.

Drag opposes the motion on both legs, so $W_{\text{drag}} \lt 0$ going up and coming down.

$\Delta K_{\text{sys}} + \Delta U_{\text{sys}} = W_{\text{drag}} = -E_{\text{thermal}} \lt 0$ : that share is spread over countless molecules and no rerun recovers it.

💡 The recoverable share earns a name of its own: $\Delta U_{\text{sys}} \equiv -W_c$ , the internal potential energy. The next video computes it directly from the force.

VQuiz Questions

Problem 1 · Reversing the Leg

Given: A conservative force does $W_1 = -45\ \text{J}$ on an object carried from $A$ to $B$ along path 1. The object is then carried from $B$ back to $A$ along a different route, path 2 — find the work $W_2$ done by that same force on the return leg.

✅ Correct! Path independence fixes

\int_A^B \vec{F}_c\cdot d\vec{r} = -45\ \text{J}

on every route, and reversing the limits flips the sign — so the loop closes at

-45 + 45 = 0

❌ Check the direction. Path independence makes the

A \to B

integral

-45\ \text{J}

on path 2 as well, but the object travels

B \to A

: every

d\vec{r}

reverses, so the sign flips.

❌ That is the whole loop, not one leg.

\oint \vec{F}_c\cdot d\vec{r} = 0

describes the round trip

W_1 + W_2

; each individual leg is generally nonzero.

❌ The route is exactly what does not matter. A conservative force has the same

A \to B

work on every path — that is the definition.

Show solution

Conservative means the $A \to B$ integral is route-blind:

\int_A^B \vec{F}_c(2)\cdot d\vec{r}_2 = \int_A^B \vec{F}_c(1)\cdot d\vec{r}_1 = -45\ \text{J}

The return leg is that same integral run backwards, so reversing the limits flips its sign:

W_2 = \int_B^A \vec{F}_c(2)\cdot d\vec{r}_2 = -\int_A^B \vec{F}_c(2)\cdot d\vec{r}_2 = +45\ \text{J}

Check the loop: $W = W_1 + W_2 = -45 + 45 = 0$ , as $\oint \vec{F}_c\cdot d\vec{r} = 0$ requires.

Problem 2 · Gravity's Books with Drag Present

Given: A ball is thrown straight up and caught at the launch height, and air resistance is not negligible — the ball is noticeably slower on the catch. Find the work done by gravity over the complete flight.

✅ Correct! Gravity is conservative and the flight is a closed loop in height, so

\oint \vec{F}^{\,g}\cdot d\vec{r} = -mg\,\Delta y = 0

— drag is irrelevant to gravity's ledger.

❌ Right deficit, wrong bookkeeper. The missing kinetic energy is drag's doing:

W^g + W^{\text{drag}} = \Delta K

, with

W^g = 0

and

W^{\text{drag}} \lt 0

on both legs.

❌ Drag changes the trajectory, not gravity's total.

W^g = -mg\,\Delta y

depends only on the net height change, which is zero however slowly or quickly the ball travels.

❌ Not quite. Gravity does

-mgh

climbing and

+mgh

falling back to the same height.

Show solution

For near-earth gravity $\vec{F}^{\,g} = -mg\,\hat{j}$ and $\vec{F}^{\,g}\cdot d\vec{r} = -mg\,dy$ , so on any route

W^g = \int -mg\,dy = -mg\,(y_f - y_i)

Up to the apex $h$ : $W^g_{\text{up}} = -mgh$ . Back down to the launch height: $W^g_{\text{down}} = +mgh$ . Total:

W^g = -mgh + mgh = \oint \vec{F}^{\,g}\cdot d\vec{r} = 0

Drag shortens $h$ and slows the ball, but it cannot put a nonzero number in gravity's column. The energy audit for the flight reads $\Delta K = W^g + W^{\text{drag}} = 0 + W^{\text{drag}} \lt 0$ — every joule missing on the catch was dissipated by drag.

Problem 3 · Auditing a Real Throw

Given: A ball of mass $m = 0.50\ \text{kg}$ leaves a hand straight up at $12\ \text{m/s}$ and returns to the launch height moving at $10\ \text{m/s}$ — find gravity's work over the round trip and the thermal energy generated.

Work done by gravity over the round trip

Thermal energy generated by air resistance

✅ Correct! Gravity's column closes at zero and the entire

11\ \text{J}

shortfall in kinetic energy is the dissipated share.

❌ Check gravity's column.

W^g = -mg\,\Delta y

, and the ball ends at the height it started, so

\Delta y = 0

for the full flight.

❌ Check the kinetic energies. Subtract

\tfrac{1}{2}mv^2

values — never square the difference of the speeds, and do not drop the factor

\tfrac{1}{2}m = 0.25\ \text{kg}

Show solution

Step 1 — gravity. The flight starts and ends at the same height, so $\Delta y = 0$ and

W^g = -mg\,\Delta y = \oint \vec{F}^{\,g}\cdot d\vec{r} = 0\ \text{J}

Step 2 — the kinetic energy audit.

K_i = \tfrac{1}{2}(0.50)(12)^2 = 36\ \text{J}, \qquad K_f = \tfrac{1}{2}(0.50)(10)^2 = 25\ \text{J}

\Delta K = 25 - 36 = -11\ \text{J}

Step 3 — assign the loss. The only forces are gravity and drag, so $\Delta K = W^g + W^{\text{drag}}$ :

-11\ \text{J} = 0 + W^{\text{drag}} \;\Longrightarrow\; W^{\text{drag}} = -11\ \text{J}

E_{\text{thermal}} = -W^{\text{drag}} = 11\ \text{J}

Common mistakes:

$\tfrac{1}{2}m(v_i - v_f)^2 = 0.25(2)^2 = 1.0\ \text{J}$ — squaring the difference instead of differencing the squares.
$\tfrac{1}{2}(v_i^2 - v_f^2) = 22\ \text{J}$ — forgetting the mass.
$v_i^2 - v_f^2 = 44\ \text{J}$ — dropping $\tfrac{1}{2}m$ altogether.

Problem 4 · A Loop That Does Not Close at Zero

Given: A block is pushed once around a closed rectangular loop on a horizontal tabletop, total perimeter $4.0\ \text{m}$ , against a kinetic friction force of constant magnitude $5.0\ \text{N}$ — find the work done around the loop by friction and by gravity.

Work done by friction around the loop

Work done by gravity around the loop

✅ Correct! Two forces, one closed path, opposite verdicts: friction bills the distance travelled, gravity bills only the net height change.

❌ The theorem has a hypothesis.

\oint \vec{F}\cdot d\vec{r} = 0

holds for conservative forces; friction is path-dependent, and a closed path is precisely where that shows up.

❌ Check the magnitude and sign. Friction opposes the motion on every segment, so each contributes

-f\,ds

and the total is

-f

times the full path length.

❌ Check gravity's geometry. The table is horizontal, so

\vec{F}^{\,g}

is perpendicular to every displacement — and the net height change is zero besides.

Show solution

Friction. Kinetic friction points opposite $d\vec{r}$ at every instant, so $\vec{f}\cdot d\vec{r} = -f\,ds$ and the contributions add instead of cancelling:

W^f = \oint \vec{f}\cdot d\vec{r} = -f \oint ds = -(5.0\ \text{N})(4.0\ \text{m}) = -20\ \text{J}

Reversing the direction of travel does not help — the sign flips on $d\vec{r}$ and on $\vec{f}$ , so $W^f$ is negative either way. That path dependence is exactly why friction is non-conservative.

Gravity. On a horizontal table $\vec{F}^{\,g} = -mg\,\hat{j}$ is perpendicular to every $d\vec{r}$ , so each leg contributes zero; equivalently $\Delta y = 0$ around the loop:

W^g = -mg\,\Delta y = 0\ \text{J}

The lesson. A closed path guarantees zero work only for a conservative force. The $20\ \text{J}$ friction removed became thermal energy in the block and tabletop, and no number of extra laps recovers it.

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