Classical-Mechanics · Unit 14 · Video 1 · Interactive Practice

Energy Transforms, Never Vanishes: Conservation, Systems, and Surroundings

IKey Formulas

Formula	Name	What you need
$\Delta E_j \equiv E_{j,\text{final}} - E_{j,\text{initial}}$	Change in form $j$	An initial and a final state
$\sum_{j=1}^{N} \Delta E_j = 0$	Conservation of energy	Every channel counted
$\Delta E = \Delta E_{\text{sys}} + \Delta E_{\text{sur}} = 0$	The energy principle	A system, a boundary, surroundings
$\Delta E_{\text{sur}} = 0 \;\Rightarrow\; \Delta E_{\text{sys}} = 0$	Unchanged surroundings	No energy crosses the boundary

Key Insight: Energy physics tracks changes, never absolute totals — and the ledger closes only when every channel is counted.

IIForms Change, the Total Does Not

100 J of gravitational energy falls, spins a turbine, and ends as electrical and thermal energy.

Process time 0.00

💡 The 15 J that ends up as thermal energy is the channel the next few units deliberately ignore — friction, and later the First Law of Thermodynamics, put it back into the ledger.

IIIThe Ledger Only Closes If Every Channel Is Counted

Drop one channel from the sum and the books refuse to balance — that residual is the missing physics.

Process channel 1 channel 2 channel 3 channel 4

IVSystem, Boundary, Surroundings

The same dam, four different boundaries: the physics is fixed, the bookkeeping is your choice.

Inside the boundary: Reservoir Turbine Alternator House heater

💡 Challenge: find a boundary with $\Delta E_{\text{sys}} = 0$ that energy still crosses in both directions.

VQuiz Questions

Problem 1 · Close the Sum

Given: A process has exactly three energy channels. Measurement gives $\Delta E_{\text{grav}} = -250\ \text{J}$ and $\Delta E_{\text{kin}} = +180\ \text{J}$ — find $\Delta E_{\text{therm}}$ .

✅ Correct! The two known changes leave a deficit of

70\ \text{J}

, and thermal energy is the channel that absorbs it.

❌ Close, but check the sign. The known changes sum to

-70\ \text{J}

, so the missing channel must be

+70\ \text{J}

for the total to vanish.

❌ Not quite. Conservation is a sum of signed changes: add them, do not subtract magnitudes.

Show solution

Conservation of energy says the changes sum to zero:

\sum_{j=1}^{3} \Delta E_j = \Delta E_{\text{grav}} + \Delta E_{\text{kin}} + \Delta E_{\text{therm}} = 0

(-250\ \text{J}) + (+180\ \text{J}) + \Delta E_{\text{therm}} = 0

-70\ \text{J} + \Delta E_{\text{therm}} = 0 \quad\Longrightarrow\quad \Delta E_{\text{therm}} = +70\ \text{J}

Gravitational energy lost $250\ \text{J}$ ; only $180\ \text{J}$ of it showed up as motion, so the remaining $+70\ \text{J}$ went into random molecular motion.

Problem 2 · Closed, Open, or Isolated?

Given: A sealed thermos of hot coffee cools on a table. No matter crosses the thermos wall, but the wall leaks a little energy into the room. Take the coffee plus the thermos as the system — which statement is correct?

✅ Correct! Energy crosses the boundary but matter does not — the defining case of a closed system, and whatever the coffee loses the room gains.

❌ Close, but not isolated. "Isolated" requires that no energy cross the boundary; the leaking wall rules it out, and

\Delta E_{\text{sys}} = 0

with it.

❌ Not quite. "Open" is about matter crossing, and conservation applies to system plus surroundings together, never to the system alone.

Show solution

Classify by what crosses the boundary:

Isolated — neither energy nor matter crosses.
Closed — energy crosses, matter does not.
Open — both energy and matter cross.

The thermos is sealed (no matter) but leaky to energy, so it is closed. The energy principle then gives

\Delta E_{\text{sys}} + \Delta E_{\text{sur}} = 0,

and since the coffee cools, $\Delta E_{\text{sys}} \lt 0$ , forcing $\Delta E_{\text{sur}} = -\Delta E_{\text{sys}} \gt 0$ . Energy is conserved; it simply moved across the boundary into the room.

Problem 3 · Same Chain, Your Boundary

Given: In the dam chain the reservoir's gravitational energy drops by $100\ \text{J}$ , the turbine and the alternator each pass their energy straight through ( $\Delta E = 0$ for both), and the house heater's thermal energy rises by $100\ \text{J}$ . Take the reservoir alone as the system.

What is $\Delta E_{\text{sys}}$ ?

For that same boundary, what is $\Delta E_{\text{sur}}$ ?

✅ Correct! The system loses

100\ \text{J}

, everything outside the boundary gains exactly that much, and

\Delta E_{\text{sys}} + \Delta E_{\text{sur}} = 0

❌ Check the system. The boundary encloses only the reservoir, so add up only the reservoir's own energy changes.

❌ That is the answer for a different boundary.

\Delta E_{\text{sys}} = 0

would hold if you enclosed the whole chain; the reservoir alone genuinely loses energy.

❌ Check the surroundings. The turbine, alternator and heater are all outside this boundary, and their changes are

0

0

and

+100\ \text{J}

Show solution

Step 1 — Add up what is inside the boundary. Only the reservoir is inside:

\Delta E_{\text{sys}} = \Delta E_{\text{grav}} = -100\ \text{J}

Step 2 — Add up what is outside. Turbine, alternator and house heater:

\Delta E_{\text{sur}} = 0 + 0 + (+100\ \text{J}) = +100\ \text{J}

Step 3 — Check the principle.

\Delta E_{\text{sys}} + \Delta E_{\text{sur}} = (-100\ \text{J}) + (+100\ \text{J}) = 0 \;\checkmark

Redrawing the boundary around all four stages would give $\Delta E_{\text{sys}} = 0$ and $\Delta E_{\text{sur}} = 0$ — the same physics, different bookkeeping.

Problem 4 · A Ledger That Refuses to Balance

Given: In the fusion reaction $p + d \rightarrow {}^{3}\text{He} + \text{photons}$ , an experimenter tallies only the nuclear internal energies and finds $\sum_j \Delta E_j = -5.5\ \text{MeV} \neq 0$ — what does the energy principle require?

✅ Correct! A nonzero sum never means energy vanished — it means an energy expression is wrong or a channel is uncounted, and here the photons are that channel.

❌ Not quite. The internal energy of the reactants exceeds that of the helium-3, and the difference is carried off by radiation at the instant of the reaction.

Show solution

The tally over nuclear internal energy alone gives

\Delta E_{\text{internal}} = E_{{}^{3}\text{He}} - \big(E_{p} + E_{d}\big) = -5.5\ \text{MeV}.

A nonzero total flags one of exactly two possibilities: an energy expression is wrong, or a channel has not been counted. Here it is the second — the reaction also emits photons, so

\Delta E_{\text{internal}} + \Delta E_{\text{rad}} = (-5.5\ \text{MeV}) + (+5.5\ \text{MeV}) = 0 \;\checkmark

The same logic runs at the frontier: the ledger of the accelerating universe currently refuses to balance, and the entry proposed to close it is dark energy.

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