Classical-Mechanics · Unit 16 · Video 3 · Interactive Practice

Defining Potential Energy from Newton's Third Law

IKey Formulas

Formula	Name	What you need
$\vec{F}_{2,1} = -\vec{F}_{1,2}$	Newton's Third Law	One internal pair; the first index names the source, the second the receiver
$W_c = \displaystyle\int_A^B \vec{F}_{2,1}\cdot d\vec{r}_{2,1}$	Two-body work, collapsed	The pair force and the relative displacement $d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2$ — neither individual path
$\Delta U_{\text{sys}} \equiv -W_c$	Change in internal potential energy	A conservative force, so the integral is path-independent and depends only on states $A$ and $B$
$\Delta K_{\text{sys}} + \Delta U_{\text{sys}} = 0$	The closed-system ledger	No external work and no non-conservative internal forces

Key Insight: The work of an internal pair depends only on how the separation changes, so no share of $U$ can be handed to either object. Potential energy is a property of the interaction — of the pair — never of the book or of the earth alone.

IIVisualization 1 — Two Work Integrals, One Relative Displacement

Each object collects its own work, yet the total depends only on how the separation changes.

Interaction

💡 Challenge: find a pair of moves that leaves the separation along the force line unchanged — each object is worked on, yet $W_c = 0$ and $\Delta U_{\text{sys}} = 0$ .

IIIVisualization 2 — Kinetic and Potential Energy Trade Exactly

Whatever kinetic energy the pair loses, its potential energy gains — and a round trip zeroes both.

Time 0.00 s

IVVisualization 3 — One Effective Particle of Mass $\mu$

The pair's relative motion obeys $\vec{F}_{2,1} = \mu\,d\vec{v}_{2,1}/dt$ — a single particle of blended mass.

Partner mass (m₂) 3.00

💡 When $m_2 \gg m_1$ the reduced mass approaches $m_1$ and the heavy partner barely moves — which is why lifting a book looks as though the book alone holds the energy.

VQuiz Questions

Problem 1 · Collapsing One Step

Given: Two objects interact through a single conservative internal force. Over one small step the force on object 1 due to object 2 is $\vec{F}_{2,1} = (4\hat{\imath} - 3\hat{\jmath})\ \text{N}$ , object 1 moves through $d\vec{r}_1 = (0.5\,\hat{\imath})\ \text{m}$ and object 2 through $d\vec{r}_2 = (0.2\,\hat{\imath} + 0.4\,\hat{\jmath})\ \text{m}$ — find the work $W_c$ done by the pair.

✅ Correct!

d\vec{r}_{2,1} = (0.3\,\hat{\imath} - 0.4\,\hat{\jmath})\ \text{m}

, and one dot product with

\vec{F}_{2,1}

delivers both objects' work at once.

❌ You added the displacements. Newton's Third Law puts a minus sign in front of the second integral:

\vec{F}_{1,2} = -\vec{F}_{2,1}

, so the two displacements subtract. Using

d\vec{r}_1 + d\vec{r}_2

gives

1.6\ \text{J}

; the relative displacement is

d\vec{r}_1 - d\vec{r}_2

❌ That is object 1's share only.

\vec{F}_{2,1}\cdot d\vec{r}_1 = 2.0\ \text{J}

\Delta K_1

. Object 2 also has work done on it,

\vec{F}_{1,2}\cdot d\vec{r}_2 = 0.4\ \text{J}

, and the pair's work is the sum.

❌ That is $\Delta U_{\text{sys}}$ , not $W_c$ . The minus sign belongs to the definition

\Delta U_{\text{sys}} \equiv -W_c

, not to the work integral itself.

❌ Not quite. Build

d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2

first, then take a single dot product with

\vec{F}_{2,1}

Show solution

Add the two work-energy statements and use the Third Law on the second one:

W_c = \vec{F}_{2,1}\cdot d\vec{r}_1 + \vec{F}_{1,2}\cdot d\vec{r}_2 = \vec{F}_{2,1}\cdot d\vec{r}_1 - \vec{F}_{2,1}\cdot d\vec{r}_2 = \vec{F}_{2,1}\cdot\big(d\vec{r}_1 - d\vec{r}_2\big)

The relative displacement is

d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2 = (0.5 - 0.2)\hat{\imath} + (0 - 0.4)\hat{\jmath} = (0.3\,\hat{\imath} - 0.4\,\hat{\jmath})\ \text{m}

W_c = (4)(0.3) + (-3)(-0.4) = 1.2 + 1.2 = 2.4\ \text{J}

Check the long way: $\vec{F}_{2,1}\cdot d\vec{r}_1 = 2.0\ \text{J}$ and $\vec{F}_{1,2}\cdot d\vec{r}_2 = (-4)(0.2) + (3)(0.4) = 0.4\ \text{J}$ , and $2.0 + 0.4 = 2.4\ \text{J}$ ✓. The collapsed form never needed the two separate paths.

Problem 2 · Carrying the Whole Pair

Given: Two objects interacting through a conservative internal force with $\vec{F}_{2,1} = (5\,\hat{\imath})\ \text{N}$ are each carried by external hands through the same displacement $d\vec{r}_1 = d\vec{r}_2 = (2\,\hat{\imath})\ \text{m}$ , so their separation never changes — find $\Delta U_{\text{sys}}$ .

✅ Correct! The relative displacement is zero, so the pair's work is zero — the

+10\ \text{J}

delivered to object 1 is exactly the

10\ \text{J}

taken from object 2.

❌ You kept only one object's work. Each object does have work done on it,

\pm 10\ \text{J}

, but the two terms are equal and opposite because

\vec{F}_{1,2} = -\vec{F}_{2,1}

acts through the same displacement.

❌ The two works subtract, not add. Adding

|W_1| + |W_2|

ignores the Third Law's minus sign — the second term is

-10\ \text{J}

, not

+10\ \text{J}

❌ Not quite. Compute

d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2

before doing anything else.

Show solution

The relative displacement vanishes:

d\vec{r}_{2,1} = d\vec{r}_1 - d\vec{r}_2 = (2\,\hat{\imath}) - (2\,\hat{\imath}) = \vec{0}

Therefore

W_c = \vec{F}_{2,1}\cdot d\vec{r}_{2,1} = 0 \qquad\Longrightarrow\qquad \Delta U_{\text{sys}} = -W_c = 0

Object by object: $\vec{F}_{2,1}\cdot d\vec{r}_1 = (5)(2) = +10\ \text{J}$ and $\vec{F}_{1,2}\cdot d\vec{r}_2 = (-5)(2) = -10\ \text{J}$ ; the pair transfers $10\ \text{J}$ from object 2 to object 1 and stores nothing.

The external hands may well change the system's kinetic energy, but $\Delta U_{\text{sys}}$ answers only to the internal pair, and that force sees only the separation — which never moved.

Problem 3 · Reduced Mass and the Ledger

Given: A closed system of $m_1 = 2.0\ \text{kg}$ and $m_2 = 6.0\ \text{kg}$ interacts only through one conservative internal force. Their relative speed grows from $v_A = 1.0\ \text{m/s}$ to $v_B = 3.0\ \text{m/s}$ — find the reduced mass and $\Delta U_{\text{sys}}$ .

What is the reduced mass?

What is the change in potential energy?

✅ Correct!

\mu = 1.5\ \text{kg}

gives

W_c = \Delta K_{\text{sys}} = +6.0\ \text{J}

, so the interaction spent

6.0\ \text{J}

of its stored energy.

❌ That is the arithmetic mean. The reduced mass blends the reciprocals:

1/\mu = 1/m_1 + 1/m_2

, so

\mu

is always smaller than the lighter mass.

❌ That is $m_1 + m_2$ . The sum is the denominator of

\mu = m_1 m_2/(m_1+m_2)

, not the answer.

❌ Check the combination.

\mu = m_1 m_2/(m_1+m_2)

, and it can never exceed the smaller of the two masses.

❌ Check the sign. The relative speed grew, so

\Delta K_{\text{sys}} > 0

; since

\Delta U_{\text{sys}} = -\Delta K_{\text{sys}}

, the potential energy must fall.

❌ You squared the difference. The formula carries

v_B^2 - v_A^2 = 9 - 1 = 8\ \text{m}^2/\text{s}^2

, not

(v_B - v_A)^2 = 4\ \text{m}^2/\text{s}^2

❌ The factor $\tfrac{1}{2}$ is missing.

W_c = \tfrac{1}{2}\mu(v_B^2 - v_A^2)

, exactly like

K = \tfrac{1}{2}mv^2

for one particle.

❌ Not quite. Compute

W_c = \tfrac{1}{2}\mu(v_B^2 - v_A^2)

first, then negate it.

Show solution

Step 1 — reduced mass. Subtracting the two Second-Law statements gives $d\vec{v}_{2,1}/dt = (1/m_1 + 1/m_2)\vec{F}_{2,1}$ , so the bracket is $1/\mu$ :

\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{(2.0)(6.0)}{2.0 + 6.0} = \frac{12}{8} = 1.5\ \text{kg}

Step 2 — work from the relative speeds. With $d\vec{r}_{2,1} = \vec{v}_{2,1}\,dt$ the integrand is an exact time derivative, and only the endpoints survive:

W_c = \int_A^B \mu\,\frac{d\vec{v}_{2,1}}{dt}\cdot\vec{v}_{2,1}\,dt = \tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right) = \tfrac{1}{2}(1.5)\big(3.0^2 - 1.0^2\big) = \tfrac{1}{2}(1.5)(8.0) = 6.0\ \text{J}

Step 3 — the ledger. $W_c = \Delta K_{\text{sys}} = +6.0\ \text{J}$ , and by definition

\Delta U_{\text{sys}} = -W_c = -6.0\ \text{J}, \qquad \Delta K_{\text{sys}} + \Delta U_{\text{sys}} = 0\ \checkmark

Note that $\mu = 1.5\ \text{kg}$ is smaller than either mass — the relative motion always responds like a particle lighter than both partners.

Problem 4 · Several Interactions, One Ledger

Given: A closed system of three objects has two conservative internal interactions and no non-conservative ones. Going from state $A$ to state $B$ , interaction 1 does $W_{c,1} = +8.0\ \text{J}$ and interaction 2 does $W_{c,2} = -3.0\ \text{J}$ — find the two energy changes, then the result of a round trip back to $A$ .

From state A to state B

Now back from B to A by any route

✅ Correct! Works add, so

W_c = +5.0\ \text{J}

and the pair energies trade one for one; and because

U

is a state quantity, returning to configuration

A

restores it exactly.

❌ The signs are swapped. Positive total work means the interactions released stored energy:

\Delta U_{\text{sys}} = -W_c < 0

while

\Delta K_{\text{sys}} = +W_c > 0

❌ You added the magnitudes. Interaction 2 does negative work, so the total is

8.0 - 3.0 = 5.0\ \text{J}

, not

8.0 + 3.0

❌ They cannot both fall. In a closed system with no non-conservative forces,

\Delta K_{\text{sys}} + \Delta U_{\text{sys}} = 0

— one rises exactly as much as the other falls.

❌ Not quite. Add the works first:

W_c = W_{c,1} + W_{c,2}

, then apply

\Delta U_{\text{sys}} = -W_c = -\Delta K_{\text{sys}}

❌ Conservative forces have no memory of the route. Each

\Delta U_{\text{sys},i}

depends only on the initial and final states, so a closed excursion returns every term to its starting value.

❌ Check what a round trip means. The final state is state

A

, so every

U

term ends where it began.

Show solution

Step 1 — one term per interaction. Work is additive, so the potential energy changes add too:

\Delta U_{\text{sys}} = \Delta U_{\text{sys},1} + \Delta U_{\text{sys},2} = -W_{c,1} - W_{c,2} = -(8.0 - 3.0) = -5.0\ \text{J}

Step 2 — the ledger. With no external work and no non-conservative internal forces,

\Delta K_{\text{sys}} = W_c = +5.0\ \text{J}, \qquad \Delta K_{\text{sys}} + \Delta U_{\text{sys}} = 5.0 - 5.0 = 0\ \checkmark

Step 3 — the round trip. Each force is conservative, so its work around a closed path vanishes:

\oint \vec{F}_{c,i}\cdot d\vec{r}_i = 0 \qquad\Longrightarrow\qquad \Delta U_{\text{sys}} = 0 \quad\text{and}\quad \Delta K_{\text{sys}} = 0

On the way back the interactions must do $-5.0\ \text{J}$ of work in total, cancelling the $+5.0\ \text{J}$ of the outbound leg — the energy the system spent is handed back in full.

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Defining Potential Energy from Newton's Third Law

IKey Formulas

IIVisualization 1 — Two Work Integrals, One Relative Displacement

IIIVisualization 2 — Kinetic and Potential Energy Trade Exactly

IVVisualization 3 — One Effective Particle of Mass μ\muμ

VQuiz Questions

IVVisualization 3 — One Effective Particle of Mass $\mu$