AP Physics C: Mechanics · Topic 3.3

Topic 3.3: Potential Energy

Unit 3: Work, Energy, and Power15-25% of the multiple-choice section

Potential energy is a scalar belonging to a system whose parts interact through conservative forces. AP Physics C adds three things AP Physics 1 never states: get the potential energy by integrating the force, get the force back by taking the slope, and read equilibrium and stability off the graph.

AP Physics: Unit 3 (topics 3.3 Potential Energy). Topic 3.3 of the current AP Physics C: Mechanics course and exam description (Unit 3, weighted 15 to 25% of the multiple-choice section at about 12 to 17 class periods). One learning objective, 3.3.A, with eight numbered essential knowledge statements and seven sub-statements, the largest required-content block in Unit 3. No boundary statement, in either course. Suggested skills 1.C, 2.A, 2.B and 3.B. This is the topic where the two courses genuinely diverge: AP Physics 1's Topic 3.3 has five numbered statements to this course's eight, and the three extra ones are exactly the calculus content. 3.3.A.4 gives Delta U as the negative integral of the conservative force over the path, 3.3.A.5 gives F_x = -dU(x)/dx together with the statement that conservative forces point in the direction of decreasing potential energy, and 3.3.A.6 with its four sub-statements covers potential energy graphs and defines stable and unstable equilibrium (3.3.A.6.i and .ii by the direction of the force after a small displacement, 3.3.A.6.iii and .iv as local minima and local maxima of U). Both 3.3.A.4 and 3.3.A.5 are printed on the C: Mechanics sheet and appear on neither the AP Physics 1 sheet nor in the AP Physics 1 framework; the phrase stable equilibrium appears nowhere in the AP Physics 1 CED. Everything else is carried over word for word and renumbered: AP Physics 1's 3.3.A.1 to .3 are this course's 3.3.A.1 to .3, its 3.3.A.4 and subs are this course's 3.3.A.7 and subs, and its 3.3.A.5 is this course's 3.3.A.8. Notation note verified against the rendered appendix at 300 dpi: both frameworks write U_g with a lowercase g at the general gravitational form while both equation sheets print U_G with a capital G, keeping lowercase on Delta U_g = mg Delta y. Learning objective 3.3.A aligns to sample free-response Question 1 (Mathematical Routines, 10 points, alongside 3.1.A and 3.4.B) and to none of the fifteen sample multiple-choice questions; that question's part B asks for U(x) from F(x) = -beta x squared with U defined zero at x = -2 m, published answer U(x) = x cubed over 8, plus 1.

What Topic 3.3 requires, in full

Topic 3.3 carries one learning objective and eight numbered essential knowledge statements, two of which have sub-statements beneath them. It is the largest required-content block in Unit 3.

Learning objective 3.3.A: describe the potential energy of a system.

Statement	What the CED says
3.3.A.1	A system composed of two or more objects has potential energy if the objects within that system only interact with each other through conservative forces
3.3.A.2	Potential energy is a scalar quantity associated with the position of objects within a system
3.3.A.3	The definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis
3.3.A.4	The relationship between conservative forces exerted on a system and the system's potential energy is $\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$
3.3.A.5	The conservative forces exerted on a system in a single dimension can be determined using the slope of the system's potential energy with respect to position in that dimension; these forces point in the direction of decreasing potential energy. Relevant equation: $F_x = -\dfrac{dU(x)}{dx}$
3.3.A.6	Graphs of a system's potential energy as a function of its position can be useful in determining physical properties of that system
3.3.A.6.i	Stable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object opposite to the direction of the small displacement, accelerating the object back toward the equilibrium position
3.3.A.6.ii	Unstable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object in the same direction as the small displacement, accelerating the object away from the equilibrium position
3.3.A.6.iii	In a given dimension, stable equilibrium positions exist at locations where the potential energy as a function of position in that dimension has a local minimum
3.3.A.6.iv	In a given dimension, unstable equilibrium positions occur at locations where the potential energy as a function of position in that dimension has a local maximum
3.3.A.7	The potential energy of common physical systems can be described using the physical properties of that system
3.3.A.7.i	The elastic potential energy of an ideal spring is $U_s = \frac{1}{2}k(\Delta x)^2$ , where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length
3.3.A.7.ii	The general form for the gravitational potential energy of a system consisting of two approximately spherical distributions of mass (for example moons, planets, or stars) is $U_g = -G\dfrac{m_1m_2}{r}$
3.3.A.7.iii	Because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system of an object of mass $m$ and a planet with gravitational field of magnitude $g$ , when the object is near the surface, may be approximated by $\Delta U_g = mg\Delta y$
3.3.A.8	The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system

Suggested skills: 1.C (create qualitative sketches of graphs), 2.A (derive a symbolic expression), 2.B (calculate or estimate an unknown quantity with units) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Topic 3.3 prints no boundary statement, in either course.

Three statements that do not exist in AP Physics 1

This is the topic where the two courses genuinely separate, and the separation is countable. AP Physics 1's Topic 3.3 has five numbered statements. AP Physics C: Mechanics has eight. Map them and the pattern is clean:

AP Physics 1	AP Physics C: Mechanics	Relationship
3.3.A.1	3.3.A.1	word for word identical
3.3.A.2	3.3.A.2	word for word identical
3.3.A.3	3.3.A.3	word for word identical
does not exist	3.3.A.4	$\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$
does not exist	3.3.A.5	$F_x = -dU(x)/dx$ and the slope statement
does not exist	3.3.A.6 and its four sub-statements	potential energy graphs, stable and unstable equilibrium
3.3.A.4, .4.i, .4.ii, .4.iii	3.3.A.7, .7.i, .7.ii, .7.iii	word for word identical, renumbered
3.3.A.5	3.3.A.8	word for word identical, renumbered

So the calculus-based course inserts exactly three numbered statements, carrying seven statements in total once the sub-statements of 3.3.A.6 are counted, and shifts everything after them down by three. Nothing that AP Physics 1 says is removed or reworded.

The insert is not a matter of emphasis. The phrase "stable equilibrium" does not appear anywhere in the AP Physics 1 course and exam description, and neither does any derivative or integral relating force to potential energy. An AP Physics 1 student can describe where a system stores energy. An AP Physics C student is expected to move between $U(x)$ and $F(x)$ in both directions and to classify what the shape of $U(x)$ implies about the motion.

If the algebra-based treatment is what you need, it is at AP Physics 1 Topic 3.3. That page is complete for a course that stops at the three shared statements plus the three named potential energies. Everything in the next three sections of this page is beyond that exam.

The two printed calculus equations, used in both directions

Statements 3.3.A.4 and 3.3.A.5 are inverses of each other, and both are printed on the AP Physics C: Mechanics equation sheet. Neither appears on the AP Physics 1 sheet.

\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r} \qquad \qquad F_x = -\frac{dU(x)}{dx}

Force to potential energy. Given a conservative force as a function of position, integrate and negate. Every question of this shape has a second half that students skip: the constant of integration. The integral fixes $U$ only up to an additive constant, and 3.3.A.3 says choosing it is your decision. Sample free-response Question 1, part B, hands you $F_{\text{BC}}(x) = -\beta x^2$ , states that the potential energy is zero at $x = -2$ m, and asks for $U(x)$ . One of its three scoring points goes for any one of: relating $\Delta U$ to $U(x) - U(x')$ , using limits that let the definite integral be evaluated by setting $U(-2\ \mathrm{m}) = 0$ , or indicating a correct integration constant. The published answer is $U(x) = \frac{x^3}{8} + 1$ , and the $+1$ is that whole scoring point.

Potential energy to force. Given $U(x)$ , differentiate and negate. Statement 3.3.A.5 also states the physical reading in words, and it is worth carrying as a sentence rather than as a formula: the conservative forces point in the direction of decreasing potential energy. A ball rolls downhill. The minus sign is that sentence.

The subscript on $\vec{F}_{\text{cf}}$ is doing real work. It means conservative force, and 3.2.A.1.iii is the reason: potential energies are associated only with conservative forces. You cannot integrate a friction force and call the result a potential energy, because the answer would depend on which path you integrated along. See Topic 3.2 for that split.

One consistency check that costs nothing and catches sign errors: after integrating a force to get $U$ , differentiate your answer and confirm you get the force back with the right sign. Worked example 1 does that explicitly.

Equilibrium and stability from the shape of U

Statement 3.3.A.6 and its four sub-statements are the single largest new idea in Unit 3 for a student arriving from the algebra-based course. Read them in the order the CED prints them, because the order is the argument.

First the physics (3.3.A.6.i and .ii). Stable equilibrium is where a small displacement produces a force back toward the equilibrium position. Unstable equilibrium is where a small displacement produces a force away from it. Both definitions are stated in terms of the force, not the graph.

Then the graph (3.3.A.6.iii and .iv). Stable equilibrium positions exist where $U(x)$ has a local minimum. Unstable equilibrium positions occur where $U(x)$ has a local maximum.

The bridge is $F_x = -dU/dx$ . Equilibrium of any kind means $F_x = 0$ , so the graph's slope is zero. Which kind depends on what the slope does either side: at a minimum it goes from negative to positive, so the force goes from pushing right to pushing left, which restores. At a maximum, the reverse.

On the graph	Slope	Force just to the right	Verdict
local minimum	zero, rising either side	back toward the point	stable (3.3.A.6.iii)
local maximum	zero, falling either side	away from the point	unstable (3.3.A.6.iv)
flat region	zero across an interval	zero everywhere in it	equilibrium everywhere, neither

A note on method. The sign of the second derivative is a quick way to tell a local minimum from a local maximum, and worked example 2 uses it. But "curvature" is not the CED's language. Statements 3.3.A.6.iii and .iv say local minimum and local maximum, and a free-response answer should too, since the graph you are given may be piecewise linear with no curvature anywhere.

Two further readings a $U$ graph supports, both from conservation of energy rather than from Topic 3.3 itself:

Where the object is fastest. With the total energy constant, $K = E - U$ , so the speed peaks wherever $U$ is smallest along the accessible region. At the bottom of a well the force is zero and the speed is greatest, at the same point. Confusing those two is the most reliable error in this topic.
Where the motion stops and reverses. The object cannot go where $U(x) > E$ , so the motion is bounded by the positions at which $U(x) = E$ , where $K = 0$ . Neither course and exam description uses a name for those positions, so if you write one on the exam, define it. Worked example 2 finds a pair of them.

Choosing the zero, and two notation traps

Statement 3.3.A.3 says the zero is yours to choose. The definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis. It is identical in both courses and it is the reason the printed gravitational equations look inconsistent: $U_g = -Gm_1m_2/r$ puts the zero at infinite separation, while $\Delta U_g = mg\Delta y$ never commits to a zero at all because it only ever gives a change.

The rule that follows: every physical answer comes out of a difference, so it cannot depend on your choice. If a result changes when you move the zero, the result is wrong. Declare the choice before the first line of algebra and keep it.

Notation trap one: the subscript on the gravitational potential energy. The framework writes $U_g$ with a lowercase $g$ at 3.3.A.7.ii. The equation sheet prints $U_G$ with a capital $G$ on the same equation, while keeping the lowercase $g$ on $\Delta U_g = mg\Delta y$ . This is true of both the AP Physics 1 and the AP Physics C: Mechanics documents, so it is a framework-against-sheet difference rather than a course-against-course one. Neither is wrong and neither will be penalised. Recognising them as the same quantity when the sheet and the question disagree is the point.

Notation trap two: which $g$ . The Table of Information for AP Physics C: Mechanics prints $g = 9.8\ \mathrm{m/s^2}$ and $g = 9.8\ \mathrm{N/kg}$ . A boundary statement under Topic 1.3 says something else: both AP Physics C courses expect that for all situations in which a numerical quantity is required for $g$ , the value $g \approx 10\ \mathrm{m/s^2}$ will be used, then adds that students will not be penalised for correctly using the more precise commonly accepted values of $g = 9.81\ \mathrm{m/s^2}$ or $g = 9.8\ \mathrm{m/s^2}$ . Both statements sit in the same document. This site uses 9.8, the sanctioned value the appendix prints. The guide on whether g is 9.8 or 10 works the discrepancy through in full.

A last point on 3.3.A.1, which students skim. Potential energy belongs to a system, not to an object. A ball alone has no gravitational potential energy; the ball-and-Earth system does. The CED's unit opener names this as a misconception the unit is built to address, asking whether a single object can "have" potential energy. In an exam answer, name the system.

The three named potential energies, and the pairwise rule

Statement 3.3.A.7 and its three sub-statements are identical to AP Physics 1's, and all three equations are printed on both courses' sheets.

System	Equation	Where the zero sits
Ideal spring (3.3.A.7.i)	$U_s = \frac{1}{2}k(\Delta x)^2$	at the spring's equilibrium length
Two spherical masses (3.3.A.7.ii)	$U_g = -G\dfrac{m_1m_2}{r}$	at infinite separation
Object near a planet's surface (3.3.A.7.iii)	$\Delta U_g = mg\Delta y$	not fixed; the equation gives only a change

Read the conditions, because they are where the marks are. $\Delta x$ in the spring equation is the distance from the spring's equilibrium length, not from the origin of your coordinate system and not from the floor. The gravitational general form is stated for approximately spherical distributions of mass, with moons, planets and stars as the examples. The near-surface form is explicitly an approximation, justified in the statement itself by the gravitational field near the surface of a planet being nearly constant.

The calculus-based course can connect the last two rather than list them. Differentiate $U_g = -Gm_1m_2/r$ with respect to $r$ per 3.3.A.5 and you get back the inverse-square attraction the sheet prints as $\left\lvert \vec{F}_g \right\rvert = Gm_1m_2/r^2$ . Take the same expression at a planet's surface and let the separation change by a small $\Delta y \ll R$ , and the difference reduces to $mg\Delta y$ with $g = GM/R^2$ . Both are the sort of thing skill 2.A asks for.

Statement 3.3.A.8 is the one that gets forgotten in a three-body question. The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system. Three objects means three pairs, four objects means six, and in general $n$ objects give $n(n-1)/2$ terms. It is a sum over pairs and not over objects, which is the mistake it exists to prevent. Worked example 3 puts a number on the difference.

How Topic 3.3 is tested

Learning objective 3.3.A appears in exactly one question in the CED's sample set, and it is the biggest one: sample free-response Question 1, the Mathematical Routines question, worth 10 points. It aligns to 3.1.A, 3.4.B and 3.3.A, with skills 2.A, 3.B and 1.C. Objective 3.3.A appears in none of the fifteen sample multiple-choice questions.

It is Topic 3.3 in both directions in one sitting. Part A, worth 7 points, gives a graph of the potential energy $U_{\text{BA}}(x)$ of a two-object system, releases Object A from rest at $x = 3x_0$ , and asks for a derived expression for its speed at $x = 7x_0$ and then for a graph of the force exerted on Object A by Object B. Part B, worth 3 points, reverses direction: a different conservative force as a function of position, a stated zero, and asks for $U(x)$ .

Details from the published scoring guidelines that tell you what the graders want:

Two parts open with the instruction to begin the derivation by writing a fundamental physics principle or an equation from the reference information. That is a stated expectation, not a stylistic preference.
In part A.i, one point goes for a multistep derivation indicating either that the total energy at the two positions is equal or that $\Delta K = -\Delta U$ ; one for stating that the total energy is $U + K$ ; one for indicating that the velocity or the kinetic energy is zero at the release point; one for correct substitution of the potential energies, $\Delta U = -5U_0$ ; and one for the answer $v = \sqrt{10U_0/m_{\text{A}}}$ .
In part A.ii the force graph earns one point for showing zero force across the flat region of the potential energy graph and one point for drawing a constant force across one interval that has twice the magnitude of a constant force across another. Flat $U$ means zero force, and the steeper the potential energy graph, the larger the force. That is $F_x = -dU/dx$ read entirely off the shape.

So the realistic preparation for Topic 3.3 is not memorising three potential energy formulas. It is sketching $F(x)$ under a given $U(x)$ , and integrating a given $F(x)$ into a $U(x)$ with the constant handled, symbolically, showing the principle you started from.

Related pages. Topic 3.4 supplies the conservation statements that turn a $U$ graph into a speed, and Topic 3.2 supplies the integral that 3.3.A.4 negates. The potential energy curve glossary entry is the short definition, kinetic versus potential energy the short comparison, and the conservation of energy guide owns the energy-accounting routine. The Unit 3 hub lists every equation in the unit against what the sheet prints.

Integrating a force into a potential energy, constant included

A conservative force acts along the $x$ axis with $F_x(x) = -kx - cx^3$ , where $k = 40\ \mathrm{N/m}$ and $c = 300\ \mathrm{N/m^3}$ . This is a stiffening spring: the cubic term makes it harder to stretch the further it goes. Define $U(0) = 0$ . Find (a) $U(x)$ , (b) the potential energy at $x = 0.20$ m and how it compares with an ideal spring of the same $k$ , and (c) the speed of a 0.50 kg object released from rest at $x = 0.20$ m as it passes $x = 0$ .

Declare the convention: positive $x$ is displacement from the natural length, and the force above is the force the spring exerts on the object.
(a) Apply 3.3.A.4 in one dimension: $\Delta U = -\int_0^x F_x(x')\, dx' = -\int_0^x (-kx' - cx'^3)\, dx' = \int_0^x (kx' + cx'^3)\, dx'$ .
Using the power rule printed in the Calculus table of the Table of Information: $U(x) - U(0) = \dfrac{kx^2}{2} + \dfrac{cx^4}{4}$ .
The choice $U(0) = 0$ from 3.3.A.3 fixes the constant at zero, so $U(x) = \dfrac{kx^2}{2} + \dfrac{cx^4}{4} = 20x^2 + 75x^4$ joules with $x$ in metres.
Check by reversing, per 3.3.A.5: $-\dfrac{dU}{dx} = -(40x + 300x^3) = -kx - cx^3$ , which is the force we started from. Sign and coefficients both confirmed.
(b) $U(0.20) = 20(0.040) + 75(0.0016) = 0.80 + 0.12 = 0.92$ J.
An ideal spring with the same $k$ would store $U_s = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(40)(0.040) = 0.80$ J, which is the first term alone. The cubic term adds 0.12 J, or 15% more.
(c) Released from rest, so $E = U(0.20) = 0.92$ J and $K_i = 0$ . At $x = 0$ the potential energy is zero by our choice, so $K = 0.92$ J.
$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(0.92)}{0.50}} = \sqrt{3.68} = 1.9183$ m/s, so 1.9 m/s.
Note what an algebra-based course could not do here. $U_s = \frac{1}{2}k(\Delta x)^2$ at 3.3.A.7.i is stated for an ideal spring. This spring is not ideal, and without 3.3.A.4 there is no route to its stored energy at all.

(a) $U(x) = \frac{1}{2}kx^2 + \frac{1}{4}cx^4 = 20x^2 + 75x^4$ J, with the constant fixed to zero by the choice $U(0) = 0$ . (b) $U(0.20\ \mathrm{m}) = 0.92$ J, which is 15% more than the 0.80 J an ideal spring of the same $k$ would store. (c) 1.9 m/s.

A double well: equilibria, stability, and where the motion stops

A 1.0 kg object moves along the $x$ axis in a system whose potential energy is $U(x) = ax^4 - bx^2$ with $a = 2.0\ \mathrm{J/m^4}$ and $b = 8.0\ \mathrm{J/m^2}$ . Find (a) the force as a function of $x$ , (b) every equilibrium position and whether each is stable or unstable, and (c) for an object released from rest at $x = 1.0$ m, the two positions where the motion reverses and the maximum speed.

Declare the convention: positive $F_x$ points toward larger $x$ .
(a) Apply 3.3.A.5: $F_x = -\dfrac{dU}{dx} = -(4ax^3 - 2bx) = 2bx - 4ax^3 = 16x - 8.0x^3$ newtons, which factors as $8.0x(2.0 - x^2)$ .
(b) Equilibrium means $F_x = 0$ , so $x = 0$ or $x^2 = 2.0$ , giving $x = 0$ and $x = \pm 1.4142$ m.
Classify by 3.3.A.6.iii and .iv, which ask whether $U$ has a local minimum or a local maximum there. $U(0) = 0$ , and $U(\pm 1.4142) = 2.0(4.0) - 8.0(2.0) = 8.0 - 16 = -8.0$ J.
Since $U$ is lower on both sides of $x = 0$ than at $x = 0$ itself, $x = 0$ is a local maximum, so it is an unstable equilibrium (3.3.A.6.iv). The two outer points are local minima, so both are stable (3.3.A.6.iii).
Confirm with the second derivative, which is a method rather than CED language: $\dfrac{d^2U}{dx^2} = 12ax^2 - 2b = 24x^2 - 16$ . At $x = 0$ it is $-16\ \mathrm{J/m^2}$ , negative, a maximum. At $x = \pm 1.4142$ m it is $24(2.0) - 16 = +32\ \mathrm{J/m^2}$ , positive, a minimum.
(c) Released from rest at $x = 1.0$ m, so $E = U(1.0) = 2.0 - 8.0 = -6.0$ J and stays there while no nonconservative force acts.
The motion reverses where $K = 0$ , that is where $U(x) = -6.0$ J: $2.0x^4 - 8.0x^2 + 6.0 = 0$ , so $x^4 - 4.0x^2 + 3.0 = 0$ , which factors as $(x^2 - 1.0)(x^2 - 3.0) = 0$ .
That gives $x = 1.0$ m and $x = \sqrt{3.0} = 1.7321$ m. One of them had to be the release point, and finding it is the check on the algebra. Verify the other: $U(1.7321) = 2.0(9.0) - 8.0(3.0) = 18 - 24 = -6.0$ J.
That gives $x = 1.0$ m and $x = \sqrt{3.0} = 1.7321$ m. One of them had to be the release point, and finding it checks the algebra. Verify the other: $U(1.7321) = 18 - 24 = -6.0$ J. The object oscillates between those two and never reaches $x = 0$ , because $U(0) = 0$ is above its total energy of $-6.0$ J. It is trapped in the right-hand well.
Note that the speed peaks exactly where the force is zero. Both statements describe the same zero-slope point on the graph, and treating them as different places is the standard error in this topic.

(a) $F_x = 8.0x(2.0 - x^2)$ N. (b) Unstable equilibrium at $x = 0$ , a local maximum of $U$ ; stable equilibria at $x = \pm 1.41$ m, local minima where $U = -8.0$ J. (c) Released from rest at $x = 1.0$ m the object oscillates between $x = 1.0$ m and $x = 1.73$ m, with a maximum speed of 2.0 m/s at $x = 1.41$ m.

Three masses, three pairs (3.3.A.8)

Three small spheres of mass 2.0 kg, 3.0 kg and 4.0 kg sit at the corners of an equilateral triangle of side 0.50 m. Take $G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$ and the zero of gravitational potential energy at infinite separation. Find the total gravitational potential energy of the system, and state how much energy would be needed to separate all three to infinity.

Apply 3.3.A.8: the total is the sum over each pair, not over each object. Three objects give $n(n-1)/2 = 3$ pairs.
Each pair contributes $U_g = -Gm_im_j/r$ from 3.3.A.7.ii, and every separation here is the same 0.50 m, so the three terms share a factor.
Sum the mass products: $(2.0)(3.0) + (2.0)(4.0) + (3.0)(4.0) = 6.0 + 8.0 + 12 = 26\ \mathrm{kg^2}$ .
$U_{\text{total}} = -\dfrac{G}{r}\sum_{\text{pairs}} m_im_j = -\dfrac{(6.67 \times 10^{-11})(26)}{0.50}$ .
$(6.67 \times 10^{-11})(26) = 1.7342 \times 10^{-9}$ , and dividing by 0.50 gives $3.4684 \times 10^{-9}$ , so $U_{\text{total}} = -3.47 \times 10^{-9}$ J.
The energy needed to separate all three to infinity is the amount that brings the total to zero, which is $+3.47 \times 10^{-9}$ J.
Check the size of the error 3.3.A.8 exists to prevent. Counting only the heaviest pair gives $-(6.67 \times 10^{-11})(12)/0.50 = -1.60 \times 10^{-9}$ J, which is 46% of the correct value. Counting only the lightest pair gives $-8.00 \times 10^{-10}$ J, 23% of it.
Note the sign convention that comes with the choice of zero. Every pair term is negative because gravitation is attractive and the zero is at infinite separation, so a bound system always has negative total gravitational potential energy. Statement 3.3.A.3 permits a different zero, but this one is the one the printed equation assumes.

$U_{\text{total}} = -3.47 \times 10^{-9}$ J, the sum of three pair terms with mass products 6.0, 8.0 and 12 kg squared over the same 0.50 m separation. Separating all three to infinity requires $3.47 \times 10^{-9}$ J. Using one pair instead of all three would understate the magnitude by more than half.

Frequently asked questions

What is the formula for force from potential energy in AP Physics C?

The force is the negative derivative of the potential energy with respect to position, written F_x equals minus dU(x)/dx. Essential knowledge 3.3.A.5 of the AP Physics C: Mechanics course and exam description gives it as a relevant equation and it is printed on the AP Physics C: Mechanics equation sheet. The same statement puts the idea in words: the conservative forces exerted on a system in a single dimension can be determined using the slope of the system's potential energy with respect to position, and those forces point in the direction of decreasing potential energy. Neither the equation nor the statement exists in AP Physics 1.

How do you find potential energy from a force in AP Physics C?

Integrate the conservative force along the path and negate it. Essential knowledge 3.3.A.4 gives the change in potential energy as minus the integral from a to b of the conservative force dotted with the displacement element, and that equation is printed on the AP Physics C: Mechanics sheet. The integral fixes the potential energy only up to an additive constant, so you also have to use the problem's stated zero to pin that constant down. Essential knowledge 3.3.A.3 says choosing where the zero sits is the observer's decision. The CED's own sample free-response Question 1 awards a scoring point specifically for handling that constant correctly.

How do you find stable and unstable equilibrium from a potential energy graph?

Find where the slope is zero, then look at the shape. Since the force is the negative slope of the potential energy, equilibrium is wherever the graph is flat. Essential knowledge 3.3.A.6.iii says stable equilibrium positions exist where the potential energy as a function of position has a local minimum, and 3.3.A.6.iv says unstable equilibrium positions occur at a local maximum. The underlying definitions come first: 3.3.A.6.i says a small displacement from stable equilibrium produces a force opposite to the displacement, accelerating the object back, and 3.3.A.6.ii says a small displacement from unstable equilibrium produces a force in the same direction, accelerating it away.

Is AP Physics C Topic 3.3 harder than AP Physics 1 Topic 3.3?

It is genuinely larger. AP Physics 1's Topic 3.3 has five numbered essential knowledge statements and AP Physics C: Mechanics has eight. The three extra ones are the calculus content: the integral relating a conservative force to a potential energy at 3.3.A.4, the derivative relating a potential energy back to the force at 3.3.A.5, and potential energy graphs with stable and unstable equilibrium at 3.3.A.6 and its four sub-statements. Everything the algebra-based course states is carried over word for word and simply renumbered. The phrase stable equilibrium does not appear anywhere in the AP Physics 1 course and exam description.

Why is gravitational potential energy negative?

Because of where the zero was placed, not because of anything physical about the system. The equation printed on the sheet, U equals minus G m one m two over r, defines the potential energy to be zero when the two masses are infinitely far apart. Bringing them closer releases energy, so any finite separation gives a value below zero. Essential knowledge 3.3.A.3 states that the definition of zero potential energy is a decision made by the observer to simplify or assist in analysis, so a different choice would shift every value by a constant. Since only differences in potential energy ever appear in a physical answer, that shift cancels out.

Is the potential energy of three objects just the sum of three formulas?

It is a sum over pairs, which for three objects means three terms rather than three objects' worth of terms. Essential knowledge 3.3.A.8 states that the total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system. Three objects give three pairs, four objects give six pairs, and in general n objects give n times n minus one, all over two. Each pair term uses that pair's own two masses and its own separation, so identical spacing does not mean identical terms unless the masses match too.