AP Physics C: Mechanics · Unit 3 of 7

Unit 3: Work, Energy, and Power

15-25% of the multiple-choice section5 topics

Topics in this unit

3.1Translational Kinetic Energy
3.2Work
3.3Potential Energy
3.4Conservation of Energy
3.5Power

Work, Energy, and Power is Unit 3 of AP Physics C: Mechanics, 15 to 25 percent of the multiple-choice section over about 12 to 17 class periods. Five topics, seven objectives. Work is an integral here, and the two formulas AP Physics 1 prints for work and power are demoted to derived cases.

AP Physics: Unit 3 (topics 3.1 Translational Kinetic Energy, 3.2 Work, 3.3 Potential Energy, 3.4 Conservation of Energy, 3.5 Power). Unit 3 of the current AP Physics C: Mechanics course and exam description, weighted 15 to 25% of the multiple-choice section (the widest range in the course) at about 12 to 17 class periods. Five topics and seven learning objectives (3.1.A; 3.2.A; 3.3.A; 3.4.A, 3.4.B, 3.4.C; 3.5.A). Two boundary statements, under Topics 3.2 and 3.4; Topics 3.1, 3.3 and 3.5 print none. Topic 3.2's says the course only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound. Topic 3.4's says the course expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. The calculus differentiator is that the sheet prints work as the integral of force dotted with displacement (3.2.A.3) and instantaneous power as dW/dt (3.5.A.4), while W = F_parallel d = Fd cos theta (3.2.A.3.iii) and P_inst = F_parallel v = Fv cos theta (3.5.A.5) are labelled derived equations and are not printed, even though both are printed on the AP Physics 1 sheet. Unit 3 also adds two equations absent from the algebra-based course, both printed: Delta U as the negative integral of the conservative force (3.3.A.4) and F_x as the negative derivative of U with respect to x (3.3.A.5). Of fifteen equation entries in the unit, three are not printed: the two derived special cases and Delta E_mech = F_f d cos theta at 3.2.A.4.iii. Objective 3.4.B appears in three of the four sample free-response questions, and sample free-response Question 1 (Mathematical Routines, 10 points) is a potential-energy-curve question built on 3.3.A.5 and 3.3.A.4 used in both directions.

What calculus changes: two formulas get demoted

The AP Physics 1 equation sheet prints work as $W = F_\parallel d = Fd\cos\theta$ and instantaneous power as $P = F_\parallel v = Fv\cos\theta$ . Neither appears on the AP Physics C: Mechanics equation sheet. What appears instead is this:

W = \int_a^b \vec{F} \cdot d\vec{r} \qquad P_{\text{inst}} = \frac{dW}{dt}

Essential knowledge 3.2.A.3 introduces the first as the work done on an object by a variable force, and adds that the integral is taken over the path from point $a$ to point $b$ . Statement 3.5.A.4 gives the second.

The two familiar formulas have not disappeared. They have been reclassified. Statement 3.2.A.3.iii says that if the component of the force parallel to the displacement is constant, the work is given by the derived equation $W = F_\parallel d = Fd\cos\theta$ . Statement 3.5.A.5 says the instantaneous power delivered by the component of a constant force parallel to the velocity can be described with the derived equation $P_{\text{inst}} = F_\parallel v = Fv\cos\theta$ . The CED's Required Equations page defines a derived equation as one provided to demonstrate the final results of derivations expected of students on the exam, and neither is on the sheet.

So the relationship between the two courses in this unit is not that one is harder. It is that the algebra-based course is handed the special case and the calculus-based course is handed the general one and expected to produce the special case from it.

Potential energy moves the same way. Statement 3.3.A.4 gives the relationship between conservative forces and potential energy as an integral, and 3.3.A.5 inverts it as a derivative:

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

Both are printed. Statement 3.3.A.5 states the physical reading of the derivative in words as well: 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, and these forces point in the direction of decreasing potential energy.

That pair is the reason this unit exists in calculus form. Given a force law, you integrate to get a potential energy. Given a potential energy curve, you differentiate to get the force at every point. An algebra-based course can read a bar chart; this one can move between $U(x)$ and $F(x)$ in either direction.

What the CED requires across Unit 3

Unit 3 of AP Physics C: Mechanics is Work, Energy, and Power. The course and exam description weights it at 15 to 25% of the multiple-choice section, the widest range of any unit in the course, and suggests about 12 to 17 class periods. Only Unit 2 is weighted higher, at 20 to 25%. Unit 4 sits at 10 to 20% and Units 1, 5, 6 and 7 at 10 to 15% each.

Five topics and seven learning objectives.

Topic	Learning objectives	Suggested skills
3.1 Translational Kinetic Energy	3.1.A	1.C, 2.C, 3.B, 3.C
3.2 Work	3.2.A	1.A, 1.C, 2.A, 2.D, 3.B
3.3 Potential Energy	3.3.A	1.C, 2.A, 2.B, 3.B
3.4 Conservation of Energy	3.4.A, 3.4.B, 3.4.C	1.B, 2.A, 2.C, 3.A, 3.C
3.5 Power	3.5.A	1.A, 2.B, 2.C, 3.B

Four of the five topics carry exactly one learning objective, and Topic 3.4 carries three. Skill 2.A, deriving a symbolic expression, is listed for Topics 3.2, 3.3 and 3.4, the three topics in the middle of the unit.

The CED's framing is that students are introduced to the idea of conservation as a foundational principle of physics, along with the concept of work as the primary agent of change for energy, and that students will be encouraged to call on their knowledge of Units 1 and 2 to determine the most appropriate technique for approaching a problem and will be challenged to understand the limiting factors of each technique. That last clause is the honest description of what this unit is for: knowing when to use energy rather than forces, and when energy will not answer the question.

The unit's "Building the Science Practices" page names four skills, 1.A, 1.C, 2.A and 3.C, and names two misconceptions it is meant to dispel: whether a force does work on an object that does not move, and whether a single object can "have" potential energy. Statement 3.3.A.1 settles the second, since potential energy requires a system of two or more objects interacting only through conservative forces.

The "Preparing for the AP Exam" note ties Unit 3 to the first free-response question, Mathematical Routines, and then adds a caution worth repeating: while Unit 3 offers content perfect for practicing the Mathematical Routines question, that question can pull content from any of the seven units of the course.

The Unit 3 equations, and what the sheet prints

Unit 3 is the unit where the AP Physics C: Mechanics equation sheet is most generous. Almost everything the framework marks as a relevant equation for this unit is printed.

Equation	Where	Printed
$K = \frac{1}{2}mv^2$	3.1.A.1	yes
$W = \int_a^b \vec{F}(r) \cdot d\vec{r}$	3.2.A.3	yes
$\vec{A} \cdot \vec{B} = AB\cos\theta$	3.2.A.3.i	yes, Vectors table
$W = F_\parallel d = Fd\cos\theta$	3.2.A.3.iii, derived	no
$\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$	3.2.A.4	yes
$\Delta E_{\text{mech}} = F_f d\cos\theta$	3.2.A.4.iii	no
$\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$	3.3.A.4	yes
$F_x = -\dfrac{dU(x)}{dx}$	3.3.A.5	yes
$U_s = \frac{1}{2}k(\Delta x)^2$	3.3.A.7.i	yes
$U_g = -\dfrac{Gm_1m_2}{r}$	3.3.A.7.ii	yes
$\Delta U_g = mg\Delta y$	3.3.A.7.iii	yes
$P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$	3.5.A.2	yes
$P_{\text{avg}} = \dfrac{W}{\Delta t}$	3.5.A.3	yes
$P_{\text{inst}} = \dfrac{dW}{dt}$	3.5.A.4	yes
$P_{\text{inst}} = F_\parallel v = Fv\cos\theta$	3.5.A.5, derived	no

Three of the fifteen are absent, and two of those three are the derived special cases from the section above. The third, $\Delta E_{\text{mech}} = F_f d\cos\theta$ at 3.2.A.4.iii, is the energy dissipated by friction, and the framework hedges it in words: the energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted. That word is doing work, and it is why friction problems ask for the path length rather than the displacement.

Two notes on where things live. The dot product is not in the Mechanics table; it is in the Table of Information's separate Vectors table, alongside the cross product, unit-vector notation and the vector sum. And the Table of Information carries a Calculus table with the power rule in both directions, including $\int x^n\, dx = \frac{1}{n+1}x^{n+1}$ for $n \neq -1$ , plus the derivatives and integrals of $e^{ax}$ , $\ln(ax)$ , $\sin(ax)$ and $\cos(ax)$ . A student who works only from the Mechanics table is leaving printed tools on the desk.

Potential energy curves, and the traps around them

Learning objective 3.3.A has eight essential knowledge statements under it, and four are about reading a graph of $U$ against position. Statement 3.3.A.6 says such graphs can be useful in determining physical properties of the system. Then:

3.3.A.6.i defines stable equilibrium as a location where a small displacement produces a force opposite to that displacement, accelerating the object back toward equilibrium.
3.3.A.6.ii defines unstable equilibrium as a location where a small displacement produces a force in the same direction as the displacement, accelerating the object away.
3.3.A.6.iii says stable equilibrium positions occur where $U(x)$ has a local minimum.
3.3.A.6.iv says unstable equilibrium positions occur where $U(x)$ has a local maximum.

Read those next to $F_x = -dU/dx$ and the whole toolkit is one derivative and one second derivative. Equilibrium is where the slope is zero. Stability is the sign of the curvature.

The traps in this unit cluster here and in Topic 3.2.

Zero displacement means zero work, no matter how large the force. This is the misconception the CED names on the unit opener. Holding a heavy box still does no work on it, because the integral $\int \vec{F} \cdot d\vec{r}$ has zero path.

A perpendicular force does no work and still changes the motion. Statement 3.2.A.3.iv says the component of a force perpendicular to the displacement of the system's centre of mass can change the direction of the system's motion without changing its kinetic energy. That single sentence is why uniform circular motion needs a force and no work.

The zero of potential energy is your choice, and the answer must not depend on it. Statement 3.3.A.3 says the definition of zero potential energy for a given system is a decision made by the observer to simplify or otherwise assist in analysis. Choose it, declare it, and note that every physical answer comes out of a difference.

Where $U$ is a minimum, the speed is a maximum, and the force is zero. All three statements say the same thing about a curve with zero slope. Confusing zero force with zero speed at the bottom of a well is the most reliable error available in Topic 3.3.

Friction wants the path length, not the displacement. Statement 3.2.A.4.iii equates the dissipated energy to the friction force times the length of the path over which it is exerted. A round trip has zero displacement and a nonzero path, which is 3.2.A.1.iv's statement that the work done by a nonconservative force is path-dependent, with numbers attached.

Unit 3's two boundary statements, both about dissipation

Unit 3 prints two boundary statements, under Topics 3.2 and 3.4. Topics 3.1, 3.3 and 3.5 print none. Both of the two are about the same question, which is how far the course follows energy once it leaves mechanical form, and they need to be read together because neither says the whole thing.

Under Topic 3.2, quoted whole:

"AP Physics C: Mechanics only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound."

Under Topic 3.4, quoted whole:

"AP Physics C: Mechanics expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces."

Together they draw a clear line. Analysis stops at mechanical energy. Awareness extends one step past it, to the fact that the missing energy became thermal energy or sound, and to the fact that nonconservative forces are what did it. What is not asked for is any thermodynamic treatment of where that energy went, no temperature change, no specific heat, no entropy. If a problem's mechanical energy has decreased, the expected answer names the nonconservative force and says the energy was dissipated as thermal energy or sound. That is the whole of it.

The rest of the unit's fencing is not in boundary statements at all, it is in the derived-equation labels covered above and in the careful wording of the conservation statements. Statement 3.4.B.3 says a system may be selected so that the total energy of that system is constant, and 3.4.C.1 through 3.4.C.3 turn that into a procedure: energy is conserved in all interactions; if the work done on a selected system is zero and there are no nonconservative interactions within it, the total mechanical energy of the system is constant; and if the work done on a selected system is nonzero, energy is transferred between the system and the environment. Conservation of mechanical energy is a conclusion you earn by choosing the system, not an assumption you start from.

If you want the algebra-based treatment of this unit

AP Physics 1 has a Unit 3 with the same title and the same five topic titles. If that is your course, the page you want is AP Physics 1 Unit 3: Work, Energy, and Power. If you are in AP Physics C: Mechanics, this is the page.

	AP Physics 1 Unit 3	AP Physics C: Mechanics Unit 3
Multiple-choice weighting	18 to 23%	15 to 25%
Topic titles 3.1 to 3.5	identical	identical
Work on the equation sheet	$W = Fd\cos\theta$	$W = \int_a^b \vec{F} \cdot d\vec{r}$
Instantaneous power on the sheet	$P = Fv\cos\theta$	$P_{\text{inst}} = dW/dt$
Force from potential energy	not in the course	$F_x = -dU(x)/dx$ , printed
Potential energy from force	not in the course	$\Delta U = -\int_a^b \vec{F}_{\text{cf}} \cdot d\vec{r}$ , printed
Variable forces	not examinable	the default case

This is the unit where the two courses share the most titles and the least mathematics. All five topic titles match exactly, so a search engine has no way to tell the two pages apart from their headings, and yet the AP Physics 1 sheet and this one disagree on the two most-used equations in the unit. If you are in the algebra-based course, everything on this page about integrals and derivatives is beyond your exam, and the Physics 1 unit page is the correct one for you.

The shared ground is real for the conservation half of the unit, so the site's work-energy theorem guide and conservation of energy guide serve both courses, as do the kinetic energy calculator and work and power calculator. All of them assume a constant force, which is exactly the regime where $W = Fd\cos\theta$ is valid and where the two courses agree.

How Unit 3 is assessed

The AP Physics C: Mechanics exam is 3 hours long. Section I is 42 multiple-choice questions in 85 minutes for 50% of the score. Section II is 4 free-response questions in 95 minutes for the other 50%, one of each type in a fixed order: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. A four-function, scientific, or graphing calculator is allowed on both sections. Unit 3's Progress Check runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

The CED's own sample questions put Unit 3 in front of you more than any other unit. Three of the fifteen sample multiple-choice questions align to it: Questions 1 and 8 to learning objective 3.4.B and essential knowledge 3.4.B.2, with skills 2.B and 3.B respectively, and Question 11 to 3.2.A and 3.2.A.4 with skill 2.D. More striking is the free-response set. Objective 3.4.B appears in three of the four sample free-response questions, in Question 1, Question 3 and Question 4. No other objective in the course appears in three.

Sample free-response Question 1, the Mathematical Routines question worth 10 points, is a Unit 3 question end to end, and it is worth knowing its shape. It gives a graph of the potential energy $U_{BA}(x)$ of a two-object system as a function of position, releases an object from rest at one position, and then asks students to derive an expression for the speed at a later position, to draw the corresponding graph of the force on the grid provided, and then, for a different conservative force given as a function of position, to derive an equation for that system's potential energy given where it is defined to be zero. Two of its parts open with the instruction to begin the derivation by writing a fundamental physics principle or an equation from the reference information.

That is $F_x = -dU/dx$ and $\Delta U = -\int \vec{F}_{\text{cf}} \cdot d\vec{r}$ used in both directions in one question.

Five optional sample instructional activities are listed: one on Topic 3.3, three on Topic 3.4, one on Topic 3.5, and none on 3.1 or 3.2. The Topic 3.5 one asks students to graph the power delivered to a car as it accelerates from rest to full speed and then describe how that graph could give the car's velocity as a function of time, which is $P_{\text{inst}} = dW/dt$ read as an integral.

Work done by a variable force, and why Fd fails

A 3.0 kg block starts at rest at $x = 0$ on a frictionless horizontal surface. A single horizontal force $F_x(x) = ax - bx^2$ acts on it, with $a = 12 \ \mathrm{N/m}$ and $b = 2.0 \ \mathrm{N/m^2}$ . Find (a) the work done from $x = 0$ to $x = 3.0$ m, (b) the speed there, (c) the position where the block is moving fastest and its speed there, and (d) what $W = Fd$ would have given.

Declare the convention: positive $x$ is the direction of motion, and the force is along $x$ so the dot product of 3.2.A.3 reduces to an ordinary integral.
(a) $W = \int_0^{3.0} (ax - bx^2)\, dx = \left[\dfrac{a x^2}{2} - \dfrac{b x^3}{3}\right]_0^{3.0}$ .
With $a = 12$ and $b = 2.0$ : $\dfrac{12(9.0)}{2} - \dfrac{2.0(27)}{3} = 54 - 18 = 36$ J.
(b) The block starts at rest, so the work-energy theorem of 3.2.A.4 gives $\Delta K = W = 36$ J with $K_i = 0$ : $v = \sqrt{\dfrac{2W}{m}} = \sqrt{\dfrac{2(36)}{3.0}} = \sqrt{24} = 4.8990$ m/s, so 4.9 m/s.
(c) The block speeds up while the force is positive and slows once it turns negative, so the fastest point is where $F_x = 0$ : $ax = bx^2$ gives $x = a/b = 12/2.0 = 6.0$ m.
$W$ out to there is $\dfrac{12(36)}{2} - \dfrac{2.0(216)}{3} = 216 - 144 = 72$ J, so $v = \sqrt{\dfrac{2(72)}{3.0}} = \sqrt{48} = 6.9282$ m/s, or 6.9 m/s.
(d) Now the wrong way. The force at $x = 3.0$ m is $F_x = 12(3.0) - 2.0(9.0) = 36 - 18 = 18$ N, and $Fd = 18 \times 3.0 = 54$ J, which is 50% too large. Using the force at $x = 0$ instead gives zero, which is 100% too small.
The honest constant-force equivalent is the average force over the path, $W/d = 36/3.0 = 12$ N, which happens to equal $F_x$ at $x = 2.0$ m and is not the force anywhere you would have guessed.
This is exactly why 3.2.A.3.iii attaches the condition "if the component of the force exerted on a system that is parallel to the displacement is constant" to the derived equation $W = F_\parallel d$ . The condition is not decoration.

(a) $W = 36$ J. (b) $v = 4.9$ m/s. (c) Fastest at $x = 6.0$ m, where $F_x = 0$ , with $v = 6.9$ m/s. (d) $Fd$ using the force at $x = 3.0$ m gives 54 J, half again too large; the average force over the path is 12 N.

Reading a potential energy curve in both directions

Two objects interact through a conservative force. The system's potential energy is $U(x) = \dfrac{A}{x^2} - \dfrac{B}{x}$ with $A = 4.0 \ \mathrm{J \cdot m^2}$ , $B = 8.0 \ \mathrm{J \cdot m}$ and $x > 0$ in metres. A 2.0 kg object is released from rest at $x = 2.0$ m. Find (a) $F_x(x)$ , (b) the equilibrium position and whether it is stable, (c) the maximum speed and where it occurs, and (d) the two turning points of the motion.

Declare the convention: $x$ is the separation, positive $F_x$ pushes toward larger $x$ .
(a) Differentiate, per 3.3.A.5: $F_x = -\dfrac{dU}{dx} = -\left(-\dfrac{2A}{x^3} + \dfrac{B}{x^2}\right) = \dfrac{2A}{x^3} - \dfrac{B}{x^2}$ . With the numbers, $F_x = \dfrac{8.0}{x^3} - \dfrac{8.0}{x^2}$ in newtons, which factors as $\dfrac{8.0(1 - x)}{x^3}$ .
(b) Equilibrium is where $F_x = 0$ , so $x = 1.0$ m. Check stability with 3.3.A.6.iii by looking at the curvature: $\dfrac{d^2U}{dx^2} = \dfrac{6A}{x^4} - \dfrac{2B}{x^3}$ , which at $x = 1.0$ m is $24 - 16 = +8.0 \ \mathrm{J/m^2}$ . Positive curvature means a local minimum, so the equilibrium is stable.
(c) Released from rest, so the total energy is $E = U(2.0) = \dfrac{4.0}{4.0} - \dfrac{8.0}{2.0} = 1.0 - 4.0 = -3.0$ J. Statement 3.4.C.2 licenses treating the mechanical energy as constant, since no work is done on the system from outside and the interaction is conservative.
The speed is greatest where $U$ is least, which is the equilibrium at $x = 1.0$ m: $U(1.0) = 4.0 - 8.0 = -4.0$ J.
$K_{\max} = E - U_{\min} = -3.0 - (-4.0) = 1.0$ J, so $v_{\max} = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(1.0)}{2.0}} = 1.0$ m/s.
(d) Turning points are where $K = 0$ , so $U(x) = -3.0$ J: $\dfrac{4.0}{x^2} - \dfrac{8.0}{x} = -3.0$ . Multiply through by $x^2$ and rearrange: $3.0x^2 - 8.0x + 4.0 = 0$ .
$x = \dfrac{8.0 \pm \sqrt{64 - 48}}{6.0} = \dfrac{8.0 \pm 4.0}{6.0}$ , giving $x = 2.0$ m and $x = \dfrac{2}{3} = 0.67$ m. One of them had to be 2.0 m, since that is where it was released from rest, and finding it is the check on the algebra.
So the object oscillates between $x = 0.67$ m and $x = 2.0$ m, passing through $x = 1.0$ m at 1.0 m/s. Reverse the first step to confirm the integral form: $-\int F_x\, dx = -\int \left(\dfrac{8.0}{x^3} - \dfrac{8.0}{x^2}\right) dx = \dfrac{4.0}{x^2} - \dfrac{8.0}{x} + C$ , which is $U(x)$ with $C = 0$ . Statement 3.3.A.3 is what lets you set $C$ , and 3.3.A.4 is the integral you just did.

(a) $F_x = 8.0(1-x)/x^3$ N. (b) Stable equilibrium at $x = 1.0$ m, since the curvature of $U$ there is $+8.0 \ \mathrm{J/m^2}$ . (c) $v_{\max} = 1.0$ m/s at $x = 1.0$ m. (d) Turning points at $x = 0.67$ m and $x = 2.0$ m. Integrating the force back returns $U(x)$ , which is 3.3.A.4 and 3.3.A.5 as inverses.

Frequently asked questions

How much of the AP Physics C Mechanics exam is Unit 3?

Unit 3, Work, Energy, and Power, is weighted at 15 to 25% of the multiple-choice section of the AP Physics C: Mechanics exam, the widest range of any unit in the course, and the course and exam description suggests about 12 to 17 class periods for it. Only Unit 2 is weighted higher, at 20 to 25%. Unit 4 sits at 10 to 20% and Units 1, 5, 6 and 7 at 10 to 15% each. Unit 3 also appears more often than any other unit in the CED's own sample questions: three of the fifteen sample multiple-choice questions align to it, and learning objective 3.4.B appears in three of the four sample free-response questions.

Is W = Fd cos theta on the AP Physics C Mechanics equation sheet?

No. The AP Physics C: Mechanics equation sheet prints work as the integral of the force dotted with the displacement, taken over the path from one point to another. The familiar product of force, distance and the cosine of the angle appears in the framework at essential knowledge 3.2.A.3.iii, but it is labelled a derived equation and carries the condition that the component of the force parallel to the displacement is constant. The course and exam description defines derived equations as results provided to demonstrate the final outcomes of derivations expected of students on the exam, so producing that product from the integral is a step you perform rather than a formula you are given. The same is true of instantaneous power: the sheet prints the derivative of work with respect to time, and force times velocity times cosine is the derived special case at 3.5.A.5.

What is the difference between AP Physics C Unit 3 and AP Physics 1 Unit 3?

All five topic titles are identical, so the difference is entirely in the mathematics. AP Physics 1 is given work as force times distance times a cosine and instantaneous power as force times velocity times a cosine, both printed on its equation sheet, and it treats forces as constant. AP Physics C: Mechanics is given work as an integral and instantaneous power as a derivative, and demotes both algebra-based formulas to derived special cases that are not printed. Physics C also adds two equations that do not exist in the algebra-based course at all: the force as the negative derivative of the potential energy with respect to position, and the change in potential energy as the negative integral of the conservative force. Weightings are 18 to 23% for Physics 1 and 15 to 25% for Physics C.

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

Look at the slope and then the curvature. Essential knowledge 3.3.A.5 gives the force as the negative derivative of the potential energy with respect to position, so equilibrium is where the graph's slope is zero. Statement 3.3.A.6.iii then says stable equilibrium positions occur 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 physical 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.

Does AP Physics C Mechanics cover thermal energy or entropy?

No, and two boundary statements say where the line is. The one under Topic 3.2 reads that AP Physics C: Mechanics only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound. The one under Topic 3.4 reads that AP Physics C: Mechanics expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. So you are expected to name thermal energy or sound as the destination of missing mechanical energy and to attribute the loss to a nonconservative force, and nothing beyond that. There is no thermodynamics in the course: no temperature change, no specific heat, no entropy.