AP Physics C: Mechanics · Topic 3.2

Topic 3.2: Work

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

In AP Physics C, work is the integral of the force dotted with the displacement along the path. Force times distance times cosine is the special case that applies only when the parallel component of the force is constant, and unlike in AP Physics 1 it is not printed on your equation sheet.

AP Physics: Unit 3 (topics 3.2 Work). Topic 3.2 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, 3.2.A, with thirteen essential knowledge statements counting sub-statements (3.2.A.1 and five subs, 3.2.A.2, 3.2.A.3 and four subs, 3.2.A.4 and three subs, 3.2.A.5). Suggested skills 1.A, 1.C, 2.A, 2.D and 3.B. Boundary statement, 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. The AP Physics 1 version of that boundary statement adds a parenthetical defining mechanical energy by reference to its Unit 3 Topic 4 and a closing sentence about AP Physics 2 covering thermal transfer; the C version has neither. The calculus differentiator is a swap of general and special case: 3.2.A.3 gives the variable-force integral W = integral of F(r) dot dr over the path (printed on the C sheet, absent from the AP Physics 1 sheet), 3.2.A.3.i adds the dot product A dot B = AB cos theta (printed in the C Table of Information's Vectors table), and W = F_parallel d = Fd cos theta is demoted to 3.2.A.3.iii and labelled a derived equation with the condition that F_parallel is constant, so it is NOT on the C sheet although it IS on the AP Physics 1 sheet. The AP Physics 1 framework instead carries that product at 3.2.A.3.i as a Relevant equation. Delta E_mech = F_f d cos theta (3.2.A.4.iii) is on neither course's sheet. The work-energy theorem sentence at 3.2.A.4 is identical in both frameworks; the relevant equation differs by one subscript (d_i in C, bare d in AP Physics 1), and both courses' sheets print the subscripted form. In the CED's sample set, 3.2.A aligns to exactly one question, sample multiple-choice question 11 (skill 2.D, EK 3.2.A.4, published answer four times the work), and to none of the four sample free-response questions.

What Topic 3.2 requires, in full

Topic 3.2 carries one learning objective and, counting sub-statements, seventeen essential knowledge entries.

Learning objective 3.2.A: describe the work done on an object or system by a given force or collection of forces.

Statement	What the CED says
3.2.A.1	Work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance
3.2.A.1.i	The work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system
3.2.A.1.ii	The work done by a conservative force on a system, or the change in the potential energy of the system, will be zero if the system returns to its initial configuration
3.2.A.1.iii	Potential energies are associated only with conservative forces
3.2.A.1.iv	The work done by a nonconservative force is path-dependent
3.2.A.1.v	The most common nonconservative forces are friction and air resistance
3.2.A.2	Work is a scalar quantity that may be positive, negative, or zero
3.2.A.3	The work done on an object by a variable force is calculated using $W = \int_a^b \vec{F}(r) \cdot d\vec{r}$ , where the integral is taken over the path from point $a$ to point $b$
3.2.A.3.i	The dot product between two vectors, $\vec{A}$ and $\vec{B}$ , results in a scalar quantity of magnitude $\vec{A} \cdot \vec{B} = AB\cos\theta$
3.2.A.3.ii	Only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy
3.2.A.3.iii	If the component of the force exerted on a system that is parallel to the displacement is constant, the work done on the system by the force is given by the derived equation $W = F_\parallel d = Fd\cos\theta$
3.2.A.3.iv	The component of the force exerted on a system perpendicular to the direction of the displacement of the system's center of mass can change the direction of the system's motion without changing the system's kinetic energy
3.2.A.4	The work-energy theorem states that the change in an object's kinetic energy is equal to the sum of the work (net work) being done by all forces exerted on the object. Relevant equation: $\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$
3.2.A.4.i	An external force may change the configuration of a system. The component of the external force parallel to the displacement times the displacement of the point of application of the force gives the change in kinetic energy of the system
3.2.A.4.ii	If the system's center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as an object, and only the system's kinetic energy can change
3.2.A.4.iii	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. $\Delta E_{\text{mech}} = F_f d\cos\theta$
3.2.A.5	Work is equal to the area under the curve of a graph of $F_\parallel$ as a function of displacement

Suggested skills: 1.A (create diagrams, tables, charts, or schematics), 1.C (create qualitative sketches of graphs), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.D (predict new values or factors of change of physical quantities using functional dependence between variables) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Topic 3.2's boundary statement, 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."

The general case and the special case have swapped places

Both courses have a Topic 3.2 called Work. Line the two frameworks up statement by statement and the structure is the same until 3.2.A.3, where AP Physics C: Mechanics inserts two statements and demotes one.

Number	AP Physics 1	AP Physics C: Mechanics
3.2.A.1 and its five sub-statements	conservative and nonconservative work	same, except 3.2.A.1.v reads "the most common nonconservative forces" rather than "examples of nonconservative forces"
3.2.A.2	work is a scalar	same
3.2.A.3	"The amount of work done on a system by a constant force is related to the components of that force and the displacement of the point at which that force is exerted"	"The work done on an object by a variable force is calculated using $W = \int_a^b \vec{F}(r) \cdot d\vec{r}$ "
3.2.A.3.i	parallel-component statement, with $W = F_\parallel d = Fd\cos\theta$ as a Relevant equation	the dot product, $\vec{A} \cdot \vec{B} = AB\cos\theta$
3.2.A.3.ii	perpendicular-component statement	parallel-component statement, with no equation attached
3.2.A.3.iii	does not exist	$W = F_\parallel d = Fd\cos\theta$ , labelled the derived equation, with the condition that $F_\parallel$ is constant
3.2.A.3.iv	does not exist	perpendicular-component statement
3.2.A.4 to 3.2.A.5	work-energy theorem, configuration, friction, area under the curve	same statements, same order

Read the third and fourth rows together. The equation AP Physics 1 hands you as the definition of work is in AP Physics C moved two sub-statements down the page, given an explicit "if" clause, and relabelled.

That relabelling is not decorative. Both course and exam descriptions carry an identical Required Equations note: not all equations in the framework appear on the equation sheet, many are provided for reference and guidance or to demonstrate the final results of derivations expected of students on the exam, and those are denoted as "Derived Equations." So in the calculus-based course $W = Fd\cos\theta$ is a result you produce, not a tool you are handed.

The two versions answer different questions. $Fd\cos\theta$ answers how much work a steady push did. The integral answers how much work a push did while it was changing, which is the only version that reaches a spring, a gravitational field far from a planet, a drag force, or a graph that is not a rectangle. If the algebra-based treatment is what you need, it is at AP Physics 1 Topic 3.2. That page is for students whose sheet supplies $Fd\cos\theta$ and whose exam never asks for the integral. This page is for students whose sheet supplies the integral and whose exam expects $Fd\cos\theta$ as an output.

What each course's sheet actually prints

Checked against the Table of Information printed in each course and exam description, not recalled:

Equation	AP Physics 1 sheet	AP Physics C: Mechanics sheet
$W = \int_a^b \vec{F} \cdot d\vec{r}$	no	yes
$W = F_\parallel d = Fd\cos\theta$	yes	no
$\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$	yes	yes
$\vec{A} \cdot \vec{B} = AB\cos\theta$	no	yes, in the Vectors table
$\Delta E_{\text{mech}} = F_f d\cos\theta$	no	no
$\left\lvert \vec{F}_f \right\rvert \leq \left\lvert \mu \vec{F}_N \right\rvert$	yes	yes
$\int x^n\, dx = \frac{1}{n+1}x^{n+1},\ n \neq -1$	no	yes, in the Calculus table

Two things in that table are worth stating out loud.

First, the swap is exact. Each course's sheet prints one of the two work equations and not the other, and each prints the one its own framework treats as primary. No course's sheet prints both.

Second, $\Delta E_{\text{mech}} = F_f d\cos\theta$ is on neither sheet, even though it appears in both frameworks at 3.2.A.4.iii. The framework hedges it in words too: the energy dissipated by friction is typically equated to the force of friction times the path length. It is a rule of thumb rather than a printed law.

The dot product's location matters as well. It is not in the Mechanics table. It sits in a separate Vectors table on the same appendix page as the Geometry, Calculus and Identities tables, alongside $\left\lvert \vec{A} \times \vec{B} \right\rvert = AB\sin\theta$ and the unit-vector form. That page also prints the power rule in both directions and the integrals of $e^{ax}$ , $\sin(ax)$ , $\cos(ax)$ and $\frac{1}{x+a}$ , so every integral a Topic 3.2 question is likely to need is printed. Working only from the Mechanics column leaves those unused. Full transcription: the AP Physics C: Mechanics formula sheet.

Evaluating the integral: dot products and areas

Statement 3.2.A.3.i introduces the dot product, and it is the mechanism that turns a vector integral into a number. You will meet it in three forms, and they are one operation.

Component form, for when force and displacement are given as components:

\vec{F} \cdot d\vec{r} = F_x\, dx + F_y\, dy + F_z\, dz

Magnitude-and-angle form, which is what the sheet prints: $\vec{A} \cdot \vec{B} = AB\cos\theta$ .

Parallel-component form, which is 3.2.A.3.ii read as arithmetic: take the part of $\vec{F}$ along the path and multiply by the path length.

The angle $\theta$ is between the force and the displacement, not between the force and any axis. Worked example 1 evaluates one work both ways and gets 18 J twice. The sign of $\cos\theta$ carries the sign of the work, which is 3.2.A.2 in operation: below 90 degrees is positive work and energy in, exactly 90 degrees is zero work (3.2.A.3.iv, which is why a normal force on a horizontal surface and the tension in a conical pendulum string do none), and above 90 degrees is negative work and energy out, with kinetic friction the standard case at 180 degrees. The Table of Information prints trigonometric values for 0, 30, 37, 45, 53, 60 and 90 degrees, and the 37 and 53 entries exist because the 3-4-5 triangle keeps appearing.

The graphical route is the same integral. Statement 3.2.A.5, identical in both courses, says work is the area under the curve of a graph of $F_\parallel$ against displacement. In AP Physics C that is not a separate technique; it is the printed equation drawn. Three rules a graph question tests:

Area below the axis is negative work. The areas subtract. Worked example 2 has one.
The object is fastest where the curve crosses zero going downward, not where the curve peaks. At the peak the force is largest and the object is still speeding up. This is the graph-reading trap of Topic 3.2, and it is the same idea Topic 3.3 states as a potential energy minimum.
The axis label matters. The statement says $F_\parallel$ , not $F$ . If a graph plots the total force magnitude while the force is at an angle to the motion, the area is not the work.

Two of this topic's five suggested skills, 1.A and 1.C, belong to Science Practice 1, which is not assessed on the multiple-choice section at all and carries 20 to 35% of the free-response section. The graph work in Topic 3.2 points at the free-response paper.

Path dependence, friction, and where the course stops

The first six statements of Topic 3.2 are about one split: conservative or not.

Conservative (3.2.A.1.i and .ii): the work depends only on the initial and final configurations, so any two paths between the same endpoints give the same work and a closed loop gives exactly zero.

Nonconservative (3.2.A.1.iv and .v): the work depends on the path, so a longer route costs more. The CED names friction and air resistance as the most common cases.

3.2.A.1.iii links this to the next topic: potential energies are associated only with conservative forces. There is no friction potential energy, which is why Topic 3.3's machinery does not apply to a rough surface and why friction enters as a work term instead.

For friction specifically, 3.2.A.4.iii equates the dissipated energy to the friction force times the length of the path over which the force is exerted. Path length, not displacement. A round trip has zero displacement and a nonzero path, so a conservative force does zero work over it and friction does not. That contrast, with numbers, is worked example 3.

Then the boundary statement stops you. 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."

The AP Physics 1 version says the same of its own course and adds two clauses this one lacks: a parenthetical defining mechanical energy by reference to Unit 3 Topic 4, and a sentence that in AP Physics 2 students will also study how thermal energy can be transferred between systems through heating or cooling. AP Physics C: Mechanics has no successor in that sequence, so the pointer is dropped.

In practice, for both: name thermal energy or sound as the destination of the missing mechanical energy, attribute the loss to a nonconservative force, and stop. No temperature change, no specific heat, no entropy. Topic 3.4 carries the other half of this fence.

How Topic 3.2 is tested

In the CED's own sample set, one question aligns to learning objective 3.2.A: sample multiple-choice question 11, tagged to essential knowledge 3.2.A.4 and skill 2.D. Two identical blocks are dropped from rest from different heights, the second reaching the floor with twice the speed of the first, and the question asks for the ratio of the work done by gravity. The published answer is four times, because the work-energy theorem makes the work proportional to $v^2$ . No integral is required, and no potential energy is required either.

Learning objective 3.2.A appears in none of the four sample free-response questions, though the sample free-response Question 1 does ask students to move between a force and a potential energy in both directions, which is Topic 3.3's version of the same calculus.

That is a fair picture of the topic. It is foundational rather than headline: the integral turns up inside questions about springs, gravitation, potential energy and power more often than as the subject of one. Skill 2.A is suggested for Topics 3.2, 3.3 and 3.4 and is the most heavily weighted skill on the multiple-choice section at 25 to 30%.

The unit's Preparing for the AP Exam note ties Unit 3 to the first free-response question, Mathematical Routines, worth 10 points at a suggested 20 to 25 minutes, and warns that although Unit 3 offers ideal content for practising it, that question can pull from any of the seven units.

Related pages. The work-energy theorem guide and work and power calculator own the constant-force routine both courses share, the regime where 3.2.A.3.iii holds. The conservative versus nonconservative force comparison is the short version of the split above, the glossary entry on work is the one-paragraph definition, and work, energy and power practice has problems in the shared regime. The Unit 3 hub lists every equation in the unit against what the sheet prints.

The same work two ways: components and the cosine

A constant force $\vec{F} = (6.0\hat{i} - 3.0\hat{j})$ N acts on a crate while the crate's point of application moves in a straight line from the origin to the point $(4.0, 2.0)$ m. Find the work done (a) using the component form of the dot product and (b) using $W = Fd\cos\theta$ , and (c) state the angle between the force and the displacement.

Declare the convention: $\hat{i}$ is the $+x$ direction and $\hat{j}$ is the $+y$ direction, and the displacement vector is $\vec{d} = (4.0\hat{i} + 2.0\hat{j})$ m.
The force is constant, so 3.2.A.3.iii applies and the integral collapses: $W = \int_a^b \vec{F} \cdot d\vec{r} = \vec{F} \cdot \vec{d}$ .
(a) Component form: $W = F_x d_x + F_y d_y = (6.0)(4.0) + (-3.0)(2.0) = 24 - 6.0 = 18$ J.
(b) Magnitudes: $F = \sqrt{(6.0)^2 + (-3.0)^2} = \sqrt{45} = 6.7082$ N, and $d = \sqrt{(4.0)^2 + (2.0)^2} = \sqrt{20} = 4.4721$ m.
Their product is exact: $\sqrt{45}\sqrt{20} = \sqrt{900} = 30.0$ N m. So $W = 30.0\cos\theta$ .
(c) Setting the two results equal, $\cos\theta = 18/30.0 = 0.600 = 3/5$ , so $\theta = 53$ degrees. That value is printed in the Table of Information's table of trigonometric functions for common angles, which lists 0, 30, 37, 45, 53, 60 and 90 degrees with $\cos 53^\circ = 3/5$ .
Note what did not matter. Nothing in the problem said the path was straight, only that the displacement was. For a constant force the two calculations above are valid for any path between those endpoints, because $\int \vec{F} \cdot d\vec{r} = \vec{F} \cdot \int d\vec{r} = \vec{F} \cdot \vec{d}$ when $\vec{F}$ can come out of the integral. That is 3.2.A.1.i for the special case of a uniform force field.

(a) and (b) both give $W = 18$ J. (c) The angle between the force and the displacement is 53 degrees, since $\cos\theta = 18/30.0 = 3/5$ .

Reading work off a force against position graph

A 3.0 kg block starts at rest at $x = 0$ on a frictionless horizontal surface. The only horizontal force acting is along the direction of motion, and a graph of $F_\parallel$ against $x$ is piecewise linear: it rises from 0 at $x = 0$ to 12 N at $x = 3.0$ m, holds at 12 N until $x = 5.0$ m, then falls linearly to $-6.0$ N at $x = 8.0$ m. Find (a) the total work done from $x = 0$ to $x = 8.0$ m, (b) the speed there, and (c) the position and value of the maximum speed.

Set the convention: positive $F_\parallel$ is in the direction of motion, so the areas above the axis are positive work and areas below are negative, per 3.2.A.2 and 3.2.A.5.
Segment 1, the triangle from $x = 0$ to $x = 3.0$ m: $W_1 = \frac{1}{2}(3.0)(12) = 18$ J.
Segment 2, the rectangle from $x = 3.0$ to $x = 5.0$ m: $W_2 = (2.0)(12) = 24$ J.
Segment 3 runs from $(5.0, 12)$ to $(8.0, -6.0)$ , a slope of $-18/3.0 = -6.0$ N/m, so $F_\parallel(x) = 12 - 6.0(x - 5.0)$ . It crosses zero at $x = 7.0$ m.
The positive part, $x = 5.0$ to 7.0 m: $W_3 = \frac{1}{2}(2.0)(12) = 12$ J. The negative part, $x = 7.0$ to 8.0 m: $W_4 = \frac{1}{2}(1.0)(-6.0) = -3.0$ J.
(a) $W_{\text{total}} = 18 + 24 + 12 - 3.0 = 51$ J.
(b) The block started at rest, so the work-energy theorem of 3.2.A.4 gives $\Delta K = 51$ J with $K_i = 0$ : $v = \sqrt{2W/m} = \sqrt{2(51)/3.0} = \sqrt{34} = 5.8310$ m/s, so 5.8 m/s.
(c) The block speeds up wherever the work is still accumulating, which is wherever $F_\parallel > 0$ . That stops at $x = 7.0$ m. Work out to there is $18 + 24 + 12 = 54$ J, so $v_{\max} = \sqrt{2(54)/3.0} = \sqrt{36} = 6.0$ m/s exactly.
Check the trap: the force is largest at $x$ between 3.0 and 5.0 m, but the speed is largest at $x = 7.0$ m, two metres past the point where the force starts falling. Peak force and peak speed are different places.

(a) 51 J. (b) 5.8 m/s at $x = 8.0$ m. (c) The speed peaks at $x = 7.0$ m, where the force crosses zero, at exactly 6.0 m/s.

A closed loop: zero for gravity, not zero for friction

A 4.0 kg block is pushed around a closed loop on a rough horizontal tabletop and returns to exactly where it started. The total path length is 2.40 m and the coefficient of kinetic friction is 0.30. Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) the work done by gravity over the loop, (b) the work done by the normal force, (c) the energy dissipated by friction, and (d) show symbolically that the answer to (c) does not depend on the shape of the loop.

(a) Gravity points down and every displacement on the tabletop is horizontal, so $\theta = 90$ degrees at every point of the path and the work done by gravity is zero. This is also 3.2.A.1.ii read directly: gravity is conservative and the system returned to its initial configuration.
(b) The normal force is also perpendicular to the motion everywhere, so it too does zero work. Statement 3.2.A.3.iv is the general form: a perpendicular force component can change the direction of motion without changing the kinetic energy.
(c) Friction is nonconservative, so 3.2.A.1.iv applies and the path matters. On a horizontal surface with no vertical applied component, $F_N = mg = (4.0)(9.8) = 39.2$ N.
The kinetic friction force is $F_f = \mu_k F_N = (0.30)(39.2) = 11.76$ N.
Per 3.2.A.4.iii the energy dissipated is the friction force times the path length: $\Delta E_{\text{mech}} = (11.76)(2.40) = 28.224$ J, so 28 J to two significant figures. The applied push had to supply that same 28 J for the block to return at its original speed.
(d) Symbolically, from the integral. Kinetic friction on a sliding block always opposes the velocity, so $\vec{F}_f \cdot d\vec{r} = -\left\lvert \vec{F}_f \right\rvert ds$ along the path, giving $W_f = -\int \mu_k mg\, ds = -\mu_k mg \int ds = -\mu_k mg L$ .
$\mu_k$ , $m$ and $g$ are constants here, so they come out of the integral and only the total path length $L$ is left. Nowhere in that argument did the shape of the loop appear.
Contrast the two results. Over the same closed loop the conservative force did exactly zero work and the nonconservative force did $-\mu_k mg L$ . Doubling the path length with the endpoints fixed leaves the first at zero and doubles the second. That is 3.2.A.1.i against 3.2.A.1.iv, with numbers.

(a) Zero. (b) Zero. (c) 28 J of mechanical energy is dissipated, from $\mu_k mg L = (0.30)(4.0)(9.8)(2.40) = 28.224$ J. (d) The integral gives $W_f = -\mu_k mg L$ for any shape of loop, because kinetic friction is antiparallel to $d\vec{r}$ everywhere and the constants come out of the integral, leaving only the total path length.

Frequently asked questions

Why is W = Fd cos theta not on the AP Physics C equation sheet when it is on the AP Physics 1 sheet?

Because the two courses treat it as a different kind of statement. The AP Physics 1 framework attaches it to essential knowledge 3.2.A.3.i as a Relevant equation and prints it on the AP Physics 1 sheet. The AP Physics C: Mechanics framework moves it to 3.2.A.3.iii, adds the condition that the component of the force parallel to the displacement is constant, and calls it a derived equation. Both course and exam descriptions define derived equations identically, as results provided to demonstrate the final outcomes of derivations expected of students, and those are exactly the ones that do not appear on the sheet. The AP Physics C: Mechanics sheet prints the integral of the force dotted with the displacement instead.

How do you find the work done by a variable force in AP Physics C?

Integrate the force along the path. Essential knowledge 3.2.A.3 gives the work done on an object by a variable force as the integral from a to b of the force dotted with the displacement element, with the integral taken over the path from point a to point b, and that equation is printed on the AP Physics C: Mechanics equation sheet. In one dimension with the force along the motion it reduces to an ordinary integral of the force with respect to position. If the force is given as a graph rather than a function, essential knowledge 3.2.A.5 gives you the same answer as the signed area under a plot of the parallel force component against displacement, counting area below the axis as negative.

Is the work-energy theorem different in AP Physics C than in AP Physics 1?

The statement is not. Essential knowledge 3.2.A.4 is word for word identical in both course and exam descriptions: the change in an object's kinetic energy equals the sum of the work being done by all forces exerted on the object. The equation printed under it differs by one subscript, since the AP Physics C: Mechanics framework writes the sum with a path length subscripted to each force while the AP Physics 1 framework writes a single unsubscripted path length; both courses' equation sheets print the subscripted form. What genuinely differs is how you get each work term, since in AP Physics C each one may be an integral.

Does friction use the distance travelled or the displacement?

The path length. Essential knowledge 3.2.A.4.iii states that 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 is the practical face of 3.2.A.1.iv, which says the work done by a nonconservative force is path-dependent. A block pushed around a closed loop on a rough table has zero displacement, so gravity and the normal force do zero work, while friction has dissipated the friction force times the whole distance travelled. Take a longer route between the same two points and the friction loss grows, while every conservative force does exactly the same work as before.

Can work be negative in AP Physics C?

Yes. Essential knowledge 3.2.A.2 states that work is a scalar quantity that may be positive, negative, or zero. The sign comes from the cosine of the angle between the force and the displacement. An angle less than 90 degrees gives positive work and puts energy into the system, exactly 90 degrees gives zero work, and an angle greater than 90 degrees gives negative work and takes energy out. Kinetic friction on a sliding object is the standard negative case, at 180 degrees. Negative work does not mean negative kinetic energy: kinetic energy is never negative, and negative work simply reduces it.

What is the difference between AP Physics C Topic 3.2 and AP Physics 1 Topic 3.2?

The titles match and about two thirds of the essential knowledge is identical, but the two courses insert their equations at opposite ends. AP Physics C: Mechanics adds two statements that AP Physics 1 does not have, the variable-force integral at 3.2.A.3 and the dot product at 3.2.A.3.i, and demotes force times distance times cosine to a conditional derived equation at 3.2.A.3.iii. AP Physics 1 gives that product as its primary definition and never mentions an integral. The boundary statements also differ: the AP Physics 1 version adds a parenthetical pointing at its own Topic 3.4 and a closing sentence about AP Physics 2 covering thermal transfer, neither of which appears in the AP Physics C version.