AP Physics 1 · Topic 3.2

Topic 3.2: Work

Unit 3: Work, Energy, and Power18-23% of the multiple-choice section

Work is the energy a force transfers into or out of a system as the point where the force acts moves through a distance. Only the component of the force lying along that displacement counts, which is what the cosine in the formula picks out. Work is a scalar and may be positive, negative, or zero.

AP Physics: Unit 3 (topics 3.2 Work). AP Physics 1 Unit 3, Topic 3.2. Learning objective 3.2.A asks students to describe the work done on an object or system by a given force or collection of forces, across five essential knowledge statements covering the definition of work, its sign, the parallel-component equation, the work-energy theorem, and work as the area under a graph of the parallel force against displacement. A boundary statement limits AP Physics 1 to the transfer of mechanical energy. Unit 3 carries 18 to 23 percent of the multiple-choice section, and the suggested skills for this topic are 1.B, 2.B, 2.D, 3.A, and 3.B.

What Topic 3.2 requires

Topic 3.2 has one learning objective, 3.2.A: describe the work done on an object or system by a given force or collection of forces. It is the longest essential knowledge list in Unit 3, and it splits into five ideas.

3.2.A.1 defines work as energy transferred by a force over a distance, with five sub-statements separating conservative from nonconservative forces.
3.2.A.2 says work is a scalar quantity that may be positive, negative, or zero.
3.2.A.3 relates the work done by a constant force to the components of that force and the displacement of the point at which the force is exerted.
3.2.A.4 states the work-energy theorem.
3.2.A.5 says work is equal to the area under the curve of a graph of $F_{\parallel}$ as a function of displacement.

The CED's suggested skills here are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities; 2.D, predict new values or factors of change using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Topic 3.2 is one of only two topics in Unit 3 that list 3.A, the experimental design skill; Topic 3.5 is the other.

A boundary statement closes the topic: AP Physics 1 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. Heating and cooling between systems belong to AP Physics 2.

The definition, and the equation that implements it

Essential knowledge 3.2.A.1 defines work as the amount of energy transferred into or out of a system by a force exerted on that system over a distance. Read that as a bookkeeping statement rather than a formula: work is energy crossing a boundary, and the force is the mechanism.

3.2.A.3 narrows it to the calculable case. 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. 3.2.A.3.i then gives the rule and the equation: 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.

W = F_{\parallel}d = Fd\cos\theta

Read the symbols off the CED wording: $F$ is how hard the force pushes or pulls, $d$ is how far the point of application travels, and $\theta$ opens between those two directions. The unit is the newton meter, renamed the joule, which is the same joule that measures kinetic energy in Topic 3.1. One force gives one work value; a collection of forces gives one value each, and you add them.

The phrase point of application is deliberate. For a rigid crate sliding across a floor, that point moves exactly as far as the crate does, so the distinction never bites. It matters when it does not, which the section below on naming the system covers.

The angle is between the force and the displacement

$\theta$ in $W = Fd\cos\theta$ is measured from the force vector to the displacement vector. It is not the angle to any coordinate axis you happened to draw, and on an incline it is not the incline angle. Get that wrong and the cosine silently returns the wrong number.

$\theta$	$\cos\theta$	What the force does
$0^\circ$	1	adds the most energy possible
$30^\circ$	0.866	adds energy
$60^\circ$	0.500	adds energy
$90^\circ$	0	transfers none
$120^\circ$	$-0.500$	removes energy
$180^\circ$	$-1$	removes the most energy possible

Work through the incline case once and it stops being a trap. A box slides a distance $d$ down a ramp inclined at $30^\circ$ . Its weight points straight down; its displacement points down the slope. The angle between those two vectors is $90^\circ - 30^\circ = 60^\circ$ , so gravity does $W_g = mgd\cos 60^\circ = 0.500mgd$ . That matches the height check, since the box drops $d\sin 30^\circ = 0.500d$ and gravity's work is $mg$ times the drop. The inclined plane guide sets up the geometry in full, and the inclined plane simulator lets you vary the angle and watch the components move.

Positive, negative, and zero work (3.2.A.2)

Work is a scalar quantity that may be positive, negative, or zero. Each of the three cases has a physical reading.

Positive means the parallel component of the force points along the displacement, so energy flows into the system. A tow rope pulling a sled forward does positive work on the sled.

Negative means the parallel component points against the displacement, so energy flows out. Kinetic friction on a sliding block is the standard case, at $\theta = 180^\circ$ .

Zero arrives three different ways: the force is perpendicular to the displacement, the displacement is zero, or the force itself is zero. The first is 3.2.A.3.ii, which says the component of a force 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. A satellite on a circular orbit is the cleanest illustration: gravity points at the center, the velocity is at right angles to it the whole way round, and the orbital speed never changes even though gravity is the only force acting. See the centripetal force guide for the force side of that picture.

The second is the misconception the CED's Unit 3 overview names outright: whether a force does work on an object even though the object does not move. It does not. Hold a heavy box motionless for ten minutes and you have done zero work on the box, whatever your arms report.

Name the force, and name the system

A work value without two labels is unfinished. Say which force did the work and what it did the work on. *Friction does $-250 \, \mathrm{J}$ of work on the crate is a complete statement; the work is $-250 \, \mathrm{J}$ * is not, because the second one leaves the reader unable to check it.

The choice of system changes the answer, not just the wording. Take a crate sliding to a stop on a rough floor. If the system is the crate alone, friction is an external force that does negative work on it and its kinetic energy falls. If the system is the crate plus the floor, friction is internal: no energy crosses the boundary, and the lost kinetic energy reappears as thermal energy inside the system. Both descriptions are correct, and they are not interchangeable mid-problem.

Essential knowledge 3.2.A.4.ii sets the condition for the simple treatment: 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. A rigid crate passes that test. A skater shoving off a wall does not: the wall's push accelerates her, but the point of contact stays put on the wall while her arms extend, so that force transfers no energy across the boundary and the kinetic energy comes from stored chemical energy inside. Her center of mass moves and the point of application does not, so say which distance your $d$ refers to before you multiply.

Conservative and nonconservative forces

Five sub-statements under 3.2.A.1 draw the line between two kinds of force, and the line is about paths.

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 Examples of nonconservative forces are friction and air resistance.

3.2.A.1.iii is the statement that sets up Topic 3.3. Gravity and an ideal spring force each get a potential energy because their work depends only on where the system started and finished. Friction gets none, so there is no such thing as frictional potential energy.

For the energy friction removes, 3.2.A.4.iii gives the working rule: 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, written $\Delta E_{\mathrm{mech}} = F_f d \cos\theta$ . The $d$ there is path length, not net displacement, and that single word is what path-dependence means in practice. Shove a book up a ramp and let it slide back to the exact spot it started from. Gravity did negative work on the way up and an equal amount of positive work on the way down, netting zero, because the configuration ended where it began. Friction did negative work on both legs and netted twice the loss of one leg, because it charges by path length and the book covered the slope twice.

Work is the area under a force versus position graph (3.2.A.5)

Essential knowledge 3.2.A.5 states that work is equal to the area under the curve of a graph of $F_{\parallel}$ as a function of displacement. Two details in that sentence decide whether you get the right number.

The vertical axis is $F_{\parallel}$ , the component of the force parallel to the displacement, not the full force magnitude. If a graph hands you the total force at an angle, take the parallel component before you measure any area.

The horizontal axis is displacement, that is, position. Area under a force versus time graph is impulse, a completely different quantity that lives in Unit 4. Check the axis label every time; the two graphs look identical.

Beyond that the method is arithmetic. Cut the shaded region at every kink so what remains is rectangles and triangles, work out each piece with its units still attached, and treat anything under the horizontal axis as a subtraction. Skill 1.B asks for quantitative graphs with appropriate scales and units, so label newtons up and meters across: a grid square 1 N tall and 1 m wide is then worth exactly 1 J.

This rule also derives a formula you will meet in the next topic. An ideal spring pulls back with a force of magnitude $k\Delta x$ , so its graph is a straight line through the origin, and the area out to a stretch $\Delta x$ is the triangle $\frac{1}{2}(\Delta x)(k\Delta x) = \frac{1}{2}k(\Delta x)^2$ . That is the elastic potential energy on the equation sheet, obtained purely from the area rule.

The work-energy theorem, and what to read next

Essential knowledge 3.2.A.4 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.

\Delta K = \sum_i W_i = \sum_i F_{\parallel,i} d_i

Two routes reach the same number when the forces are constant. Compute each force's work separately and add the signed values, which is what the middle expression says, or find the net force first and take its parallel component once. The first route is safer on a crowded free-body diagram, because it forces you to account for every force including the ones that contribute zero.

Statement 3.2.A.4.i adds the case where a force changes a system's configuration: 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.

For the full solving procedure, including how to pick between an energy route and a kinematics route, work through the work-energy theorem guide and the conservation of energy guide. To check your own numbers, the work and power calculator handles both the $Fd\cos\theta$ form and the power forms that Topic 3.5 adds. Then take the sign conventions into Topic 3.3, where the work done by a conservative force becomes a stored potential energy.

Work done by each force on a pushed crate

A 25 kg crate is pushed 6.0 m in a straight line across a level floor by a 90 N force directed $20^\circ$ below the horizontal. The coefficient of kinetic friction between crate and floor is 0.15. Taking the crate alone as the system, find the work done on it by each of the four forces acting on it, and the net work.

Declare the positive direction: the direction the crate travels, horizontally. The crate's displacement is $d = 6.0 \, \mathrm{m}$ along that direction, and it has no vertical displacement at all. The crate is rigid and slides, so the point of application of every force moves with it, which is condition 3.2.A.4.ii.
The push. The angle between the 90 N force and the horizontal displacement is $20^\circ$ , so $W_{\mathrm{push}} = Fd\cos\theta = (90)(6.0)\cos 20^\circ = (90)(6.0)(0.9397) = 507 \, \mathrm{J}$ . Positive, because the parallel component points with the motion.
Gravity and the normal force. Both are vertical and the displacement is horizontal, so $\theta = 90^\circ$ for each and $\cos 90^\circ = 0$ . $W_{\mathrm{gravity}} = 0$ and $W_{\mathrm{normal}} = 0$ . They set the vertical force balance without moving any energy into or out of the crate.
Friction needs the normal force first. The push has a downward component $F\sin 20^\circ = (90)(0.3420) = 30.8 \, \mathrm{N}$ , which presses the crate harder into the floor, so $F_N = mg + F\sin 20^\circ = (25)(9.8) + 30.8 = 245 + 30.8 = 275.8 \, \mathrm{N}$ .
On the value of $g$ : the number printed in the CED's table of information is $g = 9.8 \, \mathrm{m/s^2}$ , while a Unit 1 boundary statement reads 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, and that students will not be penalized for correctly using the more precise commonly accepted values of 9.81 or 9.8. Every calculation on this site runs on 9.8, which keeps the pages consistent with each other.
Kinetic friction is then $f_k = \mu_k F_N = (0.15)(275.8) = 41.4 \, \mathrm{N}$ , pointing opposite the motion.
Friction's work. $\theta = 180^\circ$ and $\cos 180^\circ = -1$ , so $W_{\mathrm{friction}} = (41.4)(6.0)(-1) = -248 \, \mathrm{J}$ .
Net work. Add the signed values: $W_{\mathrm{net}} = 507 + 0 + 0 - 248 = 259 \, \mathrm{J}$ . Every input carries two significant figures, so round at the end: $+510 \, \mathrm{J}$ , $-250 \, \mathrm{J}$ , and $+260 \, \mathrm{J}$ .
Sanity check with the work-energy theorem. If the crate started from rest, $\Delta K = 259 \, \mathrm{J}$ gives $v = \sqrt{2(259)/25} = 4.6 \, \mathrm{m/s}$ . A force-based route agrees: the net horizontal force is $84.6 - 41.4 = 43.2 \, \mathrm{N}$ , so $a = 43.2/25 = 1.73 \, \mathrm{m/s^2}$ and $v = \sqrt{2(1.73)(6.0)} = 4.6 \, \mathrm{m/s}$ .

$W_{\mathrm{push}} = +510 \, \mathrm{J}$ , $W_{\mathrm{gravity}} = 0$ , $W_{\mathrm{normal}} = 0$ , $W_{\mathrm{friction}} = -250 \, \mathrm{J}$ , and net work $+260 \, \mathrm{J}$ . Note that pushing downward on the crate raised the normal force above $mg$ , so it cost more friction work than a horizontal push of the same magnitude would have.

Reading work off a force versus position graph

A 2.0 kg cart starts from rest at $x = 0$ on a frictionless horizontal track and moves in the $+x$ direction. The only horizontal force on it is parallel to the track, and a graph of $F_{\parallel}$ against position shows three straight segments: a constant $+12 \, \mathrm{N}$ from $x = 0$ to $x = 3.0 \, \mathrm{m}$ ; a straight-line fall from $+12 \, \mathrm{N}$ at $x = 3.0 \, \mathrm{m}$ to zero at $x = 7.0 \, \mathrm{m}$ ; and a constant $-4.0 \, \mathrm{N}$ from $x = 7.0 \, \mathrm{m}$ to $x = 10.0 \, \mathrm{m}$ . Find the work over each segment, the total work, and the cart's speed at $x = 10.0 \, \mathrm{m}$ .

Positive direction is $+x$ , the direction the cart moves throughout. By essential knowledge 3.2.A.5 the work over any stretch is the area under the $F_{\parallel}$ versus position graph on that stretch, with area above the axis counting as positive and area below it as negative.
Segment 1, a rectangle. Height $12 \, \mathrm{N}$ , width $3.0 \, \mathrm{m}$ , so $W_1 = (12)(3.0) = +36 \, \mathrm{J}$ .
Segment 2, a triangle. Base $7.0 - 3.0 = 4.0 \, \mathrm{m}$ , height $12 \, \mathrm{N}$ , so $W_2 = \frac{1}{2}(4.0)(12) = +24 \, \mathrm{J}$ .
Segment 3, a rectangle below the axis. The force is $-4.0 \, \mathrm{N}$ over $10.0 - 7.0 = 3.0 \, \mathrm{m}$ , so $W_3 = (-4.0)(3.0) = -12 \, \mathrm{J}$ . The force now points against the motion and removes energy.
Total. $W = 36 + 24 - 12 = +48 \, \mathrm{J}$ .
Gravity and the normal force are vertical while the displacement is horizontal, so each does zero work, and the track is frictionless. That makes $48 \, \mathrm{J}$ the net work, so $\Delta K = 48 \, \mathrm{J}$ . The cart started at rest, so $K_f = 48 \, \mathrm{J}$ and $v = \sqrt{2K_f/m} = \sqrt{2(48)/2.0} = \sqrt{48} = 6.9 \, \mathrm{m/s}$ .
One extra reading the graph gives free: the cart is fastest at $x = 7.0 \, \mathrm{m}$ , where the force changes sign. Up to that point $K = 36 + 24 = 60 \, \mathrm{J}$ , so $v = \sqrt{2(60)/2.0} = \sqrt{60} = 7.7 \, \mathrm{m/s}$ , and the final segment slows it back down.

$W_1 = +36 \, \mathrm{J}$ , $W_2 = +24 \, \mathrm{J}$ , $W_3 = -12 \, \mathrm{J}$ , total $+48 \, \mathrm{J}$ , and the cart reaches $x = 10.0 \, \mathrm{m}$ at $6.9 \, \mathrm{m/s}$ after peaking at $7.7 \, \mathrm{m/s}$ . No equation for the varying force was ever needed, only the area.

Frequently asked questions

What is the formula for work in AP Physics 1?

The equation sheet prints W = F-parallel times d = Fd cos theta. Multiply how hard the force pushes by how far its point of application travels, then by the cosine of the angle opening between those two directions. The unit is the joule, which is one newton meter.

What makes the work done by a force negative?

The parallel component of the force points opposite the displacement, so the angle between them is greater than 90 degrees and the cosine is negative. Physically, that force is taking energy out of the system rather than putting it in. Kinetic friction on a sliding object is the standard example, at 180 degrees with a cosine of exactly -1.

Can a force change an object's motion without doing any work?

Yes. Essential knowledge 3.2.A.3.ii says a force component perpendicular to the displacement of the system's center of mass can change the direction of the system's motion without changing its kinetic energy. String tension on a ball moving in a horizontal circle is the classic case: it turns the ball continuously and does zero work, so the speed never changes.

What is the difference between a conservative and a nonconservative force?

The work done by a conservative force is path-independent and depends only on the initial and final configurations, so it is zero around any closed path. The work done by a nonconservative force is path-dependent. The CED names friction and air resistance as nonconservative examples, and states that potential energies are associated only with conservative forces.

How do you find work from a graph?

Take the area under a graph of the parallel force component against displacement, counting area below the axis as negative. Split the region into rectangles and triangles and add the signed areas. Check the horizontal axis first: area under a force versus time graph is impulse, not work.