AP Physics C: Mechanics · Topic 3.5

Topic 3.5: Power

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

Power is the rate at which energy changes with time. AP Physics C gives instantaneous power as the derivative of work with respect to time and prints that on the sheet. Force times velocity times cosine is the derived special case for a constant force, and it is not printed here.

AP Physics: Unit 3 (topics 3.5 Power). Topic 3.5 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, 3.5.A, and five essential knowledge statements with no sub-statements: 3.5.A.1 defines power as the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within it; 3.5.A.2 gives P_avg = Delta E / Delta t; 3.5.A.3 gives P_avg = W / Delta t; 3.5.A.4 gives P_inst = dW/dt; and 3.5.A.5 gives P_inst = F_parallel v = Fv cos theta as a derived equation for the component of a CONSTANT force parallel to the velocity. No boundary statement, in either course. Suggested skills 1.A, 2.B, 2.C and 3.B. AP Physics 1's Topic 3.5 has four statements: its 3.5.A.1 to .3 are identical to this course's, it has no dW/dt statement, and its 3.5.A.4 is this course's 3.5.A.5, also word for word identical. VERIFIED AND NON-OBVIOUS: both frameworks label Fv cos theta a derived equation in identical wording, yet the AP Physics 1 sheet PRINTS it and the AP Physics C: Mechanics sheet does not, so the derived label alone does not determine what reaches a sheet. Both sheets print P_avg = W/Delta t = Delta E/Delta t as a single line; only the C sheet prints P_inst = dW/dt. The calculus differentiator is the two-line derivation P_inst = dW/dt = F dot (dr/dt) = F dot v, which needs no constant-force assumption, and its inverse, W = integral of P dt, so the area under a power-time graph is an energy. Learning objective 3.5.A appears in NONE of the fifteen sample multiple-choice questions and NONE of the four sample free-response questions, the only Unit 3 objective absent from both lists. Unit 3's fifth sample instructional activity is the only one for this topic: graph the power delivered to a car as a function of time as it accelerates from rest to full speed, then describe how to use that graph to determine the car's velocity as a function of time. Suggested-skill inversion worth noting: AP Physics 1's Topic 3.5 lists 2.A (derive a symbolic expression) while AP Physics C: Mechanics lists 2.B (calculate an unknown quantity) and not 2.A.

What Topic 3.5 requires, in full

Topic 3.5 carries one learning objective and five essential knowledge statements, none of which has sub-statements.

Learning objective 3.5.A: describe the transfer of energy into, out of, or within a system in terms of power.

Statement	What the CED says
3.5.A.1	Power is the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within a system
3.5.A.2	Average power is the amount of energy being transferred or converted, divided by the time it took for that transfer or conversion to occur. Relevant equation: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
3.5.A.3	Because work is the change in energy of an object or system due to a force, average power is the total work done, divided by the time during which that work was done. Relevant equation: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
3.5.A.4	The instantaneous power delivered to an object by a force is given by the equation $P_{\text{inst}} = \dfrac{dW}{dt}$
3.5.A.5	The instantaneous power delivered to an object by the component of a constant force parallel to the object's velocity can be described with the derived equation $P_{\text{inst}} = F_\parallel v = Fv\cos\theta$

Suggested skills: 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.B (calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway), 2.C (compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Topic 3.5 prints no boundary statement, in either course. Unit 3's two boundary statements sit under Topic 3.2 and Topic 3.4, and both are about dissipation.

Notice that 3.5.A.1 is broader than the equations under it. Power is the rate at which energy changes either by transfer across a system boundary or by conversion from one type to another inside it. Both of those are power, and only the first is captured by thinking of power as "work over time."

The one statement AP Physics 1 does not have

AP Physics 1's Topic 3.5 has four essential knowledge statements. AP Physics C: Mechanics has five. The insert is exactly one:

AP Physics 1	AP Physics C: Mechanics	Relationship
3.5.A.1	3.5.A.1	word for word identical
3.5.A.2	3.5.A.2	word for word identical, same equation
3.5.A.3	3.5.A.3	word for word identical, same equation
does not exist	3.5.A.4	$P_{\text{inst}} = dW/dt$
3.5.A.4	3.5.A.5	word for word identical, renumbered

The calculus-based course inserts the derivative as the definition of instantaneous power and pushes the constant-force result down one place. Nothing is removed and nothing is reworded.

Here is the part that is not obvious, and it is worth getting right. Both courses call $P_{\text{inst}} = F_\parallel v = Fv\cos\theta$ a derived equation, in identical wording, and both course and exam descriptions define derived equations the same way: results provided to demonstrate the final results of derivations expected of students on the exam, denoted "Derived Equations," and not among the equations available on the sheet.

And yet the AP Physics 1 equation sheet prints $P_{\text{inst}} = F_\parallel v = Fv\cos\theta$ . The AP Physics C: Mechanics sheet does not.

So the "derived" label is not by itself what determines whether an equation reaches the sheet, and a page that told you it was would be wrong. In AP Physics 1 this equation is labelled derived and printed anyway; in AP Physics C: Mechanics it is labelled derived and left off, because the course expects you to reach it from $dW/dt$ in two lines. Do not reason from the label to the sheet. Check the sheet.

If the algebra-based treatment is what you need, it is at AP Physics 1 Topic 3.5. That page is for students who are handed $Fv\cos\theta$ on their sheet and never meet $dW/dt$ . This page is for students in the reverse position.

What each sheet prints

Checked against the Table of Information printed in each course and exam description:

Equation	AP Physics 1 sheet	AP Physics C: Mechanics sheet
$P_{\text{avg}} = \dfrac{W}{\Delta t} = \dfrac{\Delta E}{\Delta t}$	yes, as one line	yes, as one line
$P_{\text{inst}} = \dfrac{dW}{dt}$	no	yes
$P_{\text{inst}} = F_\parallel v = Fv\cos\theta$	yes	no

Three observations.

Average power is common ground. Both sheets print the two forms of average power combined on a single line, with the work version and the energy version separated by an equals sign, matching 3.5.A.2 and 3.5.A.3 taken together. There is nothing calculus-based about average power and this page will not pretend otherwise.

Instantaneous power is where the courses split, and the split is complete: each sheet prints exactly one instantaneous-power line, and they are different lines.

The watt appears in the Unit Symbols box, not in the equations. The AP Physics C: Mechanics Table of Information lists hertz, joule, kilogram, meter, newton, second and watt in that box. There is no separate conversion or definition printed for the watt; it is one joule per second, which follows from any of the equations above.

The full transcription of both sheets lives on the AP Physics C: Mechanics formula sheet page and the AP Physics 1 formula sheet page.

From dW/dt to F dot v, in two lines

This is the derivation the AP Physics C course expects and the algebra-based course cannot perform. It takes two lines and it is worth being able to write from memory.

Start from what the sheet prints, $P_{\text{inst}} = \dfrac{dW}{dt}$ , and from Topic 3.2's definition of work, $W = \int \vec{F} \cdot d\vec{r}$ , whose integrand is $dW = \vec{F} \cdot d\vec{r}$ . Then:

P_{\text{inst}} = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}

Write the dot product out with the angle between the force and the velocity, per essential knowledge 3.2.A.3.i, and it becomes $P_{\text{inst}} = Fv\cos\theta$ , which is 3.5.A.5.

Two things to notice about what that argument used and did not use.

It never assumed the force was constant. The relation $P_{\text{inst}} = \vec{F} \cdot \vec{v}$ holds instant by instant whatever the force is doing, because it is a statement about one moment. Statement 3.5.A.5 states the constant-force case specifically, because that is the case in which the same expression also gives the average power over an interval. When the force varies, $\vec{F} \cdot \vec{v}$ still gives the power at each instant and the average has to come from 3.5.A.2 or 3.5.A.3 instead. Worked example 2 has exactly that situation.

It runs backwards too. Since $P = dW/dt$ , the work done over an interval is $W = \int P\, dt$ , so the area under a power-against-time graph is an energy. The CED's own Unit 3 sample instructional activity for this topic is built on that: it asks students to construct a graph of the power delivered to a car as a function of time as the car accelerates from rest to full speed, and then to describe how they could use that graph to determine the velocity of the car as a function of time. The route is $\int P\, dt = W = \Delta K$ , then solve for $v$ . Worked example 1 does that with numbers.

$\cos\theta$ carries the sign here as it does in work. A force perpendicular to the velocity delivers zero power, which is why the tension in a string during uniform circular motion never changes the speed. A force opposing the motion delivers negative power, which is what a brake does.

Average against instantaneous, and when they differ

Statements 3.5.A.2 and 3.5.A.3 are about an interval; 3.5.A.4 and 3.5.A.5 are about an instant. Keeping them apart is most of the marks in this topic.

P_{\text{avg}} = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t} \qquad \qquad P_{\text{inst}} = \frac{dW}{dt}

The two agree only when the power is constant over the interval. Three cases worth having in mind:

Constant power. Average equals instantaneous at every moment. A machine rated at a fixed output is this case, and it has an odd consequence: since $P = Fv$ , the force must fall as the speed rises, without limit as $v \to 0$ . Worked example 2 works through it.
Power rising linearly with time. The average over an interval starting from zero is exactly half the final instantaneous value, because the area under a straight line through the origin is half the enclosing rectangle. Worked example 1 gets 24 kW against 48 kW.
Constant force and constant velocity. Everything collapses: the force is constant, the speed is constant, so the power is constant, average equals instantaneous, and 3.5.A.5's derived equation applies exactly. This is the regime the work and power calculator assumes, and it is the only regime AP Physics 1 can pose. Worked example 3 is one.

A trap worth naming. Average power is not the power at the average speed, and it is not the average of the starting and ending powers unless the power happens to be linear in time. Compute it as total energy divided by total time, which is what 3.5.A.2 says.

A second trap, from the wording of 3.5.A.1. Power is the rate at which energy changes either by transfer into or out of a system or by conversion from one type to another within a system. So a system can be delivering power internally with no energy crossing its boundary at all. A block sliding down a frictionless ramp converts gravitational potential energy into kinetic energy at a rate that is a genuine power, even though the block-and-Earth system exchanges nothing with its surroundings.

How Topic 3.5 is tested

Learning objective 3.5.A appears in none of the fifteen sample multiple-choice questions and in none of the four sample free-response questions in the AP Physics C: Mechanics course and exam description. It is the only learning objective in Unit 3 that appears in neither list.

That is a fact about the CED's sample set, not a prediction about your exam, and the framework says as much on its exam-weighting page: required course content, meaning the learning objectives and essential knowledge, can be assessed with any skill. Unit 3 as a whole is 15 to 25% of the multiple-choice section.

What the CED does provide for this topic is a sample instructional activity, the fifth and last listed for Unit 3. Students construct a graph of the power delivered to a car as a function of time as the car accelerates from rest and reaches full speed, then describe how they could use that graph to determine the velocity of the car as a function of time. Of the five activities Unit 3 lists, one is for Topic 3.3, three are for Topic 3.4, and this is the only one for Topic 3.5. Topics 3.1 and 3.2 have none.

The suggested skills point the same way as that activity. This topic lists 2.B, calculate an unknown quantity, and 2.C, compare quantities between scenarios, but not 2.A, derive a symbolic expression, which Topics 3.2, 3.3 and 3.4 all list. AP Physics 1's Topic 3.5, by contrast, does list 2.A along with 3.A and 3.C. On the suggested skills alone, the calculus-based course frames Power as more computational than its algebra-based counterpart, which inverts what you might expect. The framework's caveat above applies to that too.

Related pages. The work and power calculator handles the constant-force, constant-speed regime, work compared with power is the short version of the distinction that trips students up, and the glossary entry on power is the one-paragraph definition. Topic 3.2 supplies the $W$ that this topic differentiates, and Topic 3.1 supplies the kinetic energy that most power problems end up solving for. The Unit 3 hub lists every equation in the unit against what the sheet prints.

Power that ramps with time, and the velocity it implies

A 1200 kg car starts from rest, and its drivetrain delivers power to the car that increases linearly with time, $P(t) = bt$ with $b = 6.0$ kW/s. Assume all of that energy goes into the car's kinetic energy. Find (a) the work done in the first 8.0 s, (b) the speed at 8.0 s, (c) the speed as a function of time and what that implies about the acceleration, and (d) the average power over the 8.0 s compared with the instantaneous power at the end.

Convert once: $b = 6.0\ \mathrm{kW/s} = 6000\ \mathrm{W/s}$ , so $P(t) = 6000t$ watts with $t$ in seconds.
(a) Invert 3.5.A.4. Since $P = dW/dt$ , the work is $W = \int_0^t P\, dt' = \int_0^t bt'\, dt' = \frac{1}{2}bt^2$ .
At $t = 8.0$ s: $W = \frac{1}{2}(6000)(64) = 192000$ J, or 192 kJ.
(b) The car started from rest, so by the work-energy theorem $\Delta K = W$ with $K_i = 0$ : $v = \sqrt{2W/m} = \sqrt{2(192000)/1200} = \sqrt{320} = 17.889$ m/s, or 17.9 m/s.
(c) Symbolically, $\frac{1}{2}mv^2 = \frac{1}{2}bt^2$ , so $v(t) = t\sqrt{b/m}$ .
$\sqrt{b/m} = \sqrt{6000/1200} = \sqrt{5.0} = 2.2361$ , so $v(t) = 2.24t$ m/s. The speed is linear in time, which means the acceleration is constant at $2.24\ \mathrm{m/s^2}$ .
Check at $t = 8.0$ s: $v = 2.2361(8.0) = 17.889$ m/s, matching part (b).
Cross-check with 3.5.A.5. The force here is constant, $F = ma = (1200)(2.2361) = 2683.3$ N, and it acts along the velocity, so the derived equation applies: $P_{\text{inst}} = Fv = (2683.3)(17.889) = 48000$ W at $t = 8.0$ s. Direct from the given function, $P(8.0) = 6000(8.0) = 48000$ W. The two routes agree.
(d) Average power over the interval, from 3.5.A.3: $P_{\text{avg}} = W/\Delta t = 192000/8.0 = 24000$ W, or 24 kW.
So the average power is exactly half the final instantaneous power of 48 kW. That factor of two is general for any power rising linearly from zero, because the area under a straight line through the origin is half the rectangle around it.
The physical reading is neat: a power output that ramps in proportion to time is exactly what constant acceleration requires, because the force stays fixed while the speed grows in proportion to time.

(a) 192 kJ. (b) 17.9 m/s. (c) $v(t) = t\sqrt{b/m} = 2.24t$ m/s, which is linear in time, so the acceleration is constant at $2.24\ \mathrm{m/s^2}$ . (d) The average power is 24 kW, exactly half the 48 kW instantaneous power at 8.0 s.

Constant power, and the force that cannot be constant

A 1400 kg car accelerates from rest along a level road while its engine delivers a constant 60 kW to it, all of which goes into kinetic energy. Find (a) the speed after 5.0 s, (b) the driving force and acceleration at that moment, and (c) explain why the constant-force version of the power equation does not apply here and what happens as the speed approaches zero.

(a) The power is constant, so the work done is simply $W = P\Delta t = (60000)(5.0) = 300000$ J, which is 3.5.A.3 rearranged.
From rest, $\frac{1}{2}mv^2 = 300000$ J, so $v = \sqrt{2(300000)/1400} = \sqrt{428.57} = 20.702$ m/s, or 20.7 m/s.
Symbolically, $v(t) = \sqrt{2Pt/m}$ : the speed grows as the square root of time, not linearly.
(b) Use the instantaneous relation $P = Fv$ , valid at each moment: $F = P/v = 60000/20.702 = 2898.3$ N.
Then $a = F/m = 2898.3/1400 = 2.0702\ \mathrm{m/s^2}$ .
Confirm by differentiating: $a = \dfrac{dv}{dt} = \dfrac{d}{dt}\sqrt{2Pt/m} = \dfrac{1}{2}\sqrt{\dfrac{2P}{mt}} = \dfrac{1}{2}\sqrt{\dfrac{120000}{(1400)(5.0)}} = \dfrac{1}{2}\sqrt{17.143} = 2.0702\ \mathrm{m/s^2}$ . The two agree.
(c) The force is not constant. It was 2898 N at 5.0 s; at 2.0 s the speed is $\sqrt{2(60000)(2.0)/1400} = 13.093$ m/s and the force is $60000/13.093 = 4583$ N. So 3.5.A.5, which is stated for the component of a constant force parallel to the velocity, does not describe this situation.
What does still hold at every instant is $P_{\text{inst}} = \vec{F} \cdot \vec{v}$ , derived from $dW/dt$ in the section above without any assumption that $\vec{F}$ is constant. That derivation is why part (b) was legitimate.
As $v \to 0$ , $F = P/v \to \infty$ . A real car cannot start from rest at constant power, which is why this model is only ever used above some starting speed. The mathematics is telling you something true about the model rather than about cars.
Contrast with worked example 1: there the power ramped and the acceleration was constant; here the power is constant and the acceleration falls. Constant power is not constant force, and it is not constant acceleration.

(a) 20.7 m/s, from $v = \sqrt{2Pt/m}$ . (b) $F = 2898$ N and $a = 2.07\ \mathrm{m/s^2}$ , confirmed both from $P = Fv$ and by differentiating $v(t)$ . (c) The force falls as the speed rises, so 3.5.A.5's constant-force condition fails; the general relation $P = \vec{F} \cdot \vec{v}$ still holds instant by instant, and it forces $F$ to diverge as $v$ approaches zero.

The case where the derived equation applies exactly

A 30 kg crate is dragged across a level floor at a constant 1.5 m/s by a rope pulling at 37 degrees above the horizontal. The coefficient of kinetic friction is 0.25. Take $g = 9.8\ \mathrm{m/s^2}$ , $\sin 37^\circ = 3/5$ and $\cos 37^\circ = 4/5$ from the trigonometric table printed in the Table of Information. Find the tension and the power the rope delivers, and verify that answer a second way.

Declare the convention: the direction of motion is positive $x$ , upward is positive $y$ , and the crate moves at constant velocity so it is in equilibrium in both directions.
Vertical equilibrium: $F_N + T\sin 37^\circ - mg = 0$ , so $F_N = mg - 0.600T = (30)(9.8) - 0.600T = 294 - 0.600T$ newtons.
Horizontal equilibrium: $T\cos 37^\circ = \mu_k F_N$ , so $0.800T = 0.25(294 - 0.600T)$ .
Expand: $0.800T = 73.5 - 0.150T$ , so $0.950T = 73.5$ and $T = 77.368$ N.
Every quantity is now constant in time: the tension, the angle and the speed. So 3.5.A.5 applies exactly, with $\theta$ the 37 degrees between the rope and the velocity.
$P = Tv\cos\theta = (77.368)(1.5)(0.800) = 92.842$ W, or 93 W.
Verification a second way. The crate's kinetic energy is constant, so by 3.4.B the energy the rope puts in must equal the energy friction takes out, at every instant.
The normal force is $F_N = 294 - 0.600(77.368) = 294 - 46.421 = 247.58$ N, so the friction force is $\mu_k F_N = 0.25(247.58) = 61.895$ N.
Friction is antiparallel to the velocity, so the power it removes is $(61.895)(1.5) = 92.842$ W. That matches the rope's input to the digit.
Because everything is constant, the average power over any interval is the same 93 W, and 3.5.A.2, 3.5.A.3 and 3.5.A.5 all give the same number. This is the one regime where the distinction between average and instantaneous power carries no information.
Note where the angle went. Only $T\cos 37^\circ = 61.9$ N of the 77.4 N tension is parallel to the motion, and only that component appears in the power. The vertical component does no work at all, but it is not idle: by lifting some of the crate's weight it reduced the normal force from 294 N to 248 N and so reduced the friction the rope has to overcome.

The tension is 77 N and the rope delivers 93 W. Friction removes the same 92.8 W, as it must for a crate whose kinetic energy is constant. Because the force, angle and speed are all constant, this is a case where 3.5.A.5's derived equation is exactly valid and average and instantaneous power coincide.

Frequently asked questions

What is the formula for instantaneous power in AP Physics C?

Instantaneous power is the derivative of work with respect to time, written P_inst equals dW/dt. Essential knowledge 3.5.A.4 of the AP Physics C: Mechanics course and exam description gives it in exactly that form and it is printed on the AP Physics C: Mechanics equation sheet. That statement does not exist in AP Physics 1. From it you can derive the more familiar force-times-velocity form in two lines, since dW equals the force dotted with the displacement element, so dW/dt equals the force dotted with the velocity, which written out with the angle between them is F v cos theta.

Is P = Fv cos theta on the AP Physics C Mechanics equation sheet?

No. The AP Physics C: Mechanics sheet prints instantaneous power only as the derivative of work with respect to time. Force times velocity times cosine appears in the framework at essential knowledge 3.5.A.5, where it is labelled a derived equation and stated specifically for the component of a constant force parallel to the object's velocity. Interestingly, the AP Physics 1 framework labels the same equation a derived equation in identical wording at its own 3.5.A.4, and the AP Physics 1 sheet prints it anyway. So the derived label does not by itself determine what reaches a sheet, and each sheet has to be checked directly.

What is the difference between average and instantaneous power?

Average power is about an interval and instantaneous power is about a moment. Essential knowledge 3.5.A.2 defines average power as the energy transferred or converted divided by the time it took, and 3.5.A.3 gives the equivalent form as total work divided by the time during which the work was done. Both appear on a single printed line of the AP Physics C: Mechanics equation sheet. Instantaneous power at 3.5.A.4 is the derivative of work with respect to time. The two are equal only when the power is constant over the interval. For a power that rises linearly from zero, the average over the interval is exactly half the final instantaneous value.

How do you find velocity from a power against time graph?

Integrate the graph, then use the work-energy theorem. Since instantaneous power is the derivative of work with respect to time, the area under a power against time curve is the work done. If the object started from rest and all of that work went into kinetic energy, then one half m v squared equals that area, so the speed is the square root of twice the area divided by the mass. The AP Physics C: Mechanics course and exam description sets exactly this task as its sample instructional activity for Topic 3.5, asking students to graph the power delivered to a car as a function of time and then describe how to use that graph to determine the car's velocity as a function of time.

Is AP Physics C Topic 3.5 different from AP Physics 1 Topic 3.5?

AP Physics 1's Topic 3.5 has four essential knowledge statements and AP Physics C: Mechanics has five. The first three are word for word identical in both courses, including both average power equations. The calculus-based course inserts instantaneous power as the derivative of work with respect to time at 3.5.A.4 and renumbers the constant-force result to 3.5.A.5. Neither course prints a boundary statement for this topic. The suggested skills also differ, and not in the direction most people expect: AP Physics 1 lists 2.A, derive a symbolic expression, while AP Physics C: Mechanics lists 2.B, calculate an unknown quantity, instead.

Why does the force decrease as a car speeds up at constant power?

Because power is the product of the force and the velocity at every instant, so at fixed power the two are inversely related. Once the speed has doubled, the same engine output can only supply half the driving force. This follows from the general relation between instantaneous power and velocity, which is derived from the printed equation for instantaneous power as the derivative of work with respect to time and does not require the force to be constant. It also means that a constant-power model breaks down at very low speed, since the force it demands grows without limit as the speed approaches zero. Essential knowledge 3.5.A.5, by contrast, is stated only for a constant force.

What units is power measured in on the AP Physics C exam?

Watts. The AP Physics C: Mechanics Table of Information lists the watt in its Unit Symbols box alongside the hertz, joule, kilogram, meter, newton and second, and gives no separate definition for it because one watt is one joule per second, which follows from any of the topic's equations. Since power is energy divided by time, an answer in joules per second is the same thing and will be marked correct. Kilowatts appear in real-world figures and need converting before they meet a mass in kilograms and a speed in metres per second.