Work vs Power: What Is the Difference?

Work is an amount of energy transferred by a force, measured in joules. Power is the rate at which that energy moves, measured in watts, where one watt is one joule per second. Work contains no reference to time, so the same work can be done at any power at all by taking longer or shorter over it.

AP Physics: Unit 3 (topics 3.2 Work, 3.5 Power). Both halves sit in AP Physics 1 Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section over an estimated 22 to 27 class periods. EK 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; EK 3.2.A.2 makes it a scalar that may be positive, negative or zero; EK 3.2.A.3.i gives W = F_parallel d = F d cos theta and restricts the transfer to the parallel component; EK 3.2.A.3.ii notes that the perpendicular component can change the direction of motion without changing the kinetic energy; EK 3.2.A.4 states the work-energy theorem; EK 3.2.A.4.iii equates the energy dissipated by friction to the friction force times the length of the path; EK 3.2.A.5 makes work the area under a graph of F_parallel against displacement. The Topic 3.2 boundary statement limits AP Physics 1 to the transfer of mechanical energy while expecting awareness that mechanical energy may be dissipated as thermal energy or sound, and defers heating and cooling to AP Physics 2. Topic 3.5 has one learning objective, 3.5.A, describe the transfer of energy into, out of, or within a system in terms of power, supported by EK 3.5.A.1 (power as the rate at which energy changes with respect to time), EK 3.5.A.2 (P_avg = delta E / delta t), EK 3.5.A.3 (P_avg = W / delta t) and EK 3.5.A.4 (P_inst = F_parallel v = F v cos theta, labelled a derived equation although the sheet prints it). The CED lists no boundary statement under Topic 3.5. Suggested skills are 1.B, 2.B, 2.D, 3.A and 3.B for Topic 3.2, and 1.B, 2.A, 2.C, 3.A and 3.C for Topic 3.5.

The distinction, stated once

Work is an amount. Power is a rate. Work answers how much energy moved; power answers how fast it moved.

The AP Physics 1 CED writes them in exactly those terms, three topics apart in the same unit. EK 3.2.A.1 says work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance. EK 3.5.A.1 says 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.

Look at what is present in each definition. Work mentions a force and a distance and no time whatsoever. Power mentions time and does not mention a force. So the two are not two sizes of the same quantity, and "more powerful" is not a stronger version of "more work".

The link between them is a division. EK 3.5.A.3 makes it explicit: 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.

P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}

That single division is the whole relationship, and it runs one way only. Knowing the work tells you nothing about the power until somebody gives you a time. Knowing the power tells you nothing about the work until somebody gives you a time. The time is not optional information; it is half the question.

Each side on its own sits in the work and power glossary entries. The procedure for computing work in a given geometry belongs to the work-energy theorem guide, and this page is about telling the two quantities apart.

Work vs power, side by side

	Work	Power
What it is	The amount of energy transferred by a force acting over a distance	The rate at which energy is transferred or converted
CED source	EK 3.2.A.1	EK 3.5.A.1
Symbol	$W$	$P$
SI unit	joule, J	watt, W, which is one joule per second
Scalar or vector	Scalar, and it may be positive, negative or zero, by EK 3.2.A.2	Scalar
Contains time	No, not at all	Yes, it is the whole point
Defining relation	$W = F_{\parallel} d = F d \cos\theta$	$P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$
Instantaneous form	Not a separate quantity	$P_{\text{inst}} = F_{\parallel} v = F v \cos\theta$ , from EK 3.5.A.4
Depends on how long the job took	No	Yes, inversely
Depends on how fast the object is moving	No	Yes, directly, through $P = Fv$
Zero when	The force is perpendicular to the displacement, or nothing moved	The energy of the system is not changing
Read off a graph as	The area under a graph of $F_{\parallel}$ against displacement, from EK 3.2.A.5	The slope of a graph of energy against time
Links to kinetic energy by	$\Delta K = \sum W_i$ , the work-energy theorem in EK 3.2.A.4	No direct counterpart
On the AP Physics 1 equation sheet	$W = F_{\parallel} d = F d \cos\theta$	Both $P_{\text{avg}}$ and $P_{\text{inst}}$ are printed
Everyday word that misleads	"Work" as effort, which can be nonzero when the physics work is zero	"Powerful" as strong, which describes a force

Read the time rows together. Work has one row saying no and power has two rows saying yes, and every worked example below is built by holding the work fixed and varying the time, or holding the force fixed and varying the speed.

The case that separates them: one job, three powers

Lift a 20 kg box 1.5 m onto a shelf, steadily, three times at three different paces. Take up as positive.

The work done by the lifting force is the same every time. At constant speed the lifting force balances the weight, so its magnitude is $mg = (20)(9.8) = 196\ \text{N}$ , and it acts through 1.5 m in the same direction as the motion:

W = F_{\parallel} d = (196\ \text{N})(1.5\ \text{m}) = 294\ \text{J}

That number does not know how long you took. Now divide it by three different times:

How long the lift took	Work done	Average power
$0.50\ \text{s}$	$294\ \text{J}$	$588\ \text{W}$
$2.0\ \text{s}$	$294\ \text{J}$	$147\ \text{W}$
$6.0\ \text{s}$	$294\ \text{J}$	$49\ \text{W}$
$60\ \text{s}$	$294\ \text{J}$	$4.9\ \text{W}$

One column is constant and the other spans a factor of 120. The box cannot tell the difference between the four lifts: it gains the same $294\ \text{J}$ of gravitational potential energy each time, since $\Delta U_g = mg\Delta y = (20)(9.8)(1.5) = 294\ \text{J}$ contains no time either.

Run it the other way and the asymmetry is just as sharp. A $100\ \text{W}$ output for $10\ \text{s}$ delivers $1000\ \text{J}$ ; a $100\ \text{W}$ output for $1000\ \text{s}$ delivers $100{,}000\ \text{J}$ . Same power, a hundredfold difference in work.

So the two quantities are independent until a time is supplied, and the practical reading is:

A question that gives you a force and a distance and no time can only be asking for work.
A question that mentions a rate, a duration, a speed, or a per-second anything is asking for power.
If a question gives you all three of force, distance and time, it wants both, and the order is work first.

The instantaneous form, and why speed buys power

Average power divides a whole job by a whole duration. That is fine when the rate is steady and useless when it is not, which is why the CED gives an instantaneous version. EK 3.5.A.4 says 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 = F v \cos\theta

and this relation is printed on the AP Physics 1 equation sheet, even though the CED labels it derived.

Read its structure, because it explains most of what people find counterintuitive about power. A given force delivers more power at higher speed, in direct proportion. A tractor pulling with 5000 N at 2 m/s delivers 10 kW; the same tractor pulling with 5000 N at 8 m/s delivers 40 kW. Nothing about the force changed. The energy is simply being handed over four times as fast because the point of application is moving four times as fast.

Three consequences worth having:

A stalled machine delivers zero power however hard it pushes. With $v = 0$ , $P_{\text{inst}} = 0$ . It is also doing zero work, since nothing is moving through any distance, which is the same fact seen from the other side.
The $\cos\theta$ is the same projection as in the work relation. A force perpendicular to the velocity delivers no power at all, which is why the normal force on a sliding block and the tension on an object in uniform circular motion transfer no energy. EK 3.2.A.3.ii says the perpendicular component can change the direction of the motion without changing the kinetic energy.
Power can be negative. If $\theta$ exceeds $90^\circ$ the cosine turns negative, and the force is removing energy rather than supplying it. Braking and friction are the standard cases. This follows from $P = W/\Delta t$ with a negative work and a positive time, and EK 3.2.A.2 already allows work to be negative.

The relation between the average and the instantaneous forms is not always simple, but there is one clean case. For a constant force acting along the motion of an object starting from rest, the average power is exactly half the final instantaneous power, because $P_{\text{avg}} = F v_{\text{avg}}$ and, under constant acceleration, $v_{\text{avg}}$ is half the final speed. Worked example three shows $141\ \text{W}$ against $283\ \text{W}$ for the same push.

Units, and the letter W that means two things

The units are the fastest way to catch a work-and-power mix-up, and they are also where a notation collision lives.

A joule is a unit of energy and therefore of work. Its dimensions come straight from $W = Fd$ : one newton times one metre.

A watt is a unit of power, and one watt is one joule per second. That definition is the whole distinction expressed in units. Multiply a watt by a second and you get a joule back; divide a joule by a second and you get a watt.

Both appear in the unit-symbols table on the AP Physics 1 equation sheet, which lists eight units in total, joule (J) and watt (W) among them. And here is the collision: the sheet's symbol key uses $W$ for work while its unit table uses W for the watt. So $W = 294\ \text{J}$ is a perfectly sensible statement in which the letter on the left and the unit on the right belong to different quantities. The sheet also lists $P$ = power and $E$ = energy in the same key.

One unit check that catches nearly everything: if your answer to a power question is in joules, or your answer to a work question is in watts, stop. The task verb Calculate is defined in the CED as performing mathematical steps to arrive at a final answer including algebraic expressions, properly substituted numbers, and correct labeling of units and significant figures, so a right number with the wrong unit is not a right answer.

Two further unit facts, clearly labelled as outside the AP sheets. The kilowatt hour is a unit of energy, not of power, despite containing the word watt: it is a power multiplied by a time, and one kilowatt hour is $3.6 \times 10^6\ \text{J}$ . And horsepower is a unit of power. Neither is printed on the AP Physics 1 equation sheet, whose unit-symbols table carries only the eight SI units, and no exam question will require them, but both are common enough outside class to be worth classifying correctly.

Where each one is read off a graph

Both quantities have a graphical form, and they live on different graphs, which is a useful thing to have straight before a Translation Between Representations question.

Work is an area. EK 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. That is why a variable force is still tractable in AP Physics 1: you never need to integrate, you need to find the area of the triangles and rectangles under the graph. A spring is the standard case, where the linear $F$ against $\Delta x$ graph gives a triangular area of $\frac{1}{2}k(\Delta x)^2$ , which is the printed $U_s$ .

The sign is in the geometry. Area below the horizontal axis counts as negative work, exactly as area below the axis on a velocity graph counts as negative displacement.

Power is a slope. It follows from EK 3.5.A.1, which calls power the rate at which energy changes with respect to time: plot the system's energy against time and the slope at any instant is the instantaneous power, while the slope of the straight line joining two points is the average power over that interval. A flat energy graph means zero power, and a downward-sloping one means negative power, energy leaving the system.

So the pair of readings is:

Force against displacement: the area is the work.
Energy against time: the slope is the power.

And one more that combines them, from the instantaneous relation. Since $P_{\text{inst}} = F_{\parallel} v$ , a graph of force against time and a graph of velocity against time can be multiplied point by point to build a power against time graph, and the area under that returns you to the work. The two operations, area and slope, run in opposite directions between the same pair of quantities, which is worth knowing purely as a check: if you find a power by taking a slope, you can recover the work by taking an area, and the two answers must agree.

Where it costs a mark

Reporting a work in watts or a power in joules. The fastest error to make and the fastest to catch.
Dividing by time when the question asked for work. Work has no time in it. If a problem gives a duration and asks for the work done, the duration is either irrelevant or is there to let you find a distance.
Forgetting to divide by time when the question asked for power. The mirror image, and it usually shows up as an answer that is numerically the work.
Using $P = Fv$ with the average speed and calling the result an instantaneous power. $P_{\text{inst}} = F_{\parallel} v$ needs the instantaneous speed at that moment. Substituting an average gives an average power, which for a constant force from rest is half the final instantaneous value.
Dropping the $\cos\theta$ from either relation. Both $W = Fd\cos\theta$ and $P_{\text{inst}} = Fv\cos\theta$ project onto the direction of motion, and $\theta$ is the angle between the force and the displacement or velocity, not the angle to the horizontal unless those happen to coincide.
Claiming work is done when nothing moves. Holding a heavy box motionless at arm's length does zero work on the box, because the displacement is zero. You will still get tired, and that physiological cost is outside the mechanical accounting a Topic 3.2 boundary statement sets: 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.
Using the net displacement of a round trip in the work relation. Apply it leg by leg. This is the same trap as in distance vs displacement, where the letter $d$ carries a displacement in the work relation and a path length in the friction-dissipation relation of EK 3.2.A.4.iii.
Assuming a larger force means a larger power. Power is a product of force and speed. A very large force at a very small speed can deliver less power than a small force at high speed.
Treating power as a property of a machine rather than of a situation. The same engine delivers different power at different speeds, which is what $P = Fv$ says.

What the equation sheets give you, and what they do not

This is a well-served corner of the AP Physics 1 sheet, which prints four relevant lines and hides nothing.

For work it prints

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

and the work-energy theorem in the summed form

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

and for power it prints both the average and the instantaneous versions:

P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}

P_{\text{inst}} = F_{\parallel} v = F v \cos\theta

That is a contrast worth registering against the kinematics section of the same sheet, which prints no average-velocity definition at all. The sheet carries an explicit average for power and not for velocity, so do not reason from a supposed rule about what sheets print averages for. Check the sheet.

One curiosity in the CED's own labelling. EK 3.5.A.2 and EK 3.5.A.3 present the two average-power forms as relevant equations, while EK 3.5.A.4 introduces $P_{\text{inst}} = F_{\parallel} v = F v \cos\theta$ as a derived equation, and the sheet prints all three anyway. The labels are not a reliable guide to what is printed, which is why any claim about the sheet should be checked against the sheet itself.

In AP Physics C: Mechanics the same pair appears in calculus form. That sheet prints work as $W = \int_a^b \vec{F} \cdot d\vec{r}$ , keeps $P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$ unchanged, and replaces the instantaneous form with

P_{\text{inst}} = \frac{dW}{dt}

which is the clearest statement of the whole distinction on any of the four sheets: power is the derivative of work with respect to time. It also prints $F_x = -\frac{dU(x)}{dx}$ and $\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$ , so on that sheet the area-and-slope pairing from the previous section is written out as an integral and a derivative.

Both sheets are here: AP Physics 1 and C: Mechanics. For the CED framing, Topic 3.2 covers work and Topic 3.5 covers power. The solving routines are in the work-energy theorem and conservation of energy, the work and power calculator will run both for a given force and time, and the work, energy and power practice set has problems to sort.

The same 294 J at four different powers

A 20 kg box is lifted steadily through a vertical height of $1.5\ \text{m}$ . Using $g = 9.8\ \text{m/s}^2$ , find the work done by the lifting force and the box's gain in gravitational potential energy, then find the average power for lift times of $0.50\ \text{s}$ , $2.0\ \text{s}$ , $6.0\ \text{s}$ and $60\ \text{s}$ .

Convention: up is positive, and the box moves up at constant speed, so its acceleration is zero throughout.
Lifting force. With zero acceleration the vertical forces balance, so the lifting force has magnitude $mg = (20\ \text{kg})(9.8\ \text{m/s}^2) = 196\ \text{N}$ , directed up.
Work done by the lifting force. It acts in the same direction as the displacement, so $\theta = 0$ and $W = F d \cos 0 = (196\ \text{N})(1.5\ \text{m}) = 294\ \text{J}$ . No time appears anywhere in that line, which is the point of the example.
Cross-check through the energy. $\Delta U_g = mg\Delta y = (20)(9.8)(1.5) = 294\ \text{J}$ , matching the work done by the lifting force. It matches because the box gained no kinetic energy, so all the energy the lift supplied went to potential energy. Gravity meanwhile did $-294\ \text{J}$ , and the total work on the box is zero, consistent with $\Delta K = 0$ .
Average power at $0.50\ \text{s}$ : $P_{\text{avg}} = \frac{294\ \text{J}}{0.50\ \text{s}} = 588\ \text{W}$ .
At $2.0\ \text{s}$ : $\frac{294}{2.0} = 147\ \text{W}$ . At $6.0\ \text{s}$ : $\frac{294}{6.0} = 49\ \text{W}$ . At $60\ \text{s}$ : $\frac{294}{60} = 4.9\ \text{W}$ .
Check by multiplying back. $(588)(0.50) = 294\ \text{J}$ , $(147)(2.0) = 294\ \text{J}$ , $(49)(6.0) = 294\ \text{J}$ , $(4.9)(60) = 294\ \text{J}$ . Every power times its own time returns the same work, which is EK 3.5.A.3 read backwards.
Now the speeds, to connect this to the instantaneous form. The $2.0\ \text{s}$ lift moves at $1.5 / 2.0 = 0.75\ \text{m/s}$ , and $P_{\text{inst}} = F v = (196)(0.75) = 147\ \text{W}$ , matching the average power because the speed was steady. The $0.50\ \text{s}$ lift moves at $3.0\ \text{m/s}$ and gives $(196)(3.0) = 588\ \text{W}$ . Same force in both, and the power tracked the speed.

Work 294 J in every case, and the box gains 294 J of gravitational potential energy in every case. Average power 588 W, 147 W, 49 W and 4.9 W for the four lift times, a range of a factor of 120 for an unchanged job. The instantaneous form $P = Fv$ reproduces each figure from the same 196 N force and the lift speed.

A car at steady speed: power from force and speed

A car travels at a constant $25\ \text{m/s}$ along a level road against a total resistive force of $600\ \text{N}$ . Find the power the driving force must deliver, the work it does in $60\ \text{s}$ , and the distance covered. Then find the power needed if, at $50\ \text{m/s}$ , the total resistive force is measured to be $2400\ \text{N}$ .

Convention: the direction of travel is positive. The speed is constant, so the acceleration is zero and the driving force balances the resistive force at $600\ \text{N}$ .
Power at $25\ \text{m/s}$ . The driving force is along the velocity, so $\theta = 0$ and $P_{\text{inst}} = F v \cos 0 = (600\ \text{N})(25\ \text{m/s}) = 15{,}000\ \text{W}$ , which is $15\ \text{kW}$ .
Work in $60\ \text{s}$ . The power is steady, so $W = P \Delta t = (15{,}000\ \text{W})(60\ \text{s}) = 9.0 \times 10^5\ \text{J}$ , or $900\ \text{kJ}$ .
Distance covered: $d = v \Delta t = (25\ \text{m/s})(60\ \text{s}) = 1500\ \text{m}$ .
Independent check on the work, through the force and the distance rather than the power and the time: $W = F d = (600\ \text{N})(1500\ \text{m}) = 9.0 \times 10^5\ \text{J}$ . The two routes agree, which they must, since one is the other multiplied and divided by the same $\Delta t$ .
Note where the energy went. The car's kinetic energy is unchanged, since its speed is constant, so all $9.0 \times 10^5\ \text{J}$ was removed again by the resistive forces, which did $-9.0 \times 10^5\ \text{J}$ of work. The net work is zero and $\Delta K = 0$ , consistent with EK 3.2.A.4.
At $50\ \text{m/s}$ with a measured resistive force of $2400\ \text{N}$ : $P = (2400\ \text{N})(50\ \text{m/s}) = 120{,}000\ \text{W}$ , which is $120\ \text{kW}$ .
Compare the two. Doubling the speed doubled one factor and quadrupled the other, so the power went up by a factor of eight: $120 / 15 = 8$ . The force required grew, and the rate at which the same force has to deliver energy grew as well, and power multiplies the two. That is why the top speed of a vehicle is a power question rather than a force question.

At $25\ \text{m/s}$ : power 15 kW, work in 60 s 9.0 x 10^5 J, distance 1500 m, confirmed two ways. At $50\ \text{m/s}$ against $2400\ \text{N}$ : power 120 kW, eight times as much for twice the speed.

Average power against final power for a constant push

An $8.0\ \text{kg}$ cart, initially at rest on a frictionless horizontal surface, is pushed by a constant horizontal force of $40\ \text{N}$ through $5.0\ \text{m}$ . Find the work done, the cart's final speed, the time taken, the average power, and the instantaneous power at the start and at the end.

Convention: the direction of the push is positive $x$ , and the surface is frictionless so the push is the only horizontal force.
Work done by the push: $W = F d \cos 0 = (40\ \text{N})(5.0\ \text{m}) = 200\ \text{J}$ .
Final speed, from the work-energy theorem in EK 3.2.A.4. The cart starts at rest, so $\Delta K = K_f = 200\ \text{J}$ , and $\frac{1}{2}(8.0)v^2 = 200$ gives $v^2 = 50$ and $v = \sqrt{50} = 7.071\ \text{m/s}$ .
Time taken. The acceleration is $a = F/m = 40/8.0 = 5.0\ \text{m/s}^2$ , constant, so from $x = \frac{1}{2}a t^2$ : $5.0 = \frac{1}{2}(5.0)t^2$ , giving $t^2 = 2.0$ and $t = 1.414\ \text{s}$ . Cross-check with $v = at$ : $(5.0)(1.414) = 7.07\ \text{m/s}$ , matching the speed found from the energy.
Average power: $P_{\text{avg}} = \frac{W}{\Delta t} = \frac{200\ \text{J}}{1.414\ \text{s}} = 141.4\ \text{W}$ .
Instantaneous power at the start: $P_{\text{inst}} = F v = (40)(0) = 0\ \text{W}$ . The full 40 N is acting and the power is zero, because nothing is yet moving.
Instantaneous power at the end: $P_{\text{inst}} = F v = (40)(7.071) = 282.8\ \text{W}$ .
Compare. $282.8 / 141.4 = 2.00$ exactly, and the reason is visible: $P_{\text{avg}} = F v_{\text{avg}}$ for a constant force along the motion, and under constant acceleration from rest the average velocity is half the final velocity. Check it: $(40)(7.071/2) = (40)(3.536) = 141.4\ \text{W}$ .
So the power climbed linearly from $0$ to $282.8\ \text{W}$ while the force never changed by a newton, and the total work was $200\ \text{J}$ whichever way you compute it. The force says how much work per metre; the speed says how fast those metres arrive.

Work 200 J, final speed 7.07 m/s, time 1.414 s, average power 141.4 W. Instantaneous power 0 W at the start and 282.8 W at the end, exactly twice the average, because for a constant force from rest the average velocity is half the final velocity.

Frequently asked questions

What is the difference between work and power?

Work is an amount of energy and power is a rate. The AP Physics 1 CED defines work at essential knowledge 3.2.A.1 as the amount of energy transferred into or out of a system by a force exerted on that system over a distance, and power at essential knowledge 3.5.A.1 as the rate at which energy changes with respect to time. Work is measured in joules and contains no reference to time. Power is measured in watts, where one watt is one joule per second, and is the work divided by the time it took. The same job can therefore be done at any power at all by taking longer or shorter over it.

Can the same work be done at different powers?

Yes, and that is the clearest way to see that they are different quantities. Lifting a 20 kilogram box 1.5 metres transfers 294 joules of energy to it, whether you take half a second or a full minute. Take half a second and the average power is 588 watts. Take a minute and it is 4.9 watts. The box gains the same 294 joules of gravitational potential energy either way, because that energy change depends only on the mass, the field strength and the height. Power divides that fixed amount by whatever time you chose.

What is the unit of power, and is a watt the same as a joule?

The SI unit of power is the watt, and it is not the same as a joule. One watt is one joule per second, so a watt is an energy divided by a time. Both units appear in the unit-symbols table on the AP Physics 1 equation sheet. A useful check on any answer: work and energy come out in joules, and power comes out in watts. Note that the kilowatt hour, despite its name, is a unit of energy rather than power, because it is a power multiplied by a time; one kilowatt hour is 3.6 million joules. It does not appear on any AP equation sheet.

What does P equals Fv mean and when can you use it?

It gives the instantaneous power delivered by a force to an object moving at speed v, where only the component of the force parallel to the velocity counts, so the full form is F times v times the cosine of the angle between them. The AP Physics 1 CED introduces it at essential knowledge 3.5.A.4 for the component of a constant force parallel to the object's velocity, and the equation sheet prints it. The key reading is that a fixed force delivers power in direct proportion to speed: the same 5000 newton pull delivers four times the power at four times the speed. It also means a stalled machine delivers zero power however hard it pushes, since v is zero.

Can power be negative?

Yes, whenever energy is leaving the system rather than entering it. Power is the work divided by the time interval, and the AP Physics 1 CED states at essential knowledge 3.2.A.2 that work is a scalar quantity that may be positive, negative or zero. A positive time interval divided into a negative work gives a negative power. In the instantaneous form the sign comes from the cosine: if the angle between the force and the velocity is greater than 90 degrees the cosine is negative, which is the case for braking and for friction on a sliding object. A negative power is not a direction, since power is a scalar.

Is any work done if you hold a heavy object still?

No work is done on the object. Work requires a force acting over a distance, and if the object does not move, the distance is zero and so is the work, however large the force. Your muscles do consume energy, because they are contracting and relaxing internally, but that is a physiological cost rather than mechanical work on the object. A Topic 3.2 boundary statement in the AP Physics 1 CED keeps this outside the course, saying that 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. Since no work is done, the average power delivered to the object is also zero.

Does the AP Physics 1 equation sheet give both average and instantaneous power?

Yes, both are printed. The sheet carries average power as the work divided by the time interval and equally as the energy change divided by the time interval, and it carries instantaneous power as the parallel force component times the speed, or F times v times cosine theta. It also prints the work relation and the work-energy theorem in its summed form. This is worth contrasting with the kinematics section of the same sheet, which prints no definition of average velocity at all, so there is no general rule about which averages a sheet carries. On the AP Physics C: Mechanics sheet the instantaneous form is printed as the derivative of work with respect to time.