Topic 3.5: Power

What Topic 3.5 requires

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. Four essential knowledge statements sit under it, and three of them carry an equation.

• 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. The CED gives the relevant equation as $P_{\text{avg}} = \Delta E/\Delta t$.
• 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. The relevant equation is $P_{\text{avg}} = W/\Delta t$.
• 3.5.A.4 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 $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$.

The topic carries no boundary statement. The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Unit 3 is weighted at 18 to 23 percent of the multiple-choice section and runs about 22 to 27 class periods.

Every one of those equations is printed on the AP Physics 1 formula sheet, and the two average-power forms share a single line there: $P_{\text{avg}} = W/\Delta t = \Delta E/\Delta t$, with $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$ on the line below.

Power is a rate, and the CED counts two kinds of change (3.5.A.1)

Power is not an amount of energy. It is how fast an amount is changing, so its unit is joules per second, given the name watt.

Statement 3.5.A.1 is careful to name two different things that both count as a change:

• Transfer into or out of a system. Energy crosses the boundary. A motor lifting a load, a rope hauling a cart, friction draining a sliding block: each carries a power, because each moves energy across the line you drew.
• Conversion from one type to another within a system. Nothing crosses the boundary at all, and there is still a rate. A falling rock inside a rock-and-Earth system converts gravitational potential energy into kinetic energy at some number of joules per second, even though that system's total energy never changes.

The second case follows straight from Topic 3.4. A system whose total energy is constant can have a great deal going on inside it, and power measures the pace of it. So the total energy is constant and the power is zero are not the same claim.

Power therefore inherits the system dependence of everything else in Unit 3. Redraw the boundary and a quantity that was a transfer becomes a conversion, or the reverse. Say whose power you mean and for which system: the power delivered by the rope, the power dissipated by friction, the net power on the cart. On a free response, an unlabeled $P$ is an invitation to be marked down.

Average power: one idea, two printed equations (3.5.A.2 and 3.5.A.3)

The CED lists these as two statements because they answer two different questions, and it is worth keeping them apart.

3.5.A.2 is the general one. Average power is the amount of energy transferred or converted divided by the time it took. Any energy change counts: potential to kinetic, mechanical to thermal, whatever the situation contains.

3.5.A.3 is the specialization to work. Because work is the change in energy of an object or system due to a force, the average power delivered by a force is the total work that force did divided by the time it took to do it.

Reach for $\Delta E/\Delta t$ when the question describes an energy change, and for $W/\Delta t$ when it names a force and a displacement. The two agree whenever the work you divide by the time is the whole energy change, and they come apart only when you are holding part of it. A motor that lifts a load and speeds it up at the same time is filling two stores at once, so either printed form works provided you count both: the total energy change, or the total work the motor did. The first worked example below is that case, and taking only the gravitational part returns the wrong number.

Two mechanical points. First, $\Delta t$ is the duration of the transfer, not the duration of the whole problem, so a machine that works for 3.0 s and then idles for 7.0 s has two defensible average powers depending on which interval the question means. Second, average power is blind to everything inside its interval: it sees totals only, which is what the next section is about.

Instantaneous power: P = F v cos theta (3.5.A.4)

Statement 3.5.A.4 is precise about what this covers: the instantaneous power delivered to an object by the component of a constant force parallel to the object's velocity. The CED calls it a derived equation rather than a definition, and the derivation is one line. Over a short enough interval the force does work $W = F_{\parallel}d$, so

because displacement over time is the speed at that moment. That is skill 2.A in miniature, and it is worth being able to reproduce rather than memorize.

The angle $\theta$ is between the force and the velocity. It is not the angle to the ground and not the angle of a ramp. That is the same rule as $W = Fd\cos\theta$ in Topic 3.2, with velocity standing in for displacement, and the sign behaves identically.

• $\theta$ less than $90^\circ$: positive power, the force is feeding energy in.
• $\theta$ equal to $90^\circ$: zero power. The normal force on a level floor and the string tension in uniform circular motion each deliver zero power at every instant, however large they are.
• $\theta$ greater than $90^\circ$: negative power. Kinetic friction and air resistance remove energy, and writing the negative sign is better physics than writing a magnitude and calling it a loss.

The reading of $P = Fv$ worth carrying into the exam is that a constant force delivers power in proportion to the speed. The same push delivers twice the power at twice the speed, because its point of application now moves through twice the distance every second.

Average versus instantaneous, and when they agree

The middle row is the one worth practising, because it is where the two quantities differ by a fixed factor. A constant net force on an object starting from rest makes the speed rise linearly with time, so $P_{\text{inst}} = Fv$ rises linearly too, from zero at the start to $Fv_f$ at the end. A quantity rising linearly from zero averages to half its final value, so the average power over that interval is exactly half the instantaneous power at the end, and it is reached at the midpoint of the interval. The second worked example checks that against both definitions rather than assuming it.

The trap runs the other way. You cannot recover an instantaneous value from an average. A machine averaging 400 W over 10 s might have run steadily at 400 W, or delivered 4000 W for one second and nothing for nine; the average keeps no record. If a question says at that instant, you need a force and a speed at that instant, and if it says over the interval, you need totals.

Compare quantities carefully too, which is skill 2.C for this topic. Same work in half the time is double the average power. Same power moving twice the load means half the speed, because $P = Fv$ fixes the product and not either factor.

The area under a power versus time graph is energy

Skill 1.B asks for quantitative graphs with appropriate scales and units, and the graph to be fluent with here is power against time.

Rearrange 3.5.A.2 into $\Delta E = P_{\text{avg}}\Delta t$ and the graph reads itself. A rectangle of height $P$ and width $\Delta t$ on those axes is an energy, so adding the rectangles up gives the whole transfer.

• The area between the curve and the time axis is the energy transferred or converted, in joules when power is in watts and time in seconds.
• The height at a point is the instantaneous power at that moment, which is a rate and not an energy.
• The average power over an interval is the area divided by the width of that interval, which is the height of the rectangle with the same area.

This is the same move you already made in Topic 3.2, where 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. The status of the two readings is not identical, though, and it is worth being straight about that: the force-versus-position area is written into the CED as its own essential knowledge statement, while the power-versus-time area is not printed anywhere. It follows from 3.5.A.2. Treat it as a result you can derive in one line, not as a line you can quote.

Area below the time axis counts as negative, meaning energy leaving the system. A curve that dips below the axis is describing an interval in which the force opposed the motion, and that area is subtracted rather than added. The third worked example takes a piecewise graph apart end to end.

Measuring power in the lab (skill 3.A)

Suggested skill 3.A asks you to create experimental procedures appropriate for a given scientific question, and power is easy to measure badly, because it needs two quantities that come from different instruments.

For the question what average power does this motor deliver while lifting a load?, a defensible procedure is short:

1. Measure the mass of the load on a balance, so the energy change is computed rather than assumed.
2. Measure the vertical rise with a meter stick, and measure the final speed if the load is still moving at the end, since $\Delta E$ has to include any kinetic energy gained.
3. Time the lift with a photogate or a video frame count rather than a hand-held stopwatch. Reaction time is a systematic error here, and it shortens $\Delta t$, which inflates the power.
4. Compute $\Delta E = \Delta U_g + \Delta K$ and divide by the measured time.
5. Repeat, average, and state plainly whether the number is an average over the whole lift or a peak.

For the graphical version, which is where skill 1.B earns its place, vary the load and plot the energy transferred against the time taken. If the motor delivers a steady power, the points fall on a straight line through the origin whose slope is that power. A slope drawn through several points is stronger evidence for a claim, which is skill 3.C, than a single ratio computed once.

On units, one watt is one joule per second, so power multiplied by time is energy. If an answer to a power question comes out in joules, a division by time went missing somewhere; if an answer to an energy question comes out in watts, a multiplication did.

Where Topic 3.5 leads

Power closes Unit 3 by putting time back into an account that deliberately left it out. Topic 3.4 tells you how much energy moved and where it went. Topic 3.5 tells you how fast. Nothing in the balance itself changes, which is why almost every power question is an energy question with one extra division at the end.

Practice the underlying work calculation with the work-energy theorem guide, and check any $W$, $P_{\text{avg}}$, or $P_{\text{inst}}$ arithmetic against the work and power calculator, which runs all three sheet equations. The conservation of energy guide supplies the $\Delta E$ that most power questions need before they can begin.

The definition reappears later rather than being replaced. Unit 6 puts rotational work on the sheet as $W = \tau\Delta\theta$, and dividing any work by the time it took is still average power. In AP Physics 2 the electrical power of a circuit element is $P = I\Delta V$, which is still joules per second. Everything on this page about the gap between an average and an instant carries straight across. The Unit 3 overview shows how the five topics fit together.