Energy is the amount. Power is the rate

Two space heaters run for the same hour. The 1500 W heater delivers three times the energy of the 500 W one. But run the 1500 W for one minute and the 500 W for three minutes, and they deliver the same amount of energy. Energy is how much. Power is how fast. Lifting a box up the same stairs slowly or quickly takes the same energy, but very different power.

§1

What power is.

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Power is the rate at which energy is transferred. Two forms appear on the AP equation sheet. Same idea, different inputs.

Average power

$P_\text{avg} = \dfrac{W}{\Delta t} = \dfrac{\Delta E}{\Delta t}$

Instantaneous power

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

The first form divides total energy by the time it took. The second multiplies the force by the component of velocity along that force, or equivalently $Fv\cos\theta$. Either way, the answer comes out in joules per second.

The unit is the watt: $1\ \text{W} = 1\ \text{J/s}$. A kilowatt is $1000$ W. One horsepower is about $746$ W, handy for everyday comparisons but rarely used on a physics problem.

Power is a scalar. Both forms give one number. The vectors $\vec F$ and $\vec v$ go into the second form, but their dot product is itself just a number. Two cars driving in opposite directions with the same engine produce the same power.

The sign of power tells direction of energy flow, not magnitude. A positive $P$ means energy is entering the system; a negative $P$ means energy is leaving. A motor whose power on a falling weight is $-200$ W is not a smaller motor than one delivering $+200$ W. Same magnitude, energy flowing the other way.

The angle $\theta$ between $\vec F$ and $\vec v$ matters. When $\theta = 0$, force and motion line up and $P = Fv$. When $\theta = 90^\circ$, force and motion are perpendicular and $P = 0$, even though both $F$ and $v$ are nonzero. A satellite in a circular orbit feels gravity always perpendicular to its motion, so gravity does zero power on the satellite.

§2

Two routes, one number.

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A 60 kg climber goes up a 3 m flight of stairs in 4.0 seconds, moving at a constant vertical speed. What is the average power output of the climber's legs during the climb? Take $g = 10\ \text{N/kg}$.

Both equation-sheet forms apply here. They have to give the same answer.

Setup

$m = 60\ \text{kg}$, $h = 3\ \text{m}$, $\Delta t = 4.0\ \text{s}$, $g = 10\ \text{N/kg}$. The climb is at constant velocity, so the climber's kinetic energy does not change. All of the work the legs do raises the gravitational potential energy.

Route 1 · via energy and time

Find the energy going into gravitational potential, then divide by the time:

$$\Delta E = mgh = (60)(10)(3) = 1800\ \text{J}$$ $$P_\text{avg} = \dfrac{\Delta E}{\Delta t} = \dfrac{1800}{4.0} = 450\ \text{W}$$

Route 2 · via force and velocity

At constant vertical velocity, the upward force from the legs equals the weight. The velocity is the height per time:

$$v = \dfrac{h}{\Delta t} = \dfrac{3}{4.0} = 0.75\ \text{m/s}$$ $$F = mg = (60)(10) = 600\ \text{N}$$ $$P = Fv = (600)(0.75) = 450\ \text{W}$$

Why they match

Both routes give 450 W. They had to: $P = Fv = (mg)(h/\Delta t) = mgh/\Delta t = \Delta E/\Delta t$. The two forms are one equation written two ways.

These are not two formulas. They are two paths to one number. Use whichever fits what you are given: energy and time go through $\Delta E/\Delta t$; force and velocity go through $Fv\cos\theta$.

§3

Where students go wrong.

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Three traps come up again and again. Each one treats a rate as if it were a quantity, a sign as if it were a size, or a scalar as if it were a vector.

Pitfall · 01

"A 60 W bulb stores 60 W of energy inside it."

This treats a watt rating as a quantity of energy sitting in the bulb. The intuition comes from everyday phrasing like "this battery has a lot of watts," which mixes up power with capacity.

Fix. A watt is a rate, $1\ \text{W} = 1\ \text{J/s}$. The bulb does not contain joules; it converts electrical energy into light and heat at a rate of 60 joules every second. Run it for 1 second, you have delivered 60 J. Run it for 1 hour, you have delivered $(60)(3600) = 216{,}000\ \text{J}$. The wattage tells you the rate of delivery, not a stash.

Pitfall · 02

"The motor's power is $-300$ W, so the motor must be off or broken."

Two errors get rolled into one here: reading a negative power as a smaller power, and reading it as evidence the device has stopped. Both ignore what the sign actually means.

Fix. The magnitude is 300 W; the sign tells the direction of energy flow. A negative $P$ means energy is leaving the system through that force. For an elevator descending at constant velocity, the cable pulls up while the car moves down, so $P_\text{T} = Fv\cos 180^\circ = -Fv$. Energy is leaving the car along the cable, back to the motor. The motor is fully active; the sign just tells you which way the joules are moving.

Pitfall · 03

"Power is a vector, since $P = Fv$ and both $F$ and $v$ have direction."

Force and velocity are vectors, so their product seems like it should be one too. The cosine in $P = Fv\cos\theta$ adds to that impression by looking like a direction factor.

Fix. Power is a scalar. The form $P = F v \cos\theta$ is the dot product $\vec F \cdot \vec v$, which always returns one number. That number can be positive or negative, but it never points in a direction. Two cars driving opposite ways with the same engine produce the same power.

§4

Skill Check.

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Ten scenarios. Each one targets one of the traps above, or applies the equation-sheet forms in a slightly different setup. Progress is saved as you go.