\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\)
Chapter 6: Energy
10.

Power

Power
\(P\)
W
When it comes to transferring energy, the total amount of energy is not the only consideration: it also matters how fast the energy is transferred. For example, I might be able to pick up a heavy weight just like Superman could, and if we hold it over our heads we will be doing the same amount of work. But Superman can do it a lot faster than I can. We call the rate of energy flow power
$$P={\Delta E\over\Delta t}$$

which we measure in units of watts (or joules per second: $1\u{W} = 1\u{J/s}$.

If the energy transfer in question is the work done by a force moving an object, then we can write this equation as $$P={Fd\cos\theta\over\Delta t}=F{d\over\Delta t}\cos\theta$$ but $d\over\Delta t$ is the speed of the object the work is done on, so we could also write the work done by a force F as $$P=Fv\cos\theta$$ where $v$ is the speed of the object, and $\theta$ is the angle between the force and the motion.

You are probably more familiar with watts in the context of electricity, when the power in question is related to the flow of electrical energy into a house (on an electrical bill) or the output of heat and light energy from a light bulb (although these days, a "60W light bulb" is more likely to mean "an LED or CFL light bulb that has the same brightness as a 60W incandescent bulb").

Electric consumption is usually listed in kilowatt-hours, or $\u{kW\cdot hr}$, which is a unit of energy equal to 3.6 million joules.