\subsection{Power and Instantaneous Power}

This subsection introduces power as the rate at which work is done. In AP mechanics, the key idea is local: the instantaneous power delivered by a force depends on the component of that force along the motion, so the sign of the power tells whether the force is adding energy to the object or removing it.

\dfn{Average and instantaneous power}{Let $\Delta W$ denote the work done by a force during a time interval $\Delta t>0$. The \emph{average power} over that interval is
\[
P_{\text{avg}}=\frac{\Delta W}{\Delta t}.
\]
The \emph{instantaneous power} at time $t$ is the limit of the average power over shorter and shorter time intervals:
\[
P=\lim_{\Delta t\to 0}\frac{\Delta W}{\Delta t}=\frac{dW}{dt}.
\]
Power is a scalar quantity. Its SI unit is the watt:
\[
1\,\mathrm{W}=1\,\mathrm{J/s}.
\]
If $P>0$, the force is doing positive work and transferring energy to the object. If $P<0$, the force is doing negative work and removing mechanical energy from the object.}

\thm{Mechanical power from force and velocity}{Let $\vec{F}$ denote the force acting on a particle, let $\vec{v}=d\vec{r}/dt$ denote the particle's velocity, let $F=|\vec{F}|$ denote the magnitude of the force, let $v=|\vec{v}|$ denote the speed, let $\theta$ denote the angle between $\vec{F}$ and $\vec{v}$, and let $F_{\parallel}$ denote the component of the force parallel to the motion. Then the instantaneous power delivered by the force is
\[
P=\vec{F}\cdot \vec{v}=Fv\cos\theta=F_{\parallel}v.
\]
Therefore only the component of the force along the motion contributes to power. If $\theta<90^\circ$, then $P>0$; if $\theta>90^\circ$, then $P<0$; and if $\theta=90^\circ$, then $P=0$. In one-dimensional motion along the $x$-axis,
\[
P=F_xv_x.
\]}

\pf{Short derivation from $P=dW/dt$}{Let $d\vec{r}$ denote the particle's infinitesimal displacement. From the local definition of work,
\[
dW=\vec{F}\cdot d\vec{r}.
\]
Divide by $dt$ to obtain the instantaneous rate at which work is done:
\[
P=\frac{dW}{dt}=\vec{F}\cdot \frac{d\vec{r}}{dt}.
\]
Since
\[
\frac{d\vec{r}}{dt}=\vec{v},
\]
it follows that
\[
P=\vec{F}\cdot \vec{v}.
\]
If $\theta$ is the angle between $\vec{F}$ and $\vec{v}$, then the dot-product form gives
\[
P=Fv\cos\theta=F_{\parallel}v.
\]}

\cor{Perpendicular forces do zero instantaneous power}{If a force is perpendicular to the velocity at a given instant, then
\[
\vec{F}\cdot \vec{v}=0,
\]
so
\[
P=0.
\]
Thus a force can change the direction of motion without transferring energy through work at that instant. Common AP examples are a normal force on frictionless motion along a surface and the centripetal force in uniform circular motion.}

\qs{Worked example}{A student pulls a sled across level snow with a rope of tension magnitude $T=85\,\mathrm{N}$ at an angle $\theta=25^\circ$ above the horizontal. The sled moves horizontally with constant speed $v=1.8\,\mathrm{m/s}$ for a time interval $\Delta t=40\,\mathrm{s}$. Let $\vec{T}$ denote the tension force and let $\vec{v}$ denote the sled's velocity.

(a) Find the instantaneous power delivered by the tension.

(b) Find the work done by the tension during the $40\,\mathrm{s}$ interval.

(c) Find the average power delivered by the tension over that interval.}

\sol Because the sled's velocity is horizontal, only the horizontal component of the tension contributes to the power and the work. The component of the tension along the motion is
\[
T_{\parallel}=T\cos\theta=(85\,\mathrm{N})\cos 25^\circ\approx 77.0\,\mathrm{N}.
\]

For part (a), the instantaneous power is
\[
P=\vec{T}\cdot \vec{v}=Tv\cos\theta.
\]
Substitute the given values:
\[
P=(85\,\mathrm{N})(1.8\,\mathrm{m/s})\cos 25^\circ\approx 1.39\times 10^2\,\mathrm{W}.
\]
So the instantaneous power delivered by the tension is
\[
P\approx 1.4\times 10^2\,\mathrm{W}.
\]
The power is positive because the tension has a component in the same direction as the motion.

For part (b), the sled's horizontal displacement during the interval is
\[
\Delta x=v\Delta t=(1.8\,\mathrm{m/s})(40\,\mathrm{s})=72\,\mathrm{m}.
\]
The work done by the tension is
\[
W=T\Delta x\cos\theta.
\]
Therefore,
\[
W=(85\,\mathrm{N})(72\,\mathrm{m})\cos 25^\circ\approx 5.55\times 10^3\,\mathrm{J}.
\]
So the tension does
\[
W\approx 5.6\times 10^3\,\mathrm{J}
\]
of positive work on the sled. The vertical component of the tension does no work because the displacement is horizontal.

For part (c), the average power is
\[
P_{\text{avg}}=\frac{\Delta W}{\Delta t}=\frac{5.55\times 10^3\,\mathrm{J}}{40\,\mathrm{s}}\approx 1.39\times 10^2\,\mathrm{W}.
\]
Thus,
\[
P_{\text{avg}}\approx 1.4\times 10^2\,\mathrm{W}.
\]
This matches the instantaneous power because the force magnitude, the angle, and the speed are all constant throughout the motion.
