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concepts/mechanics/u3/m3-5-power.tex

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+\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.