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\subsection{Work as a Line Integral}
This subsection develops work from the local relation between force and an infinitesimal displacement. In AP mechanics, this gives a clean calculus-based way to handle variable forces and to track when a force helps, opposes, or does no work on the motion.
\dfn{Infinitesimal work and total work along a path}{Let a particle move along a path $C$. Let $d\vec{r}$ denote an infinitesimal displacement vector of the particle, let $ds=|d\vec{r}|$ denote the corresponding infinitesimal path length, and let $\vec{F}$ denote the force acting on the particle at that location.
The \emph{infinitesimal work} done by the force during that displacement is
\[
dW=\vec{F}\cdot d\vec{r}.
\]
If $\hat{t}$ denotes the unit tangent to the path and $F_{\parallel}=\vec{F}\cdot \hat{t}$ denotes the component of the force parallel to the motion, then
\[
dW=F_{\parallel}\,ds.
\]
Therefore the \emph{total work} done by the force as the particle moves along the path $C$ is
\[
W=\int_C \vec{F}\cdot d\vec{r}=\int_C F_{\parallel}\,ds.
\]}
\thm{Line-integral form of work and the constant-force special case}{Let a particle move from an initial point to a final point along a path $C$ under a force $\vec{F}$. Then the work done by that force is
\[
W=\int_C \vec{F}\cdot d\vec{r}.
\]
Only the component of $\vec{F}$ parallel to the instantaneous displacement contributes, so equivalently
\[
W=\int_C F_{\parallel}\,ds.
\]
If the force is constant in magnitude and direction, then
\[
W=\vec{F}\cdot \Delta \vec{r},
\]
where $\Delta \vec{r}$ is the total displacement from the initial point to the final point. In particular, if the motion is along a straight line parallel to the constant force, then
\[
W=F\,\Delta s.
\]}
\nt{Work is positive when the force has a component in the same direction as the displacement, negative when that component is opposite the displacement, and zero when the force is perpendicular to the displacement. For a general force, the value of $W=\int_C \vec{F}\cdot d\vec{r}$ can depend on the path $C$, not just on the endpoints. In AP problems, this often appears when the force changes with position or when different paths make the parallel component $F_{\parallel}$ different. Typical zero-work cases include a normal force on motion along a surface or a centripetal force in uniform circular motion, because those forces are perpendicular to the instantaneous displacement.}
\pf{Why the line integral gives total work}{Break the path into many small displacement vectors $\Delta \vec{r}_1,\Delta \vec{r}_2,\dots,\Delta \vec{r}_n$. Over each small segment, the work is approximately
\[
\Delta W_k\approx \vec{F}_k\cdot \Delta \vec{r}_k.
\]
Summing over the path gives
\[
W\approx \sum_{k=1}^n \vec{F}_k\cdot \Delta \vec{r}_k.
\]
In the limit as the segments become infinitesimal, this Riemann sum becomes
\[
W=\int_C \vec{F}\cdot d\vec{r}.
\]
If $\vec{F}$ is constant, then it can be taken outside the integral:
\[
W=\vec{F}\cdot \int_C d\vec{r}=\vec{F}\cdot \Delta \vec{r}.
\]}
\qs{Worked example}{A cart moves along a straight horizontal track from $x_i=0$ to $x_f=5.0\,\mathrm{m}$. Let $x$ denote the cart's position coordinate, and let the applied force on the cart be
\[
\vec{F}(x)=(6.0-2.0x)\hat{\imath}\,\mathrm{N}.
\]
Find the work done by this force on the cart over the interval from $x=0$ to $x=5.0\,\mathrm{m}$. State where the force does negative work.}
\sol Because the motion is along the $x$-axis, the displacement element is
\[
d\vec{r}=dx\,\hat{\imath}.
\]
Therefore,
\[
dW=\vec{F}\cdot d\vec{r}=[(6.0-2.0x)\hat{\imath}]\cdot (dx\,\hat{\imath})=(6.0-2.0x)\,dx.
\]
So the total work is
\[
W=\int_0^{5.0} (6.0-2.0x)\,dx.
\]
Evaluate the integral:
\[
W=\left[6.0x-x^2\right]_0^{5.0}.
\]
Substitute the limits:
\[
W=(6.0)(5.0)-(5.0)^2-\left[(6.0)(0)-0^2\right]=30.0-25.0=5.0\,\mathrm{J}.
\]
Thus the force does
\[
5.0\,\mathrm{J}
\]
of net work on the cart.
To identify where the force does negative work, find where the force component along the motion becomes negative:
\[
6.0-2.0x<0.
\]
This occurs when
\[
x>3.0\,\mathrm{m}.
\]
So from $x=3.0\,\mathrm{m}$ to $x=5.0\,\mathrm{m}$, the force points opposite the displacement and does negative work. From $x=0$ to $x=3.0\,\mathrm{m}$, it does positive work.

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\subsection{Kinetic Energy and the Work-Energy Theorem}
This subsection introduces kinetic energy as the energy of motion and the work-energy theorem as the main AP bridge from force and displacement to speed without solving for time explicitly.
\dfn{Kinetic energy and net work}{Let $m$ denote the mass of a particle or body, let $\vec{v}$ denote its velocity, and let $v=|\vec{v}|$ denote its speed. The \emph{kinetic energy} of the body is
\[
K=\tfrac12 mv^2.
\]
If forces $\vec{F}_1$, $\vec{F}_2$, $\dots$, and $\vec{F}_n$ act on the body while it undergoes an infinitesimal displacement $d\vec{r}$, then the differential work done by force $i$ is
\[
dW_i=\vec{F}_i\cdot d\vec{r}.
\]
Let the net force be
\[
\vec{F}_{\text{net}}=\sum_{i=1}^n \vec{F}_i.
\]
Then the \emph{net work} is the sum of the works done by all forces:
\[
dW_{\text{net}}=\sum_{i=1}^n dW_i=\vec{F}_{\text{net}}\cdot d\vec{r}.
\]
Over a finite motion,
\[
W_{\text{net}}=\int \vec{F}_{\text{net}}\cdot d\vec{r}=\sum_{i=1}^n W_i.
\]
The SI unit of both work and kinetic energy is the joule, where $1\,\mathrm{J}=1\,\mathrm{N\,m}$.}
\thm{Work-energy theorem}{Let $m$ denote the mass of a body, let $\vec{v}$ denote its velocity, let $v=|\vec{v}|$ denote its speed, and let $d\vec{r}$ denote an infinitesimal displacement of the body. Then
\[
dK=\vec{F}_{\text{net}}\cdot d\vec{r}.
\]
Integrating from an initial state to a final state gives the work-energy theorem:
\[
W_{\text{net}}=\Delta K=K_f-K_i=\tfrac12 mv_f^2-\tfrac12 mv_i^2.
\]
Thus the net work done on a body equals the change in its kinetic energy.}
\nt{The work-energy theorem is often more efficient than combining Newton's second law with kinematics when the problem asks for a speed after a known displacement or after a known amount of work. It avoids solving for time and often avoids solving for acceleration explicitly. In AP mechanics, \emph{net work} means the algebraic sum of the work done by all forces on the chosen system. A force parallel to the displacement does positive work, a force opposite the displacement does negative work, and a force perpendicular to the displacement does zero work.}
\pf{Short derivation from Newton II}{Let $m$ denote the constant mass of the body, let $\vec{v}$ denote its velocity, and let $d\vec{r}$ denote its infinitesimal displacement. Start with Newton's second law,
\[
\vec{F}_{\text{net}}=m\frac{d\vec{v}}{dt}.
\]
Since
\[
d\vec{r}=\vec{v}\,dt,
\]
dot both sides of Newton's second law with $d\vec{r}$:
\[
\vec{F}_{\text{net}}\cdot d\vec{r}=m\frac{d\vec{v}}{dt}\cdot (\vec{v}\,dt)=m\vec{v}\cdot d\vec{v}.
\]
Now use
\[
v^2=\vec{v}\cdot \vec{v},
\]
so
\[
d(v^2)=2\vec{v}\cdot d\vec{v}.
\]
Therefore,
\[
\vec{F}_{\text{net}}\cdot d\vec{r}=m\vec{v}\cdot d\vec{v}=d\!\left(\tfrac12 mv^2\right)=dK.
\]
Integrating gives
\[
W_{\text{net}}=\int \vec{F}_{\text{net}}\cdot d\vec{r}=\int dK=K_f-K_i=\Delta K.
\]}
\qs{Worked example}{Choose the positive $x$-axis to the right. A crate of mass $m=4.0\,\mathrm{kg}$ moves to the right on a horizontal floor with initial speed $v_i=3.0\,\mathrm{m/s}$. A constant applied force of magnitude $F_A=20.0\,\mathrm{N}$ acts to the right while the crate moves a horizontal distance $\Delta x=6.0\,\mathrm{m}$. Kinetic friction of magnitude $f_k=4.0\,\mathrm{N}$ acts to the left. The normal force and the weight act vertically.
Find the crate's final speed $v_f$.}
\sol Use the work-energy theorem:
\[
W_{\text{net}}=\Delta K=\tfrac12 mv_f^2-\tfrac12 mv_i^2.
\]
Compute the work done by each force.
The applied force is parallel to the displacement, so its work is positive:
\[
W_A=F_A\Delta x=(20.0\,\mathrm{N})(6.0\,\mathrm{m})=120\,\mathrm{J}.
\]
The friction force is opposite the displacement, so its work is negative:
\[
W_f=-f_k\Delta x=-(4.0\,\mathrm{N})(6.0\,\mathrm{m})=-24\,\mathrm{J}.
\]
The normal force and the weight are perpendicular to the horizontal displacement, so each does zero work:
\[
W_N=0,
\qquad
W_g=0.
\]
Therefore the net work is
\[
W_{\text{net}}=W_A+W_f+W_N+W_g=120\,\mathrm{J}-24\,\mathrm{J}=96\,\mathrm{J}.
\]
Now find the initial kinetic energy:
\[
K_i=\tfrac12 mv_i^2=\tfrac12 (4.0\,\mathrm{kg})(3.0\,\mathrm{m/s})^2=18\,\mathrm{J}.
\]
So the final kinetic energy is
\[
K_f=K_i+W_{\text{net}}=18\,\mathrm{J}+96\,\mathrm{J}=114\,\mathrm{J}.
\]
Use $K_f=\tfrac12 mv_f^2$:
\[
\tfrac12 (4.0\,\mathrm{kg})v_f^2=114\,\mathrm{J}.
\]
Thus
\[
2.0\,v_f^2=114,
\qquad
v_f^2=57,
\qquad
v_f=\sqrt{57}\,\mathrm{m/s}\approx 7.5\,\mathrm{m/s}.
\]
Therefore the crate's final speed is
\[
v_f\approx 7.5\,\mathrm{m/s}.
\]
This method is shorter than solving for the acceleration and then using a kinematics equation, because the theorem connects net work directly to the change in speed.

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\subsection{Conservative Forces and Potential Energy}
This subsection introduces conservative forces through path-independent work and uses that idea to define potential energy differences for an interacting system.
\dfn{Conservative force and potential energy difference}{Let $\vec{F}_c$ denote a force associated with some interaction, let $i$ and $f$ denote initial and final positions, and let $C$ denote a path from $i$ to $f$. The force $\vec{F}_c$ is called \emph{conservative} if the work
\[
W_c(i\to f)=\int_C \vec{F}_c\cdot d\vec{r}
\]
depends only on the endpoints $i$ and $f$, not on the path $C$.
For a conservative force, the corresponding \emph{potential energy difference} of the interacting system is defined by
\[
\Delta U=U_f-U_i=-\int_i^f \vec{F}_c\cdot d\vec{r}.
\]
Thus the work done by the conservative force is
\[
W_c=-\Delta U.
\]}
\thm{Equivalent conservative-force relations}{Let $\vec{F}_c$ denote a conservative force and let $d\vec{r}$ denote an infinitesimal displacement. Then the following relations hold:
\[
\oint \vec{F}_c\cdot d\vec{r}=0,
\]
so the work done by $\vec{F}_c$ around any closed loop is zero.
Equivalently, for any two paths $C_1$ and $C_2$ connecting the same endpoints,
\[
\int_{C_1} \vec{F}_c\cdot d\vec{r}=\int_{C_2} \vec{F}_c\cdot d\vec{r}.
\]
The local potential-energy relation is
\[
dU=-\vec{F}_c\cdot d\vec{r}.
\]
For one-dimensional motion along the $x$-axis,
\[
F_x=-\frac{dU}{dx}.
\]
More generally, one may write lightly
\[
\vec{F}_c=-\nabla U.
\]}
\nt{Potential energy is a property of a \emph{system}, not of a single isolated object. For example, gravitational potential energy belongs to the Earth-object system, and spring potential energy belongs to the block-spring system. A conservative force can transfer energy between kinetic and potential forms without making the potential difference depend on the path. By contrast, nonconservative forces such as kinetic friction and air resistance have path-dependent work, so a single-valued potential-energy function for that interaction is not defined in this AP sense.}
\pf{Why zero closed-loop work gives a well-defined $\Delta U$}{Assume that for every closed path,
\[
\oint \vec{F}_c\cdot d\vec{r}=0.
\]
Take two paths $C_1$ and $C_2$ from the same initial point $i$ to the same final point $f$. Traverse $C_1$ from $i$ to $f$ and then traverse $C_2$ backward from $f$ to $i$. This makes a closed loop, so
\[
\int_{C_1} \vec{F}_c\cdot d\vec{r}+\int_{f\to i\text{ on }C_2} \vec{F}_c\cdot d\vec{r}=0.
\]
Reversing the limits changes the sign of the second integral, giving
\[
\int_{C_1} \vec{F}_c\cdot d\vec{r}=\int_{C_2} \vec{F}_c\cdot d\vec{r}.
\]
So the work depends only on the endpoints. Therefore the quantity
\[
U_f-U_i=-\int_i^f \vec{F}_c\cdot d\vec{r}
\]
is path independent and is a well-defined potential-energy difference.}
\qs{Worked example}{A block of mass $m=0.50\,\mathrm{kg}$ is attached to an ideal horizontal spring of spring constant $k=200\,\mathrm{N/m}$ on a frictionless track. Let $x$ denote the block's displacement from the spring's equilibrium position, with positive $x$ to the right. Initially the block is held at rest at $x_i=+0.15\,\mathrm{m}$ and then released. Find:
\begin{enumerate}
\item the change in spring potential energy $\Delta U_s$ as the block moves to $x_f=0$,
\item the work done by the spring during that motion, and
\item the block's speed $v_f$ when it passes through equilibrium.
\end{enumerate}}
\sol For an ideal spring, the spring force is
\[
\vec{F}_s=-kx\,\hat{\imath}.
\]
Since the motion is one-dimensional, the potential-energy relation gives
\[
F_x=-\frac{dU_s}{dx}.
\]
So
\[
-kx=-\frac{dU_s}{dx}
\qquad \Rightarrow \qquad
\frac{dU_s}{dx}=kx.
\]
Integrate with respect to $x$:
\[
U_s(x)=\int kx\,dx=\tfrac12 kx^2+C.
\]
Choose the usual reference $U_s=0$ at $x=0$, so $C=0$ and
\[
U_s(x)=\tfrac12 kx^2.
\]
At the initial position,
\[
U_{s,i}=\tfrac12 (200)(0.15)^2=100(0.0225)=2.25\,\mathrm{J}.
\]
At the final position $x_f=0$,
\[
U_{s,f}=\tfrac12 (200)(0)^2=0.
\]
Therefore the change in spring potential energy is
\[
\Delta U_s=U_{s,f}-U_{s,i}=0-2.25\,\mathrm{J}=-2.25\,\mathrm{J}.
\]
Because the spring force is conservative,
\[
W_s=-\Delta U_s=+2.25\,\mathrm{J}.
\]
So the spring does positive work on the block as the spring relaxes toward equilibrium.
The track is frictionless, so the spring is the only force doing work on the block in the horizontal direction. Thus
\[
W_{\text{net}}=\Delta K.
\]
Since the block starts from rest,
\[
K_i=0,
\qquad
K_f=W_{\text{net}}=2.25\,\mathrm{J}.
\]
Then
\[
\tfrac12 mv_f^2=2.25.
\]
Substitute $m=0.50\,\mathrm{kg}$:
\[
\tfrac12 (0.50)v_f^2=2.25
\qquad \Rightarrow \qquad
0.25v_f^2=2.25
\qquad \Rightarrow \qquad
v_f^2=9.0.
\]
Hence
\[
v_f=3.0\,\mathrm{m/s}.
\]
So the results are
\[
\Delta U_s=-2.25\,\mathrm{J},
\qquad
W_s=+2.25\,\mathrm{J},
\qquad
v_f=3.0\,\mathrm{m/s}.
\]

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\subsection{Mechanical Energy Conservation}
This subsection packages work and potential energy into an energy-accounting method. In AP mechanics, the key step is to choose a system first, then decide which interactions are represented by potential energy and which must be tracked as nonconservative work.
\dfn{Mechanical energy and nonconservative work}{Let a chosen system move between an initial state and a final state. Let $K$ denote the total kinetic energy of the system, and let $U$ denote the total potential energy associated with all conservative interactions included in the system, such as gravitational and spring interactions. The \emph{mechanical energy} of the system is
\[
E_{\mathrm{mech}}=K+U.
\]
Let $W_{\mathrm{nc}}$ denote the total work done on the system by forces or processes that are not represented by a potential-energy function in the chosen model. With this sign convention,
\[
W_{\mathrm{nc}}>0 \text{ increases } E_{\mathrm{mech}},
\qquad
W_{\mathrm{nc}}<0 \text{ decreases } E_{\mathrm{mech}}.
\]
Typical examples include kinetic friction, air drag, and an external applied force not absorbed into $U$.}
\thm{Mechanical energy equation}{Let $K_i$ and $U_i$ denote the initial kinetic and potential energies of a chosen system, and let $K_f$ and $U_f$ denote the corresponding final quantities. If $W_{\mathrm{nc}}$ is the total nonconservative work done on the system, then
\[
\Delta E_{\mathrm{mech}}=\Delta(K+U)=W_{\mathrm{nc}}.
\]
Equivalently,
\[
K_i+U_i+W_{\mathrm{nc}}=K_f+U_f.
\]
If the motion is governed only by conservative forces already accounted for in $U$, then $W_{\mathrm{nc}}=0$ and mechanical energy is conserved:
\[
\Delta(K+U)=0,
\qquad
K_i+U_i=K_f+U_f.
\]}
\nt{Choose the system before writing any energy equation. If the system is \emph{object + Earth}, then gravitational potential energy belongs in $U$ and gravity should not also be counted as separate work. If the system is \emph{object + spring}, then spring potential energy belongs in $U$. If the system is \emph{object + Earth + spring}, then both $U_g$ and $U_s$ belong in $U$. Mechanical energy is conserved only when no nonconservative work changes $K+U$ for that chosen system. When friction, drag, or an external agent transfers energy into or out of the system, use
\[
\Delta(K+U)=W_{\mathrm{nc}}
\]
instead of setting $\Delta(K+U)$ equal to zero. Total energy is still conserved overall; it is specifically \emph{mechanical} energy that may change.}
\pf{Derivation from the work-energy theorem}{Let $W_{\mathrm{net}}$ denote the net work done on the chosen system. By the work-energy theorem,
\[
\Delta K=W_{\mathrm{net}}.
\]
Split the net work into conservative and nonconservative parts:
\[
W_{\mathrm{net}}=W_c+W_{\mathrm{nc}}.
\]
For the conservative forces represented by the potential energy $U$,
\[
W_c=-\Delta U.
\]
Therefore,
\[
\Delta K=-\Delta U+W_{\mathrm{nc}}.
\]
Rearranging gives
\[
\Delta K+\Delta U=W_{\mathrm{nc}},
\]
so
\[
\Delta(K+U)=W_{\mathrm{nc}}.
\]
If $W_{\mathrm{nc}}=0$, then
\[
\Delta(K+U)=0,
\]
which is the conservation of mechanical energy.}
\qs{Worked example}{A block of mass $m=2.0\,\mathrm{kg}$ is released from rest at a height $h=1.20\,\mathrm{m}$ above the bottom of a ramp. The ramp is frictionless. After reaching the bottom, the block crosses a rough horizontal surface of length $d=0.80\,\mathrm{m}$ with coefficient of kinetic friction $\mu_k=0.25$. The block then compresses a horizontal spring of spring constant $k=400\,\mathrm{N/m}$ on a frictionless section of track and momentarily comes to rest at maximum compression. Let $x$ denote the maximum spring compression.
Find $x$.}
\sol Choose the system to be \emph{block + Earth + spring}. Then gravity and the spring are accounted for through potential energy, and the only nonconservative work is the work done by kinetic friction on the rough horizontal section.
Take the initial state to be the release point and the final state to be the instant of maximum compression. Let the gravitational potential energy be zero at the bottom of the ramp, and let the spring potential energy be zero when the spring is uncompressed.
Use the mechanical energy equation
\[
K_i+U_i+W_{\mathrm{nc}}=K_f+U_f.
\]
At the initial state, the block is released from rest, so
\[
K_i=0.
\]
Its gravitational potential energy is
\[
U_i=mgh=(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})(1.20\,\mathrm{m})=23.52\,\mathrm{J}.
\]
At the final state, the block momentarily stops at maximum compression, so
\[
K_f=0.
\]
Its gravitational potential energy is zero because it is at the bottom level, and its spring potential energy is
\[
U_f=\tfrac12 kx^2.
\]
Now compute the nonconservative work. On the rough horizontal section, the kinetic friction force has magnitude
\[
f_k=\mu_k mg=(0.25)(2.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=4.9\,\mathrm{N}.
\]
Because friction opposes the motion over the distance $d=0.80\,\mathrm{m}$, the work done by friction is
\[
W_{\mathrm{nc}}=-f_k d=-(4.9\,\mathrm{N})(0.80\,\mathrm{m})=-3.92\,\mathrm{J}.
\]
Substitute into the energy equation:
\[
0+23.52\,\mathrm{J}-3.92\,\mathrm{J}=0+\tfrac12 (400\,\mathrm{N/m})x^2.
\]
So
\[
19.60\,\mathrm{J}=200x^2.
\]
Hence
\[
x^2=0.0980,
\qquad
x=\sqrt{0.0980}\,\mathrm{m}\approx 0.313\,\mathrm{m}.
\]
Therefore the maximum spring compression is
\[
x\approx 0.31\,\mathrm{m}.
\]
If the rough section were absent, then $W_{\mathrm{nc}}=0$ and mechanical energy would be conserved exactly. Here the negative friction work reduces the mechanical energy before the block reaches the spring.

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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.