\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.}

\wc{Net work is the algebraic sum}{Net work $W_{\mathrm{net}}=W_1+W_2+\cdots$ includes the \emph{signs} of individual works. If one force does $+120\,\mathrm{J}$ and another does $-24\,\mathrm{J}$, the net work is $+96\,\mathrm{J}$, \emph{not} $144\,\mathrm{J}$.}

\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.
