---
title: Work and Kinetic Energy
module: Energy
moduleNumber: 3
lessonNumber: 1
order: 301
summary: >
  A constant push along a straight path is trivial to score, but real forces
  vary and bend along curved trajectories, and only the component along the
  motion transfers any energy. Work captures exactly that transfer as the line
  integral $W=\int\vec F\cdot\d\vec r$, and the work-kinetic-energy theorem turns
  it into a statement about speed: the net work on a particle equals the change
  in its $\tfrac12 mv^2$. We build work up from the dot product to the signed
  area under a force curve, derive the theorem from Newton's second law, and read
  power as its instantaneous rate $P=\vec F\cdot\vec v$.
topics: [Energy]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 6 — Work and Kinetic Energy; §§6-1–6-4"
---

## Work and the work-energy theorem

Work measures energy transferred to or from a particle by a force while the point
of application is displaced. For a constant force acting during a straight-line
displacement, the definition is

$$
W=\vec F\cdot\Delta\vec r
 =F\Delta r\cos\theta.
$$

The dot product selects the component parallel to the displacement. If
$\hat s=\Delta\vec r/\Delta r$, then

$$
W=(\vec F\cdot\hat s)\Delta r=F_{\parallel}\Delta r.
$$

Work is a scalar; its SI unit is the joule, $1\ \mathrm J=1\ \mathrm{N\,m}$. Force
remains a vector measured in newtons.

> **Definition (positive, negative, and zero work).** A force does positive work
> when its component along the displacement is positive, negative work when that
> component is negative, and zero work when it is perpendicular to the displacement
> or when the displacement is zero.

The sign follows from the geometry.

- **Tow rope** pulling a sled forward: positive work.
- **Kinetic friction** on a sliding crate: negative work.
- **Normal force** on an object crossing a smooth horizontal surface: zero work, though it still enforces the constraint.
- **Pendulum tension**: zero work, because the instantaneous displacement is tangential and the tension radial.

$$
% caption: A constant force pulls a block at an angle across a level surface;
% only its horizontal component does work over the horizontal displacement,
% while the normal force and weight stay perpendicular to the motion.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (0,0) -- (6.5,0);
  \draw[thick] (1.6,0) rectangle (3.0,0.9);
  \draw[->, black, thick] (2.3,0.9) -- (2.3,1.85) node[above] {$N$};
  \draw[->, black, thick] (2.3,0) -- (2.3,-0.9) node[right] {$mg$};
  \draw[dashed, black] (4.905,1.55) -- (4.905,0.45);
  \draw[->, acc, very thick] (3.0,0.45) -- (4.905,1.55) node[above right] {$\mathbf F$};
  \draw[->, acc, thick] (3.0,0.45) -- (4.905,0.45) node[midway, below] {$F_x$};
  \draw[black] (3.7,0.45) arc (0:30:0.7);
  \draw[->, black, thick] (1.6,-1.35) -- (3.7,-1.35) node[midway, below] {$dx$};
\end{tikzpicture}
$$

The scalar-product form remains valid in three dimensions. For components,

$$
\vec F\cdot\Delta\vec r
=F_x\Delta x+F_y\Delta y+F_z\Delta z.
$$

On an incline, each force contributes through its component along the displacement;
the magnitudes alone do not fix the work. A vertical rise has $\Delta x=0$ but
nonzero work by weight; horizontal travel has $\Delta y=0$ but nonzero work by a
horizontal applied force. When a force changes direction along the path, the
constant-force formula no longer applies.

**Variable force and the work integral.**

Over a sufficiently short displacement $\d\vec r$, the force is approximately
constant, giving

$$
\d W=\vec F(\vec r)\cdot\d\vec r.
$$

Adding the segments and taking the limiting sum gives the line integral

$$
W_C=\int_C\vec F\cdot\d\vec r.
$$

The subscript $C$ matters. For a general force field, work can depend on the path
as well as on the endpoints. Along one-dimensional motion on the $x$ axis this
reduces to

$$
W_{x_i\to x_f}=\int_{x_i}^{x_f}F_x(x)\,\d x.
$$

The signed area under an $F_x$ versus $x$ graph is therefore work. Areas above the
$x$ axis are positive; areas below are negative. The graph need not describe a
force proportional to position. A constant force produces a rectangle, a linear
spring produces a triangle over an interval beginning at equilibrium, and arbitrary
curves require integration or a numerical area estimate.

$$
% caption: Work is the signed area under a force-position curve. The area from
% $x_a$ to $x_b$ is positive and the area from $x_b$ to $x_c$ is negative, so the
% two regions enter the work with opposite signs.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.0,0) node[right] {$x$};
  \draw[->, black] (0,-2.1) -- (0,3.1) node[above] {$F_x$};
  \fill[acc!12] (0.8,0) -- plot[domain=0.8:3.5,samples=80]
    (\x,{1.6*sin(360*(\x-0.8)/5.4)}) -- (3.5,0) -- cycle;
  \draw[very thick, domain=0.8:6.2, samples=160]
    plot (\x,{1.6*sin(360*(\x-0.8)/5.4)});
  \node at (2.15,0.8) {$W>0$};
  \node at (4.85,-0.8) {$W<0$};
  \node[anchor=north] at (0.8,-0.12) {$x_a$};
  \node[anchor=north] at (3.5,-0.12) {$x_b$};
  \node[anchor=north] at (6.2,-0.12) {$x_c$};
\end{tikzpicture}
$$

For measured force-position data, register force and position to the same physical
configuration before estimating an area. A sensor offset creates a work error that
grows with the displacement interval; position lag shifts a sharp force feature into
the wrong interval.

A spring obeying Hooke's law, $F_x=-kx$, does work from $x_i$ to $x_f$ of

$$
W_s=\int_{x_i}^{x_f}(-kx)\,\d x
=-\frac12k(x_f^2-x_i^2).
$$

The minus sign reflects the restoring direction. Stretching the spring from zero
to positive $x$ requires an external force to do positive work; the spring does
negative work over that displacement. Releasing the stretched spring reverses the
sign of its work as it moves toward $x=0$.

**Kinetic energy and the work-energy theorem.**

Newton's second law gives a direct connection between net work and speed. For
one-dimensional motion,

$$
F_{\rm net}=ma,
\qquad
a=\frac{\d v}{\d t}=\frac{\d v}{\d x}\frac{\d x}{\d t}
=v\frac{\d v}{\d x}.
$$

Multiplication by $\d x$ and integration yields

$$
\int_{x_i}^{x_f}F_{\rm net}\,\d x
=\int_{v_i}^{v_f}mv\,\d v
=\frac12mv_f^2-\frac12mv_i^2.
$$

Thus

$$
W_{\rm net}=\Delta K,
\qquad
K=\frac12mv^2.
$$

> **Theorem (work-kinetic-energy theorem).** In an inertial frame, the net work
> done on a particle equals its change in kinetic energy.

The theorem sums work from every force: applied forces, gravity, friction, normal
forces, tension, and any other interaction. Begin with $W_{\rm net}$, then substitute
the individual work contributions:

$$
\sum_j W_j=K_f-K_i.
$$

The speed rather than the velocity appears in $K$. Reversing a particle's direction
at unchanged speed leaves its kinetic energy unchanged, even though its velocity
changes. A net force perpendicular to the velocity changes direction but performs
no instantaneous work. Uniform circular motion provides the standard example:
centripetal force is perpendicular to $\vec v$, so $\d W=0$ and the speed is
constant.

$$
% caption: On a circular path the velocity is tangent to the circle while the
% centripetal force points to the centre. That force stays perpendicular to the
% displacement, so it does no work and the speed is unchanged.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (3,1.6) circle (1.4);
  \draw[->, black, thick] (3,3.0) -- (3,1.78) node[midway, right] {$F$};
  \draw[->, acc, very thick] (3,3.0) -- (4.4,3.0) node[above] {$v$};
  \draw[->, acc, thick, dashed] (4.4,3.0) -- (5.2,3.0) node[right] {$dr$};
  \draw[black] (3.24,3.0) -- (3.24,2.76) -- (3,2.76);
  \draw[fill=black, draw=black, thick] (3,3.0) circle (1.9pt);
  \draw[fill=white, draw=black, thick] (3,1.6) circle (1.6pt);
  \node[black, anchor=north] at (3,1.5) {center};
\end{tikzpicture}
$$

> **Worked example (stopping distance).** A $1200\ \mathrm{kg}$ car travels at
> $25\ \mathrm{m\,s^{-1}}$ on a level road. A braking force of magnitude
> $7.5\times10^3\ \mathrm N$ opposes its motion. Neglecting air resistance, the
> braking work is $W_b=-F_b d$ and the initial kinetic energy is
>
> $$
> K_i=\frac12(1200)(25)^2=3.75\times10^5\ \mathrm J.
> $$
>
> At rest $K_f=0$, so
>
> $$
> -F_b d=0-K_i,
> \qquad
> d=\frac{K_i}{F_b}=50\ \mathrm m.
> $$

No constant-acceleration assumption enters: the result holds whenever the total
braking work is known, including position-dependent forces. Doubling the initial
speed quadruples both the required work and the stopping distance.

## Power and multiple forces

Power is the rate at which work is done:

$$
P_{\rm av}=\frac{\Delta W}{\Delta t},
\qquad
P=\frac{\d W}{\d t}.
$$

Using $\d W=\vec F\cdot\d\vec r$ gives the instantaneous mechanical-power
relation

$$
P=\vec F\cdot\vec v.
$$

One watt is one joule per second. For a force parallel to velocity, $P=Fv$; a motor
that exerts the same tractive force requires more power at greater speed. For a
force perpendicular to velocity, $P=0$. This is consistent with the work-energy
theorem because

$$
P_{\rm net}=\frac{\d K}{\d t}.
$$

Power is often quoted in kilowatts; the engineering horsepower conversion is
$1\ \mathrm{hp}\approx746\ \mathrm W$. Efficiency compares output power with input
power, but no efficiency factor enters a work-energy calculation unless the system
boundary is stated.

Power measurements require force and velocity from the same instant and frame. A
delayed velocity channel can flip the sign during a reversal. For rotating
machinery the analogous pair is torque and angular velocity, with the same timing
requirement.

**Several forces and constraints.**

A particle subject to several forces has additive work because the dot product and
integral are linear:

$$
W_{\rm net}=\int_C\left(\sum_j\vec F_j\right)\cdot\d\vec r
=\sum_j\int_C\vec F_j\cdot\d\vec r.
$$

Compute net work either by adding forces before integration or by adding the work of
each force afterward. The latter form is often clearer when gravity, friction, and an
applied force have distinct physical roles. A sign convention is still required for
every scalar component. Along an incline at angle $\alpha$, the
work of gravity over a displacement $s$ upward along the surface is

$$
W_g=(-mg\sin\alpha)s=-mg(s\sin\alpha)=-mg\Delta y.
$$

The normal force is perpendicular to the surface displacement and contributes zero
work. In force analysis it still fixes the contact constraint and sets the
kinetic-friction magnitude $f_k=\mu_kN$ when sliding occurs. A constraint force can
have nonzero work if the point of application has displacement along the force, as in
a moving support or a deforming system.

The work-energy theorem also applies when a particle travels on a curved path. In a
short interval, $\d\vec r=\vec v\,\d t$, so

$$
\d W=\vec F\cdot\vec v\,\d t,
\qquad
\frac{\d K}{\d t}=\vec F_{\rm net}\cdot\vec v.
$$

Tangential force changes speed; perpendicular force changes direction. Decomposing
$\vec F=F_t\hat t+F_n\hat n$ gives

$$
\frac{\d K}{\d t}=F_tv.
$$

The normal component has no instantaneous power. In circular motion, a purely
centripetal force can be large while its work is identically zero.

> **Worked example (position-dependent resistance).** A $2.0\ \mathrm{kg}$ cart
> starts from rest at $x=0$. A horizontal applied force $F_a=(12-2x)\ \mathrm N$
> acts for $0\le x\le4.0\ \mathrm m$, against a constant opposing force of
> $3.0\ \mathrm N$ over the same interval. The individual works are
>
> $$
> W_a=\int_0^4(12-2x)\,\d x=[12x-x^2]_0^4=32\ \mathrm J,
> $$
>
> $$
> W_r=\int_0^4(-3.0)\,\d x=-12\ \mathrm J.
> $$
>
> Then $W_{\rm net}=20\ \mathrm J=\tfrac12mv_f^2$, giving
>
> $$
> v_f=\sqrt{\frac{2(20)}{2.0}}=4.47\ \mathrm{m\,s^{-1}}.
> $$

Substituting the initial value $12\ \mathrm N$ for the variable force gives the
wrong answer; its average over the interval is $8\ \mathrm N$, which the area
calculation accounts for automatically.

**Diagnostic checks.**

- **Units:** a force law integrated over distance must retain a factor of length and yield joules.
- **Sign:** $W_{\rm net}>0$ raises $v^2$, $W_{\rm net}<0$ lowers it. A force perpendicular to the motion contributes no work however large its magnitude.
- **Direction:** the theorem predicts speed only; the direction of the final velocity needs a separate force or constraint analysis.
- **Path vs. displacement:** net displacement suffices only when the force is truly constant in magnitude and direction. Kinetic friction follows the local motion and needs a path integral proportional to distance travelled.
- **System boundary:** it fixes whether a force is an internal conversion or an external transfer; the numerical work is unchanged.

## Work along changing paths

The line integral is needed whenever the force direction changes along a path, even
at constant magnitude. A force tangent to the path does work equal to its magnitude
times path length; a force radial from a fixed centre, acting on fixed-radius
circular motion, stays perpendicular to every local displacement and does zero work.
Each short segment contributes a scalar product, and the full work is their signed
sum.

Whether tension or contact does work depends on the displacement of its point of
application. A taut string does zero work on a mass on a fixed-radius arc because the
tension is radial, but does work if its length changes, since the application point
then moves partly along the tension. A normal force on a fixed smooth track does zero
work; a moving support transfers energy through the same contact.

> **Worked example (arc work).** A $15\ \mathrm N$ tangential force acts through a quarter-circle arc of radius
> $2.0\ \mathrm m$. The arc length is $(\pi/2)(2.0)=3.14\ \mathrm m$, so the work
> is $47.1\ \mathrm J$. A simultaneous $15\ \mathrm N$ radial force does zero work
> on the same fixed-radius arc. Adding magnitudes and multiplying by arc length would
> incorrectly attribute work to the radial constraint force.

**Power as a component measurement.**

Instantaneous power is the force component along the current velocity times speed.
A traction force parallel to motion gives positive power; a braking or drag force,
negative power; a sideways constraint force, zero. The same force flips from
positive to negative power if the motion reverses while the force direction holds.

Power is not energy. A large power over a short time can transfer less energy than a
smaller power sustained longer. The area under a power-time graph is work, as is the
area under a force-position graph; for recorded motion, work follows from integrating
either force along the path or power over time.

$$
% caption: The area under a power-time history is the work transferred. Positive
% power adds kinetic energy and negative power removes it, so the signed areas, not
% the peak power alone, set the net work over the interval shown.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.1,0) node[right] {time};
  \draw[->, black] (0,-2.0) -- (0,3.2) node[above] {power};
  \fill[acc!12] (0.5,0) -- (1.0,1.55) -- (2.4,1.55) -- (3.1,0) -- cycle;
  \draw[very thick] (0.5,0) -- (1.0,1.55) -- (2.4,1.55) -- (3.1,0) -- (3.9,-1.2) -- (5.1,-1.2) -- (5.6,0);
  \node[anchor=south] at (1.7,1.55) {energy added};
  \node[anchor=north] at (4.5,-1.2) {energy removed};
\end{tikzpicture}
$$

> **Worked example (power-time).** A motor supplies $2.0\ \mathrm{kW}$ for $8.0\ \mathrm s$, then regenerative
> braking absorbs $0.50\ \mathrm{kW}$ for $4.0\ \mathrm s$. The corresponding works
> are $+16\ \mathrm{kJ}$ and $-2.0\ \mathrm{kJ}$, so the net work is
> $+14\ \mathrm{kJ}$. Power without a stated interval does not specify an energy
> transfer.

**Work-energy checks and stopping feasibility.**

A brake, surface, or resisting field removes kinetic energy through the negative
work available over the stated distance. If the initial kinetic energy exceeds that
capacity, the object cannot stop within the region, and a signed energy balance shows
this without a constant-acceleration assumption.

Write each work source separately before forming the net: an uphill displacement
gives negative work by weight, kinetic friction gives negative work while sliding, a
motor gives positive work. The final kinetic energy must stay nonnegative; a negative
value signals that the presumed direction, interval, or stopping location is not
reachable under the model.

> **Worked example (stopping capacity).** A $900\ \mathrm{kg}$ vehicle enters a $30\ \mathrm m$ escape lane at
> $20\ \mathrm{m\,s^{-1}}$. Braking and rolling resistance together provide a
> constant opposing force of $5.0\ \mathrm{kN}$. Their available negative work is
> $-(5000)(30)=-150\ \mathrm{kJ}$, while the initial kinetic energy is
> $180\ \mathrm{kJ}$. The vehicle leaves the lane with $30\ \mathrm{kJ}$ of kinetic
> energy. The work-energy theorem identifies the shortfall directly; solving for a
> negative stopping distance would obscure the physical conclusion.

**Signed line-integral work on a path with reversals.**

The differential displacement carries the direction of motion in a line integral.
In one dimension, a path from $x=0$ to $x=4$ and back to $x=1$ requires separate
oriented segments. Forward and return segments have opposite $d x$, so a fixed
positive $x$ force component does positive work outward and negative work on return.
With a constant force, net work depends on final displacement; signed segment
contributions show transfer along the full path.

Parameterizing a path makes the orientation explicit. If a curve is described by
$\vec r(u)$ as parameter $u$ increases, then

$$
W=\int_{u_i}^{u_f}\vec F[\vec r(u)]\cdot
\frac{\d\vec r}{\d u}\,\d u.
$$

Reversing the path reverses the sign of the integral for the same force field. A
constant force does zero work on a closed path, since its net displacement is zero,
even though the outward and return segments carry opposite signed works.

$$
% caption: A constant force does positive work on the outward leg of a path and
% negative work on the return, because the local displacement reverses. Over the
% closed excursion shown the oriented contributions cancel and the net work is
% zero despite the nonzero path length.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.2,0) node[right] {position};
  \draw[->] (1.1,1.35) -- (2.6,1.35) node[midway, above] {constant force};
  \draw[->, acc, very thick] (0.7,0.5) -- (4.7,0.5) node[midway, above] {outward displacement};
  \draw[->, black, thick] (4.7,-0.5) -- (1.7,-0.5) node[midway, below] {return displacement};
  \draw[fill=white, draw=black, thick] (0.7,0) circle (1.9pt);
  \draw[fill=white, draw=black, thick] (4.7,0) circle (1.9pt);
  \draw[fill=white, draw=black, thick] (1.7,0) circle (1.9pt);
\end{tikzpicture}
$$

> **Worked example (return path).** A constant horizontal force of $6.0\ \mathrm N$ acts while a cart moves from
> $x=1.0$ to $x=5.0\ \mathrm m$, then returns to $x=2.0\ \mathrm m$. The outward
> work is $(6.0)(4.0)=+24\ \mathrm J$, and the return work is
> $(6.0)(-3.0)=-18\ \mathrm J$. Net work is $+6.0\ \mathrm J$, equal to the force
> times the overall displacement of $+1.0\ \mathrm m$. The total distance of
> $7.0\ \mathrm m$ is not substituted into a signed-work calculation.

**Variable-force graphs: signed areas and interval boundaries.**

An $F_x$--$x$ graph must be read as a definite integral. A force that changes sign at
an interior point contributes positive area on one side and negative area on the
other, and a region below the axis removes kinetic energy rather than being discarded.

Piecewise records integrate exactly with rectangles, triangles, and trapezoids;
curved experimental data give a trapezoidal estimate whose precision matches the
graph resolution. The horizontal axis must be displacement, not time: the area under
a force-time graph is impulse, while the area under a force-position graph is work.

Under a position-only force law, reversing the direction of travel over the same
coordinate reverses the sign of the work. A velocity-dependent force such as kinetic
friction or drag is not a single-valued $F(x)$ curve through a forward-and-return
cycle, so the graph must name the segment or carry the velocity dependence.

> **Worked example (piecewise area).** A force is $10\ \mathrm N$ for the first
> $2.0\ \mathrm m$, then decreases linearly from $10\ \mathrm N$ to zero over the
> next $3.0\ \mathrm m$, and finally equals $-4.0\ \mathrm N$ for $1.5\ \mathrm m$.
> The work is a rectangle plus a triangle plus a negative rectangle:
>
> $$
> W=(10)(2.0)+\frac12(10)(3.0)+(-4.0)(1.5)=29\ \mathrm J.
> $$

Treating the final segment as a positive area would give $41\ \mathrm J$ and
overpredict the kinetic-energy increase. The negative force component matters even
though the object continues in the positive coordinate direction.

## Instantaneous power and conservative forces

The dot-product formula $P=\vec F\cdot\vec v$ is local: it uses force and
velocity at the same instant and stays valid when either varies. Decomposing force
into tangent and normal components gives $P=F_tv$, so only the tangent component
contributes. A cornering vehicle can carry a large sideways tire force that changes
no kinetic energy while its forward component controls the speed.

A motor at maximum output power has available tractive force $F_\parallel=P/v$, which
falls as speed rises. At low speed the expression breaks down without a separate
torque or traction limit, since real drives have finite force capacity. At high speed
drag power grows quickly, leaving less net power for acceleration even at constant
motor output.

The sign of power follows the force component along velocity. A braking force gives
negative power and removes kinetic energy. A force perpendicular to velocity gives
zero power even if it changes heading. The total instantaneous power of all forces
equals the rate of kinetic-energy change, so a zero net power reading implies
constant speed, not necessarily zero acceleration.

$$
% caption: The force and velocity components set the instantaneous power. The
% forward part of the drive force adds power, the opposing drag removes power, and
% the perpendicular steering force changes heading while adding no power.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, black] (3.95,2.4) -- (3.95,1.35);
  \draw[thick] (2.3,1.35) circle (0.32);
  \draw[->, acc, very thick] (2.62,1.35) -- (4.8,1.35) node[right] {motion $v$};
  \draw[->, black, thick] (2.57,1.52) -- (3.95,2.4) node[above right] {thrust};
  \draw[->, black, thick] (1.98,1.35) -- (0.8,1.35) node[left] {drag};
  \draw[->, black, thick] (2.3,1.67) -- (2.3,3.05) node[above] {steering force};
\end{tikzpicture}
$$

> **Worked example (force-power).** A motor supplies $24\ \mathrm{kW}$ of mechanical power while a vehicle travels
> at $16\ \mathrm{m\,s^{-1}}$ on level ground. If all of that output is tangential,
> the corresponding drive force is $24\,000/16=1500\ \mathrm N$. At twice the speed,
> the same power would correspond to $750\ \mathrm N$ before accounting for the
> larger drag force. The calculation identifies an instantaneous force capability;
> the work over a route still requires integration over time or displacement.

**Work by gravity and by a spring.**

Near Earth's surface, gravity is approximately a constant vector. Its work depends
only on the vertical displacement between the endpoints, even when the actual path
is curved or contains horizontal segments. With upward positive,

$$
W_g=-mg(y_f-y_i).
$$

Moving upward gives negative work by gravity, moving downward positive; path length
does not enter. Carrying a box up a long ramp and lifting it vertically through the
same height give the same gravitational work under the constant-$g$ model, though
other forces can do different work along the two routes.

A spring is a position-dependent force whose restoring force grows linearly with
extension or compression from the unstretched reference. Its work follows from
integrating the changing force, not from multiplying a final force by the full
displacement: negative while the spring stretches away from equilibrium, positive
while it returns.

$$
% caption: Gravity and a linear spring are both position-dependent forces. Near the
% ground the gravity force is constant with height, so its force-height graph is
% flat, while the spring force grows in proportion to extension and its triangular
% area gives the work.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->, black] (0,0) -- (4.2,0) node[right] {height};
    \draw[->, black] (0,0) -- (0,3.2) node[above] {weight};
    \draw[dashed, black] (0,1.8) -- (0.4,1.8);
    \draw[very thick] (0.4,1.8) -- (3.7,1.8);
  \end{scope}
  \begin{scope}[shift={(5.6,0)}]
    \draw[->, black] (0,0) -- (4.2,0) node[right] {extension};
    \draw[->, black] (0,0) -- (0,3.2) node[above] {spring force};
    \fill[acc!12] (0,0) -- (3.4,2.6) -- (3.4,0) -- cycle;
    \draw[very thick] (0,0) -- (3.4,2.6);
  \end{scope}
\end{tikzpicture}
$$

> **Worked example (gravity and spring).** A $0.80\ \mathrm{kg}$ mass is raised
> $1.50\ \mathrm m$, then a spring of constant $120\ \mathrm{N\,m^{-1}}$ is
> stretched from zero to $0.20\ \mathrm m$. During the lift gravity does
> $-(0.80)(9.8)(1.50)=-11.8\ \mathrm J$. The spring does
>
> $$
> W_s=-\frac12(120)(0.20)^2=-2.40\ \mathrm J
> $$
>
> while stretching.

Each work comes from its own force law and displacement; they are not yet summed
into a conservation statement.

**Constraint-force work and moving contacts.**

A constraint force does no work only when its point of application moves
perpendicular to it. A normal force on a fixed smooth track meets this because the
permitted displacement is tangent to the track; tension in a fixed-length pendulum
string stays radial while the bob moves tangent to the arc. These zero-work results
follow from geometry, not from the labels "normal" or "tension."

When the constraint itself moves or deforms, it can do work. A moving conveyor's
normal force does positive work on a package; a winch that changes a string's length
does work through its tension. The work is evaluated from the actual motion of the
application point in the chosen frame, so treating every constraint as zero-work would
discard real transfers from motors, moving supports, and driven strings.

In a multi-body system a contact force can do negative work on one body and equal
positive work on another. Compute each with its own point motion first; for an ideal
internal constraint the two cancel in the combined-system balance. The force arrow
alone means nothing without the receiving body, its displacement, and the system
boundary.

$$
% caption: A fixed smooth guide carries a normal force perpendicular to the
% particle's tangent displacement, so it does zero work. A moving support instead
% pushes along the object's displacement, transferring energy through the same kind
% of contact constraint.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[black, thick] (1.079,1.145) arc (250:290:2.4);
    \draw[->, black, thick] (2.109,1.009) -- (2.022,2.005) node[above] {normal};
    \draw[->, acc, thick] (2.109,1.009) -- (3.205,1.105) node[right] {tangent displacement};
    \draw[black] (2.328,1.028) -- (2.309,1.247) -- (2.090,1.228);
    \draw[fill=black, draw=black, thick] (2.109,1.009) circle (2pt);
  \end{scope}
  \begin{scope}[shift={(6.0,0)}]
    \draw[black, thick] (0.2,0) -- (4.1,0);
    \draw[thick] (1.3,0) rectangle (2.5,0.85);
    \draw[->, black, thick] (2.5,0.42) -- (3.75,0.42) node[right] {support force};
    \draw[->, acc, thick] (1.3,-0.45) -- (3.0,-0.45) node[midway, below] {moving contact};
  \end{scope}
\end{tikzpicture}
$$

> **Worked example (moving support).** A horizontal conveyor exerts a $40\ \mathrm N$ forward contact force on a package
> while the package moves $3.0\ \mathrm m$ forward. The conveyor's contact force does
> $+120\ \mathrm J$ of work on the package. Calling it a normal support force would
> not make the work vanish; the relevant displacement is parallel to the actual force
> in this model. A fixed vertical normal force on the same package would instead do
> zero work during horizontal travel.

## Detailed work integrals

When a force law is given algebraically, work follows from integrating its component
along the displacement coordinate over the stated limits. Units give a dimensional
check: each term in $F_x(x)$ must have units of force, and integrating over position
must produce joules. A constant term is a rectangle, a term proportional to $x$ a
triangle, higher powers curved areas.

The limits carry the physical interval. Integrating from zero when the object begins
at a nonzero coordinate adds work from a segment that never occurred. A negative
integral does not mean a negative force magnitude; it means the force component
opposed the oriented displacement on balance. Several variable forces can be summed
separately or added first and integrated once.

> **Worked example (polynomial force).** A cart moves from $x=1.0$ to $x=4.0\ \mathrm m$ under a horizontal force
> $F_x(x)=6.0+2.0x-x^2$, in newtons when $x$ is in metres. The work is
>
> $$
> W=\int_1^4(6.0+2.0x-x^2)\,\d x
> =\left[6.0x+x^2-\frac{x^3}{3}\right]_1^4
> =12\ \mathrm J.
> $$

Using a final force times the displacement would ignore the curvature of the force
law. Using limits from zero to four metres would include an extra, untravelled
segment. The integral states both the force model and the actual path interval.

**Integral audit.**

A variable-force integral needs the force-law coordinate, the actual endpoints, and
the sign of the force component along the increasing coordinate; a graph shows the
same audit, with positive area raising kinetic energy and negative area lowering it.
When forces are given in component form, integrate only the component parallel to the
differential displacement, and integrate each oriented segment in its actual
direction so a return segment reverses the limits.

A measured force law gives an area estimate rather than an exact value. A dense set
of force-position points supports a trapezoidal or fitted-curve integral; averaging
the first and last force values is justified only when the graph is linear between
them.

## Frame dependence and force direction

Kinetic energy depends on the inertial frame because speed depends on the observer.
A package resting on a train floor has zero kinetic energy in the train frame and
nonzero kinetic energy in the ground frame when the train moves; both descriptions
are valid. The theorem holds in each frame provided force, displacement, velocity,
work, and kinetic energy are all evaluated in that same frame. Mixing frames is an
accounting error.

The change in kinetic energy also transforms between frames. If a frame moves at
constant velocity $\vec V$ relative to another, then

$$
\vec v'=\vec v-\vec V.
$$

The kinetic-energy difference is not generally the same number in the two frames,
because the squared speed changes. Work by a constant force can likewise differ:
the force acts through the displacement measured in the selected frame. A force that
does no work on an object stationary in one frame can do work on the same object in
another frame where it moves.

Frame selection specifies the meaning of an energy statement. A stopping distance
measured relative to a road is naturally analysed in the road frame; a package
sliding inside a moving carriage can be analysed in the carriage frame if the
carriage moves at constant velocity. State the chosen frame before assigning
numerical kinetic energies or work values, especially when a system includes moving
supports.

$$
% caption: The same block and force are described from two inertial frames. In the
% ground frame the block moves through a nonzero displacement, while in the train
% frame it stays put, so the work and kinetic energy differ between frames even
% though the force arrow is identical.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->, black] (0,0) -- (4.9,0) node[right] {ground position};
    \draw[thick] (1.5,0) rectangle (2.5,0.78);
    \draw[->, acc, thick] (2.5,0.39) -- (3.7,0.39) node[right] {force};
    \draw[->, black, dashed] (1.1,-0.5) -- (2.9,-0.5) node[midway, below] {ground displacement};
  \end{scope}
  \begin{scope}[shift={(7.4,0)}]
    \draw[->, black] (0,0) -- (4.9,0) node[right] {train position};
    \draw[thick] (1.5,0) rectangle (2.5,0.78);
    \draw[->, acc, thick] (2.5,0.39) -- (3.7,0.39) node[right] {same force};
    \draw[fill=black!55, draw=black] (2.0,-0.5) circle (1.6pt);
    \node[anchor=north] at (2.0,-0.68) {no displacement here};
  \end{scope}
\end{tikzpicture}
$$

> **Worked example (two-frame work).** A constant $10\ \mathrm N$ force acts for $3.0\ \mathrm s$ on a cart moving at
> $4.0\ \mathrm{m\,s^{-1}}$ in the ground frame. In that frame the displacement is
> $12\ \mathrm m$ and the force does $120\ \mathrm J$ of work. In a frame moving
> with the cart at the start, the cart's displacement is not generally zero over the
> full interval because the force changes its velocity; its work must be calculated
> from the frame-specific displacement, not assumed from the initial relative rest.
> The two kinetic-energy changes differ by the corresponding frame transformation and
> each equals the work calculated within its own frame.

**Net work, speed change, and force direction.**

Net work fixes the change in speed squared: positive net work raises the speed,
negative net work lowers it, zero net work leaves it unchanged. The direction of the
final velocity and the force responsible still need a vector analysis. A force
component parallel to velocity changes speed; a perpendicular component changes
heading only. A general force has both, curving the path while speeding the object up
or slowing it down, and the net-work theorem sums their energy effects without
reconstructing the trajectory.

A velocity reversal occurs at a turning point where the speed reaches zero. The
work-energy equation locates it through $K=0$ once the force work is known, but
cannot select the direction afterward without the force past the turn. Uniform
circular motion is the opposite case: nonzero net force, zero net work, since the
force stays perpendicular to velocity.

> **Worked example (directional-work check).** A $2.0\ \mathrm{kg}$ puck has speed $6.0\ \mathrm{m\,s^{-1}}$ on a curved guide.
> Over one interval, a tangential resisting force does $-20\ \mathrm J$ while the
> normal guide force does zero work. The final kinetic energy is
> $36-20=16\ \mathrm J$, giving final speed $4.0\ \mathrm{m\,s^{-1}}$. The normal
> force may still turn the puck substantially during that interval; the work result
> reports the speed change, not a straight-line final velocity.

**Variable-force stopping distance.**

Position-dependent braking stops the object when the accumulated negative work equals
the initial kinetic energy. Write the signed resistive component along the direction
of travel and integrate from the initial position to the unknown stopping position;
the resulting equation may be linear, quadratic, or numerical, but the energy logic
is unchanged.

Force variation can arise from a changing road grade, a spring buffer, a magnetic
brake, or an energy-absorbing barrier. The sign follows the motion coordinate: for a
vehicle moving in the positive direction a resistive force is negative, and a slope
rising with distance adds a further negative gravitational component. An algebraic
stopping root is physical only if it lies in the stated force-law interval and leaves
nonnegative kinetic energy up to the stop.

> **Worked example (variable-force stop).** A $1000\ \mathrm{kg}$ cart moves at
> $12\ \mathrm{m\,s^{-1}}$ on a level track. Over the braking region the opposing
> force is
>
> $$
> F_x(x)=-(2000+100x)\ \mathrm N,
> $$
>
> with $x$ in metres from brake application. The initial kinetic energy is
> $72\,000\ \mathrm J$, so stopping requires
>
> $$
> \int_0^{d}-(2000+100x)\,\d x=-72\,000.
> $$
>
> Thus $2000d+50d^2=72\,000$, giving the positive root $d=22.9\ \mathrm m$, where the
> force law has removed exactly the initial kinetic energy.

A constant-force estimate using the initial $2000\ \mathrm N$ would predict
$36\ \mathrm m$, missing the increasing braking force by a wide margin.

**Piecewise stopping models.**

Real stopping regions often contain more than one force regime: a brake applies a
constant force, then a spring buffer or gravel bed adds a position-dependent
resistance. Work is computed in the physical order of travel. Integrate the force
over the first segment, subtract that work from the initial kinetic energy, and carry
any positive remainder into the next segment as its starting value. If kinetic energy
reaches zero before a boundary, the later force law never engages. Each boundary is
an event in the energy account; record the energy carried into it before selecting
the next force law.

| Event | Energy condition | Consequence |
| --- | --- | --- |
| Entry to a new region | $K_{\rm in}=K_0+\sum W_{\rm prior}$ | initialize the next integral |
| Stop before boundary | $K_{\rm in}+W_{\rm segment}=0$ | later region is not used |
| Exit with residual | $K_{\rm out}>0$ | report speed and remaining path |

A stop inside the first segment uses that segment's root, not a later formula; a root
beyond the available barrier length leaves the object outside the region with residual
kinetic energy, which the account states without assuming later motion. Every segment
must share one inertial frame: a road-frame stopping distance uses the road-frame
speed and displacement, while a moving walkway or vehicle floor changes both.

> **Worked example (residual-speed check).** If a variable brake can remove only $50\ \mathrm{kJ}$ before the end of its available
> path and a cart enters with $68\ \mathrm{kJ}$ of kinetic energy, it cannot stop in
> that path. Its remaining kinetic energy is $18\ \mathrm{kJ}$. For a
> $900\ \mathrm{kg}$ cart, the exit speed is
> $\sqrt{2(18\,000)/900}=6.3\ \mathrm{m\,s^{-1}}$. Reporting a stopping distance
> longer than the actual brake region would conceal the model limitation rather than
> solve it.

**Stopping-model audit.**

Before accepting a stopping root, check that its distance lies in the interval where
the force law applies, that each work term has the correct sign, and that one
inertial frame is used throughout. Audit a piecewise stop as an energy ledger indexed
by location: the entry energy of each segment is the remaining kinetic energy from the
previous one, not the initial energy copied into every force law.

| Segment | Work evaluation | Physical check |
| --- | --- | --- |
| Entry region | $W_1=\int_{x_0}^{x_1}F_1(x)\,\d x$ | root lies inside $[x_0,x_1]$ |
| Later region | $W_2=\int_{x_1}^{x_2}F_2(x)\,\d x$ | use only if $K_0+W_1>0$ |
| Exit | $K_f=K_0+W_1+W_2$ | report residual speed if $K_f>0$ |

**Reporting a work-energy result.**

Report a work or kinetic-energy change with its system, reference frame, travel
interval, force model, and sign convention. A number such as $-18\ \mathrm{kJ}$ can
mean work by a brake on a cart, work by the cart on the brake under the opposite
convention, or one contribution inside a larger balance; the sentence must say which.
Give force parameters and distances with units, and round only after the final
balance is checked.

For measured force-position data, a constant sensor offset produces an area error
proportional to path length, and a short unmeasured spike can dominate the work even
when most points look smooth. Comparing against the measured change in kinetic energy
gives an independent closure check when mass and speed are known. Any residual then
belongs to one of:

- a stated external work,
- a changed potential energy,
- measurement uncertainty,
- a limitation of the chosen system boundary.

[^tipler6]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §§6-1–6-4.