Unit 3: Work, Energy, and Power

Work as Energy Transfer

What “work” means (and what it does not mean)

In mechanics, work is a precise way to quantify energy transferred into or out of a system by a force acting through a displacement. The key idea is that work connects a cause (a force) to an effect (a change in energy). If a force acts but nothing moves, then no mechanical work is done on the object, no matter how tired you feel.

A common confusion is mixing up “effort” with physical work. Holding a book motionless requires metabolic energy, but since the book’s displacement is zero, the mechanical work done on the book is zero.

The general (calculus) definition of work

If a force is $\vec{F}$ and $d\vec{r}$ is an infinitesimal displacement along the path, then

$dW=\vec{F}\cdot d\vec{r}$

and over a path from point 1 to point 2,

$W=\int_1^2 \vec{F}\cdot d\vec{r}$

This matters whenever the force changes in magnitude or direction, or the path is curved.

Work by a constant force (dot product form)

When a constant force acts over a straight-line displacement,

$W=\vec{F}\cdot \Delta \vec{r}$

If $\theta$ is the angle between the force and the displacement, then

$W=F\Delta r\cos\theta$

Only the component parallel to the displacement contributes.

• If $\theta=0$, the force helps the motion and work is positive.
• If $\theta=180^\circ$, the force opposes the motion and work is negative.
• If $\theta=90^\circ$, the force is perpendicular to the displacement and

$W=0$

If the force is constant, it can be factored out of the integral, which is another way to see why the constant-force formula works.

Units

Work is measured in joules. A newton-meter is the same unit:

$1\ \mathrm{J}=1\ \mathrm{N\,m}=1\ \mathrm{kg\,m^2/s^2}$

Work from a varying force (1D form)

If motion is along the $x$-axis and the force component along the motion is $F_x(x)$,

$W=\int_{x_1}^{x_2} F_x(x)\,dx$

This integral form is especially important in AP Physics C because you’re expected to be comfortable using calculus to compute work.

Work from graphs: area under a force–position curve

If you’re given a graph of $F_x$ versus $x$,

$W=\int F_x\,dx$

Geometrically, this is the signed area between the curve and the $x$-axis.

• Area above the axis contributes positive work.
• Area below the axis contributes negative work.

Work depends on force component and displacement, not time

Work is not directly about how long it takes. A slow push and a fast push can do the same work if the force and displacement details match. Time enters later through power.

Common misconception: “work done equals force times distance”

That shortcut only works when the force is constant and parallel to the displacement. In general you must use the dot product (component along motion), and if the force varies you must integrate. If the force is perpendicular, work is zero no matter how far the object travels.

Example 1: Constant-force work at an angle

A box is pulled across a horizontal floor by a force of magnitude $F=50\ \mathrm{N}$ at an angle $\theta=30^\circ$ above horizontal. The box moves $\Delta r=4\ \mathrm{m}$ horizontally. Only the horizontal component contributes:

$W=F\Delta r\cos\theta$

$W=(50)(4)\cos 30^\circ$

$W=200\left(\frac{\sqrt{3}}{2}\right)$

$W\approx 173\ \mathrm{J}$

A frequent mistake here is using the full force magnitude without the cosine factor.

Sample Problem 1: Lifting a book at constant velocity

You slowly lift a book of mass $2\ \mathrm{kg}$ at constant velocity through a distance of $3\ \mathrm{m}$. How much work do you do on the book?

Because the velocity is constant, the net force is zero, so your upward force balances weight:

$F=mg$

Using the common approximation $g\approx 10\ \mathrm{m/s^2}$,

$F=(2)(10)=20\ \mathrm{N}$

Force and displacement are parallel, so

$W=Fd$

$W=(20)(3)=60\ \mathrm{J}$

Sample Problem 2: Crate pulled with a rope at an angle

A $15\ \mathrm{kg}$ crate is moved along a horizontal floor by a worker pulling with a rope that makes a $30^\circ$ angle with the horizontal. The tension is $200\ \mathrm{N}$ and the crate slides $10\ \mathrm{m}$. How much work is done on the crate by the rope?

Only the component of the tension along the direction of motion does work:

$W=Fd\cos\theta$

$W=(200)(10)\cos 30^\circ$

$W=2000\left(\frac{\sqrt{3}}{2}\right)$

$W=1000\sqrt{3}\ \mathrm{J}$

$W\approx 1.73\times 10^3\ \mathrm{J}$

Example 2: Work from a variable force (spring-like)

Suppose

$F_x(x)=3x^2$

acts along the $x$ direction as an object moves from $x=0$ to $x=2\ \mathrm{m}$.

$W=\int_0^2 3x^2\,dx$

$W=\left[x^3\right]_0^2$

$W=8\ \mathrm{J}$

The key skill is setting up the integral with correct limits and using the correct component of force.

Exam Focus

• Typical question patterns:
  • Compute work from a force–position graph (area under curve, including negative regions).
  • Use the integral definition of work for a force that varies with position.
  • Determine which forces do zero work (normal force in many situations, centripetal force in uniform circular motion).
• Common mistakes:
  • Forgetting the cosine factor (using force magnitude instead of the component along displacement).
  • Confusing the integral for work with “area under velocity–time” or other graphs.
  • Using the wrong limits or integrating with respect to the wrong variable.
  • Dropping signs by assuming “distance is always positive so work is positive.”

Kinetic Energy and the Work–Energy Theorem

Kinetic energy: energy of motion

Kinetic energy is the energy associated with an object’s motion. For a particle of mass $m$ and speed $v$,

$K=\frac{1}{2}mv^2$

Kinetic energy is scalar and nonnegative because it depends on $v^2$. The SI unit is the joule, but kinetic energy may be expressed in other units (for example calories, electronvolts, or foot-pounds) depending on context.

It is not generally true that “kinetic energy is conserved” in every closed system. Kinetic energy stays constant only in special situations (for example, when the net work on the system is zero, or in perfectly elastic collisions). In general, kinetic energy can transform into potential energy or internal (thermal) energy.

Why kinetic energy matters

Kinetic energy lets you connect forces and motion without directly solving differential equations in time. Instead of tracking acceleration at every moment, you can track how energy changes between two positions.

The work–energy theorem

The total work done on an object, equivalently the work done by the net force, equals the change in kinetic energy:

$W_{\mathrm{net}}=\Delta K$

with

$\Delta K=K_f-K_i=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

This can be derived from Newton’s second law and the definition of work. Conceptually, forces acting through distances change speed in a way captured by kinetic energy.

What counts as “net work”

Net work is the sum of the work done by all forces acting on the object:

$W_{\mathrm{net}}=\sum W_i$

A strong habit is to list forces, decide which do work, compute each work with sign, then sum.

Special but common cases

1. If displacement is tangent to a surface, the normal force is often perpendicular to displacement, so

$W_N=0$

2. In uniform circular motion, the centripetal force is perpendicular to instantaneous displacement, so it does no work. That is why speed can remain constant even with a nonzero net force.

3. Kinetic friction typically does negative work, transferring mechanical energy into thermal energy.

Sample Problem 3: Kinetic energy from mass and speed

What is the kinetic energy of a ball of mass $0.10\ \mathrm{kg}$ moving with speed $30\ \mathrm{m/s}$?

$K=\frac{1}{2}mv^2$

$K=\frac{1}{2}(0.10)(30^2)$

$K=0.05\cdot 900$

$K=45\ \mathrm{J}$

Sample Problem 4: Tennis ball launched straight up (finding maximum height)

A tennis ball of mass $0.06\ \mathrm{kg}$ is hit straight upward with initial speed $50\ \mathrm{m/s}$. How high does it go if air resistance is negligible?

This could be done with kinematics (“the Big Five”), but energy is quicker. As the ball rises, gravity does negative work (because gravity opposes the upward motion). At the top, speed is zero, so all the initial kinetic energy becomes gravitational potential energy:

$\frac{1}{2}mv_i^2=mgh$

Mass cancels:

$h=\frac{v_i^2}{2g}$

Using $g\approx 10\ \mathrm{m/s^2}$,

$h=\frac{50^2}{2\cdot 10}=125\ \mathrm{m}$

(Using $g=9.8\ \mathrm{m/s^2}$ gives a very similar value.)

Example 1: Stopping distance using work–energy

A car of mass $m$ moves at speed $v_i$ on a level road. Brakes provide an approximately constant friction force magnitude $F_k$ opposite motion. Find stopping distance $d$.

The only horizontal force doing work is friction:

$W_{\mathrm{net}}=-F_k d$

The car stops, so

$\Delta K=0-\frac{1}{2}mv_i^2$

Set equal:

$-F_k d=-\frac{1}{2}mv_i^2$

$d=\frac{mv_i^2}{2F_k}$

Doubling speed quadruples stopping distance if braking force stays the same.

Example 2: Work–energy with multiple forces on an incline

A block slides $5\ \mathrm{m}$ down a rough incline. Along the incline, gravity component is $mg\sin\theta$ downhill and kinetic friction is $f_k=\mu_k mg\cos\theta$ uphill. Starting from rest, find speed after distance $L=5\ \mathrm{m}$.

Net work:

$W_{\mathrm{net}}=(mg\sin\theta)L-(\mu_k mg\cos\theta)L$

Starting from rest means

$W_{\mathrm{net}}=\frac{1}{2}mv^2$

Cancel $m$ and solve:

$gL(\sin\theta-\mu_k\cos\theta)=\frac{1}{2}v^2$

$v=\sqrt{2gL(\sin\theta-\mu_k\cos\theta)}$

Exam Focus

• Typical question patterns:
  • Use the work–energy theorem to find speed after moving through a region with forces (inclines, friction, applied forces).
  • Identify which forces do zero work due to perpendicularity.
  • Combine multiple work contributions, including negative work by friction.
• Common mistakes:
  • Mixing up “net force” with “net work” (work depends on displacement and direction).
  • Forgetting that work can be negative, leading to smaller final speeds.
  • Using velocity components instead of speed in kinetic energy.

Conservative Forces and Potential Energy

Why potential energy exists

Sometimes energy is stored in a configuration (height, compression, stretching) and later recovered as kinetic energy. This is captured by potential energy, which is defined for conservative forces.

A conservative force is one for which the work done between two points depends only on endpoints, not on path. This lets you compute energy changes without solving the detailed motion.

Equivalent tests for a conservative force

1. Path independence.
2. Zero work in a closed loop:

$\oint \vec{F}\cdot d\vec{r}=0$

3. A potential energy function exists such that

$W_{\mathrm{cons}}=-\Delta U$

Potential energy definition (work-based)

For a conservative force,

$\Delta U=U_f-U_i=-\int_i^f \vec{F}_{\mathrm{cons}}\cdot d\vec{r}$

If the conservative force does positive work, potential energy decreases. If you do work against it, potential energy increases.

Types of potential energy (common examples)

• Gravitational potential energy: energy due to position in a gravitational field.
• Elastic potential energy: energy due to deformation (stretching/compression).
• Chemical potential energy: energy stored in chemical bonds (important broadly in science, though not typically modeled in AP Mechanics calculations).

Gravitational potential energy (near Earth)

Near Earth, where $g$ is approximately constant,

$\Delta U_g=mg\Delta y$

A common reference choice is

$U_g=mgy$

The absolute value depends on your reference level, but changes do not.

Spring potential energy

For an ideal spring obeying Hooke’s law,

$F_s=-kx$

and the associated potential energy is

$U_s=\frac{1}{2}kx^2$

where $x$ is displacement from equilibrium.

Conservative vs nonconservative forces

• Conservative: gravity, ideal spring force, electrostatic force.
• Nonconservative: kinetic friction, air resistance, many applied forces that depend on time or path.

Friction is nonconservative because the work it does depends on path length, not just endpoints.

Force and potential energy relationship in 1D

In one dimension,

$F_x=-\frac{dU}{dx}$

This relationship is central for reading potential-energy graphs, finding equilibria, and discussing stability.

Example 1: Gravitational potential energy change

A mass $m=2\ \mathrm{kg}$ is raised by $\Delta y=3\ \mathrm{m}$.

$\Delta U_g=mg\Delta y=(2)(9.8)(3)=58.8\ \mathrm{J}$

If lifted slowly at constant speed, your work is approximately $+58.8\ \mathrm{J}$ and gravity’s work is $-58.8\ \mathrm{J}$.

Example 2: Spring potential energy and work

A spring with $k=200\ \mathrm{N/m}$ is compressed by $x=0.10\ \mathrm{m}$.

$U_s=\frac{1}{2}kx^2=\frac{1}{2}(200)(0.10)^2=1.0\ \mathrm{J}$

Sample Problem 5: Stuntwoman’s gravitational potential energy

A stuntwoman of mass $60\ \mathrm{kg}$ scales a $40\ \mathrm{m}$ rock face. What is her gravitational potential energy relative to the ground?

Taking ground as $h=0$,

$U_g=mgh$

Using $g\approx 10\ \mathrm{m/s^2}$,

$U_g=(60)(10)(40)=2.4\times 10^4\ \mathrm{J}$

Sample Problem 6: Stuntwoman jumps (final speed, no air resistance)

If she jumps off the cliff, what is her final speed as she lands on a large air-filled cushion on the ground (neglect air resistance)?

Gravitational potential energy transforms into kinetic energy:

$mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$

Using $g\approx 10\ \mathrm{m/s^2}$ and $h=40\ \mathrm{m}$,

$v=\sqrt{2(10)(40)}$

$v=\sqrt{800}$

$v\approx 28.3\ \mathrm{m/s}$

Exam Focus

• Typical question patterns:
  • Determine $\Delta U$ from displacement in gravity or from spring compression/extension.
  • Use $W_{\mathrm{cons}}=-\Delta U$ to relate force work and potential energy.
  • Interpret conservative-force criteria conceptually (path independence, closed-loop work).
• Common mistakes:
  • Sign errors: positive work by a conservative force means potential energy decreases.
  • Confusing displacement from equilibrium in the spring potential energy formula.
  • Treating potential energy as absolute rather than reference-dependent.

Conservation of Mechanical Energy and Energy Accounting with Nonconservative Work

Mechanical energy and when it is conserved

Mechanical energy is

$E_{\mathrm{mech}}=K+U$

If only conservative forces do work (or nonconservative forces do no net work), mechanical energy is conserved:

$K_i+U_i=K_f+U_f$

This connects two positions or moments without requiring time.

The more general energy equation (including nonconservative work)

When nonconservative forces do work,

$W_{\mathrm{nc}}=\Delta K+\Delta U$

Equivalently,

$K_i+U_i+W_{\mathrm{nc}}=K_f+U_f$

Mechanical energy isn’t “destroyed”; it’s transferred into other forms (thermal, sound, internal energy) not included in $U$ unless you model them.

Choosing the system

Your energy equation depends on what you define as the system.

• If the system is “block + Earth,” gravity is internal and you include gravitational potential energy.
• If the system is just the “block,” gravity is external and you include work by gravity instead.

Both are valid if you’re consistent. Many errors come from double-counting (including both $U_g$ and work by gravity).

Strategy for energy problems

Energy methods are accounting:

1. Identify initial and final states.
2. Decide which energy terms matter (kinetic, gravitational, spring).
3. Decide whether nonconservative work is present.
4. Write one consistent equation connecting states.

Example 1: Speed at bottom of a frictionless ramp

A block starts from rest at height $h$ and slides down a frictionless track. Take bottom as $U=0$.

$mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$

Mass cancels.

Sample Problem 7: Dropped ball (conservation of mechanical energy)

A ball of mass $2\ \mathrm{kg}$ is dropped from a height of $5.0\ \mathrm{m}$ above the floor. Find its speed as it strikes the floor (ignore air resistance). Taking the floor as $h=0$,

$mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$

Using $g=9.8\ \mathrm{m/s^2}$,

$v=\sqrt{2(9.8)(5.0)}$

$v\approx 9.9\ \mathrm{m/s}$

This illustrates the core idea: potential energy decreases while kinetic energy increases.

Example 2: Ramp with friction (nonconservative work)

A block slides down a ramp of length $L$ with kinetic friction coefficient $\mu_k$ and vertical drop $h$. On an incline, $N=mg\cos\theta$, so

$f_k=\mu_k mg\cos\theta$

Friction work is

$W_f=-f_k L=-\mu_k mg\cos\theta\,L$

Energy equation:

$mgh-\mu_k mg\cos\theta\,L=\frac{1}{2}mv^2$

Solve:

$v=\sqrt{2g\left(h-\mu_k\cos\theta\,L\right)}$

A classic mistake is to treat friction work as depending only on vertical drop; kinetic friction work depends on path length.

Example 3: Spring launcher with a friction patch

A block of mass $m$ is launched by a spring compressed by $x$ with spring constant $k$, then crosses a rough horizontal patch of length $d$ with coefficient $\mu_k$. Starting from rest at compression and ending on level ground:

$\frac{1}{2}kx^2-\mu_k mgd=\frac{1}{2}mv^2$

$v=\sqrt{\frac{kx^2}{m}-2\mu_k gd}$

This treats friction as an “energy cost” subtracted from stored spring energy.

Exam Focus

• Typical question patterns:
  • Use $K_i+U_i=K_f+U_f$ for frictionless motion involving height and springs.
  • Use $W_{\mathrm{nc}}=\Delta K+\Delta U$ when friction or drag is present.
  • Multi-stage energy problems (spring launch then incline, drop then spring compression).
• Common mistakes:
  • Double-counting gravity (including both $U_g$ and work by gravity).
  • Using the wrong distance for friction work (distance along the surface where friction acts).
  • Assuming mechanical energy is conserved even when friction/drag is explicitly present.

Power: The Rate of Doing Work

What power measures

Power tells you how fast work is done or how fast energy is transferred. Two devices can do the same work, but the one that does it faster has greater power.

Average power:

$P_{\mathrm{avg}}=\frac{W}{\Delta t}$

Instantaneous power:

$P=\frac{dW}{dt}$

Using $dW=\vec{F}\cdot d\vec{r}$ and $\vec{v}=\frac{d\vec{r}}{dt}$,

$P=\vec{F}\cdot \vec{v}$

Power depends on the component of force along the velocity.

Units:

$1\ \mathrm{W}=1\ \mathrm{J/s}$

Interpreting the dot product

• If force is along motion, power is

$P=Fv$

• If force is opposite velocity, power is negative.
• If force is perpendicular to velocity, power is zero at that instant.

Example 1: Lifting at constant speed

A motor lifts a mass $m$ vertically upward at constant speed $v$. Then tension equals weight, and

$P=mgv$

Sample Problem 8: Mover pushing a crate (average power)

A mover pushes a crate from the inside of a truck to the back end, a distance of $6\ \mathrm{m}$, exerting a steady push of $300\ \mathrm{N}$. He moves the crate this distance in $20\ \mathrm{s}$. What is his power output during this time?

Work:

$W=Fd$

$W=(300)(6)=1800\ \mathrm{J}$

Average power:

$P_{\mathrm{avg}}=\frac{W}{t}$

$P_{\mathrm{avg}}=\frac{1800}{20}=90\ \mathrm{W}$

Sample Problem 9: Rocket engine power at constant speed

What must be the power output of a rocket engine that moves a $1000\ \mathrm{kg}$ rocket at a constant speed of $8.0\ \mathrm{m/s}$ (using $g\approx 10\ \mathrm{N/kg}$)?

Using

$P=Fv$

and taking the needed force as the weight,

$F=mg$

then

$P=mgv$

$P=(1000)(10)(8.0)=80000\ \mathrm{W}$

$P=80\ \mathrm{kW}$

Example 2: Maximum power and steady speed on level ground (linear drag model)

If resistive force is proportional to speed,

$F_r=bv$

At top speed on level ground, acceleration is zero, so driving force matches resistive force, and the power needed to overcome drag is

$P=F_r v$

$P=bv^2$

Solve:

$v=\sqrt{\frac{P}{b}}$

Even if this exact model is rare, the reasoning pattern is common: relate power to force and speed, then use “steady speed” conditions.

Exam Focus

• Typical question patterns:
  • Compute power from work over time or from force and speed.
  • Use motor power limits to find speed while lifting or climbing.
  • Interpret negative power for braking or drag forces.
• Common mistakes:
  • Treating power as a force (mixing up units and meaning).
  • Forgetting the dot product and using $P=Fv$ when force is not parallel to velocity.
  • Using average power when instantaneous power is required (or vice versa).

Potential Energy Curves, Equilibrium, and Stability (1D)

Why potential energy graphs are so useful

When motion is constrained to one dimension, a graph of $U(x)$ can reveal where the particle can move, where it speeds up or slows down, and where equilibrium points exist.

The key connection is

$F_x=-\frac{dU}{dx}$

So the slope of $U(x)$ gives force direction and magnitude, and zero slope indicates equilibrium.

Total mechanical energy, allowed regions, and turning points

If only conservative forces act,

$E=K+U$

and total energy $E$ is constant. Because kinetic energy cannot be negative,

$K=E-U(x)\ge 0$

So motion is allowed only where

$U(x)\le E$

Turning points occur where speed is zero, so

$U(x)=E$

On a $U(x)$ graph, a horizontal line at $E$ intersects the curve at turning points.

Equilibrium points and stability

Equilibrium occurs where

$F_x(x_0)=0$

equivalently

$\frac{dU}{dx}\bigg|_{x_0}=0$

• Stable equilibrium corresponds to a local minimum of $U(x)$.
• Unstable equilibrium corresponds to a local maximum of $U(x)$.

If using the second derivative test,

$\frac{d^2U}{dx^2}\bigg|_{x_0}>0$

means stable, and

$\frac{d^2U}{dx^2}\bigg|_{x_0}<0$

means unstable.

Example 1: Reading force direction from $U(x)$

If $U(x)$ is increasing with $x$ at some location, then $\frac{dU}{dx}$ is positive there, so

$F_x=-\frac{dU}{dx}$

and the force is negative (toward decreasing $x$). This sign reasoning is a frequent exam target.

Example 2: Finding speed from a potential energy curve

If total energy is $E$ and you know $U(x)$ at a position, then

$K=E-U(x)$

and

$\frac{1}{2}mv^2=E-U(x)$

so

$v=\sqrt{\frac{2(E-U(x))}{m}}$

Potential energy curves beyond mechanics contexts

Potential energy curves are also used to describe interactions between particles (atoms and molecules). The shape depends on the interaction. A simple harmonic oscillator has a parabolic potential-energy curve. A typical diatomic molecular potential has a minimum at the equilibrium bond length and approaches a dissociation limit as separation becomes large. Polyatomic systems can have many minima and maxima corresponding to different conformations.

Sample Problem 10: Equilibrium points and motion from a given potential function

An object of mass $m=4\ \mathrm{kg}$ has potential energy

$U(x)=(x-2)-(2x-3)^3$

where $x$ is in meters and $U$ is in joules.

(a) Determine the positions of points A and B, the equilibrium points.

Equilibrium occurs when

$\frac{dU}{dx}=0$

Compute the derivative:

$\frac{dU}{dx}=1-6(2x-3)^2$

Set to zero:

$1-6(2x-3)^2=0$

$6(2x-3)^2=1$

$2x-3=\pm \frac{1}{\sqrt{6}}$

So

$x=\frac{3\pm \frac{1}{\sqrt{6}}}{2}$

Numerically, these are about $x\approx 1.3\ \mathrm{m}$ and $x\approx 1.7\ \mathrm{m}$. One is a local minimum (stable) and the other is a local maximum (unstable).

(b) If released from rest at point B, can it reach point A or C?

Released from rest means total energy equals the potential energy at B:

$E=U(B)$

Motion is only possible where $U(x)\le E$. In this setup, the total mechanical energy at B is negative because all of it is potential energy there. The object cannot reach point C if $U(C)$ is above that energy level, but it can reach point A if $U(A)$ is less than the starting energy. As it moves from B to A, potential energy decreases (becomes more negative) and kinetic energy increases.

(c) Released from rest at point C, determine its speed as it passes point A.

First determine the total mechanical energy from point C:

$E=U(C)$

Then at point A,

$\frac{1}{2}mv^2=E-U(A)$

so

$v=\sqrt{\frac{2(E-U(A))}{m}}$

This procedure matches the general turning-point and allowed-region reasoning: you compare $U(x)$ to a fixed total energy $E$.

Exam Focus

• Typical question patterns:
  • Given $U(x)$, determine where motion is allowed (regions where $U\le E$) and locate turning points.
  • Use $F_x=-\frac{dU}{dx}$ to infer force direction and relative magnitude from slopes.
  • Identify stable vs unstable equilibrium from minima/maxima of $U(x)$.
  • Use derivatives of $U(x)$ to find equilibrium points.
• Common mistakes:
  • Forgetting the negative sign in the force–potential relationship.
  • Claiming motion is possible where $U>E$ (that would require negative kinetic energy).
  • Confusing equilibrium (slope zero) with turning points (where $U=E$).

Advanced Work Integrals and Common AP Physics C Setups

Work along curved paths and dot products

The most general definition of work is

$W=\int \vec{F}\cdot d\vec{r}$

This matters when the path is not straight or the force changes direction along the path. A common approach is to focus on the component of the force tangent to the path (the component parallel to $d\vec{r}$).

Notation you may see (and how to interpret it)

Example 1: Work done by a non-constant applied force

An applied force varies with position as

$F(x)=ax$

along the direction of motion from $x=0$ to $x=L$.

$W=\int_0^L ax\,dx$

$W=a\left[\frac{x^2}{2}\right]_0^L$

$W=\frac{1}{2}aL^2$

On an $F$ vs $x$ graph, this is the area of a triangle under a line from $0$ to $aL$.

Example 2: Comparing work by gravity using a force integral vs potential energy

A mass moves upward a vertical distance $h$. Gravity is downward with magnitude $mg$, with upward taken as positive.

Using the work integral,

$W_g=\int_0^h (-mg)\,dy$

$W_g=-mgh$

Using potential energy,

$\Delta U_g=mgh$

and for a conservative force,

$W_g=-\Delta U_g$

The methods agree, which is a useful self-check.

Exam Focus

• Typical question patterns:
  • Set up and evaluate work integrals for forces that depend on position.
  • Translate between a force function, a force graph, and the computed work.
  • Use both the integral method and the potential-energy method as consistency checks for conservative forces.
• Common mistakes:
  • Integrating the magnitude of force instead of the component along displacement.
  • Dropping signs by assuming work must be positive.
  • Treating variable-force work as if a single constant force acts over the whole distance.

Putting It All Together: Multi-Concept Energy Problems

How AP problems combine ideas

Many AP Physics C questions are not “one-formula” problems. They often combine applied-force work, potential energy changes, nonconservative work (friction), and power constraints. The unifying skill is telling a consistent energy story: where energy starts, where it goes, and what mechanisms transfer it.

Example 1: Block pulled up an incline by a rope

A block of mass $m$ is pulled up an incline of angle $\theta$ over distance $L$ by a rope with tension magnitude $T$ parallel to the incline. Kinetic friction coefficient is $\mu_k$. Starting from rest, find final speed.

Work and energy accounting:

• Tension does positive work:

$W_T=TL$

• Gravity increases potential energy by

$\Delta U_g=mgL\sin\theta$

• Friction does negative work. With

$f_k=\mu_k mg\cos\theta$

its work is

$W_f=-\mu_k mg\cos\theta\,L$

Use

$W_{\mathrm{nc}}=\Delta K+\Delta U$

Here the nonconservative work is

$W_{\mathrm{nc}}=W_T+W_f$

so

$TL-\mu_k mg\cos\theta\,L=\frac{1}{2}mv^2+mgL\sin\theta$

Solve:

$v=\sqrt{\frac{2}{m}\left(TL-\mu_k mg\cos\theta\,L-mgL\sin\theta\right)}$

Sign discipline is everything: each term is energy added or removed.

Example 2: Power-limited climb at constant speed

A vehicle of mass $m$ climbs a hill at constant speed $v$ on an incline angle $\theta$. Ignore drag and rolling resistance. What power must the engine provide?

Constant speed means $\Delta K=0$. The required uphill force balances gravity’s component:

$F=mg\sin\theta$

Power:

$P=Fv$

$P=mgv\sin\theta$

Common error pattern: mixing “conserved energy” with “constant speed”

Constant speed means

$\Delta K=0$

but it does not imply mechanical energy is conserved. Climbing at constant speed increases gravitational potential energy, so energy must come from engine work. Likewise, moving at constant speed on a rough surface requires continuous power because friction removes mechanical energy.

Exam Focus

• Typical question patterns:
  • Multi-step energy transfer with friction and springs (find speed, compression, or distance).
  • Combine constant-speed or equilibrium conditions with power or work relations.
  • Explain qualitatively where energy goes when friction is present.
• Common mistakes:
  • Writing both a full work–energy equation and a full mechanical-energy conservation equation simultaneously without reconciling them.
  • Treating applied work as conservative (it generally is not unless explicitly derived from a potential).
  • Neglecting that friction converts mechanical energy into thermal energy, so “lost” mechanical energy is transformed, not missing.