Friction and Inclined Planes

Static vs Kinetic Friction

Part 1 of 7 — Friction & Inclines

Friction is one of the most common forces you'll encounter on the AP Physics C exam. Understanding the distinction between static and kinetic friction — and when each applies — is essential.

The Nature of Friction

Friction arises from microscopic interactions between surfaces in contact. At the AP Physics C level, we model friction with two simple laws:

Type	Symbol	Condition	Formula
Static	$f_s$	Object at rest	$f_s \leq \mu_s N$
Kinetic	$f_k$	Object sliding	$f_k = \mu_k N$

Key Differences

• Static friction is a variable force: it adjusts to match the applied force, up to a maximum $f_{s,\max} = \mu_s N$.
• Kinetic friction is a constant force (for a given normal force): $f_k = \mu_k N$.
• Typically $\mu_s > \mu_k$; it takes more force to start motion than to maintain it.

The Normal Force

The normal force $N$ is perpendicular to the contact surface. On a flat surface with no vertical acceleration:

$N = mg - F\sin\theta$

where $F$ is an applied force at angle $\theta$ above the horizontal. If pushing downward at angle $\theta$ below horizontal:

$N = mg + F\sin\theta$

Free Body Diagrams with Friction

When drawing FBDs involving friction, always:

1. Identify the contact surface — friction acts along it.
2. Determine the normal force — perpendicular to the surface.
3. Check if the object moves — this determines static vs kinetic.
4. Direction of friction — opposes motion (kinetic) or opposes tendency of motion (static).

Worked Example

A 5 kg block is pushed along a floor by a force $F = 40$ N at $30°$ below the horizontal. $\mu_k = 0.25$.

Step 1: Normal force (vertical equilibrium): $N = mg + F\sin 30° = 5(10) + 40(0.5) = 70 \text{ N}$

Step 2: Kinetic friction: $f_k = \mu_k N = 0.25 \times 70 = 17.5 \text{ N}$

Step 3: Horizontal acceleration: $F\cos 30° - f_k = ma$ $40(0.866) - 17.5 = 5a$ $a = \frac{34.64 - 17.5}{5} = 3.43 \text{ m/s}^2$

Key Insight: Pushing downward increases the normal force and thus the friction force. Pulling upward decreases it.

The Threshold of Motion

A critical AP Physics C skill is determining whether a block moves under a given force. The procedure:

1. Assume static: Calculate the force needed to keep the block stationary.
2. Compare to $f_{s,\max}$: If the required friction $\leq \mu_s N$, the block stays at rest.
3. If exceeded: The block accelerates. Switch to kinetic friction for the dynamics.

The Transition: Calculus Perspective

At the instant motion begins, the friction force drops from $f_{s,\max} = \mu_s N$ to $f_k = \mu_k N$. This creates an instantaneous jump in acceleration:

$a(t = 0^+) = \frac{F_{\text{applied}} - \mu_k N}{m}$

This discontinuity in friction is why objects often "jerk" when they start sliding.

Optimal Angle for Pulling

To minimize the pulling force $F$ needed to move a block, differentiate with respect to $\theta$:

$F = \frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}$

Setting $\frac{dF}{d\theta} = 0$:

$\sin\theta - \mu_s \cos\theta = 0 \implies \tan\theta = \mu_s$

$\theta_{\text{opt}} = \arctan(\mu_s)$

Part 1 Summary

Concept	Key Formula
Static friction	$f_s \leq \mu_s N$
Kinetic friction	$f_k = \mu_k N$
Normal force (pull up)	$N = mg - F\sin\theta$
Normal force (push down)	$N = mg + F\sin\theta$
Optimal pull angle	$\theta = \arctan(\mu_s)$

Next up: Part 2 — Inclined Planes (No Friction), where we analyze motion on smooth ramps using calculus.

Inclined Planes (No Friction)

Part 2 of 7 — Friction & Inclines

Before adding friction, let's master the frictionless incline — the foundation for all ramp problems.

Setting Up Coordinates

For an incline at angle $\theta$:

Direction	Axis	Forces
Along the incline	$x$-axis	$mg\sin\theta$ (down the ramp)
Perpendicular to incline	$y$-axis	$N - mg\cos\theta = 0$

Newton's Second Law on an Incline

Along the incline (taking down-the-ramp as positive):

$ma = mg\sin\theta$ $a = g\sin\theta$

This is constant acceleration, so all kinematics equations apply:

$v(t) = v_0 + (g\sin\theta)\,t$ $x(t) = x_0 + v_0 t + \frac{1}{2}(g\sin\theta)\,t^2$ $v^2 = v_0^2 + 2(g\sin\theta)\Delta x$

Energy Methods on Inclines

For a frictionless incline, mechanical energy is conserved:

$\frac{1}{2}mv_0^2 + mgh_0 = \frac{1}{2}mv^2 + mgh$

The height change on a ramp of length $L$ at angle $\theta$:

$\Delta h = L\sin\theta$

Worked Example: Launched Up a Ramp

A block is launched up a frictionless $45°$ ramp with initial speed $v_0 = 10$ m/s.

How far up the ramp does it travel?

Using energy conservation: $\frac{1}{2}mv_0^2 = mg(L\sin\theta)$ $L = \frac{v_0^2}{2g\sin\theta} = \frac{100}{2(10)(\sin 45°)} = \frac{100}{14.14} = 7.07 \text{ m}$

Time to reach the top:

Decelerating at $a = -g\sin 45° = -7.07$ $m/s^{2}$: $v = v_0 - (g\sin\theta)t = 0$ $t = \frac{v_0}{g\sin\theta} = \frac{10}{7.07} = 1.41 \text{ s}$

Velocity as a Function of Position

Using $v^2 = v_0^2 - 2g\sin\theta \cdot x$:

$v(x) = \sqrt{v_0^2 - 2gx\sin\theta}$

This is valid until $v = 0$ at $x_{\max} = \frac{v_0^2}{2g\sin\theta}$.

Calculus Approach: Variable Angle Ramps

What if the ramp angle changes? Consider a curved ramp where $\theta = \theta(x)$.

For a small displacement $dx$ along the ramp, the height change is $dh = \sin\theta(x)\,dx$.

The equation of motion becomes: $m\frac{dv}{dt} = mg\sin\theta(x)$

Using the chain rule: $\frac{dv}{dt} = v\frac{dv}{dx}$:

$mv\frac{dv}{dx} = mg\sin\theta(x)$

$\int_{v_0}^{v} v'\,dv' = g\int_0^x \sin\theta(x')\,dx'$

$\frac{v^2 - v_0^2}{2} = g\int_0^x \sin\theta(x')\,dx'$

Example: Parabolic Ramp

For a ramp shaped like $y = \frac{x^2}{2L}$, the slope at position $x$ is:

$\tan\theta = \frac{dy}{dx} = \frac{x}{L}$

For small angles: $\sin\theta \approx \tan\theta = x/L$.

$\frac{v^2}{2} = g\int_0^x \frac{x'}{L}\,dx' = \frac{gx^2}{2L}$

$v(x) = x\sqrt{\frac{g}{L}}$

The speed increases linearly with position — a result unique to this geometry.

Part 2 Summary

Concept	Formula
Acceleration on incline	$a = g\sin\theta$
Normal force	$N = mg\cos\theta$
Speed at bottom	$v = \sqrt{2gL\sin\theta} = \sqrt{2gh}$
Height change	$\Delta h = L\sin\theta$
Chain rule substitution	$a = v\frac{dv}{dx}$

Next up: Part 3 — Inclined Planes with Friction, combining everything from Parts 1 and 2.

Inclined Planes with Friction

Part 3 of 7 — Friction & Inclines

Now we combine friction with inclined planes — the bread and butter of AP Physics C mechanics problems.

Forces on a Rough Incline

For a block on an incline of angle $\theta$ with friction:

Perpendicular to incline ($y$-direction): $N = mg\cos\theta$

Along the incline ($x$-direction):

Scenario	Net Force (down ramp +)
Sliding down	$mg\sin\theta - \mu_k mg\cos\theta = ma$
Sliding up	$-mg\sin\theta - \mu_k mg\cos\theta = ma$
At rest (checking)	$mg\sin\theta \leq \mu_s mg\cos\theta$?

Critical Angle

A block starts sliding when the component of gravity along the ramp exceeds maximum static friction:

$mg\sin\theta_c = \mu_s mg\cos\theta_c$ $\tan\theta_c = \mu_s$ $\theta_c = \arctan(\mu_s)$

This gives a clean way to measure $\mu_s$ experimentally.

Dynamics on Rough Inclines

Sliding Down

Acceleration (taking down-ramp as positive): $a_{\text{down}} = g(\sin\theta - \mu_k \cos\theta)$

The block slides down only if $\tan\theta > \mu_s$ (i.e., $\theta > \theta_c$).

Sliding Up (Launched Upward)

When a block is launched up a rough incline, both gravity and friction decelerate it:

$a_{\text{up}} = -g(\sin\theta + \mu_k \cos\theta)$

Caution: After the block stops, it may or may not slide back down. It stays at rest if $\tan\theta \leq \mu_s$.

Worked Example

A block is launched up a $37°$ rough incline ($\mu_k = 0.3$) with $v_0 = 12$ m/s.

Deceleration going up: $a_{\text{up}} = -g(\sin 37° + 0.3\cos 37°) = -10(0.6 + 0.24) = -8.4 \text{ m/s}^2$

Distance traveled up: $v^2 = v_0^2 + 2a\Delta x = 0$ $\Delta x = \frac{v_0^2}{2|a|} = \frac{144}{16.8} = 8.57 \text{ m}$

Does it slide back? $\tan 37° = 0.75 > \mu_s$ (assuming $\mu_s \approx 0.35$), so yes.

Acceleration sliding back down: $a_{\text{down}} = g(\sin 37° - 0.3\cos 37°) = 10(0.6 - 0.24) = 3.6 \text{ m/s}^2$

Speed at the bottom (returning): $v = \sqrt{2(3.6)(8.57)} = \sqrt{61.7} = 7.85 \text{ m/s}$

Notice: $v_{\text{return}} < v_0$ because friction dissipated energy.

Energy Methods with Friction

When friction is present, mechanical energy is not conserved. The work-energy theorem gives:

$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = W_{\text{net}} = -mgh - \mu_k N \cdot d$

For a ramp of length $d$ at angle $\theta$:

$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 + mgd\sin\theta - \mu_k mgd\cos\theta$

The energy dissipated by friction:

$\Delta E_{\text{thermal}} = \mu_k mgd\cos\theta = f_k \cdot d$

Round Trip Energy Loss

For a block launched up and sliding back to the start:

$\Delta E_{\text{total}} = 2\mu_k mgd\cos\theta$

Since $\frac{1}{2}mv_{\text{return}}^2 = \frac{1}{2}mv_0^2 - 2\mu_k mgd\cos\theta$:

$v_{\text{return}} = \sqrt{v_0^2 - 4\mu_k gd\cos\theta}$

where $d = \frac{v_0^2}{2g(\sin\theta + \mu_k\cos\theta)}$ is the distance traveled up.

Part 3 Summary

Concept	Formula
Critical angle	$\theta_c = \arctan(\mu_s)$
Accel. sliding down	$a = g(\sin\theta - \mu_k\cos\theta)$
Decel. sliding up	$a = -g(\sin\theta + \mu_k\cos\theta)$
Energy loss (one way)	$\Delta E = \mu_k mgd\cos\theta$

Next up: Part 4 — Velocity-Dependent Friction, where we use differential equations to handle drag and other calculus-based friction models.

Friction with Calculus (Velocity-Dependent)

Part 4 of 7 — Friction & Inclines

In many real-world situations, the resistive force depends on velocity. This is the hallmark of AP Physics C — using differential equations to solve dynamics problems.

Linear Drag: $f = bv$

A common model for low-speed drag:

$m\frac{dv}{dt} = F_{\text{applied}} - bv$

Terminal Velocity

At terminal velocity, $a = 0$:

$v_T = \frac{F_{\text{applied}}}{b}$

For an object falling under gravity with linear drag:

$v_T = \frac{mg}{b}$

Solving the ODE

$m\frac{dv}{dt} = mg - bv$

Separation of variables:

$\frac{dv}{mg - bv} = \frac{dt}{m}$

$-\frac{1}{b}\ln\left(\frac{mg - bv}{mg}\right) = \frac{t}{m}$

$v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right) = v_T\left(1 - e^{-t/\tau}\right)$

where $\tau = m/b$ is the time constant.

Quadratic Drag: $f = cv^2$

At higher speeds, drag is proportional to $v^2$:

$m\frac{dv}{dt} = mg - cv^2$

Terminal Velocity

$v_T = \sqrt{\frac{mg}{c}}$

Terminal speed always comes from setting the net force to zero, whatever the drag law. You can also find the acceleration at any speed directly from Newton's second law:

$a = g - \frac{c}{m}v^2 = g\left(1 - \frac{v^2}{v_T^2}\right)$

For example, at half the terminal speed the drag is one quarter of the weight, so $a = \tfrac{3}{4}g$.

The AP Physics C exam solves for $v(t)$ only with the linear model $f = bv$; for quadratic drag you need the terminal speed and force-balance reasoning, not a full solution.

Velocity-Dependent Friction on Inclines

Consider a block sliding down an incline with velocity-dependent friction $f = bv$:

$m\frac{dv}{dt} = mg\sin\theta - bv$

This has the same form as free fall with linear drag. The solution:

$v(t) = \frac{mg\sin\theta}{b}\left(1 - e^{-bt/m}\right)$

Terminal velocity on the incline:

$v_T = \frac{mg\sin\theta}{b}$

Worked Example

A 0.5 kg block slides down a $30°$ incline with velocity-dependent friction $f = 2v$ (in SI units). Find:

Terminal velocity: $v_T = \frac{mg\sin\theta}{b} = \frac{0.5 \times 10 \times 0.5}{2} = 1.25 \text{ m/s}$

Time constant: $\tau = \frac{m}{b} = \frac{0.5}{2} = 0.25 \text{ s}$

Speed at $t = 0.5$ s: $v(0.5) = 1.25\left(1 - e^{-0.5/0.25}\right) = 1.25(1 - e^{-2}) = 1.25(0.865) = 1.08 \text{ m/s}$

Acceleration at $t = 0.5$ s: $a(t) = g\sin\theta \cdot e^{-bt/m} = 5 \cdot e^{-2} = 0.677 \text{ m/s}^2$

Part 4 Summary

Model	ODE	Solution	$v_T$
Linear drag ($f=bv$)	$m\dot{v} = mg - bv$	$v_T(1-e^{-t/\tau})$	$mg/b$
Quadratic drag ($f=cv^2$)	$m\dot{v} = mg - cv^2$	Force balance only	$\sqrt{mg/c}$
Linear drag, no driving	$m\dot{v} = -bv$	$v_0 e^{-t/\tau}$	$0$

AP Tip: The AP Physics C exam frequently tests your ability to set up and solve first-order ODEs with separation of variables.

Next up: Part 5 — Systems on Inclines (Atwood machines, connected blocks).

Systems on Inclines

Part 5 of 7 — Friction & Inclines

Many AP Physics C problems involve multiple objects connected by strings, often on inclines. The key: apply Newton's second law to each object separately, then combine.

Atwood Machine on an Incline

A classic setup: mass $m_1$ on a rough incline (angle $\theta$, coefficient $\mu_k$) connected by a massless string over a frictionless pulley to a hanging mass $m_2$.

Free Body Diagrams

Block on incline ($m_1$): $T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a$

Hanging block ($m_2$, assuming it descends): $m_2 g - T = m_2 a$

Solving for $a$ and $T$

Adding the equations: $m_2 g - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = (m_1 + m_2)a$

$a = \frac{m_2 g - m_1 g(\sin\theta + \mu_k\cos\theta)}{m_1 + m_2}$

$T = m_2(g - a) = \frac{m_1 m_2 g(1 + \sin\theta + \mu_k\cos\theta)}{m_1 + m_2}$

Two Blocks on Different Inclines

Consider masses $m_1$ and $m_2$ connected by a string over a pulley, each on different inclines at angles $\theta_1$ and $\theta_2$.

Assuming $m_1$ moves up its incline and $m_2$ moves down its incline:

For $m_1$ (moves up, friction opposes): $T - m_1 g\sin\theta_1 - \mu_1 m_1 g\cos\theta_1 = m_1 a$

For $m_2$ (moves down, friction opposes): $m_2 g\sin\theta_2 - T - \mu_2 m_2 g\cos\theta_2 = m_2 a$

Adding: $a = \frac{m_2 g(\sin\theta_2 - \mu_2\cos\theta_2) - m_1 g(\sin\theta_1 + \mu_1\cos\theta_1)}{m_1 + m_2}$

Checking Direction

Before solving, determine which way the system tends to move:

1. Compare $m_1 g\sin\theta_1$ vs $m_2 g\sin\theta_2$ (gravitational components).
2. The side with greater gravitational component tends to slide down.
3. If $a < 0$ from your assumption, the system moves the other way (flip the friction directions and resolve).

Stacked Blocks on an Incline

For a block of mass $m_1$ on top of a block $m_2$ on a ramp:

• The friction between blocks is what accelerates/decelerates $m_1$.
• If the blocks move together: treat as one system to find $a$, then isolate one block to find the friction between them.
• If they slide relative to each other: apply kinetic friction at the interface.

Constraint Equations

When blocks are connected by strings through pulleys, the constraint equation relates their accelerations.

Simple Constraint (Single String)

If a single string connects two blocks, their speeds are equal: $v_1 = v_2 \implies a_1 = a_2 = a$

Pulley Ratio Constraint

If a string wraps around a movable pulley attached to block 2: $\Delta x_2 = \frac{\Delta x_1}{2} \implies a_2 = \frac{a_1}{2}$

And the tensions relate as: $T_2 = 2T_1$

Worked Example: Pulley System

A 5 kg block on a $37°$ rough incline ($\mu_k = 0.25$) is connected to a 3 kg hanging mass through a pulley attached to the incline block.

With constraint $a_{\text{hang}} = 2a_{\text{incline}}$:

Let $a$ be the incline block's acceleration (up the incline).

Incline block: $2T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a$

Hanging block: $m_2 g - T = m_2(2a)$

From the hanging block: $T = m_2 g - 2m_2 a = 3(10) - 6a = 30 - 6a$

Substituting: $2(30 - 6a) - 50\sin 37° - 0.25(50)\cos 37° = 5a$

$60 - 12a - 30 - 10 = 5a$ $20 = 17a \implies a = 1.18 \text{ m/s}^2$

Part 5 Summary

System Type	Key Approach
Atwood on incline	Separate FBDs, same $a$ and $T$
Two inclines	Compare gravitational components to find direction
Stacked blocks	Check if friction is sufficient; if not, blocks separate
Pulley constraints	Relate $a_1$ and $a_2$ via string length

Next up: Part 6 — Problem-Solving Workshop with multi-step AP-style problems.

Problem-Solving Workshop

Part 6 of 7 — Friction & Inclines

This workshop presents AP Physics C–style problems that integrate concepts from Parts 1–5. Practice the systematic approach:

Problem-Solving Framework

Step	Action
1	Draw a diagram and label all forces
2	Choose a coordinate system (tilted for inclines)
3	Write Newton's 2nd law for each object
4	Identify constraints (strings, contact)
5	Solve the system of equations
6	Check units and limiting cases

Problem 2: Block Launched Up a Ramp

A 3 kg block is launched up a $37°$ rough incline ($\mu_k = 0.3$) with initial speed $v_0 = 10$ m/s.

Solution

Phase 1: Going up

$a_{\text{up}} = -g(\sin 37° + \mu_k \cos 37°) = -10(0.6 + 0.24) = -8.4 \text{ m/s}^2$

Distance traveled up: $0 = v_0^2 + 2a_{\text{up}}d \implies d = \frac{v_0^2}{2 \times 8.4} = \frac{100}{16.8} = 5.95 \text{ m}$

Time to reach top: $t_1 = \frac{v_0}{|a_{\text{up}}|} = \frac{10}{8.4} = 1.19 \text{ s}$

Phase 2: Check if it slides back

$\tan 37° = 0.75$. If $\mu_s \approx 0.35 < 0.75$, it slides back.

$a_{\text{down}} = g(\sin 37° - \mu_k \cos 37°) = 10(0.6 - 0.24) = 3.6 \text{ m/s}^2$

Speed at bottom: $v = \sqrt{2(3.6)(5.95)} = \sqrt{42.84} = 6.55 \text{ m/s}$

Energy check: Energy dissipated = $\mu_k mg\cos\theta \times 2d = 0.3(30)(0.8)(11.9) = 85.7$ J.

Initial KE = $\frac{1}{2}(3)(100) = 150$ J. Final KE = $150 - 85.7 = 64.3$ J. $v = \sqrt{2(64.3)/3} = 6.55$ m/s. ✓

Problem 3: Velocity-Dependent Friction on a Ramp

A 1 kg block starts from rest at the top of a $45°$ incline with velocity-dependent friction $f = 4v$ (SI units). Find:

(a) Terminal velocity: $v_T = \frac{mg\sin\theta}{b} = \frac{1(10)(\frac{\sqrt{2}}{2})}{4} = \frac{7.07}{4} = 1.77 \text{ m/s}$

(b) Velocity as a function of time: $v(t) = v_T(1 - e^{-bt/m}) = 1.77(1 - e^{-4t})$

(c) Distance traveled in 1 second: $x(t) = \int_0^t v(t')\,dt' = v_T\left[t + \frac{m}{b}e^{-bt/m}\right]_0^t = v_T\left(t + \frac{1}{4}e^{-4t} - \frac{1}{4}\right)$

$x(1) = 1.77\left(1 + 0.25e^{-4} - 0.25\right) = 1.77(1 + 0.00458 - 0.25) = 1.77(0.755) = 1.34 \text{ m}$

(d) Acceleration at $t = 0.5$ s: $a(t) = g\sin\theta \cdot e^{-bt/m} = 7.07 e^{-2} = 0.957 \text{ m/s}^2$

Workshop Summary

Common AP Pitfalls

Mistake	Correction
Wrong friction direction	Always opposes motion (or tendency of motion)
Forgetting $N \neq mg$ on inclines	$N = mg\cos\theta$ on an incline
Using $\mu_s$ when sliding	Use $\mu_k$ once the object is in motion
Ignoring the "does it move?" check	Always compare applied force to $f_{s,\max}$ first
Wrong sign on $a$ when going up vs down	Friction always opposes velocity direction

Next up: Part 7 — Review & Applications, consolidating everything with real-world contexts.

Review & Applications

Part 7 of 7 — Friction & Inclines

Complete Topic Reference

Concept	Formula	Part
Static friction	$f_s \leq \mu_s N$	1
Kinetic friction	$f_k = \mu_k N$	1
Incline acceleration (no friction)	$a = g\sin\theta$	2
Incline acceleration (friction, down)	$a = g(\sin\theta - \mu_k\cos\theta)$	3
Critical angle	$\theta_c = \arctan(\mu_s)$	3
Linear drag terminal velocity	$v_T = mg/b$	4
Quadratic drag terminal velocity	$v_T = \sqrt{mg/c}$	4
Falling with linear drag	$v = v_T(1-e^{-bt/m})$	4
Atwood system on incline	$a = \frac{m_2 g - m_1 g(\sin\theta + \mu_k\cos\theta)}{m_1+m_2}$	5

Application: Braking on a Hill

A car of mass $m$ is traveling at speed $v_0$ down a hill of angle $\theta$. The brakes provide a constant friction force $F_b$.

Equation of motion: $ma = mg\sin\theta - F_b$

Stopping distance: $d = \frac{mv_0^2}{2(F_b - mg\sin\theta)}$

This is valid only if $F_b > mg\sin\theta$ (brakes can actually slow the car).

Critical Braking Condition

The minimum braking force to stop on a hill: $F_b > mg\sin\theta$

If the tires provide braking through friction: $F_b = \mu_k N = \mu_k mg\cos\theta$

The car can stop only if: $\mu_k mg\cos\theta > mg\sin\theta \implies \mu_k > \tan\theta$

This is why steep icy hills ($\mu_k$ small, $\theta$ large) are so dangerous.

With ABS (Maintaining Static Friction)

Modern ABS brakes prevent wheel lock, using $\mu_s$ instead of $\mu_k$:

$d_{\text{ABS}} = \frac{v_0^2}{2g(\mu_s\cos\theta - \sin\theta)}$

Since $\mu_s > \mu_k$, ABS gives shorter stopping distances.

Application: Drag on a Falling Object

A skydiver ($m = 80$ kg) falls with a drag force of magnitude $f = 0.25v^2$ (SI units).

Terminal velocity (net force zero): $mg = cv_T^2 \implies v_T = \sqrt{\frac{mg}{c}} = \sqrt{\frac{800}{0.25}} = \sqrt{3200} \approx 56.6 \text{ m/s}$

Acceleration at $v = 40$ m/s (Newton's second law at that instant): $f = 0.25(40)^2 = 400 \text{ N}, \qquad a = \frac{mg - f}{m} = \frac{800 - 400}{80} = 5.0 \text{ m/s}^2$

As the speed rises toward $v_T$, the drag approaches the weight and the acceleration falls toward zero.

🎉 Topic Complete: Friction & Inclines

You've mastered the full AP Physics C treatment of friction and inclines:

Part	Topic	Status
1	Static vs kinetic friction	✅
2	Inclined planes (no friction)	✅
3	Inclined planes with friction	✅
4	Velocity-dependent friction (calculus)	✅
5	Systems on inclines	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key Takeaway: On the AP exam, friction problems test your ability to (1) set up correct FBDs, (2) handle the static/kinetic transition, and (3) solve differential equations for velocity-dependent forces.