Friction vs Drag: What Is the Difference?

Friction acts between two solid surfaces and is modelled as a coefficient times the normal force, with no speed in it. Drag acts in a fluid and grows with speed, so it changes as the object moves. That is why friction brings a slider to a stop while drag produces a terminal velocity.

AP Physics: Unit 2 (topics 2.7 Kinetic and Static Friction, 2.9 Resistive Forces). Friction and drag split the two mechanics courses. Friction is Topic 2.7 in both AP Physics 1 and AP Physics C: Mechanics, worded identically in the two CEDs: LO 2.7.A on kinetic friction carries EK 2.7.A.1 through 2.7.A.2.ii, including 2.7.A.1.ii, that the force of friction between two surfaces does not depend on the size of the surface area of contact, and 2.7.A.2, that the kinetic friction magnitude is the product of the normal force and the coefficient of kinetic friction; LO 2.7.B on static friction carries EK 2.7.B.1 through 2.7.B.3, ending with the statement that the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces. Suggested skills for AP Physics 1 Topic 2.7 are 1.C, 2.B, 2.C and 3.B, and the topic has no boundary statement in either course. Drag appears only in AP Physics C: Mechanics, as Topic 2.9, Resistive Forces, which AP Physics 1 does not have. LO 2.9.A carries EK 2.9.A.1, defining a resistive force as a velocity-dependent force in the opposite direction of an object's velocity with F_r = -kv as the example; EK 2.9.A.2 and its three sub-points, which set up the differential equation, solve it by separation of variables, and state that position, velocity and acceleration are exponential with asymptotes determined by the initial conditions and the forces exerted on the object; and EK 2.9.A.3, defining terminal velocity as the maximum speed achieved under a constant force and an opposing resistive force, reached when the net force is zero. Suggested skills for Topic 2.9 are 1.B, 2.A, 2.C, 3.A and 3.C. AP Physics 1 Unit 2 is weighted at 18 to 23 percent of its multiple-choice section over about 22 to 27 class periods; C: Mechanics Unit 2 is weighted at 20 to 25 percent. Both mechanics sheets print one friction line, |F_f| <= |mu F_N|, and neither prints any resistive force or terminal velocity equation. Both courses' exam conventions state that air resistance is assumed negligible unless otherwise stated.

One model has a speed in it and the other does not

Both oppose motion, both remove kinetic energy, and both are listed together in the AP Physics 1 CED as examples of nonconservative forces at EK 3.2.A.1.v. The difference is what each force is a function of.

The AP friction model is $\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$ . Read the right-hand side: a coefficient set by the two materials, and the normal force. No velocity. Push the block faster and, in this model, the friction force does not change.

The AP resistive-force model is $\vec{F}_r = -k\vec{v}$ . Read that one: the velocity is the whole content of it. AP Physics C: Mechanics EK 2.9.A.1 defines a resistive force as a velocity-dependent force in the opposite direction of an object's velocity, and gives that expression as its example.

Everything else on this page follows from those two lines. A force that does not change with speed gives a constant acceleration, a straight-line velocity-time graph and a definite stopping time. A force that grows with speed gives a changing acceleration, an exponential velocity-time graph and an asymptote. The two produce different shapes of motion, not just different numbers.

The terminology worth fixing before going further: drag and air resistance are the physical cases, and "resistive force" is the CED's category name for any velocity-dependent opposing force. The drag force and resistive force entries define each; this page is about how either one differs from friction.

Friction vs drag, side by side

Question you are asking	Friction	Drag
What it acts between	Two solid surfaces in contact	An object and the fluid it moves through
AP model	$\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$	$\vec{F}_r = -k\vec{v}$
Depends on speed	No, there is no $v$ in the model	Yes, that is the definition
Depends on the normal force	Yes, directly	No, the surface is not involved
Depends on contact area	No, EK 2.7.A.1.ii says so outright	Not in the AP model, which has one constant $k$
The material constant	$\mu$ , dimensionless	$k$ , with units of $\text{kg/s}$
Direction	Opposite the relative motion of the surfaces	Opposite the object's velocity
Can it be zero while the object moves	Only if $\mu$ or $F_N$ is zero	Only at $v = 0$
Resulting acceleration	Constant	Changes continuously
Velocity-time graph	A straight line	An exponential curve with an asymptote
What the motion ends at	A dead stop, at a definite time	A terminal speed, approached and never reached
Maths needed	Algebra	A differential equation, EK 2.9.A.2
AP Physics 1	Topic 2.7, required	Not in the course
AP Physics C: Mechanics	Topic 2.7, required	Topic 2.9, required
Printed on any AP equation sheet	Yes, one line	No, on none of the four

Three rows are worth unpacking.

The units row is the distinction in symbols. $\mu$ is a ratio of two forces, so it is a pure number and cannot introduce a speed dependence. $k$ has units of $\text{kg/s}$ , which is what you get from dividing a newton by a metre per second, so a speed is built into it before any physics is discussed.

The contact-area row is a CED statement, not folklore. EK 2.7.A.1.ii reads: the force of friction between two surfaces does not depend on the size of the surface area of contact. The AP Physics 2 CED's appendix on the object model makes the same point at more length, arguing that two blocks can still be treated as objects because the amount of area in contact does not change the force of friction between them. Drag in the real world depends heavily on frontal area, but the AP model buries all of that in the single constant $k$ , so the CED never asks you to vary it.

The speed row needs one careful qualification. The CED does not contain a sentence saying "friction is independent of speed." What it gives is a model with no velocity in it: EK 2.7.A.2 states the magnitude of the kinetic friction force as the product of the normal force and the coefficient of kinetic friction, and EK 2.7.A.2.i says that coefficient depends on the material properties of the surfaces in contact. Nothing there can change when the block speeds up. So within the AP model kinetic friction does not vary with speed, and that is the safe way to state it.

The case that separates them: two ways to slow down

Give one object two different opposing forces and watch the motion diverge.

Under friction, an object stops. Worked example one slides a $3.0$ kg block at $9.0$ m/s across a floor with $\mu_k = 0.28$ . The friction force is $8.23$ N at $9.0$ m/s, $8.23$ N at $4.0$ m/s and $8.23$ N at $0.1$ m/s. A constant force gives a constant deceleration of $2.744\ \text{m/s}^2$ , and the block stops dead after $3.28$ s having travelled $14.8$ m. The velocity-time graph is a straight line hitting the axis.

Under a resistive force, an object never quite stops. Worked example three takes the same $9.0$ m/s start with $\vec{F}_r = -k\vec{v}$ . As the object slows, the force shrinks in exact proportion, so the deceleration shrinks with it. The velocity decays as $v = v_0 e^{-kt/m}$ : down to $1.22$ m/s after $1.0$ s, $0.16$ m/s after $2.0$ s, and never to zero. The graph flattens onto the axis instead of crossing it.

Add gravity and the divergence goes the other way. A block sliding down a slope under friction either accelerates forever or moves at whatever the force balance dictates from the start. An object falling under drag accelerates hard at first, then less and less as the drag builds, until at $v = mg/k$ the two forces match and the acceleration reaches zero. That speed is terminal velocity, and EK 2.9.A.3 defines it as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force that are exerted on the object in opposite directions, with the terminal condition reached when the net force exerted on the object is zero.

Friction has no analogue of that. A friction force cannot grow to meet a driving force it is not already matching, because nothing in $\mu F_N$ responds to how fast the object is going. Static friction does adjust itself, but it adjusts to the applied force, not to the speed, and it stops adjusting at $\mu_s F_N$ .

What each course actually asks for

This is the part most explanations get wrong by answering as general physics rather than as AP scope, so here is what is in each CED.

AP Physics 1 has friction and no drag. Topic 2.7, Kinetic and Static Friction, sits in Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. It carries LO 2.7.A on kinetic friction and LO 2.7.B on static friction, and that is the entire treatment of resistive contact forces in the course. The word drag appears in the CED only in the appendix list of modeling assumptions, where frictional and drag forces are listed among the things students may assume negligible unless otherwise stated. Air resistance appears twice: once as an example of a nonconservative force at EK 3.2.A.1.v, and once in the exam conventions box, which states that air resistance is assumed to be negligible unless otherwise stated.

AP Physics C: Mechanics has both. Its Topic 2.7 is word for word the same friction treatment, down to the same essential knowledge numbering. Then it adds Topic 2.9, Resistive Forces, which AP Physics 1 does not have, and that topic is where the difference lives. LO 2.9.A asks you to describe the motion of an object subject to a resistive force, and it carries three pieces of essential knowledge:

EK 2.9.A.1 defines the resistive force as velocity-dependent and opposite the velocity, with $\vec{F}_r = -k\vec{v}$ as the example.
EK 2.9.A.2 says applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity. Sub-point 2.9.A.2.i solves it by separation of variables, integrating over the proper limits of integration. Sub-point 2.9.A.2.ii gets acceleration or position from initial conditions and calculus once a function for velocity is determined. Sub-point 2.9.A.2.iii states that position, velocity and acceleration as functions of time under $\vec{F}_r = -k\vec{v}$ are exponential and have asymptotes determined by the initial conditions of the object and the forces exerted on it.
EK 2.9.A.3 defines terminal velocity.

Suggested skills for Topic 2.9 are 1.B, 2.A, 2.C, 3.A and 3.C. The topic carries no boundary statement.

Neither Physics 2 nor Physics C: E&M covers either force, since neither course contains a mechanics unit.

So the scope answer is specific: friction is common ground and identically worded in the two mechanics courses, and drag is the one that splits them, appearing only in C: Mechanics and only as a calculus problem. If you are sitting AP Physics 1 and a question mentions air resistance at all, it is almost certainly telling you to ignore it.

Neither is on the equation sheet in the form you expect

Two claims here, both checked against the printed appendices rather than recalled.

The friction line is an inequality, and there is only one of it. The AP Physics 1 Table of Information prints exactly $\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$ in its mechanics block, with $\mu$ defined in the symbol key as the coefficient of friction and no subscript. The AP Physics C: Mechanics sheet prints the identical line. So the sheet does not distinguish static from kinetic for you, and it does not hand you the kinetic equality $F_{f,k} = \mu_k F_N$ as a separate line: that is EK 2.7.A.2 in the CED, and you supply the equality yourself once you have decided the surfaces are sliding. Which one applies is the subject of the static vs kinetic friction guide, and that decision is not restated here.

No AP equation sheet prints a drag or resistive force equation at all. The AP Physics C: Mechanics equations appendix carries the full mechanics list, from the kinematic equations through the rotational block, and $\vec{F}_r = -k\vec{v}$ is not among them. Neither is a terminal velocity expression. So the form of the resistive force arrives with the question, or from EK 2.9.A.1, and $v_t = mg/k$ is something you derive on the page by setting the net force to zero.

That asymmetry is a fair summary of the two forces' status in the courses. Friction gets a printed line and a coefficient you may be asked to measure. Drag gets a topic, a differential equation and nothing printed.

Where the confusion costs a mark

Using $\mu$ for a fluid. There is no coefficient of friction between a ball and the air. A fluid problem needs a resistive-force model, and if the question has not supplied one and you are sitting AP Physics 1, the intended answer treats the air as absent.
Assuming a friction force falls off as the object slows. It does not, in this model. That assumption turns a straight velocity-time graph into a curve and makes the stopping distance come out too large.
Assuming drag is constant. The opposite error, and the more expensive one, because it makes the motion look like constant acceleration and invites the kinematic equations. The three kinematic equations require constant acceleration and are invalid the moment the force depends on velocity, which is exactly why C: Mechanics reaches for calculus at Topic 2.9.
Reaching for terminal velocity in AP Physics 1. It is not in that course. A Physics 1 answer that appeals to terminal velocity is answering a question the exam did not ask, and usually one it explicitly assumed away.
Saying a falling object stops accelerating because the air pushes it up harder than gravity. At terminal velocity the two are equal, not unequal. The net force is zero and the object keeps moving at constant velocity. This is a Newton's first law state, not a stopping.
Sketching a velocity-time graph that kinks onto a horizontal line. EK 2.9.A.2.iii makes the approach exponential, so the curve flattens toward the asymptote and never meets it. A corner in the graph is a wrong answer even if the final value is right.
Claiming friction depends on area because a wider tyre grips better. EK 2.7.A.1.ii is explicit that the force of friction does not depend on the size of the surface area of contact. Whatever the real tyre is doing, it is outside this model.
Mixing the two constants. $\mu$ is dimensionless and multiplies a force; $k$ carries units of $\text{kg/s}$ and multiplies a velocity. A number quoted with no units is a $\mu$ ; a number in $\text{kg/s}$ is a $k$ .

For the friction routines themselves, how to find the coefficient of friction has the measurement methods and the friction calculator checks numbers.

When they behave alike, and why that lulls you

Three situations make the two look interchangeable, and all three are common.

Both are nonconservative, and the CED names them in the same breath. EK 3.2.A.1.v lists friction and air resistance as its examples of nonconservative forces, and EK 3.2.A.1.iv says the work done by a nonconservative force is path-dependent. So for an energy-accounting question, the two behave identically: both remove mechanical energy, both make the path matter, and neither has a potential energy associated with it, since EK 3.2.A.1.iii reserves potential energies for conservative forces.

Both are zero-net-force states at steady speed. A crate dragged at constant velocity has the pull balancing kinetic friction. A falling object at terminal velocity has gravity balancing the drag. Both are $\sum \vec{F} = 0$ , both are Newton's first law, and both give an acceleration of zero. If the question only asks for the force balance at that instant, the two are the same problem.

Both point opposite the motion. EK 2.7.A.1.i puts kinetic friction opposite the motion of each surface relative to the other, and EK 2.9.A.1 puts the resistive force opposite the object's velocity. Neither ever speeds the object up along its direction of travel, so a free-body diagram treats them the same way.

The distinction reappears the moment the question asks what happens next rather than what is happening now. Any question about how the force changes as the motion changes separates them immediately, and that is the shape of nearly every AP question that involves drag at all: sketch the graph, describe the approach to terminal velocity, find the velocity as a function of time. None of those has an answer for friction, because for friction the force is a constant and the answer is a line.

What the CED asks, and how the exam frames it

Friction, AP Physics 1 and AP Physics C: Mechanics, Topic 2.7. LO 2.7.A, describe kinetic friction between two surfaces, carries EK 2.7.A.1 (kinetic friction occurs when two surfaces in contact move relative to each other), 2.7.A.1.i (direction opposite the relative motion of the surfaces), 2.7.A.1.ii (no dependence on contact area), 2.7.A.2 (the magnitude is the product of the normal force and the coefficient of kinetic friction), 2.7.A.2.i (the coefficient depends on the material properties of the surfaces in contact) and 2.7.A.2.ii (the definition of normal force). LO 2.7.B, describe static friction, carries EK 2.7.B.1, 2.7.B.2 with the relevant equation $\lvert \vec{F}_{f,s} \rvert \leq \lvert \mu_s \vec{F}_n \rvert$ , its two sub-points on slipping and on the existence of a maximum, and EK 2.7.B.3, that the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces. Suggested skills for AP Physics 1 Topic 2.7 are 1.C, 2.B, 2.C and 3.B. The topic carries no boundary statement in either course.

Drag, AP Physics C: Mechanics only, Topic 2.9, Resistive Forces. LO 2.9.A with EK 2.9.A.1 through 2.9.A.3 as set out above. Unit 2 of C: Mechanics is weighted at 20 to 25 percent of that exam's multiple-choice section, so a topic that exists in only one of the two mechanics courses still sits inside a heavily weighted unit.

The exam conventions differ between the two courses in a way worth noticing. The AP Physics 1 conventions box lists four assumptions: inertial frames, negligible air resistance, ideal springs and strings, and ideal fluids in completely filled pipes. The AP Physics C: Mechanics box lists three, dropping the fluids line, since that course has no fluids unit. Both keep the air resistance line, so even in the course that teaches drag, a question has to switch it on explicitly.

The CED framing is at Topic 2.7 in AP Physics 1 and Topic 2.9 in C: Mechanics. For problems, the friction practice set works the contact-force side.

A block stopped by friction: the force that never changes

A $3.0$ kg block slides across a level floor at $9.0$ m/s. The coefficient of kinetic friction between block and floor is $\mu_k = 0.28$ . Find the friction force, the acceleration, the time to stop and the distance travelled. Then find the friction force again at $4.0$ m/s and at $0.10$ m/s. Take the direction of motion as positive and use $g = 9.8\ \text{m/s}^2$ .

Normal force first. The floor is level and nothing pushes down or lifts up, so $F_N = mg = (3.0)(9.8) = 29.4\ \text{N}$ .
Friction force, from EK 2.7.A.2 with the surfaces sliding, so the inequality becomes an equality: $F_f = \mu_k F_N = (0.28)(29.4) = 8.232\ \text{N}$ , directed opposite the motion, so $-8.232\ \text{N}$ under the chosen sign convention.
Acceleration, from $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ : $a = \frac{-8.232}{3.0} = -2.744\ \text{m/s}^2$ . Every quantity on the right is a constant, so the acceleration is a constant, and the kinematic equations are legal.
Time to stop, from $v = v_0 + at$ with $v = 0$ : $t = \frac{9.0}{2.744} = 3.280\ \text{s}$ . A definite, finite time.
Distance, from $v^2 = v_0^2 + 2a\Delta x$ : $\Delta x = \frac{(9.0)^2}{2(2.744)} = \frac{81.0}{5.488} = 14.76\ \text{m}$ .
Now the point of the example. At $4.0$ m/s the friction force is $\mu_k F_N = (0.28)(29.4) = 8.232\ \text{N}$ . At $0.10$ m/s it is $(0.28)(29.4) = 8.232\ \text{N}$ . The speed never entered the calculation, because there is no speed in the model. The same three digits come out at every point of the slide until the instant the block stops.
A check on the shape: constant acceleration means the velocity-time graph is a straight line from $9.0$ m/s to zero over $3.28$ s, and the area under it is $\frac{1}{2}(9.0)(3.280) = 14.76\ \text{m}$ , matching the distance found from the kinematic equation.

Friction is $8.23$ N throughout, giving $a = -2.744\ \text{m/s}^2$ , a stop after $3.28$ s and a slide of $14.8$ m. The force is the same $8.23$ N at $9.0$ m/s, $4.0$ m/s and $0.10$ m/s, which is why the deceleration is constant and the block reaches a genuine stop.

A falling object under a resistive force: the force that grows

A $0.15$ kg object falls from rest through air that exerts a resistive force $\vec{F}_r = -k\vec{v}$ with $k = 0.30\ \text{kg/s}$ . Find the terminal speed, the net force and acceleration when the object is moving at $2.0$ m/s, and the speed after $0.50$ s, $1.5$ s and $3.0$ s. Take downward as positive and use $g = 9.8\ \text{m/s}^2$ .

Weight, the constant force in EK 2.9.A.3's pairing: $mg = (0.15)(9.8) = 1.47\ \text{N}$ downward, unchanging for the whole fall.
Terminal speed. The terminal condition is zero net force, so $mg = kv_t$ and $v_t = \frac{mg}{k} = \frac{1.47}{0.30} = 4.9\ \text{m/s}$ . Note this is derived, not looked up: no AP sheet prints it.
At the instant the object is doing $2.0$ m/s, the resistive force is $kv = (0.30)(2.0) = 0.60\ \text{N}$ upward. Net force $= 1.47 - 0.60 = 0.87\ \text{N}$ downward, so $a = \frac{0.87}{0.15} = 5.8\ \text{m/s}^2$ . Already well below $9.8$ , and it was $9.8$ at the moment of release when $v$ was zero.
Compare that with the friction case. There the opposing force was $8.232$ N at every speed. Here it is $0$ N at release, $0.60$ N at $2.0$ m/s and $1.47$ N at terminal speed. The force is a readout of the speed.
Velocity as a function of time. Newton's second law gives $m\frac{dv}{dt} = mg - kv$ , the differential equation EK 2.9.A.2 names. Separating variables and integrating from rest gives $v = v_t\left(1 - e^{-kt/m}\right)$ , with the time constant $\frac{m}{k} = \frac{0.15}{0.30} = 0.50\ \text{s}$ .
At $t = 0.50$ s: $v = 4.9(1 - e^{-1}) = 4.9(0.6321) = 3.10\ \text{m/s}$ . At $t = 1.5$ s: $v = 4.9(1 - e^{-3}) = 4.9(0.9502) = 4.66\ \text{m/s}$ . At $t = 3.0$ s: $v = 4.9(1 - e^{-6}) = 4.9(0.99752) = 4.888\ \text{m/s}$ .
Read the last three numbers. The gaps to $4.9$ are $1.80$ , $0.24$ and $0.012$ m/s. Each further half second closes most of what is left and none of it closes the last bit. That is EK 2.9.A.2.iii's asymptote, and it is why the graph must be drawn flattening rather than kinking.
A sanity check on the free-fall comparison: without drag, three seconds of falling from rest would give $v = gt = 29.4\ \text{m/s}$ . With this drag the object is doing $4.89$ m/s. The whole difference is a force that grew from zero to $1.47$ N as the speed rose.

Terminal speed $v_t = 4.9$ m/s. At $2.0$ m/s the net force is $0.87$ N and the acceleration $5.8\ \text{m/s}^2$ , down from $9.8\ \text{m/s}^2$ at release. The speed reaches $3.10$ m/s at $0.50$ s, $4.66$ m/s at $1.5$ s and $4.888$ m/s at $3.0$ s, closing on $4.9$ m/s without ever arriving.

Same starting speed, two ways of slowing: 9.0 m/s under each model

A $0.15$ kg object moves at $9.0$ m/s and is slowed only by a resistive force $\vec{F}_r = -k\vec{v}$ with $k = 0.30\ \text{kg/s}$ , with no gravity component along its motion. Derive the velocity as a function of time by separation of variables, and find the speed at $0.50$ s, $1.0$ s and $2.0$ s. Compare with the $3.0$ kg block of worked example one, which started at the same $9.0$ m/s and was slowed by friction alone. Take the direction of motion as positive.

Set up the differential equation. The only force along the motion is the resistive one, so $m\frac{dv}{dt} = -kv$ , which is EK 2.9.A.2's statement that applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity.
Separate the variables, the method EK 2.9.A.2.i names: $\frac{dv}{v} = -\frac{k}{m}\,dt$ . Integrating over the proper limits, from $v_0$ to $v$ on the left and $0$ to $t$ on the right, gives $\ln\frac{v}{v_0} = -\frac{k}{m}t$ , so $v = v_0 e^{-kt/m}$ .
Time constant: $\frac{m}{k} = \frac{0.15}{0.30} = 0.50\ \text{s}$ , so the exponent is $-2.0t$ with $t$ in seconds.
At $t = 0.50$ s: $v = 9.0e^{-1} = 9.0(0.36788) = 3.31\ \text{m/s}$ . At $t = 1.0$ s: $v = 9.0e^{-2} = 9.0(0.13534) = 1.218\ \text{m/s}$ . At $t = 2.0$ s: $v = 9.0e^{-4} = 9.0(0.018316) = 0.165\ \text{m/s}$ .
At $t = 3.28$ s, the moment the friction block came to rest, this object is still moving at $9.0e^{-6.56} = 0.0127\ \text{m/s}$ . Small, and not zero, and it will not be zero at any later time either. An exponential has no root.
Now the accelerations. The friction block decelerated at a flat $2.744\ \text{m/s}^2$ from start to finish. This object decelerates at $\frac{kv}{m} = 2.0v$ , so $18.0\ \text{m/s}^2$ at the start, $6.62\ \text{m/s}^2$ at $0.50$ s and $0.33\ \text{m/s}^2$ at $2.0$ s. One number against a number that tracks the speed.
Why the kinematic equations were legal for one and not the other. The three kinematic equations of EK 1.3.A.2 assume constant acceleration. The friction block satisfies that and its stopping distance came straight out of $v^2 = v_0^2 + 2a\Delta x$ . This object does not, and applying that equation to it would be a category error, which is precisely why resistive forces sit in the calculus-based course and not in the algebra-based one.

Separation of variables gives $v = v_0 e^{-kt/m}$ with a time constant of $0.50$ s, so the speed falls to $3.31$ m/s at $0.50$ s, $1.218$ m/s at $1.0$ s and $0.165$ m/s at $2.0$ s, and is still $0.0127$ m/s at $3.28$ s, when the friction block of worked example one had already been stationary. Friction gives a constant $2.744\ \text{m/s}^2$ and a real stop; the resistive force gives a deceleration that fades with the speed and no stop at all.

Frequently asked questions

What is the difference between friction and drag?

Friction acts between two solid surfaces that are in contact, and the AP model makes it a coefficient multiplied by the normal force, with no velocity anywhere in it. Drag acts on an object moving through a fluid, and it is defined by its velocity dependence: AP Physics C: Mechanics essential knowledge 2.9.A.1 calls a resistive force a velocity-dependent force in the opposite direction of an object's velocity, with force equal to minus k times velocity as its example. The consequence is that friction gives a constant deceleration and a definite stopping time, while drag gives a deceleration that shrinks as the object slows, so the motion is exponential and approaches a limit rather than reaching a stop.

Does friction depend on speed?

Not in the AP model. The equation sheet prints the friction force as a coefficient times the normal force, and neither of those changes when the object moves faster. Essential knowledge 2.7.A.2 gives the kinetic friction magnitude as the product of the normal force and the coefficient of kinetic friction, and 2.7.A.2.i says that coefficient depends on the material properties of the surfaces in contact, so there is nothing in the model that could respond to speed. A 3.0 kilogram block on a surface with a coefficient of 0.28 feels the same 8.23 newtons at 9 metres per second as at 0.1 metres per second. Real friction does vary slightly with sliding speed, but no AP course asks about that.

Does friction depend on surface area?

No, and the CED says so directly. Essential knowledge 2.7.A.1.ii reads that the force of friction between two surfaces does not depend on the size of the surface area of contact. Only the coefficient, which is set by what the two materials are, and the normal force pressing them together enter the model. The AP Physics 2 CED makes the same point in its appendix on the object model, noting that two blocks can still be treated as objects precisely because the amount of area in contact does not change the force of friction between them. Drag in the real world does depend on frontal area, but the AP resistive force model hides all of that in a single constant.

Is drag on the AP Physics 1 exam?

No. AP Physics 1 has no topic on drag or resistive forces. Its exam conventions state that air resistance is assumed to be negligible unless otherwise stated, and its appendix list of modeling assumptions includes negligible frictional and drag forces. Air resistance appears in the course only as an example of a nonconservative force, at essential knowledge 3.2.A.1.v. Resistive forces are AP Physics C: Mechanics Topic 2.9, which is not in AP Physics 1 at all, because the treatment needs a differential equation. If an AP Physics 1 question mentions air resistance, it is almost always telling you to ignore it.

Is there a drag equation on the AP equation sheet?

No, on none of the four sheets. The AP Physics C: Mechanics equations appendix carries the full mechanics list and contains no resistive force expression and no terminal velocity expression. So the form of the drag force arrives with the question, or from essential knowledge 2.9.A.1, and the terminal speed is something you derive by setting the net force to zero. Friction fares better: both mechanics sheets print one friction line, the inequality of the friction force magnitude being less than or equal to the coefficient times the normal force, with the coefficient carrying no subscript.

Why does a falling object reach terminal velocity but a sliding block just stops?

Because only one of the two opposing forces can grow to match the driving force. Drag increases with speed, so a falling object speeds up until the drag has grown to equal the weight, at which point the net force is zero and the speed stops changing. That balance point is at weight divided by the drag constant. Friction cannot do this: its size is fixed by the coefficient and the normal force, so it either exceeds the driving force, in which case the object decelerates at a constant rate to a stop, or it does not, in which case the object keeps accelerating. There is no speed at which friction happens to become the right size.

Is air resistance the same thing as friction?

They are both forces that oppose motion and remove mechanical energy, and the AP Physics 1 CED lists them together at essential knowledge 3.2.A.1.v as its two examples of nonconservative forces. For energy accounting they behave the same way, and both point opposite the motion on a free-body diagram. They are modelled quite differently, though. Air resistance is a fluid force whose size depends on how fast the object is going, while friction is a surface force whose size depends on how hard the surfaces are pressed together. Calling air resistance a kind of friction is common in everyday speech, but on an exam it will lead you to reach for a coefficient of friction that does not exist for a fluid.