Conservative vs Nonconservative Force

A conservative force does work set only by where the system starts and ends, never by the route, so a round trip nets zero and the interaction can store potential energy. A nonconservative force does path-dependent work and stores none. Gravity and springs are conservative; friction is not.

AP Physics: Unit 3 (topics 3.2 Work, 3.4 Conservation of Energy). The conservative and nonconservative classification sits inside AP Physics 1 Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. It appears under Topic 3.2, learning objective 3.2.A, describe the work done on an object or system by a given force or collection of forces, as five sub-points of essential knowledge 3.2.A.1: 3.2.A.1.i says the work done by a conservative force is path-independent and depends only on the initial and final configurations, 3.2.A.1.ii says that work, or the change in the system's potential energy, is zero if the system returns to its initial configuration, 3.2.A.1.iii says potential energies are associated only with conservative forces, 3.2.A.1.iv says the work done by a nonconservative force is path-dependent, and 3.2.A.1.v names friction and air resistance as examples. EK 3.2.A.4.iii gives the dissipation equation, energy dissipated by friction equated to the friction force times the length of the path. The consequence is in Topic 3.4 at EK 3.4.C.1, energy is conserved in all interactions, and EK 3.4.C.2, which requires no nonconservative interactions within the system for its mechanical energy to be constant. The Topic 3.4 boundary statement says AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. Suggested skills are 1.B, 2.B, 2.D, 3.A and 3.B for Topic 3.2, and 1.A, 2.A, 2.C and 3.C for Topic 3.4.

The test is the path, not the loss

Ask one question: does the work depend on the route? If it depends only on where the system started and where it ended, the force is conservative. If it depends on the path taken between those two configurations, the force is nonconservative. Nothing else classifies a force.

The AP Physics 1 CED puts both halves of the test under Topic 3.2, and the wording is worth having exactly.

EK 3.2.A.1.i: the work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system.

EK 3.2.A.1.iv: the work done by a nonconservative force is path-dependent.

EK 3.2.A.1.v: examples of nonconservative forces are friction and air resistance.

Two consequences follow immediately, and the CED states both rather than leaving you to derive them.

EK 3.2.A.1.ii gives the round-trip test: the work done by a conservative force on a system, or the change in the potential energy of the system, will be zero if the system returns to its initial configuration. Take a book to the shelf and back to the table and gravity has done exactly zero net work. Push a crate across the floor and back and friction has not.

EK 3.2.A.1.iii gives the bookkeeping consequence: potential energies are associated only with conservative forces. That is why there is a $U_g$ and a $U_s$ and there is no friction potential energy. A potential energy has to attach one number to a configuration, and a path-dependent force cannot supply one, because the answer would depend on how the system got there.

The misreading to head off now: conservative does not mean it conserves the object's energy, and nonconservative does not mean energy disappears. EK 3.4.C.1 says energy is conserved in all interactions, with no exception for friction. What a nonconservative force does is move energy out of the mechanical account into thermal energy and sound, and the Topic 3.4 boundary statement says AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. The total is still conserved. The name refers to the mechanical energy, not to energy.

Conservative vs nonconservative, side by side

Question you are asking	Conservative	Nonconservative
Work depends on	The initial and final configurations only	The path taken
Work over a closed round trip	Exactly zero	Not zero, and always negative for friction
Has a potential energy	Yes	No, ever (EK 3.2.A.1.iii)
What the work depends on geometrically	The change in position, such as $\Delta y$	The length of the path traveled
Effect on the system's mechanical energy	Moves it between $K$ and $U$	Removes it from the mechanical account
Where the energy goes	Nowhere, it stays mechanical	Thermal energy and sound
Can the work be recovered by reversing the motion	Yes, in full	No
Can the work be negative	Yes, routinely	Yes, and friction's always is
AP Physics 1 examples	Gravitational force, ideal spring force	Friction, air resistance
Mechanical energy constant	Yes, if these are the only interactions inside the system (EK 3.4.C.2)	No
CED essential knowledge	3.2.A.1.i, 3.2.A.1.ii, 3.2.A.1.iii	3.2.A.1.iv, 3.2.A.1.v, 3.2.A.4.iii

Three rows deserve a second look.

The "can the work be negative" row is the same in both columns, and that is the point. A negative sign tells you the force opposed the displacement. It tells you nothing about the classification. Gravity does negative work on every object that rises, and gravity is conservative. Worked example three below throws a ball straight up so you can watch gravity do $-12.25$ J and then hand all of it back.

The "geometrically" row is the fastest practical test there is. Gravity's work near a planet is $-mg\Delta y$ : it wants a height change and does not care how far you walked. Friction's work is $-F_f d$ where $d$ is the distance traveled along the surface: it wants a path length and does not care about your height. Ask which of those two a given force is reading off the motion and you have the answer.

The "has a potential energy" row is a hard yes or no. Not "sometimes", not "approximately". If a force has a potential energy, it is conservative, and if you cannot write a potential energy for it, it is not.

The case that separates them: two routes, same endpoints

Put a block on a horizontal table and move it from point A to point B, which are $1.50$ m apart. Do it twice, along two different routes, with $\mu_k = 0.25$ and a mass of $1.2$ kg.

	Route 1: straight, $1.50$ m	Route 2: bent, $0.90$ m then $1.20$ m
Straight-line separation of A and B	$1.50$ m	$1.50$ m
Distance actually traveled	$1.50$ m	$2.10$ m
Height change	$0$	$0$
Work done by gravity	$0$	$0$
Work done by friction	$-4.41$ J	$-6.17$ J

The two legs of route 2 are perpendicular, so $\sqrt{0.90^2 + 1.20^2} = 1.50$ m puts B in the same place both times. Gravity returns the same number for both routes. Friction does not, and it is larger on the longer route by exactly the ratio of the path lengths, $2.10/1.50 = 1.40$ . That single table is the definition made visible.

Now close the loop. Slide the block from A to B and back to A along route 1.

Gravity: the system has returned to its initial configuration, so by EK 3.2.A.1.ii the work is zero. It was zero on each leg here too, since the height never changed.
Friction: the block traveled $3.00$ m in total, so friction did $-8.82$ J. Reversing the motion did not give anything back; it took twice as much.

That asymmetry is the whole reason for two categories. A conservative force lets you forget the journey and look up the endpoints. A nonconservative force makes you measure the journey.

Worked example one runs these numbers in full, and worked example two repeats the experiment with a height change so gravity has something to do.

How friction's work is computed, and why it is not a potential energy

The CED gives friction its own accounting line. EK 3.2.A.4.iii: the energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted, with the relevant equation

\Delta E_{\text{mech}} = F_f d \cos\theta

Read the symbols carefully, because this equation looks like the ordinary work formula and one symbol means something different. In $W = F_{\parallel}d = Fd\cos\theta$ from EK 3.2.A.3.i, $d$ is the displacement of the point of application. Here, EK 3.2.A.4.iii says in words that $d$ is the length of the path. On a straight one-way slide those two agree. On the bent route above they do not, and the path length is the one that gives the right dissipation.

That difference is exactly why friction has no potential energy. Path length is not a function of position. Two arrangements of the same objects can be identical and have arrived by paths of different lengths, so there is no number to attach to the arrangement.

The friction magnitude itself comes from Unit 2. EK 2.7.A.2 gives kinetic friction as $\lvert \vec{F}_{f,k} \rvert = \lvert \mu_k \vec{F}_N \rvert$ , and the AP Physics 1 sheet prints the general relation $\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$ . So computing a friction loss is a two-step job: find $F_N$ from Newton's second law perpendicular to the surface, then multiply by $\mu_k$ and by the path length.

One thing to know about what the sheet does and does not carry: $\Delta E_{\text{mech}} = F_f d\cos\theta$ is listed as a relevant equation under EK 3.2.A.4.iii in the CED, and it is not printed on the AP Physics 1 equation sheet. Its translational energy lines are $K = \frac{1}{2}mv^2$ , $W = F_{\parallel}d = Fd\cos\theta$ , $\Delta K = \sum W_i = \sum F_{\parallel, i}d_i$ , $U_s = \frac{1}{2}k(\Delta x)^2$ , $U_G = -\frac{Gm_1m_2}{r}$ and $\Delta U_g = mg\Delta y$ , and its rotational block adds $K = \frac{1}{2}I\omega^2$ and $W = \tau\Delta\theta$ . Every one of those either is a kinetic energy, is a work, or is a potential energy for a conservative interaction. There is no printed line for a dissipation, which is consistent: the sheet has no place to put one, because a nonconservative force has no potential energy.

Which AP Physics 1 forces fall where

The CED names its nonconservative examples explicitly and identifies its conservative forces by giving them potential energies. Here is what it supports, and where it is silent.

Conservative in AP Physics 1. The gravitational force and the ideal spring force. The CED does not label either with the word conservative in a definition, but it assigns each of them a potential energy: EK 3.3.A.4.i gives $U_s = \frac{1}{2}k(\Delta x)^2$ for an ideal spring, EK 3.3.A.4.ii gives $U_g = -\frac{Gm_1m_2}{r}$ for two approximately spherical masses, and EK 3.3.A.4.iii gives the near-surface approximation $\Delta U_g = mg\Delta y$ . Combine that with EK 3.2.A.1.iii, potential energies are associated only with conservative forces, and the classification follows.

Nonconservative in AP Physics 1. Friction and air resistance, named in EK 3.2.A.1.v. Drag is the general name for the resistive force from a fluid; the CED's phrase is air resistance.

Forces the CED puts in neither list. The normal force and tension get no potential energy and no nonconservative label. That is not an oversight, and you should not invent a category for them. What matters for a calculation is how much work they do, and in the standard setups the answer is often zero:

The normal force is perpendicular to the surface by EK 2.7.A.2.ii, and an object sliding along a surface moves parallel to it, so the normal force does no work on that motion.
An ideal string does not stretch, by EK 2.3.A.3.i, so tension does no net work on the objects it connects in a standard two-body pulley system; it transfers energy from one to the other.

So when a question asks whether the mechanical energy of a system is constant, EK 3.4.C.2 is the test to run, and it names exactly one disqualifier: nonconservative interactions inside the system. A normal force or a tension doing zero work does not disqualify anything.

One further place the vocabulary appears, outside Unit 3. EK 4.4.A.4 says that in an inelastic collision some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy. That is the same idea in Unit 4 clothing, and it is why the elastic vs inelastic collision test works at all.

When you cannot tell them apart, and why that lulls you

Three situations hide the distinction, and all three are common in the first weeks of an energy unit.

One straight path in one direction. Compute the work done by gravity and by friction on a block sliding once down a single ramp. Both come out as a force times a distance, both are a single number, and nothing in the arithmetic reveals that one of them would change if you rerouted the block. The distinction only shows up when there are two routes or a return trip, which is why worked example two below uses two ramps rather than one.

A problem that never returns anything to its starting configuration. Almost every introductory problem runs one way: released from rest, slides to the bottom, stops. Path independence has nothing to bite on, so you can substitute $-F_f d$ and $-mg\Delta y$ into the same equation for weeks without noticing they are different kinds of quantity.

Friction on a fixed geometry. If the geometry is fixed, friction's loss is a fixed number, and a fixed number behaves like a potential energy in your algebra. Change the geometry, or offer two paths, and the resemblance ends.

There is also a genuine coincidence worth naming rather than mistaking. On a horizontal surface, gravity's work is zero on every path, which is path-independent in the strongest possible sense. That is not a special property of horizontal surfaces; it is $\Delta y = 0$ in $-mg\Delta y$ , and it is the reason worked example one has to look at friction to see any difference at all.

The practical reflex: when a question mentions two routes, a longer way round, a return trip, or a loop, sort the forces into the two columns before writing a single equation. When a question mentions none of those, you can compute both works the same way and the classification will not be tested.

Where the confusion costs a mark

These are specific scoring errors.

Writing conservation of mechanical energy in a problem with friction. EK 3.4.C.2 requires no nonconservative interactions inside the system. With friction present, $K_i + U_i = K_f + U_f$ is void, and you need $K_i + U_i = K_f + U_f + \lvert \Delta E_{\text{mech}} \rvert$ with the dissipation term written in. This is the expensive one, because it produces a clean number that is too large.
Calling gravity nonconservative because it does negative work. Sign is not the test. Path dependence is. Gravity does negative work on everything that rises and is conservative throughout.
Inventing a potential energy for friction. EK 3.2.A.1.iii forbids it. There is no $U_f$ .
Using the displacement instead of the path length in a friction loss. EK 3.2.A.4.iii says the length of the path over which the force is exerted. On the bent route in the table above, using the $1.50$ m displacement instead of the $2.10$ m path underestimates the loss by about $29$ percent.
Saying energy is not conserved when friction acts. EK 3.4.C.1 says energy is conserved in all interactions. Say that mechanical energy was dissipated as thermal energy and sound, which is the language the Topic 3.4 boundary statement uses.
Assuming a round trip returns an object to its original speed when air resistance acts. Worked example three has a ball that leaves at $7.0$ m/s and comes back at $6.08$ m/s once air resistance takes $3.0$ J over the round trip.
Forgetting that the normal force changes on an incline before computing friction. $F_N = mg\cos\theta$ on a ramp, so the friction force and therefore the loss depend on the angle. The how to find normal force routine is the fix.
Treating a steeper ramp as the harder one. Worked example two takes the same crate to the same height by two ramps and the steep one costs less total work, because friction acts over a shorter path. The force required is larger; the work is smaller.
Counting a conservative force twice. Once you write $\Delta U_g$ into an equation, gravity is inside the system and must not also appear as an external work. That double count is covered in kinetic vs potential energy.

What the CED asks, and how the exam frames it

This distinction lives in AP Physics 1 Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section and estimated at about 22 to 27 class periods.

The classification itself sits under Topic 3.2, Work, learning objective 3.2.A: describe the work done on an object or system by a given force or collection of forces. Everything on this page about path dependence comes from the five sub-points of EK 3.2.A.1, which is a single essential knowledge statement about what work is, with the two force types tucked underneath it. That placement tells you how the exam treats the topic: not as a vocabulary question about which forces are conservative, but as a condition you check before choosing an energy method. Suggested skills for Topic 3.2 are 1.B, 2.B, 2.D, 3.A and 3.B.

The consequence sits under Topic 3.4, Conservation of Energy, at EK 3.4.C.2 and 3.4.C.3, and the boundary statement there is the one to memorize word for word: AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. Suggested skills for Topic 3.4 are 1.A, 2.A, 2.C and 3.C.

One limit on how far the accounting goes. A boundary statement under Topic 3.2 says AP Physics 1 only expects students to analyze the transfer of mechanical energy, as defined in Unit 3 Topic 4, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound, and that in AP Physics 2 students will also study how thermal energy can be transferred between systems through heating or cooling. So you compute how much mechanical energy left; you do not compute the temperature rise it caused. That is an AP Physics 2 question.

One of Unit 3's four essential questions in the CED is why it seems easier to carry a large box up a ramp rather than up a set of stairs, and worked example two below is the answer with numbers: the ramp reduces the force you must exert while increasing the path over which friction acts.

For the routines rather than the classification, conservation of energy has the energy accounting method with a friction term, the work-energy theorem has the force-by-force route, and static vs kinetic friction decides which coefficient applies before you compute any loss. The CED framing is on Topic 3.2 and Topic 3.4.

Two routes between the same two points, and then a round trip

A $1.2$ kg block is dragged across a horizontal table from A to B, which are $1.50$ m apart, with $\mu_k = 0.25$ between block and table. Route 1 goes straight from A to B. Route 2 goes $0.90$ m, turns through a right angle, then goes $1.20$ m, arriving at the same point B. Find the work done by gravity and by friction on each route, then find both again for a round trip A to B to A along route 1. Use $g = 9.8\ \text{m/s}^2$ .

Check that the two routes really do end at the same place, because the whole comparison rests on it. The legs of route 2 are perpendicular, so the straight-line separation is $\sqrt{(0.90)^2 + (1.20)^2} = \sqrt{0.81 + 1.44} = \sqrt{2.25} = 1.50\ \text{m}$ . Same B.
Path lengths: route 1 travels $1.50$ m, route 2 travels $0.90 + 1.20 = 2.10$ m. The displacement is identical; the distance traveled is not.
Friction force, which is the same on both routes because the surface is horizontal and the block is not accelerating vertically. $F_N = mg = (1.2)(9.8) = 11.76\ \text{N}$ , so $F_f = \mu_k F_N = (0.25)(11.76) = 2.94\ \text{N}$ , opposing the motion at every instant.
Work done by gravity. The table is horizontal, so $\Delta y = 0$ on both routes and $W_g = -mg\Delta y = 0$ on both. Path-independent, and here both values are the same number for the strongest possible reason: there was no height change to be independent of.
Work done by friction, route 1. Friction opposes the displacement at every point, so $W_f = -F_f d = -(2.94\ \text{N})(1.50\ \text{m}) = -4.41\ \text{J}$ .
Work done by friction, route 2. Friction opposes the motion along each leg separately, and EK 3.2.A.4.iii tells you to use the length of the path: $W_f = -(2.94)(2.10) = -6.17\ \text{J}$ to three significant figures.
Compare. Gravity: $0$ and $0$ , identical. Friction: $-4.41$ J and $-6.17$ J, differing by the ratio of the path lengths, $\frac{2.10}{1.50} = 1.40$ . Two forces, two routes, same endpoints, and only one of the forces noticed the change.
Now the round trip, A to B to A along route 1. Gravity: the block is back where it started, so the system has returned to its initial configuration and EK 3.2.A.1.ii gives $W_g = 0$ over the loop.
Friction on the round trip: total distance $1.50 + 1.50 = 3.00\ \text{m}$ , so $W_f = -(2.94)(3.00) = -8.82\ \text{J}$ . Not zero, and not recovered by reversing the motion: it is twice the one-way loss, because friction reversed direction along with the motion and kept opposing it.
The sign is worth one final note. Friction's work came out negative on all three journeys and it cannot come out positive for a block sliding on a stationary surface, because kinetic friction on the block always opposes the block's motion relative to that surface, by EK 2.7.A.1.i.

Gravity does $0$ J on both routes and $0$ J on the round trip, because the height never changed and the block returned to its starting configuration. Friction does $-4.41$ J on the $1.50$ m route, $-6.17$ J on the $2.10$ m route, and $-8.82$ J on the $3.00$ m round trip. The work done by gravity read the endpoints; the work done by friction read the odometer.

Same crate, same height, two ramps

A $3.0$ kg crate is pushed at constant speed from the floor up to a platform $1.20$ m high, along a ramp with $\mu_k = 0.30$ . Compare a $30^\circ$ ramp with a $60^\circ$ ramp: find the work done by gravity, the work done by friction, and the total work you must do, on each ramp. Use $g = 9.8\ \text{m/s}^2$ .

Set the convention: take the positive direction as up the slope, and treat the crate as moving at constant speed so its kinetic energy does not change and the works must sum to zero.
Work done by gravity, which needs only the height. $W_g = -mg\Delta y = -(3.0)(9.8)(1.20) = -35.28\ \text{J}$ . That is the answer for the $30^\circ$ ramp and for the $60^\circ$ ramp and for a vertical lift and for a staircase, because $\Delta y$ is $1.20$ m in every one of them. Path independence in one line.
Ramp lengths, which friction does need. $L = \frac{h}{\sin\theta}$ . At $30^\circ$ : $L = \frac{1.20}{0.5000} = 2.40\ \text{m}$ . At $60^\circ$ : $L = \frac{1.20}{0.8660} = 1.386\ \text{m}$ .
Normal forces, which also change with the angle. Perpendicular to the ramp the crate does not accelerate, so $F_N = mg\cos\theta$ . At $30^\circ$ : $F_N = (29.4)(0.8660) = 25.46\ \text{N}$ . At $60^\circ$ : $F_N = (29.4)(0.5000) = 14.70\ \text{N}$ .
Friction forces and their work. At $30^\circ$ : $F_f = (0.30)(25.46) = 7.638\ \text{N}$ , so $W_f = -(7.638)(2.40) = -18.3\ \text{J}$ . At $60^\circ$ : $F_f = (0.30)(14.70) = 4.410\ \text{N}$ , so $W_f = -(4.410)(1.386) = -6.11\ \text{J}$ .
Total work you must do, since the crate's kinetic energy does not change and your push, gravity and friction are the only forces doing work: $W_{\text{you}} = 35.28 + 18.3 = 53.6\ \text{J}$ on the $30^\circ$ ramp, and $35.28 + 6.11 = 41.4\ \text{J}$ on the $60^\circ$ ramp.
Read the comparison. Gravity charged $35.28$ J on both ramps, to the digit. Friction charged three times as much on the shallow ramp as on the steep one, and the difference went straight into the total. The conservative force gave one number for both geometries; the nonconservative force gave two.
The general form is worth extracting, because it explains the pattern rather than just reporting it. $W_f = -\mu_k mg\cos\theta \cdot \frac{h}{\sin\theta} = -\mu_k mgh\frac{\cos\theta}{\sin\theta}$ . The height and the coefficient are fixed, so the loss falls as the ramp gets steeper, reaching zero for a vertical lift where there is no surface to rub along.
This is the CED's Unit 3 essential question with numbers on it. A shallow ramp is easier because the force you push with is smaller: at $30^\circ$ you push with $\frac{53.6}{2.40} = 22.3$ N against $\frac{41.4}{1.386} = 29.9$ N at $60^\circ$ . It is not cheaper. You pay $53.6$ J instead of $41.4$ J for the same $1.20$ m of height, and the extra $12.2$ J became thermal energy in the ramp and the crate.

Gravity does $-35.28$ J on both ramps, set entirely by the $1.20$ m height. Friction does $-18.3$ J on the $30^\circ$ ramp and $-6.11$ J on the $60^\circ$ ramp, so the total work required is $53.6$ J and $41.4$ J. The shallow ramp needs less force, $22.3$ N against $29.9$ N, and more total work, which is the trade a ramp actually makes.

A conservative force doing negative work, and a round trip that does not close

A $0.50$ kg ball is thrown straight up at $7.0$ m/s and returns to the height it was launched from. Part (a): ignoring air resistance, find the maximum height, the work done by gravity going up, the work done by gravity coming down, and the speed on return. Part (b): if air resistance removes $1.5$ J on the way up and $1.5$ J on the way down, find the speed on return. Take upward as positive and use $g = 9.8\ \text{m/s}^2$ .

Kinetic energy at launch: $K_i = \frac{1}{2}(0.50)(7.0)^2 = \frac{1}{2}(0.50)(49) = 12.25\ \text{J}$ .
Part (a), maximum height. With no air resistance the only force is gravity, so $K$ falls to zero exactly when gravity has done $-12.25$ J of work: $mgh = 12.25$ , giving $h = \frac{12.25}{(0.50)(9.8)} = \frac{12.25}{4.9} = 2.5\ \text{m}$ .
Work done by gravity going up. The force is down, the displacement is up, so $\cos\theta = -1$ and $W_g = -mgh = -(0.50)(9.8)(2.5) = -12.25\ \text{J}$ . Negative work, done by a conservative force, and that is entirely normal.
Work done by gravity coming down. Force and displacement now agree, so $W_g = +12.25\ \text{J}$ . Over the round trip the total is $-12.25 + 12.25 = 0$ , which is EK 3.2.A.1.ii holding: the ball has returned to its initial configuration.
Speed on return, part (a). All $12.25$ J is back in kinetic energy, so $v = \sqrt{\frac{2(12.25)}{0.50}} = \sqrt{49} = 7.0\ \text{m/s}$ , the launch speed exactly. Gravity took the energy and gave every joule back.
Part (b), with air resistance. Air resistance is nonconservative by EK 3.2.A.1.v, and it opposes the motion in both directions, so it does negative work on both legs and the two losses add rather than cancel: $-1.5 + (-1.5) = -3.0\ \text{J}$ over the round trip.
Kinetic energy on return: $K_f = 12.25 - 3.0 = 9.25\ \text{J}$ , so $v = \sqrt{\frac{2(9.25)}{0.50}} = \sqrt{37} = 6.08\ \text{m/s}$ to three significant figures.
Compare the two round trips. Gravity's contribution was $0$ J in both parts and cannot be anything else on a closed path. Air resistance's contribution was $-3.0$ J, and the observable consequence is a return speed of $6.08$ m/s instead of $7.0$ m/s. The measurable signature of a nonconservative force is that a closed path does not return you to your starting state.
A note on part (b) that is easy to get wrong: the maximum height in part (b) is lower than $2.5$ m, because air resistance took energy on the way up as well. This example gives you the round-trip loss and asks only for the return speed, so you do not need that height, but do not carry $2.5$ m into part (b) as though it still applied.

Part (a): maximum height $2.5$ m, gravity does $-12.25$ J going up and $+12.25$ J coming down for a round-trip total of zero, and the ball returns at exactly $7.0$ m/s. Part (b): air resistance takes $3.0$ J over the round trip, leaving $9.25$ J, so the ball returns at $6.08$ m/s. Negative work does not make a force nonconservative; a closed path that fails to close does.

Frequently asked questions

What is the difference between a conservative and a nonconservative force?

A conservative force does work that depends only on the initial and final configurations of the system, never on the route taken between them, so the work over any closed round trip is exactly zero and the interaction can store potential energy. A nonconservative force does work that depends on the path taken, so a round trip does not net zero and no potential energy can be defined for it. The AP Physics 1 CED states both halves at essential knowledge 3.2.A.1.i and 3.2.A.1.iv, and names friction and air resistance as the nonconservative examples at 3.2.A.1.v.

Is friction a conservative or nonconservative force?

Nonconservative. The AP Physics 1 CED names it directly at essential knowledge 3.2.A.1.v, alongside air resistance. The reason is that the work friction does depends on the length of the path traveled rather than on where the object ended up, so two routes between the same two points give different answers and a round trip loses energy rather than returning to zero. Because the work is path-dependent, there is no friction potential energy, which essential knowledge 3.2.A.1.iii rules out by restricting potential energies to conservative forces. Friction moves mechanical energy into thermal energy and sound.

Which forces are conservative in AP Physics 1?

The gravitational force and the ideal spring force. The CED identifies them by giving each a potential energy: essential knowledge 3.3.A.4.i gives the elastic potential energy of an ideal spring, and 3.3.A.4.ii and 3.3.A.4.iii give the general and near-surface gravitational potential energies. Since essential knowledge 3.2.A.1.iii says potential energies are associated only with conservative forces, having one settles the classification. The forces the CED names as nonconservative are friction and air resistance. The normal force and tension are given neither a potential energy nor a nonconservative label, and in the standard setups they do no work at all.

Does a nonconservative force violate conservation of energy?

No. Energy is conserved in every interaction, and the AP Physics 1 CED states that outright at essential knowledge 3.4.C.1. What a nonconservative force does is take energy out of the mechanical account, the sum of kinetic and potential energies, and put it into thermal energy and sound. The Topic 3.4 boundary statement says AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces. So the correct sentence is that mechanical energy was not conserved, or that mechanical energy was dissipated, never that energy disappeared.

Why do only conservative forces have potential energy?

Because a potential energy has to assign a single number to each configuration of the system, and only a path-independent force allows that. Essential knowledge 3.2.A.1.i says the work done by a conservative force depends only on the initial and final configurations, so the energy stored can be read off the arrangement alone. Essential knowledge 3.2.A.1.iv says the work done by a nonconservative force is path-dependent, so two systems in identical arrangements could have had different amounts of energy removed depending on how they got there, and no single number fits. Essential knowledge 3.2.A.1.iii states the conclusion: potential energies are associated only with conservative forces.

Does negative work mean a force is nonconservative?

No, and this is a common misreading. Negative work simply means the force had a component opposite to the displacement. Gravity does negative work on every object that rises, and gravity is conservative: throw a ball up at 7.0 meters per second and gravity does about minus 12.25 joules on the way up, then exactly plus 12.25 joules on the way down, for a round trip total of zero. The test for a conservative force is whether the work over a closed path is zero, not whether the work on one leg is negative. Friction's work is always negative for an object sliding on a stationary surface, but that is a separate fact from its path dependence.

How do you calculate the energy lost to friction?

Multiply the friction force by the length of the path over which it acted. The AP Physics 1 CED states this at essential knowledge 3.2.A.4.iii: the energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted. Find the friction force first, from the coefficient of kinetic friction multiplied by the normal force, and remember the normal force is not simply mg on an incline. Use the path length, not the straight-line displacement, because those differ whenever the route bends or doubles back, and using the displacement understates the loss.