Positive vs Negative Work: What the Sign Means

Q: What does negative work mean in physics?

It means energy is being transferred out of the object or system by that force. The AP Physics 1 CED defines work at essential knowledge 3.2.A.1 as the amount of energy transferred into or out of a system by a force exerted on that system over a distance, and the sign records which of those two happened. Negative work does not mean the object moves backwards, and it does not mean the force points backwards. A rope lowering a crate at constant speed pulls upward while the crate moves down, so the angle between them is 180 degrees, the work is negative, and the crate is neither reversing nor slowing.

Q: What determines whether work is positive, negative, or zero?

Only the angle between the force and the displacement. Work equals the force magnitude times the distance times the cosine of that angle, and since a magnitude and a distance are both positive, the cosine is the only factor that can change sign. An angle under 90 degrees gives a positive cosine and positive work, exactly 90 degrees gives zero, and an angle over 90 degrees gives negative work. The AP Physics 1 CED states the three outcomes at essential knowledge 3.2.A.2: work is a scalar quantity that may be positive, negative, or zero. The angle is measured between the force and the displacement, not between the force and the horizontal or the surface.

Q: Does friction always do negative work?

Not always, though it does in the standard AP setup. When an object slides on a stationary surface, kinetic friction opposes the object's displacement, the angle is 180 degrees and the work on that object is negative. But if the surface itself is moving, the friction can act in the direction the object travels. Drop a box onto a moving conveyor belt and the friction on the box acts forward while the box moves forward, so it does positive work on the box. What is always true is that kinetic friction between two surfaces sliding across each other dissipates mechanical energy from the system as a whole, which is why the CED classes it as a nonconservative force at essential knowledge 3.2.A.1.v.

Q: Can the normal force ever do work?

Yes, whenever it is not perpendicular to the displacement. On a level floor, or on an object sliding along an incline, the normal force is perpendicular to the motion and does exactly zero work. But stand in an elevator accelerating upward and the floor pushes you up while you move up, so the angle is zero and the normal force does positive work. For a 60 kg passenger in an elevator accelerating upward at 1.2 meters per second squared through 3.0 m, the normal force is 660 N and does plus 1980 J, gravity does minus 1764 J, and the net 216 J matches the passenger's kinetic energy. The rule that the normal force does no work is a special case, not a law.

Q: Does negative work always mean the object slows down?

No. Only the net work determines the change in kinetic energy, and one force doing negative work says nothing on its own. A car accelerating up a hill has gravity doing negative work and friction from the air doing negative work while the car speeds up, because the driving force does more positive work than those two remove. Conversely, a crate lowered at constant speed has a rope doing large negative work on it while its speed never changes, since gravity is simultaneously doing an equal amount of positive work. The AP Physics 1 work-energy theorem sums the work of every force, and it is that signed total that equals the change in kinetic energy.

Q: Why does gravity do negative work on an object moving upward?

Because gravity points down and the displacement points up, so the angle between them is 180 degrees and the cosine is minus one. Physically, energy is leaving the object's motion and being stored in the configuration of the object and Earth as gravitational potential energy. The size is the weight times the height gained. On the way back down the angle becomes zero, the work becomes positive with the same magnitude, and the round trip totals zero, which is what the CED means at essential knowledge 3.2.A.1.ii when it says the work done by a conservative force is zero if the system returns to its initial configuration.

Q: Can a force be large and still do zero work?

Yes, in two situations. If the force is perpendicular to the displacement the cosine is zero, so a string whirling a ball in a horizontal circle can pull hard while transferring no energy, and the ball's speed never changes. The CED describes this at essential knowledge 3.2.A.3.ii, saying the perpendicular component can change the direction of the system's motion without changing its kinetic energy. If nothing moves, the distance is zero and the work is zero however large the force, which is why holding a heavy box still does no work on the box. A force can be essential to the force balance and irrelevant to the energy budget at the same time.

Negative work is not work done backwards. It means energy is leaving the object through that force. The sign comes entirely from the angle between the force and the displacement: under 90 degrees gives positive work, exactly 90 degrees gives zero, and over 90 degrees gives negative work.

AP Physics: Unit 3 (topics 3.2 Work). This sits in 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. Topic 3.2, Work, carries one learning objective, 3.2.A, describe the work done on an object or system by a given force or collection of forces. The sign material runs through its essential knowledge: EK 3.2.A.1 defines work as energy transferred into or out of a system; EK 3.2.A.1.ii says a conservative force's work over a round trip is zero while EK 3.2.A.1.iv and 3.2.A.1.v make a nonconservative force's work path dependent and name friction and air resistance; EK 3.2.A.2 states that work is a scalar that may be positive, negative, or zero; EK 3.2.A.3.i gives W equals F parallel times d equals F d cosine theta; EK 3.2.A.3.ii covers the perpendicular case that changes direction without changing kinetic energy; EK 3.2.A.4 gives the work-energy theorem as a sum over all forces; EK 3.2.A.5 gives work as the area under a force against displacement graph. Suggested skills for Topic 3.2 are 1.B, 2.B, 2.D, 3.A and 3.B.

The sign says which way energy moved

Start with what the sign is not. Negative work does not mean the object moved backwards, and it does not mean the force pointed backwards. It means energy left the object through that force.

The AP Physics 1 CED sets up the arithmetic at essential knowledge 3.2.A.3.i, which says only the component of the force parallel to the displacement of the point of application changes the system's total energy, and prints the equation:

W = F_{\parallel}d = Fd\cos\theta

In that expression $F$ is a magnitude and so is positive. The distance $d$ is a magnitude and so is positive. The only factor that can be negative is $\cos\theta$ , so the sign of the work is the sign of the cosine of the angle between the force and the displacement, and nothing else contributes.

EK 3.2.A.2 states the range of outcomes: work is a scalar quantity that may be positive, negative, or zero. Three possibilities, and the CED names all three, because the cosine of an angle between $0^\circ$ and $180^\circ$ can only be positive, zero or negative.

The interpretation runs through EK 3.2.A.1, which defines work as the amount of energy transferred into or out of a system by a force exerted on that system over a distance. "Into or out of" is the sign. Positive work is energy in. Negative work is energy out. Zero work is a force that moved no energy at all, whatever else it was doing.

Three cases, and the CED names all three

The angle between force and displacement	$\cos\theta$	Sign of the work	Energy through that force	If this were the only force
$\theta = 0^\circ$	$+1$	Positive, and as large as it gets	Into the object	It speeds up fastest
$0^\circ < \theta < 90^\circ$	Between $0$ and $1$	Positive, smaller than $Fd$	Into the object	It speeds up
$\theta = 90^\circ$	$0$	Zero	None	Its speed does not change
$90^\circ < \theta < 180^\circ$	Between $-1$ and $0$	Negative	Out of the object	It slows down
$\theta = 180^\circ$	$-1$	Negative, and as large in size as it gets	Out of the object	It slows down fastest

Two notes on reading this table.

The angle is between the force and the displacement, not between the force and anything else. Not the horizontal, not the surface, not the object's initial velocity. Getting this wrong is the source of most sign errors that are not sign errors at all but geometry errors.

The last column only applies if that force acts alone. An individual force's sign tells you which way it moved energy, not what happened to the object. A crate can be speeding up while three separate forces are each doing negative work on it, as long as a fourth is doing more positive work than all three combined.

A short list of the angles you meet most often, with the sign already resolved:

Force	Situation	Angle	Work
Gravity	Object falling	$0^\circ$	Positive
Gravity	Object rising	$180^\circ$	Negative
Gravity	Horizontal motion	$90^\circ$	Zero
Gravity	Block sliding up a $30^\circ$ incline	$120^\circ$	Negative
Gravity	Block sliding down a $30^\circ$ incline	$60^\circ$	Positive
Normal force	Object sliding on a level floor	$90^\circ$	Zero
Normal force	Object sliding along an incline	$90^\circ$	Zero
Normal force	Person on the floor of an elevator moving up	$0^\circ$	Positive
Kinetic friction	Object sliding on a stationary surface	$180^\circ$	Negative
Air resistance	Any motion through air	$180^\circ$	Negative
Tension	Crate lowered at steady speed by a rope	$180^\circ$	Negative
Tension	Ball whirled in a horizontal circle	$90^\circ$	Zero
Applied push	Crate pushed along the floor	$0^\circ$	Positive

Notice that gravity and the normal force each appear more than once with different answers. No force has a fixed sign of work. The sign belongs to the situation, and specifically to the angle in that situation.

Zero work is not the same as no force

The middle row of the table is the one that gets skipped and then costs a mark, because a force doing zero work is still fully present in the free-body diagram and in Newton's second law.

EK 3.2.A.3.ii states it in the CED's own terms: the component of the force exerted on a system perpendicular to the direction of the displacement of the system's center of mass can change the direction of the system's motion without changing the system's kinetic energy.

That is a precise description of circular motion. A ball on a string moving in a horizontal circle at constant speed is being pulled hard the whole time, its velocity is changing continuously, and the tension does exactly zero work because it is always perpendicular to the motion. The kinetic energy never budges.

The two cases where work vanishes:

The angle is $90^\circ$ . The normal force on a level floor, the tension in a circular-motion problem, gravity during purely horizontal motion, and the magnetic force on a moving charge in AP Physics 2.
The displacement is zero. Holding a box still, or pushing on a wall that does not move. Here $d = 0$ kills the product whatever the angle is.

A force can be essential to the physics and irrelevant to the energy budget at the same time. The normal force on a car rounding a banked curve is what keeps the car on the road, and it contributes nothing to the car's energy accounting.

The case that separates them: an upward force doing negative work

If the sign really came from the direction of the force, an upward force could never do negative work. It can, and the example is ordinary.

Lower a $40$ kg crate slowly to the ground on a rope. The tension points up. The crate moves down. The angle between them is $180^\circ$ , so the tension does negative work, and over $3.0$ m it does $-1176$ J. The second worked example runs it in full.

That single case kills three intuitions at once.

Negative work does not require a backwards force. The tension is upward, the most forwards-looking direction available in a lowering problem.
Negative work does not require the object to slow down. The crate descends at constant speed and its kinetic energy never changes.
Negative work does not mean something went wrong. The rope is removing gravitational potential energy from the crate and Earth system in a controlled way, which is exactly what lowering something means.

Run it the other way and the same force flips. Haul the crate back up and the tension is still upward while the displacement is now upward too, so $\theta = 0^\circ$ and the tension does $+1176$ J. Same rope, same crate, same magnitude of force, opposite sign, and the only thing that changed was the direction of travel.

The reverse surprise is a normal force doing positive work. Stand in an elevator that accelerates upward: the floor pushes up on you and you move up, so $\theta = 0^\circ$ and the normal force does positive work. The first worked example computes it. Anyone who has memorised "the normal force never does work" has memorised a special case.

Signed works add, and only the total equals the change in kinetic energy

The signs are not decoration on individual answers. They are what makes the work-energy theorem an equation rather than a list.

EK 3.2.A.4 says the change in an object's kinetic energy equals the sum of the work being done by all forces exerted on the object, and the AP Physics 1 sheet prints it as

\Delta K = \sum_i W_i = \sum_i F_{\parallel, i}\, d_i

The sum is over signed quantities. Adding magnitudes instead of signed values is a routine arithmetic failure here, and it is invisible in problems where every force happens to do positive work.

The safe routine, in order:

Draw the free-body diagram and list every force.
For each force, find the angle between that force and the displacement.
Compute $W_i = F_i d\cos\theta_i$ and write down the sign with the number, every time.
Add the signed values. That total, and only that total, equals $\Delta K$ .
Check the sign of the total against the physics: a speeding-up object must come out positive.

Three working checks worth running.

If the object moves at constant speed, the signed works must sum to zero. Any nonzero total means a sign is wrong.
If the object slows to a stop, the total must be negative and equal in size to the starting kinetic energy.
A negative total from an object that clearly accelerated means you flipped a sign, not that the physics is strange.

The full procedure lives in the work-energy theorem guide; this page is about getting each sign right before you add.

Round trips, and which forces come back to zero

Send an object out and bring it back to exactly where it started, and the signs of the work done along the way decide whether that force's total is zero.

EK 3.2.A.1.i says the work done by a conservative force on a system is path independent and depends only on the initial and final configurations. EK 3.2.A.1.ii draws the conclusion: the work done by a conservative force on a system, or equivalently the change in the potential energy of the system, will be zero if the system returns to its initial configuration.

So for gravity and for spring forces, the positive and negative halves of a round trip cancel exactly. Throw a ball straight up and gravity does negative work all the way up and positive work all the way down, in equal amounts, and the ball returns to your hand at the speed it left.

Nonconservative forces do not close. EK 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. Slide a block out and back along a rough floor and friction did negative work on both legs, because friction reverses direction with the motion and therefore stays at $180^\circ$ throughout. Two negatives here do not make a positive; they add to a bigger negative. That is why a round trip on a rough surface always ends slower than it began, and it is the clearest test of what a nonconservative force is.

Force	Out	Back	Round trip total
Gravity, vertical throw	Negative	Positive	Zero
Spring, compress and release	Negative	Positive	Zero
Kinetic friction, out and back	Negative	Negative	Negative
Air resistance, out and back	Negative	Negative	Negative

The sign pattern in that table is the whole difference between the two families, worked out in more depth in conservative vs nonconservative force.

When the sign is obvious, and when it is not

In a large fraction of problems the angle is $0^\circ$ , $90^\circ$ or $180^\circ$ , the cosine is $+1$ , $0$ or $-1$ , and you can write the sign down by inspection. That fluency is worth having, and it is also why the harder cases catch people.

The cases where the sign needs a moment:

Angles strictly between $90^\circ$ and $180^\circ$ . Gravity on a block sliding up an incline is the standard one: on a $30^\circ$ incline the displacement is $30^\circ$ above horizontal and gravity points straight down, so the angle between them is $120^\circ$ and the cosine is $-0.5$ . The third worked example checks that against the height-based route and gets the same number.
Kinetic friction when the surface itself is moving. Friction on an object sliding relative to a stationary surface opposes the object's displacement and does negative work, and that is the standard AP setup. Drop an object onto a moving conveyor belt, though, and the friction on the object acts forward while the object moves forward, so it does positive work on the object. The rule that survives both is about the relative sliding, not the object's direction.
Forces on a system whose parts move differently. EK 3.2.A.4.ii says that if the system's center of mass and the point of application of the force move the same distance, the system may be modeled as an object and only its kinetic energy can change. When they do not move together, the accounting is more involved, and AP Physics 1 signals this with the phrase "displacement of the point of application" in EK 3.2.A.3.

A final sanity habit: after computing a sign, say out loud which way the energy went and check it against the picture. If your arithmetic says a rope lifting a crate removed energy from it, one of the two angles in the problem is wrong.

Where the confusion costs a mark

Each of these is a scoring event rather than general advice.

Saying negative work means the object moves backwards. It means energy left the object through that force. A crate lowered at steady speed has negative work done on it by an upward rope while moving down at constant speed.
Dropping the sign and adding magnitudes. $100$ J of push work and $47$ J of friction work give $53$ J of net work, not $147$ J. The theorem sums signed values.
Measuring the angle from the wrong reference. It is the angle between the force and the displacement. On an incline it is not the incline angle, and for gravity on a block sliding up a $30^\circ$ slope it is $120^\circ$ .
Assuming the normal force always does zero work. True whenever it is perpendicular to the motion, which covers level floors and inclines, and false in an accelerating elevator, where it does positive work on the passenger.
Assuming gravity always does negative work. It does negative work on a rising object and positive work on a falling one, and zero during horizontal motion.
Concluding an object slows down because one force did negative work. Only the net work determines the change in kinetic energy. A car accelerating uphill has gravity and friction both doing negative work while it speeds up.
Writing $\cos\theta$ with the angle in radians on a calculator set to degrees, or the reverse. AP Physics works in degrees, and $\cos(120)$ in radian mode returns a positive number, which quietly flips the sign of the whole answer.
Treating two negative works on a round trip as cancelling. For friction they add. Only conservative forces cancel over a closed path, per EK 3.2.A.1.ii.
Reporting a negative kinetic energy. If a sign error propagates that far, the mistake was upstream: kinetic energy has no sign to carry, only its change does.

What the CED asks, and what the sheet prints

This sits in 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. Topic 3.2, Work, carries one learning objective, 3.2.A: describe the work done on an object or system by a given force or collection of forces.

The sign material is spread through its essential knowledge. EK 3.2.A.1 defines work as energy transferred into or out of a system, which is what the sign records. EK 3.2.A.1.i through 3.2.A.1.v cover the conservative and nonconservative split, including the round-trip statement at 3.2.A.1.ii and the naming of friction and air resistance at 3.2.A.1.v. EK 3.2.A.2 states that work is a scalar that may be positive, negative, or zero. EK 3.2.A.3.i gives the parallel-component rule and the equation $W = F_{\parallel}d = Fd\cos\theta$ , while 3.2.A.3.ii gives the perpendicular case that changes direction without changing kinetic energy. EK 3.2.A.4 gives the work-energy theorem as a sum over forces, and 3.2.A.4.iii gives the friction dissipation estimate $\Delta E_{\text{mech}} = F_f d\cos\theta$ . EK 3.2.A.5 says work equals the area under a graph of force as a function of displacement, where area below the axis counts as negative.

Suggested skills for Topic 3.2 are 1.B, 2.B, 2.D, 3.A and 3.B. Skill 3.B is applying a definition to make a claim, and "this force does negative work because the angle exceeds ninety degrees" is exactly the shape of answer that earns it.

On the AP Physics 1 equation sheet, the two lines that carry the sign are $W = F_{\parallel}d = Fd\cos\theta$ and $\Delta K = \sum W_i = \sum F_{\parallel,i}\, d_i$ . Nothing on the sheet tells you which angle to use or how to interpret a negative result, so both of those come from you.

Going further: the work-energy theorem for the solving routine, work vs energy for why a transfer can be negative when a state cannot, conservative vs nonconservative force for the round-trip test, and inclined plane problems for the geometry that produces the awkward angles. The CED framing is on Topic 3.2.

An elevator, where the normal force does positive work

A $60$ kg passenger stands on the floor of an elevator that accelerates upward at $1.2\ \text{m/s}^2$ from rest, rising $3.0$ m. Use $g = 9.8\ \text{m/s}^2$ and take upward as positive. Find the normal force, the work done on the passenger by the normal force and by gravity, the net work, and check it against the passenger's kinetic energy.

Set the axis once: upward is positive, and both the acceleration and the displacement are upward.
Normal force. Newton's second law on the passenger: $F_N - mg = ma$ , so $F_N = m(g + a) = (60)(9.8 + 1.2) = (60)(11.0) = 660\ \text{N}$ .
Work by the normal force. The force is upward and the displacement is upward, so $\theta = 0^\circ$ and $\cos\theta = +1$ : $W_N = (660)(3.0)(+1) = +1980\ \text{J}$ . A normal force doing positive work, which the level-floor case never shows you.
Work by gravity. Downward force, upward displacement, so $\theta = 180^\circ$ and $\cos\theta = -1$ : $W_g = (60)(9.8)(3.0)(-1) = -1764\ \text{J}$ .
Net work. $\Delta K = 1980 + (-1764) = +216\ \text{J}$ . The signed sum, not the sum of magnitudes, which would have been $3744$ J.
Independent check. The passenger started from rest and accelerated at $1.2\ \text{m/s}^2$ over $3.0$ m, so $v^2 = 2ad = 2(1.2)(3.0) = 7.2\ \text{m}^2/\text{s}^2$ and $K = \frac{1}{2}(60)(7.2) = 216\ \text{J}$ . The theorem and the kinematics agree exactly.
Now change one thing. If the elevator moved up at constant speed instead, $F_N$ would be $588$ N, $W_N$ would be $+1764$ J, gravity would still do $-1764$ J, and the net work would be zero, matching a kinetic energy that does not change. The normal force still does positive work; it is just balanced.
And one more. If the passenger walked $3.0$ m across a level floor instead, the normal force would be perpendicular to the displacement, $\theta = 90^\circ$ , and $W_N = 0$ . Same force, three situations, three different answers, and the angle is the only thing that changed.

The normal force is $660$ N and does $+1980$ J of work; gravity does $-1764$ J. The net work is $+216$ J, which matches the passenger's kinetic energy of $216$ J found from kinematics. The normal force does positive work here, zero work on a level floor, and the sign came from the angle in each case.

Two ways to net zero, each with a force doing negative work

Part one: a $40$ kg crate is lowered $3.0$ m at constant speed by a vertical rope. Part two: the same crate is then pushed $3.0$ m horizontally across a level floor at constant speed, with a coefficient of kinetic friction of $0.30$ . Use $g = 9.8\ \text{m/s}^2$ . For each part, find the work done by every force and the net work.

Part one, the tension. Constant speed means zero net force, so the rope's tension balances gravity: $T = mg = (40)(9.8) = 392\ \text{N}$ , directed upward.
Part one, work by the tension. The tension is upward and the displacement is $3.0$ m downward, so $\theta = 180^\circ$ : $W_T = (392)(3.0)(-1) = -1176\ \text{J}$ . An upward force doing negative work while the crate moves down at constant speed. Nothing about it is backwards, and nothing about it is slowing the crate.
Part one, work by gravity. Downward force, downward displacement, $\theta = 0^\circ$ : $W_g = (392)(3.0)(+1) = +1176\ \text{J}$ .
Part one, net work. $-1176 + 1176 = 0\ \text{J}$ , so $\Delta K = 0$ , which is what constant speed requires. Gravity fed $1176$ J into the crate and the rope took the same amount straight back out.
Part two, the friction force. On a level floor with no vertical acceleration, $F_N = mg = 392\ \text{N}$ , so $f_k = \mu_k F_N = (0.30)(392) = 117.6\ \text{N}$ , opposing the motion.
Part two, the push. Constant speed again, so the horizontal push equals friction: $F_{\text{push}} = 117.6\ \text{N}$ forward.
Part two, the four works. Push at $\theta = 0^\circ$ : $W_{\text{push}} = (117.6)(3.0)(+1) = +352.8\ \text{J}$ . Friction at $\theta = 180^\circ$ : $W_f = (117.6)(3.0)(-1) = -352.8\ \text{J}$ . Normal force at $\theta = 90^\circ$ : $0$ J. Gravity at $\theta = 90^\circ$ : $0$ J. Net work $= 0$ J.
Compare the two parts. Both net to zero and both have a force doing hundreds of joules of negative work, but for different reasons. In part one the negative work was done by an upward tension against a downward displacement, and the energy went back up the rope. In part two it was done by friction at $180^\circ$ , and the energy became thermal energy in the crate and the floor. Same sign, same size of effect on the total, and two entirely different destinations for the energy.
Note also which forces did nothing in part two. The normal force is $392$ N, tied with gravity as the largest force in the problem, and its contribution to the energy budget is exactly zero because it is perpendicular to the motion.

Lowering: the tension does $-1176$ J, gravity does $+1176$ J, net zero. Pushing: the push does $+352.8$ J, friction does $-352.8$ J, and the normal force and gravity each do zero, net zero. In each case a force did large negative work while the crate moved at constant speed, and in neither case did negative work mean backwards motion.

A 120 degree angle, checked two ways

A $5.0$ kg block slides $4.0$ m up a frictionless incline angled at $30^\circ$ to the horizontal, and just comes to rest at the top of that slide. Use $g = 9.8\ \text{m/s}^2$ . Find the work done by gravity and by the normal force using the angle between each force and the displacement, check gravity's work against the height gained, and find the speed the block started with. Then find the works for the slide back down.

Set up the geometry. The displacement is $4.0$ m directed up the slope, which is $30^\circ$ above the horizontal. Gravity points straight down. The angle between a direction $30^\circ$ above horizontal and straight down is $90^\circ + 30^\circ = 120^\circ$ .
Gravity's work by the angle route. $F_g = mg = (5.0)(9.8) = 49.0\ \text{N}$ , and $\cos 120^\circ = -0.5$ , so $W_g = (49.0)(4.0)(-0.5) = -98.0\ \text{J}$ . Negative, because the block is climbing and gravity is taking energy out of its motion.
Gravity's work by the height route, as a check. The block rises $h = d\sin 30^\circ = (4.0)(0.5) = 2.0\ \text{m}$ , and gravity's work on a rising object is $-mgh = -(5.0)(9.8)(2.0) = -98.0\ \text{J}$ . The two routes agree exactly, which they must, because $\cos 120^\circ$ and $-\sin 30^\circ$ are the same number.
The normal force. It is perpendicular to the incline surface, and the displacement is along that surface, so $\theta = 90^\circ$ and $W_N = (F_N)(4.0)(0) = 0\ \text{J}$ . Its magnitude, $mg\cos 30^\circ = 42.4$ N, never enters the energy budget.
Net work and the starting speed. The incline is frictionless, so the only forces are gravity and the normal force and $\Delta K = -98.0 + 0 = -98.0\ \text{J}$ . The block ends at rest, so its initial kinetic energy was $98.0$ J and $v_0 = \sqrt{2(98.0)/5.0} = \sqrt{39.2} = 6.26\ \text{m/s}$ to three significant figures.
The slide back down. The displacement reverses, so the angle between gravity and the displacement becomes $180^\circ - 120^\circ = 60^\circ$ , and $\cos 60^\circ = +0.5$ : $W_g = (49.0)(4.0)(+0.5) = +98.0\ \text{J}$ . The normal force is still perpendicular and still does zero.
The round trip. Gravity did $-98.0$ J going up and $+98.0$ J coming down, totalling zero, exactly as EK 3.2.A.1.ii requires of a conservative force returning to its initial configuration. The block arrives back at the bottom at $6.26$ m/s, the speed it left with.
One thing to notice about the failure mode. A student who used the incline angle of $30^\circ$ instead of the true $120^\circ$ would compute $(49.0)(4.0)(\cos 30^\circ) = +170\ \text{J}$ , positive and larger in size than the correct answer. Both the size and the sign would be wrong, and the block would come out gaining energy while climbing.

Going up, gravity does $-98.0$ J at an angle of $120^\circ$ and the normal force does $0$ J at $90^\circ$ , so the block started at $6.26$ m/s. Coming down, gravity does $+98.0$ J at $60^\circ$ and the round trip totals zero. Using the incline's $30^\circ$ instead of the true $120^\circ$ gives $+170$ J, wrong in both sign and size.

Frequently asked questions

What does negative work mean in physics?

It means energy is being transferred out of the object or system by that force. The AP Physics 1 CED defines work at essential knowledge 3.2.A.1 as the amount of energy transferred into or out of a system by a force exerted on that system over a distance, and the sign records which of those two happened. Negative work does not mean the object moves backwards, and it does not mean the force points backwards. A rope lowering a crate at constant speed pulls upward while the crate moves down, so the angle between them is 180 degrees, the work is negative, and the crate is neither reversing nor slowing.

What determines whether work is positive, negative, or zero?

Only the angle between the force and the displacement. Work equals the force magnitude times the distance times the cosine of that angle, and since a magnitude and a distance are both positive, the cosine is the only factor that can change sign. An angle under 90 degrees gives a positive cosine and positive work, exactly 90 degrees gives zero, and an angle over 90 degrees gives negative work. The AP Physics 1 CED states the three outcomes at essential knowledge 3.2.A.2: work is a scalar quantity that may be positive, negative, or zero. The angle is measured between the force and the displacement, not between the force and the horizontal or the surface.

Does friction always do negative work?

Not always, though it does in the standard AP setup. When an object slides on a stationary surface, kinetic friction opposes the object's displacement, the angle is 180 degrees and the work on that object is negative. But if the surface itself is moving, the friction can act in the direction the object travels. Drop a box onto a moving conveyor belt and the friction on the box acts forward while the box moves forward, so it does positive work on the box. What is always true is that kinetic friction between two surfaces sliding across each other dissipates mechanical energy from the system as a whole, which is why the CED classes it as a nonconservative force at essential knowledge 3.2.A.1.v.

Can the normal force ever do work?

Yes, whenever it is not perpendicular to the displacement. On a level floor, or on an object sliding along an incline, the normal force is perpendicular to the motion and does exactly zero work. But stand in an elevator accelerating upward and the floor pushes you up while you move up, so the angle is zero and the normal force does positive work. For a 60 kg passenger in an elevator accelerating upward at 1.2 meters per second squared through 3.0 m, the normal force is 660 N and does plus 1980 J, gravity does minus 1764 J, and the net 216 J matches the passenger's kinetic energy. The rule that the normal force does no work is a special case, not a law.

Does negative work always mean the object slows down?

No. Only the net work determines the change in kinetic energy, and one force doing negative work says nothing on its own. A car accelerating up a hill has gravity doing negative work and friction from the air doing negative work while the car speeds up, because the driving force does more positive work than those two remove. Conversely, a crate lowered at constant speed has a rope doing large negative work on it while its speed never changes, since gravity is simultaneously doing an equal amount of positive work. The AP Physics 1 work-energy theorem sums the work of every force, and it is that signed total that equals the change in kinetic energy.

Why does gravity do negative work on an object moving upward?

Because gravity points down and the displacement points up, so the angle between them is 180 degrees and the cosine is minus one. Physically, energy is leaving the object's motion and being stored in the configuration of the object and Earth as gravitational potential energy. The size is the weight times the height gained. On the way back down the angle becomes zero, the work becomes positive with the same magnitude, and the round trip totals zero, which is what the CED means at essential knowledge 3.2.A.1.ii when it says the work done by a conservative force is zero if the system returns to its initial configuration.

Can a force be large and still do zero work?

Yes, in two situations. If the force is perpendicular to the displacement the cosine is zero, so a string whirling a ball in a horizontal circle can pull hard while transferring no energy, and the ball's speed never changes. The CED describes this at essential knowledge 3.2.A.3.ii, saying the perpendicular component can change the direction of the system's motion without changing its kinetic energy. If nothing moves, the distance is zero and the work is zero however large the force, which is why holding a heavy box still does no work on the box. A force can be essential to the force balance and irrelevant to the energy budget at the same time.