7.1 Work as a Line Integral Perspective in Simple Settings

Three Pushes, Three Stories

You push a box across the room with a constant horizontal force. You push a box up a ramp at an angle. And you carry a box across a flat floor at constant height.

In all three cases, you exert a force. In all three cases, the box moves. But in which cases do you do work on the box?

If your answer is "all three," you are in good company --- and you are wrong about one of them. The carrying case is the surprise. You are clearly exerting a force (holding the box up against gravity). The box is clearly moving (across the room). But the force you exert does zero work on the box. None.

This is not a technicality. It reveals something deep about what "work" actually means in physics --- and how sharply it diverges from the everyday English word. In everyday language, carrying a heavy box across a room is exhausting. It feels like work. But in physics, work has a precise geometric meaning: it depends not just on how hard you push and how far the object moves, but on the relationship between the direction of your force and the direction of the motion.

That relationship is the subject of this section.

Before you read on: You carry a box horizontally at constant height. Your hands exert an upward force on the box to support it against gravity. The box moves horizontally.

Does the upward force you exert do work on the box? Commit to a yes-or-no answer and a brief reason before continuing.

[Interactive: Predict-Then-Reveal. Student selects "Yes" or "No" and types a short justification. After submitting, the answer is revealed: No. The force is vertical (upward) and the displacement is horizontal. Because the force and displacement are perpendicular, no work is done by that force. A brief explanation follows, connecting to the cosine relationship that will be developed below.]

The Guiding Question

How does force transfer energy through displacement?

You already know that forces cause acceleration (Chapter 5) and that acceleration changes velocity (Chapter 2). But sometimes you do not care about the detailed history of the motion. You care about a bottom line: how much energy did the force deliver to the object, or take away from it?

That is what work measures. It is the mechanism by which a force transfers energy to (or from) an object through displacement. And the precise definition will hinge on a piece of mathematics you have already met: the dot product from Chapter 3.

Exploring the Angle Dependence

Before we write down a formula, let's build intuition for how the angle between force and displacement affects work.

[Interactive: Force-Displacement Angle Explorer. Two arrows are displayed on screen: a force vector $\vec{F}$ (fixed magnitude, say 10 N) and a displacement vector $\Delta\vec{r}$ (fixed magnitude, say 3 m). A slider lets students adjust the angle $\theta$ between the two vectors from $0°$ to $360°$. As the angle changes, the following update in real time:

A visual projection of $\vec{F}$ onto the direction of $\Delta\vec{r}$, showing the component $F\cos\theta$
The computed work $W = Fd\cos\theta$, displayed numerically and as a colored bar (positive = green, zero = gray, negative = red)

Guided prompts:

"Set $\theta = 0°$ (force and displacement in the same direction). What is the work? This is the maximum possible work for these magnitudes."
"Slowly increase $\theta$. What happens to the work? At what angle does the work become zero?"
"Keep increasing $\theta$ past $90°$. The work is now negative. What does that mean physically?"
"Set $\theta = 180°$ (force directly opposes displacement). What is the work now? Think of a physical scenario where this happens."]

Spend time with this interactive before reading on. The key observations are:

When the force points in the same direction as the displacement ($\theta = 0°$), all of the force contributes to the work. The work is maximized: $W = Fd$.
When the force is perpendicular to the displacement ($\theta = 90°$), none of the force contributes to the work. $W = 0$. This is the carrying-a-box case.
When the force opposes the displacement ($\theta = 180°$), the work is negative: $W = -Fd$. This is friction slowing you down, or a braking force opposing motion.
At intermediate angles, only the component of the force along the displacement contributes.

Pause and think: A person pushes a lawn mower with a force directed along the handle, which makes an angle of about $40°$ below the horizontal. The mower moves horizontally. Does the full force contribute to the work done on the mower? If not, which part does?

Check your answer

Only the horizontal component of the pushing force contributes to the work, because the displacement is horizontal. The vertical component of the push presses the mower into the ground (increasing the normal force) but does not contribute to displacing the mower forward. The work done is $W = F\cos 40° \cdot d$, where $F$ is the magnitude of the push and $d$ is the horizontal distance traveled. The factor $\cos 40° \approx 0.766$ means that about 77% of the force's magnitude actually contributes to the work.

The Concept: Work as a Dot Product

Here is the definition that captures everything the interactive just showed you.

The work done by a constant force $\vec{F}$ on an object that undergoes a displacement $\Delta\vec{r}$ is:

$$W = \vec{F} \cdot \Delta\vec{r} = F \, d \, \cos\theta$$

where $F = |\vec{F}|$ is the magnitude of the force, $d = |\Delta\vec{r}|$ is the magnitude of the displacement, and $\theta$ is the angle between the force and displacement vectors.

This is the dot product from Chapter 3. It appeared there as a mathematical operation that takes two vectors and returns a scalar. Here it appears physically: the dot product selects the component of force that actually transfers energy.

Let's unpack the three factors:

The magnitude of the force, $F$. A larger force can transfer more energy per unit displacement. This is intuitive: pushing harder moves things more.

The magnitude of the displacement, $d$. A force does more work when it acts over a longer distance. Also intuitive: pushing something farther requires more energy transfer.

The cosine of the angle, $\cos\theta$. This is the factor that makes work different from "force times distance." It captures the geometric relationship between the force direction and the motion direction:

Angle $\theta$	$\cos\theta$	Meaning
$0°$	$+1$	Force fully along displacement. Maximum positive work.
$0° < \theta < 90°$	Positive	Force has a component along displacement. Positive work.
$90°$	$0$	Force perpendicular to displacement. Zero work.
$90° < \theta < 180°$	Negative	Force has a component opposing displacement. Negative work.
$180°$	$-1$	Force fully opposes displacement. Maximum negative work.

The sign of work

The sign of the work is not a bookkeeping convention. It carries physical meaning:

Positive work ($W > 0$): The force transfers energy to the object. The object speeds up, or gains energy. Example: you push a box in the direction it moves.
Zero work ($W = 0$): The force transfers no energy. Example: the normal force on a box sliding along a flat surface (the normal force is perpendicular to the motion).
Negative work ($W < 0$): The force transfers energy away from the object. The object slows down, or loses kinetic energy. Example: friction on a sliding box.

Units

Work has units of force times distance: newtons times meters, which is called the joule (J).

$$1 \text{ J} = 1 \text{ N} \cdot \text{m}$$

Work is a scalar --- it has magnitude and sign but no direction. This is one of its great advantages over force analysis, and a theme we will develop throughout this chapter.

The Non-Example: Carrying a Box

Let's return to the scenario that started this section and examine it carefully, because it sharpens the physics meaning of work more than any formula can.

You carry a box of mass $m$ horizontally at constant height across a room, a distance $d$.

What forces act on the box? Two: gravity $m\vec{g}$ pulling straight down, and the normal force from your hands $\vec{N}$ pushing straight up. (We are ignoring air resistance and the small horizontal force needed to start and stop the box --- focus on the steady carrying phase.)

What is the displacement? Horizontal, let's say in the $\hat{x}$ direction: $\Delta\vec{r} = d\,\hat{x}$.

What is the work done by gravity? The force is $m\vec{g} = -mg\,\hat{y}$. The displacement is $d\,\hat{x}$. The angle between them is $90°$:

$$W_{\text{gravity}} = mg \cdot d \cdot \cos 90° = 0$$

What is the work done by your hands? The force is $\vec{N} = mg\,\hat{y}$ (upward, balancing gravity). The displacement is again $d\,\hat{x}$. The angle is again $90°$:

$$W_{\text{hands}} = mg \cdot d \cdot \cos 90° = 0$$

Neither force does any work. The net work on the box is zero. No energy is transferred to or from the box.

"But I'm tired!" Of course you are. Your muscles are contracting and releasing, burning chemical energy, generating heat. Your body is doing internal physiological work. But the force your hands exert on the box does no mechanical work on the box because that force is perpendicular to the box's displacement. The energy your body spends goes into your muscles, not into the box.

This is the sharpest lesson this section has to offer. Physics "work" and everyday "work" are different concepts. The everyday word describes effort, exertion, fatigue. The physics word describes energy transfer through displacement. A force can be large, a displacement can be large, and the work can still be zero --- if the two are perpendicular.

Connecting to the Dot Product

In Chapter 3, you learned that the dot product of two vectors can be computed two ways:

Geometric form: $\vec{A} \cdot \vec{B} = AB\cos\theta$

Component form: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$

Both forms apply to work. If you know the force and displacement in component form, you can compute work directly:

$$W = F_x \, \Delta x + F_y \, \Delta y + F_z \, \Delta z$$

This is often the most efficient route when you already have components. For example, if a force $\vec{F} = (3, -4, 0)$ N acts over a displacement $\Delta\vec{r} = (5, 0, 2)$ m, then:

$$W = (3)(5) + (-4)(0) + (0)(2) = 15 \text{ J}$$

No need to find the angle $\theta$ at all. The component form handles it automatically.

The connection to multivariable calculus. The dot product is not just a formula you memorized in math class. Here it appears with physical meaning: it selects the component of force that actually transfers energy. When you compute $\vec{F} \cdot \Delta\vec{r}$, the dot product is doing physical work (literally). It projects the force onto the direction of motion and multiplies by the distance traveled. Every time you see a dot product in physics, ask: "What is being projected onto what, and why?"

A First Glimpse of the Line Integral

So far, we have considered constant forces and straight-line displacements. The formula $W = \vec{F} \cdot \Delta\vec{r}$ handles these cases perfectly. But what happens when the force changes along the path, or the path curves?

You will tackle that fully in Section 7.2. But the seed of the idea is already here.

Imagine breaking a curved path into many tiny straight segments, each of length $d\vec{r}$. Along each tiny segment, the force is approximately constant. The work done over each tiny segment is $\vec{F} \cdot d\vec{r}$. The total work is the sum of all these contributions:

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

This is a line integral --- an integral along a path. You have seen the same accumulation idea before: in Section 2.3, you accumulated displacement from velocity by integrating $v(t)\,dt$. Here, you accumulate energy transfer from force by integrating $\vec{F} \cdot d\vec{r}$.

For now, we do not need the full integral machinery. The constant-force case $W = \vec{F} \cdot \Delta\vec{r}$ is the building block, and it captures the essential geometry: force, displacement, and the angle between them.

Practice

Layer 1: Concrete

Problem 1. A 50 N force is applied to a crate at an angle of $30°$ above the horizontal. The crate moves 8 m horizontally along a flat floor. Find the work done by the applied force.

Check your answer

The displacement is horizontal. The force makes an angle of $30°$ with the displacement. So: $$W = F d \cos\theta = 50 \times 8 \times \cos 30° = 50 \times 8 \times 0.866 = 346 \text{ J}$$ The vertical component of the force ($50 \sin 30° = 25$ N) does no work because it is perpendicular to the displacement. Only the horizontal component ($50 \cos 30° = 43.3$ N) contributes.

Problem 2. A force $\vec{F} = (4, -3)$ N acts on a particle as it moves from position $(1, 2)$ m to position $(6, 2)$ m. Find the work done.

Check your answer

The displacement is: $$\Delta\vec{r} = (6 - 1, \, 2 - 2) = (5, \, 0) \text{ m}$$ Using the component form of the dot product: $$W = F_x \, \Delta x + F_y \, \Delta y = (4)(5) + (-3)(0) = 20 \text{ J}$$ The $y$-component of the force ($-3$ N) does no work because the displacement has no $y$-component.

Problem 3. You pull a sled with a force of 120 N along a rope that makes an angle of $25°$ with the horizontal. The sled moves 15 m across flat ground. Meanwhile, kinetic friction exerts a force of 40 N opposing the motion. Find:

(a) The work done by the rope tension.

(b) The work done by friction.

Check your answer

**(a)** The rope force makes an angle of $25°$ with the displacement (horizontal): $$W_{\text{rope}} = 120 \times 15 \times \cos 25° = 120 \times 15 \times 0.906 = 1631 \text{ J}$$ **(b)** Friction opposes the motion, so the angle between friction and displacement is $180°$: $$W_{\text{friction}} = 40 \times 15 \times \cos 180° = 40 \times 15 \times (-1) = -600 \text{ J}$$ Negative work: friction removes energy from the sled. **(c)** The net work is the sum. (Gravity and the normal force both act perpendicular to the displacement, so they each do zero work.) $$W_{\text{net}} = 1631 + (-600) = 1031 \text{ J}$$

Layer 2: Pattern

Problem 4. Rank the following three scenarios by the amount of work done by the applied force, from greatest to least.

Scenario A: A 100 N force is applied at $0°$ to the displacement over 5 m.
Scenario B: A 150 N force is applied at $60°$ to the displacement over 5 m.
Scenario C: A 200 N force is applied at $80°$ to the displacement over 5 m.

Check your answer

Compute the work for each: - **A:** $W = 100 \times 5 \times \cos 0° = 500$ J - **B:** $W = 150 \times 5 \times \cos 60° = 150 \times 5 \times 0.5 = 375$ J - **C:** $W = 200 \times 5 \times \cos 80° = 200 \times 5 \times 0.174 = 174$ J Ranking: **A > B > C**. The pattern: a larger force does not guarantee more work. Scenario C has the largest force magnitude but the least work, because most of that force is directed nearly perpendicular to the displacement. The angle matters as much as --- sometimes more than --- the magnitude.

Layer 3: Structure

Problem 5. Can work be negative? What does negative work mean physically? Give two concrete examples of forces that do negative work on an object.

Check your answer

Yes, work can be negative. It occurs whenever the angle between the force and the displacement is greater than $90°$ --- that is, when the force has a component opposing the direction of motion. **Physical meaning:** Negative work means the force is removing energy from the object. The object loses kinetic energy due to that force. **Example 1: Kinetic friction.** When a box slides across a rough floor, the friction force points opposite to the displacement. The angle is $180°$, so $W = Fd\cos 180° = -Fd$. Friction removes kinetic energy from the box (converting it to thermal energy). **Example 2: Gravity on a rising ball.** When you throw a ball upward, gravity points downward while the displacement is upward. The angle is $180°$, so gravity does negative work on the ball. This is why the ball slows down as it rises --- gravity is removing kinetic energy from it. In both cases, "negative work" is not a mathematical curiosity. It describes a real physical process: the force takes energy away from the object's motion.

Layer 4: Debug

Problem 6. A student learns that work equals "force times distance" and consistently calculates $W = Fd$ (without the cosine factor). In what situations does this give the correct answer, and when does it fail?

Check your answer

The student's formula $W = Fd$ gives the correct answer only when $\cos\theta = 1$, which means $\theta = 0°$ --- the force is exactly parallel to the displacement. **When it works:** - Pushing a box with a horizontal force across a horizontal floor (force and displacement both horizontal). - Gravity on an object in free fall (force and displacement both vertical, downward). - Any scenario where the force is applied in exactly the same direction as the motion. **When it fails:** - Pulling a sled with a rope at an angle: the student would overestimate the work because the full force magnitude is larger than its component along the displacement. - Carrying a box horizontally: the student would calculate $W = Fd$ with a large positive value, when the correct answer is $W = 0$ (force and displacement are perpendicular). - Friction opposing motion: the student would calculate a positive value of $W$, when the correct answer is negative. The student's formula is a special case of the general formula. It works when $\theta = 0°$ and fails whenever $\theta \neq 0°$. The $\cos\theta$ factor is not an optional refinement --- it is the part of the definition that makes work a physically meaningful quantity rather than an arbitrary product of two numbers.

Spaced Retrieval

Before you move on, test your recall of earlier material.

Recall prompt 1: How do you compute the dot product of two vectors given in component form? (Section 3.4)

Recall prompt 2: What is the geometric interpretation of the dot product $\vec{A} \cdot \vec{B}$? What does it measure? (Section 3.4)

Recall prompt 3: When you decompose a force vector into components along and perpendicular to a given direction, what role does the cosine play versus the sine? (Section 3.2)

Reflection

Think back over this section.

How does the physics meaning of "work" differ from the everyday meaning?

In everyday English, "work" means effort, exertion, fatigue. You "worked hard" carrying groceries up the stairs. You "did a lot of work" holding a heavy box. But in physics, work is a precise mechanical quantity: energy transferred by a force through displacement. Holding a heavy box does zero work (no displacement). Carrying a box horizontally does zero work (force perpendicular to displacement). Your muscles burn energy, but that energy goes into your body's internal processes --- not into the box.

This disconnect between everyday language and physics language is one of the most common sources of confusion in this chapter. If the physics definition feels strange right now, that is a sign you are engaging with it honestly. The everyday meaning is deeply ingrained, and overwriting it takes deliberate practice.

Self-assessment: Can you explain, to someone who has never taken physics, why carrying a box across a room counts as "no work" in physics? If you can make that explanation clear and convincing, you understand this section.

Looking Ahead

You have seen how work is defined for a constant force acting over a straight-line displacement: $W = \vec{F} \cdot \Delta\vec{r}$. The dot product captures the essential geometry --- only the component of force along the displacement does work, and the sign of the work tells you whether energy flows into or out of the object.

But many forces are not constant. When you stretch a spring, the force grows as you pull farther. When you launch a rocket, the gravitational force weakens with altitude. In these cases, you cannot simply multiply force by distance, because the force changes as the object moves.

In the next section, you will extend the concept of work to variable forces. The key idea will be one you have already met in calculus: when a quantity changes continuously, you handle it by slicing the process into tiny pieces, computing the contribution from each piece, and summing them up. That is integration. The formula $W = \int \vec{F} \cdot d\vec{r}$ will turn the constant-force result into a general tool, and you will see that the constant-force case was just the simplest instance of a much larger idea.