Section 5.4: Work
| Course: MATH162 | Textbook: Stewart Calculus 9th Edition, Section 5.4 |
The Big Picture
How much effort does it take to lift a heavy object? To stretch a spring? To pump water out of a tank?
In physics, work measures the energy transferred when a force moves an object. When the force is constant, it’s simple: $W = Fd$ (force × distance). But what if the force varies as the object moves? That’s where calculus enters—we slice the motion into tiny pieces where force is approximately constant, then integrate.
This section brings together everything you’ve learned: Riemann sums become integrals, and the “slice and sum” strategy now applies to physics.
Key Equations
| Formula | Name | When to Use |
|---|---|---|
| $W = Fd$ | Constant Force | Force doesn’t change with position |
| $W = \displaystyle\int_a^b F(x)\,dx$ | Variable Force | Force is a function of position |
| $F(x) = kx$ | Hooke’s Law | Spring stretched $x$ beyond natural length |
| $W = \displaystyle\int (\text{weight of slice}) \cdot (\text{distance})\,dy$ | Pumping/Lifting | Lifting distributed mass (cables, water) |
Units Reference
| System | Force Unit | Distance Unit | Work Unit |
|---|---|---|---|
| SI (Metric) | Newton (N) = kg·m/s² | meter (m) | Joule (J) = N·m |
| US Customary | pound (lb) | foot (ft) | foot-pound (ft-lb) |
| Key facts: 1 J ≈ 0.74 ft-lb | Weight = mass × $g$ where $g = 9.8$ m/s² or $32$ ft/s² |
Skill Dependency Map
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E["Pumping<br/>Problems"]
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Learning Path
Phase 1: The Foundation
| Skill | What You’ll Learn | Key Takeaway |
|---|---|---|
| Work with Variable Force | $W = \int_a^b F(x)\,dx$ | Work = integral of force over distance |
Why it works: Just like area is the integral of height, work is the integral of force. Slice the motion into tiny steps where force is approximately constant.
Phase 2: Classic Applications
| Skill | What You’ll Learn | Key Takeaway |
|---|---|---|
| Spring Problems | Hooke’s Law: $F = kx$ | Find $k$ first, then integrate $kx$ |
| Cable/Chain Problems | Weight distributed along length | Each piece travels a different distance |
| Pumping Problems | Lift layers of liquid | Integrate (weight of layer) × (distance lifted) |
Skills in This Section
| Skill | Description | Difficulty |
|---|---|---|
| Pumping Work and Lifting Problems | Applications of Integration | Advanced |
| Spring Work and Hooke’s Law | Applications of Integration | Intermediate |
| Work with Constant and Variable Forces | Applications of Integration | Intermediate |
Exercise Coverage Map
| Exercises | Topic | Key Challenge |
|---|---|---|
| 1–2 | Constant force (lifting) | Use $W = Fd$; convert mass to force if needed |
| 3–4 | Variable force (basic) | Direct integration of $F(x)$ |
| 5–6 | Graph/table of force | Estimate work from data |
| 7–12 | Spring problems | Find spring constant $k$ from given data |
| 13–19 | Cables and chains | Set up Riemann sum for distributed weight |
| 20–28 | Pumping water | Layer-by-layer approach with varying radius |
| 29–30 | Gas expansion | Work = $\int P\,dV$ |
| 31–33 | Work-Energy Theorem | Connect work to kinetic energy |
| 34–36 | Challenge problems | Gravity, pyramids |
Key Problem-Solving Strategies
For Spring Problems
- Find $k$: Use given force and displacement: $k = F/x$
- Set up integral: $W = \int_{x_1}^{x_2} kx\,dx$
- Watch units: If length in cm, convert to meters for SI
For Cable/Chain Problems
- Set up coordinates: Usually $x = 0$ at top (where it’s pulled to)
- Find force on a small piece: (weight/length) · $dx$
- Distance traveled: Each piece at position $x$ travels distance $x$
- Integrate: $W = \int_0^L (\text{linear density}) \cdot x\,dx$
For Pumping Problems
- Slice horizontally: Each slice is a thin disk of liquid
- Find volume of slice: $dV = A(y)\,dy$ where $A(y)$ is cross-sectional area
- Find weight: (density) × (volume) × $g$
- Find distance: How far does this slice travel to reach the spout?
- Integrate: $W = \int (\text{weight}) \cdot (\text{distance})\,dy$
Self-Assessment Quiz
Q1: A force of 40 N stretches a spring 0.1 m. What is the spring constant $k$?
By Hooke’s Law, $F = kx$:
\[k = \frac{F}{x} = \frac{40}{0.1} = 400 \text{ N/m}\]
Q2: What integral gives the work to stretch that spring from 0.1 m to 0.2 m?
\[W = \int_{0.1}^{0.2} 400x\,dx = 400 \cdot \frac{x^2}{2}\Big\vert _{0.1}^{0.2} = 200(0.04 - 0.01) = 6 \text{ J}\]
Q3: A 50-ft cable weighing 2 lb/ft hangs from a winch. Why isn’t the work simply $W = (100 \text{ lb})(50 \text{ ft}) = 5000$ ft-lb?
Because not all of the cable travels 50 feet!
- The bottom of the cable travels 50 ft
- The top of the cable travels 0 ft
- Each piece travels a different distance
We need to integrate: $W = \int_0^{50} 2x\,dx = x^2\Big\vert _0^{50} = 2500$ ft-lb
This is half of the naïve answer—on average, each pound travels 25 ft.
Q4: When pumping water out of a cone-shaped tank, why does the radius appear in the integral?
Each horizontal slice is a circular disk. The volume of a thin slice at height $y$ is:
\[dV = \pi r(y)^2\,dy\]The radius $r(y)$ varies with $y$ (the cone gets wider or narrower), so the weight of each slice—and thus the work to lift it—depends on $y$.
Deep Connections
The Work Integral as a Riemann Sum
Every work integral comes from the same reasoning:
\[W = \lim_{n \to \infty} \sum_{i=1}^{n} F(x_i^*) \Delta x = \int_a^b F(x)\,dx\]This is identical to the logic for area and volume. The “slice, approximate, sum, limit” pattern appears throughout calculus.
Work-Energy Theorem
One of the most important results in physics:
\[W = \Delta KE = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\]The work done on an object equals its change in kinetic energy. This connects the integral definition of work to Newton’s laws.
Connection to Potential Energy
When you do work against gravity to lift an object, that energy is stored as gravitational potential energy:
\[PE = mgh = \int_0^h mg\,dy\]The work integral tells you how much energy you’ve “deposited” into the object’s position.
Common Mistakes to Avoid
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Using total weight × total distance for cables | Not all pieces travel the same distance | Integrate: each piece’s distance varies |
| Forgetting to find $k$ in spring problems | Can’t integrate without the constant | Use given data: $k = F/x$ |
| Wrong limits for pumping problems | Must match where water starts and ends | Limits are $y$-values where water is |
| Mixing mass and weight | $F = mg$, not $F = m$ | In SI: multiply mass by 9.8; in US: pounds are already force |
| Wrong distance in pumping | Distance is to spout, not to top of water | Draw a picture; distance = (spout height) − $y$ |
Key Mathematical Themes
| Theme | How It Appears | Why It Matters |
|---|---|---|
| Integration as summation | $\sum F_i \Delta x \to \int F\,dx$ | Same Riemann sum reasoning from area/volume |
| Coordinate choice matters | Origin at top for cables, at bottom for tanks | Right setup simplifies the distance expression |
| Physical reasoning guides setup | “What varies? What’s constant?” | Understanding physics helps catch errors |
| Units carry information | N·m = J; lb·ft = ft-lb | Dimensional analysis checks your work |
Looking Ahead
The “slice and integrate” strategy continues:
- Section 5.5: Average value = total integral ÷ interval length
- Chapter 8: Arc length, surface area use similar reasoning
- Physics courses: Work appears everywhere—electrostatics, thermodynamics, mechanics
| Previous | Up | Next |
|---|---|---|
| Section 5.3 | Chapter 5 | Section 5.5 |