Section 5.4: Work

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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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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

  1. Find $k$: Use given force and displacement: $k = F/x$
  2. Set up integral: $W = \int_{x_1}^{x_2} kx\,dx$
  3. Watch units: If length in cm, convert to meters for SI

For Cable/Chain Problems

  1. Set up coordinates: Usually $x = 0$ at top (where it’s pulled to)
  2. Find force on a small piece: (weight/length) · $dx$
  3. Distance traveled: Each piece at position $x$ travels distance $x$
  4. Integrate: $W = \int_0^L (\text{linear density}) \cdot x\,dx$

For Pumping Problems

  1. Slice horizontally: Each slice is a thin disk of liquid
  2. Find volume of slice: $dV = A(y)\,dy$ where $A(y)$ is cross-sectional area
  3. Find weight: (density) × (volume) × $g$
  4. Find distance: How far does this slice travel to reach the spout?
  5. 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!

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:


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