Work with Constant and Variable Forces

9 min read

Reference: Stewart §5.4

Why Force Times Distance Isn’t Enough

When you push a book across a table, the work you do is simple: force times distance. But what if the force changes as you move? A spring gets harder to stretch the farther you pull it. Lifting a heavy cable becomes easier as more of it reaches the top. Pumping water from a tank requires more effort for the water at the bottom.

In these situations, the force isn’t constant, so we can’t just multiply. Instead, we break the motion into tiny pieces where the force is approximately constant, calculate the work for each piece, and add them up. This is exactly what a definite integral does.

The key insight: Work is the accumulation of force over distance, and integration is the mathematical tool for accumulation.

Prerequisite Map

Before You Start

Quick self-check (30 seconds each). If you struggle with these, review the linked topics first.

1. Definite Integrals: Evaluate $\displaystyle\int_1^4 (2x + 3)\,dx$

Check Your Answer

$$\int_1^4 (2x + 3)\,dx = \left[x^2 + 3x\right]_1^4 = (16 + 12) - (1 + 3) = 24$$

✅ Got 24? Great: you’re ready to proceed!

❌ Stuck? Review Definite Integrals before continuing.

2. Riemann Sums: What does the sum $\displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x$ approximate as $n \to \infty$?

Check Your Answer

It approximates the definite integral $\displaystyle\int_a^b f(x)\,dx$.

Geometrically, each term $f(x_i^*)\Delta x$ is the area of a thin rectangle. Adding them approximates the area under $y = f(x)$.

✅ Makes sense? You’re ready!

❌ Fuzzy on this? Review Riemann Sums: the conceptual understanding is important for seeing why work becomes an integral.

Quick Reference

→ Jump to Practice Problems

Property	Value
Section	Chapter 5, Section 4
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

Definition: Work with Constant Force

When a constant force $F$ moves an object a distance $d$ in the direction of the force:

$$\boxed{W = Fd}$$

Units:

SI: Force in newtons (N), distance in meters (m), work in joules (J = N·m)
US: Force in pounds (lb), distance in feet (ft), work in foot-pounds (ft-lb)

Important distinction: Weight is a force (already includes gravity), while mass must be multiplied by $g = 9.8 \text{ m/s}^2$ to get force.

Visualization: From Constant to Variable Force

Constant Force:                    Variable Force:
   F                                  f(x)
   ↓↓↓↓↓↓↓↓                          ↓  ↓↓  ↓↓↓
   ────────                          ────────
   a       b                         a       b

W = F × (b-a)                      W = ∫ₐᵇ f(x) dx

Definition: Work with Variable Force

When force $f(x)$ varies with position as an object moves from $x = a$ to $x = b$:

$$\boxed{W = \int_a^b f(x) \, dx}$$

Why This Works

Divide the interval $[a, b]$ into $n$ small pieces of width $\Delta x$. On each piece, the force is approximately constant at $f(x_i^*)$, so the work on that piece is approximately:

$$W_i \approx f(x_i^*) \Delta x$$

Total work is the sum of all pieces:

$$W \approx \sum_{i=1}^{n} f(x_i^*) \Delta x$$

As $n \to \infty$ and $\Delta x \to 0$, this Riemann sum becomes the definite integral.

Still confused about Riemann sums? This is the same idea as approximating area under a curve with rectangles. Each rectangle has width $\Delta x$ and height $f(x_i^*)$. Review Riemann sums →

Physical Interpretation

Area under the force curve: The work equals the area under $y = f(x)$ from $x = a$ to $x = b$
Positive vs. negative work: If force and motion are in opposite directions, work is negative
Units check: $[f(x)] \times [dx] = \text{force} \times \text{distance} = \text{work}$ ✓

Common Force Types

Scenario	Force Function	Notes
Constant force	$f(x) = F$	Lifting at constant speed
Linear force	$f(x) = kx$	Springs (Hooke’s Law)
Inverse square	$f(x) = \frac{k}{x^2}$	Gravity at varying distances
Position-dependent	$f(x) = g(x)$	Given force function

Common Pitfalls

Mistake	Why It Happens	How to Avoid
Confusing mass and weight	Given “15 kg” and using 15 as the force	Mass (kg) needs $F = mg$. Weight (lb, N) is already a force.
Wrong limits of integration	Mixing up starting and ending positions	Draw a diagram. Label positions clearly before setting up the integral.
Forgetting units	Getting a number without checking reasonableness	Always write units. Work = force × distance, so J = N·m or ft-lb.
Using $W = Fd$ for variable force	Force changes but you multiply anyway	If $f$ depends on $x$, you must integrate: $W = \int f(x)\,dx$

Historical Note: Why "Work"?

The concept of work was formalized in the early 1800s by French mathematician Gaspard-Gustave Coriolis (yes, the Coriolis effect is named after him). He defined work as “weight lifted through a height” to analyze the efficiency of machines and waterwheels. The integral formulation came from engineers trying to calculate how much useful output they could get from steam engines: the more work per unit of fuel, the better the engine.

Practice Problems

Level 1 Lifting a Toolbox

A worker lifts a 15-kg toolbox from the ground to a shelf 1.8 m high. How much work is done? Use $g = 9.8 \text{ m/s}^2$.

Thought Process

This is a constant force problem. The force needed to lift the toolbox equals the weight (mass times gravity). Since the height is constant, we use $W = Fd$.

Steps:

Calculate force: $F = mg$
Distance is the height: $d = 1.8$ m
Apply $W = Fd$

Show Answer

Step 1: Find the force (weight of toolbox) $$F = mg = (15 \text{ kg})(9.8 \text{ m/s}^2) = 147 \text{ N}$$

Step 2: Calculate work $$W = Fd = (147 \text{ N})(1.8 \text{ m}) = 264.6 \text{ J}$$

The work done is approximately 265 joules.

Level 2 Variable Force Along a Path

A particle moves along the $x$-axis from $x = 0$ to $x = 4$ meters. A force of $f(x) = 3x + 2$ newtons acts on it. Find the work done.

Thought Process

The force varies with position, so we need the integral formula: $$W = \int_a^b f(x) \, dx$$

Here $a = 0$, $b = 4$, and $f(x) = 3x + 2$. This is a straightforward polynomial integral.

Show Answer

$$W = \int_0^4 (3x + 2) \, dx = \left[\frac{3x^2}{2} + 2x\right]_0^4$$

$$= \left(\frac{3(16)}{2} + 8\right) - (0) = 24 + 8 = 32 \text{ J}$$

The work done is 32 joules.

Level 3 Force with Inverse Relationship

A repelling magnetic force of $f(x) = \frac{12}{x^2}$ newtons acts on a charged particle, where $x$ is the distance in meters from a fixed magnet. How much work is done in moving the particle from $x = 2$ m to $x = 6$ m?

Thought Process

We need $W = \int_2^6 \frac{12}{x^2} dx$.

The antiderivative of $x^{-2}$ is $-x^{-1} = -\frac{1}{x}$.

Watch the signs carefully when evaluating the definite integral.

Show Answer

$$W = \int_2^6 \frac{12}{x^2} \, dx = 12 \int_2^6 x^{-2} \, dx$$

$$= 12 \left[-x^{-1}\right]_2^6 = 12 \left[-\frac{1}{6} - \left(-\frac{1}{2}\right)\right]$$

$$= 12 \left(-\frac{1}{6} + \frac{1}{2}\right) = 12 \left(\frac{-1 + 3}{6}\right) = 12 \cdot \frac{2}{6} = 4 \text{ J}$$

The work done is 4 joules.

Level 4 Setting Up the Work Integral

A 50-foot chain weighing 3 lb/ft hangs from the top of a building. How much work is required to pull the entire chain to the top?

Thought Process

This is tricky because different parts of the chain travel different distances. The bottom of the chain travels 50 ft, but the top barely moves.

Set up coordinates: Let $x$ = distance from the top of the building (pointing down). A small piece of chain at position $x$ has:

Length: $dx$ ft
Weight: $(3 \text{ lb/ft})(dx \text{ ft}) = 3 \, dx$ lb
Distance to lift: $x$ ft

Work for this piece: $(3 \, dx) \cdot x = 3x \, dx$

Total work: integrate from $x = 0$ to $x = 50$.

Show Answer

Set up: Place origin at top of building, $x$-axis pointing down.

A small segment of chain at depth $x$ with length $dx$:

Weight = $(3 \text{ lb/ft})(dx) = 3\,dx$ lb
Must be lifted distance $x$

Work integral: $$W = \int_0^{50} 3x \, dx = 3 \cdot \frac{x^2}{2} \Big\vert _0^{50}$$

$$= \frac{3}{2}(2500 - 0) = 3750 \text{ ft-lb}$$

The work required is 3750 ft-lb.

Level 5 Work-Energy Analysis of a Slingshot

A toy slingshot exerts a restoring force $f(x) = 80x$ N when the pouch is pulled back $x$ meters from equilibrium. A 50-gram projectile (mass = 0.05 kg) is placed in the pouch, pulled back 0.12 m, and released from rest.

Calculate the work done by the slingshot on the projectile as it accelerates from $x = 0.12$ m to $x = 0$ (equilibrium).
Derive the Work-Energy Theorem: Starting from $W = \int_{x_1}^{x_2} f(x)\,dx$ and using Newton's Second Law ($F = ma$), show that $W = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$. (Hint: Use the substitution $u = v(t)$ where $v = \frac{dx}{dt}$.)
Apply the theorem: What is the projectile's velocity at the moment it leaves the slingshot (at $x = 0$)?
If you wanted to double the launch velocity, by what factor would you need to increase the pullback distance? Explain using energy considerations.

Thought Process

This problem connects spring-like work to the Work-Energy Theorem through a concrete application.

(a) The slingshot acts like a spring with $k = 80$ N/m. Calculate work using $W = \int kx\,dx$ from $x = 0.12$ to $x = 0$. Watch the sign!

(b) The derivation requires converting the spatial integral to a velocity integral:

Start: $W = \int f(x)\,dx$
Use $f = ma = m\frac{dv}{dt}$ and $dx = v\,dt$
Substitute $u = v$ to integrate

(c) Apply $W = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$ with $v_1 = 0$ (starts at rest).

(d) Since $KE \propto v^2$ and $W \propto x^2$ (for spring-like force), think about how these scale.

Show Answer

(a) Work done by slingshot:

As the projectile moves from $x = 0.12$ m toward equilibrium, the force is in the direction of motion (accelerating the projectile).

$$W = \int_{0.12}^{0} (-80x) \, dx$$

The negative sign accounts for the force direction (pointing toward $x = 0$ when $x > 0$):

$$W = -80 \cdot \frac{x^2}{2} \Big\vert _{0.12}^{0} = -40[(0)^2 - (0.12)^2] = -40(-0.0144) = 0.576 \text{ J}$$

The slingshot does 0.576 J of positive work on the projectile.

(b) Derivation of Work-Energy Theorem:

Start with the work integral: $$W = \int_{x_1}^{x_2} f(x) \, dx$$

Let $x = s(t)$ be position at time $t$. Then $dx = v(t)\,dt$.

$$W = \int_{t_1}^{t_2} f(s(t)) \cdot v(t) \, dt$$

By Newton’s Second Law, $f = ma = m\frac{dv}{dt}$:

$$W = \int_{t_1}^{t_2} m\frac{dv}{dt} \cdot v \, dt$$

Substitute $u = v(t)$, so $du = \frac{dv}{dt}\,dt$:

$$W = \int_{v_1}^{v_2} m \cdot u \, du = \frac{mu^2}{2} \Big\vert _{v_1}^{v_2} = \boxed{\frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2}$$

(c) Launch velocity:

Apply the theorem with $v_1 = 0$ (released from rest):

$$0.576 = \frac{1}{2}(0.05)v_2^2 - 0$$

$$v_2^2 = \frac{0.576 \times 2}{0.05} = \frac{1.152}{0.05} = 23.04$$

$$v_2 = \sqrt{23.04} \approx 4.8 \text{ m/s}$$

The projectile leaves at approximately 4.8 m/s (about 10.7 mph).

(d) Doubling velocity:

To double $v$, we need to quadruple $KE$ (since $KE \propto v^2$).

This requires quadrupling the work: $W_{\text{new}} = 4W_{\text{old}}$.

Since work for a spring-like force is $W = \frac{1}{2}kx^2$, and $W \propto x^2$:

$$\frac{W_{\text{new}}}{W_{\text{old}}} = \frac{x_{\text{new}}^2}{x_{\text{old}}^2} = 4 \implies x_{\text{new}} = 2x_{\text{old}}$$

To double the launch velocity, double the pullback distance.

This is a consequence of both $W \propto x^2$ and $KE \propto v^2$, which cancel to give the simple linear relationship.

Conceptual Questions (CCI-Style)

Conceptual Interpreting Work Graphically

The graph shows force $f(x)$ (in newtons) versus position $x$ (in meters) for an object moving from $x = 0$ to $x = 5$.

f(x)
  8 |     ___________
    |    /
  4 |___/
    |
  0 +---+---+---+---+---→ x
    0   1   2   3   4   5

Which statement best describes the work done?

The work equals $8 \times 5 = 40$ J
The work equals the area of the shaded region under the curve
The work cannot be determined without knowing the mass
The work is zero because the force eventually becomes constant

Thought Process

Work is $\int f(x) \, dx$, which geometrically represents the area under the force curve. The mass doesn’t appear in the work formula directly. The force being constant later doesn’t make the total work zero.

Show Answer

Answer: (B)

Work is the definite integral $W = \int_0^5 f(x) \, dx$, which geometrically equals the area under the force curve from $x = 0$ to $x = 5$.

(A) is wrong because it treats the force as constant at 8 N throughout
(C) is wrong because work depends on force and distance, not mass directly
(D) is wrong because constant force still does work over distance

Common Misconceptions

Common misconception

work always equals force times distance, so $W = F \cdot d$ can be applied even when the force varies.

This is the rate-as-fixed-number error. The formula $W = Fd$ applies only when the force is constant throughout the displacement. When force depends on position, as with a spring governed by $f(x) = kx$ or a chain where the effective weight decreases as it is pulled up, using a single force value multiplied by total distance ignores how the force changes. For a spring with $k = 250$ N/m stretched from 0 to 0.12 m, using the maximum force $250 \times 0.12 = 30$ N gives $W = 30 \times 0.12 = 3.6$ J, which overstates the correct answer of 1.8 J because the force was 30 N only at the very end of the stretch.

Mastery Checklist

I can calculate work using $W = Fd$ for constant force problems
I distinguish between mass (kg) and weight (N or lb) in work problems
I can set up and evaluate $W = \int_a^b f(x) \, dx$ for variable force
I can explain why work equals area under the force-position curve
I understand how Riemann sums lead to the work integral
I can set up work integrals for distributed weight problems (chains/cables)

Mental Model

Work as Accumulated Effort:

Think of work like filling a bucket with water one cup at a time. Each cup is a small bit of effort (force × tiny distance). The total work is adding up all those small efforts. When force varies, some cups are heavier than others, but we still add them all up. The integral is just a precise way of doing this addition when there are infinitely many infinitely small cups.

Fun Fact: How Much Work to Climb Everest?

Mount Everest is 8,849 m tall. For a 70 kg climber, the work against gravity alone is: $$W = mgh = (70)(9.8)(8849) \approx 6.1 \times 10^6 \text{ J} = 6.1 \text{ MJ}$$

That’s equivalent to about 1,450 food Calories, roughly a day’s worth of eating. (In reality, climbers burn far more because of inefficiency, cold, and carrying gear.)

Connections

Looking back:

The work integral arises from Riemann sums, just like area under a curve
This is an application of the Fundamental Theorem of Calculus

Looking ahead:

Spring Work applies this to Hooke’s Law
Pumping Work extends to 3D tank problems

Real-world connections:

Engineers calculate work to size motors and engines
The Work-Energy Theorem is fundamental to mechanics

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Last updated: 2026-01-23