← MathScape

Spring Work and Hooke's Law

MATH162
Reference: Stewart 5.4  •  Chapter: 5  •  Section: 4

Navigation: Wiki Home > Skills > Spring Work

Spring Work and Hooke's Law

Why Springs Need Calculus

Stretch a spring a little, and it pulls back gently. Stretch it more, and it pulls back harder. This increasing resistance is described by Hooke's Law: the force is proportional to the displacement.

Because the force changes as you stretch the spring, calculating work requires integration. You can't just multiply force times distance when the force itself depends on distance.

The physics: Springs store energy as you stretch them. The work you do becomes potential energy stored in the spring.

Prerequisite Map

Prerequisites
Work with IntegralsPolynomial Integration
→
This skill
Spring Work
→
Unlocks
Pumping WorkElastic Potential Energy

Before You Start

Quick self-check. If you struggle, review the linked prerequisite first.

1. Work Integral: A force $f(x) = 5x$ newtons acts on an object moving from $x = 0$ to $x = 2$ m. What is the work done?

Check Your Answer

$$W = \int_0^2 5x\,dx = \frac{5x^2}{2}\Big\vert _0^2 = \frac{5(4)}{2} = 10 \text{ J}$$

✅ Got 10 J? You're ready for spring problems!

❌ Stuck on the integral? Review Work with Integrals first—springs are just a special case of $f(x) = kx$.

2. Unit Conversion: Convert 8 inches to feet.

Check Your Answer

$$8 \text{ in} = \frac{8}{12} \text{ ft} = \frac{2}{3} \text{ ft} \approx 0.667 \text{ ft}$$

✅ Got it? Unit conversion is critical for spring problems!

❌ Made an error? Remember: 12 inches = 1 foot. Spring problems often mix inches and feet—always convert before integrating.

Quick Reference

→ Jump to Practice Problems

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

Key Concepts

Hooke's Law

The force required to maintain a spring stretched $x$ units beyond its natural length is:

$$\boxed{f(x) = kx}$$

where:

Important: The spring constant $k$ is always positive. A larger $k$ means a stiffer spring.

Visualizing Hooke's Law

Natural length (x = 0):      Stretched by x:
┌──╔═══╗──┐                  ┌──╔═══════════╗──┐
│  ║   ║  │                  │  ║           ║  │──→ f(x) = kx
└──╚═══╝──┘                  └──╚═══════════╝──┘
                                    ←─ x ─→

The force $f(x) = kx$ acts in the opposite direction of the stretch, trying to restore the spring to its natural length.

Work Done on a Spring

To stretch (or compress) a spring from position $x = a$ to $x = b$:

$$\boxed{W = \int_a^b kx \, dx = \frac{1}{2}k(b^2 - a^2)}$$

Special case: Starting from natural length ($a = 0$):

$$W = \int_0^b kx \, dx = \frac{1}{2}kb^2$$

This is also the formula for elastic potential energy stored in a spring.

Still confused about why we integrate? The force $f(x) = kx$ changes at every position. Review Work with Integrals to see how variable force leads to integration.

Finding the Spring Constant

Given force and displacement data, solve for $k$:

$$k = \frac{F}{x}$$

Common setup: "A force of $F$ newtons is required to hold a spring stretched $x$ meters beyond its natural length."

Units Summary

Quantity SI Units US Units
Force $F$ newtons (N) pounds (lb)
Displacement $x$ meters (m) feet (ft) or inches (in)
Spring constant $k$ N/m lb/ft or lb/in
Work $W$ joules (J) foot-pounds (ft-lb)

Unit conversion: Be careful with inches vs. feet! Convert to consistent units before calculating.

Common Pitfalls

Mistake Why It Happens How to Avoid
Using total length instead of displacement Problem says "stretched to 30 cm" but you use 30 cm as $x$ $x$ = (current length) − (natural length). Always subtract!
Mixing units Problem gives inches, you integrate without converting Convert ALL lengths to the same unit (feet or meters) before setting up the integral.
Forgetting to find $k$ first Jump straight to work formula Most problems give force at ONE position. Use $k = F/x$ to find the spring constant first.
Using $W = Fx$ for springs Treating spring force as constant Spring force varies! You must integrate: $W = \int kx\,dx$
Wrong limits on non-zero start Starting at $x = 4$ in, ending at $x = 10$ in Use displacement from natural length for BOTH limits. If natural length is 0, limits are $a = 4$ in, $b = 10$ in (as displacements).
Historical Note: Robert Hooke's Discovery

Robert Hooke published his spring law in 1678 as a Latin anagram: "ceiiinosssttuv," which unscrambles to "ut tensio, sic vis" ("as the extension, so the force"). He used the anagram to establish priority while keeping the discovery secret! Hooke was famously paranoid about rivals—especially Isaac Newton—stealing his ideas. Today, Hooke's Law is one of the first physics equations students encounter, fundamental to everything from watch springs to building earthquake resistance.

Practice Problems {#practice-problems}

Level 1 Finding the Spring Constant

A spring has natural length 25 cm. A force of 20 N is required to keep it stretched to a length of 30 cm. What is the spring constant $k$?

Many students initially use 30 cm instead of 5 cm for $x$. Remember: displacement is from natural length!

Thought Process

The displacement $x$ is the stretch beyond natural length, not the total length.

  • Natural length: 25 cm
  • Stretched length: 30 cm
  • Displacement: $x = 30 - 25 = 5$ cm $= 0.05$ m

Then use $k = F/x$.

Show Answer

Step 1: Find displacement $$x = 30 \text{ cm} - 25 \text{ cm} = 5 \text{ cm} = 0.05 \text{ m}$$

Step 2: Apply Hooke's Law $$k = \frac{F}{x} = \frac{20 \text{ N}}{0.05 \text{ m}} = 400 \text{ N/m}$$

The spring constant is $k = 400$ N/m.

Level 2 Work to Stretch a Spring

A spring with spring constant $k = 250$ N/m is stretched from its natural length to 0.12 m beyond its natural length. How much work is done?

Thought Process

Starting from natural length means $a = 0$ and stretching to $b = 0.12$ m.

Use the formula: $$W = \int_0^b kx \, dx = \frac{1}{2}kb^2$$

Show Answer

$$W = \int_0^{0.12} 250x \, dx = 250 \cdot \frac{x^2}{2} \Big\vert _0^{0.12}$$

$$= 125(0.12)^2 = 125(0.0144) = 1.8 \text{ J}$$

The work done is 1.8 joules.

Level 3 Two-Step Spring Problem

A force of 8 lb is required to hold a spring stretched 6 inches beyond its natural length. How much work is done in stretching the spring from 4 inches to 10 inches beyond its natural length?

Thought Process

Two steps:

  1. Find $k$ from the given data (force = 8 lb at $x = 6$ in = 0.5 ft)
  2. Calculate work from $a = 4$ in to $b = 10$ in (convert to feet!)

Watch the unit conversion: 6 in = 0.5 ft, 4 in = 1/3 ft, 10 in = 5/6 ft.

Show Answer

Step 1: Find spring constant $$k = \frac{F}{x} = \frac{8 \text{ lb}}{0.5 \text{ ft}} = 16 \text{ lb/ft}$$

Step 2: Convert displacements to feet

  • $a = 4 \text{ in} = \frac{4}{12} = \frac{1}{3}$ ft
  • $b = 10 \text{ in} = \frac{10}{12} = \frac{5}{6}$ ft

Step 3: Calculate work $$W = \int_{1/3}^{5/6} 16x \, dx = 16 \cdot \frac{x^2}{2} \Big\vert _{1/3}^{5/6}$$

$$= 8\left[\left(\frac{5}{6}\right)^2 - \left(\frac{1}{3}\right)^2\right] = 8\left[\frac{25}{36} - \frac{1}{9}\right]$$

$$= 8\left[\frac{25}{36} - \frac{4}{36}\right] = 8 \cdot \frac{21}{36} = \frac{168}{36} = \frac{14}{3} \text{ ft-lb}$$

The work done is $\frac{14}{3} \approx 4.67$ ft-lb.

Level 4 Finding Natural Length from Work Data

It takes 5 J of work to stretch a spring from 12 cm to 15 cm, and 9 J of work to stretch it from 15 cm to 18 cm. What is the natural length of the spring?

Thought Process

Let $L$ = natural length (in meters). Then:

  • From 12 cm to 15 cm: stretch from $(0.12 - L)$ to $(0.15 - L)$
  • From 15 cm to 18 cm: stretch from $(0.15 - L)$ to $(0.18 - L)$

Work formula: $W = \frac{k}{2}(b^2 - a^2)$

Set up two equations with two unknowns ($k$ and $L$) and solve.

Show Answer

Let $L$ = natural length (in meters). Define displacements from natural length:

  • At 12 cm: $x_1 = 0.12 - L$
  • At 15 cm: $x_2 = 0.15 - L$
  • At 18 cm: $x_3 = 0.18 - L$

Equation 1: Work from 12 to 15 cm $$5 = \frac{k}{2}(x_2^2 - x_1^2) = \frac{k}{2}[(0.15-L)^2 - (0.12-L)^2]$$

Equation 2: Work from 15 to 18 cm $$9 = \frac{k}{2}(x_3^2 - x_2^2) = \frac{k}{2}[(0.18-L)^2 - (0.15-L)^2]$$

Divide equation 2 by equation 1: $$\frac{9}{5} = \frac{(0.18-L)^2 - (0.15-L)^2}{(0.15-L)^2 - (0.12-L)^2}$$

Using difference of squares: $a^2 - b^2 = (a+b)(a-b)$

Numerator: $(0.18-L + 0.15-L)(0.18-L - 0.15+L) = (0.33 - 2L)(0.03)$

Denominator: $(0.15-L + 0.12-L)(0.15-L - 0.12+L) = (0.27 - 2L)(0.03)$

$$\frac{9}{5} = \frac{(0.33 - 2L)(0.03)}{(0.27 - 2L)(0.03)} = \frac{0.33 - 2L}{0.27 - 2L}$$

Cross-multiply: $$9(0.27 - 2L) = 5(0.33 - 2L)$$ $$2.43 - 18L = 1.65 - 10L$$ $$0.78 = 8L$$ $$L = 0.0975 \text{ m} = 9.75 \text{ cm}$$

The natural length is approximately 9.75 cm (or 0.0975 m).

Level 5 Energy Storage and the Odd Number Pattern

A physics student measures that a spring stores 4.5 J of potential energy when stretched 0.15 m from its natural length. (Recall: potential energy stored = work done to stretch.)

  1. Determine the spring constant $k$.
  2. How much energy is stored when the spring is stretched only 0.05 m? Is this one-third of 4.5 J? Explain.
  3. Find the work required to stretch from 0.05 m to 0.10 m displacement, and compare it to the work for the first 0.05 m (from 0 to 0.05 m).
  4. Prove the general result: if $W_1$ is the work to stretch from 0 to $\Delta x$, then the work to stretch from $(n-1)\Delta x$ to $n\Delta x$ is $(2n-1)W_1$.
  5. Verify your formula by showing that $W_1 + W_2 + W_3 + \cdots + W_n = \frac{1}{2}k(n\Delta x)^2$. (Hint: What is $1 + 3 + 5 + \cdots + (2n-1)$?)
Thought Process

This problem connects spring work to potential energy and uncovers a beautiful mathematical pattern.

(a) Use $PE = \frac{1}{2}kx^2$ with the given data to find $k$.

(b) Energy scales as $x^2$, not linearly! So one-third the displacement gives one-ninth the energy, not one-third.

(c) Use the work formula $W = \frac{k}{2}(b^2 - a^2)$ for the interval [0.05, 0.10]. Compare to $W_1 = \frac{1}{2}k(0.05)^2$.

(d) Derive the general formula by computing $W_n = \frac{k}{2}[(n\Delta x)^2 - ((n-1)\Delta x)^2]$ and simplifying.

(e) The sum $1 + 3 + 5 + \cdots + (2n-1) = n^2$ is a famous identity. Use it to verify consistency.

Show Answer

(a) Finding $k$:

Using $PE = \frac{1}{2}kx^2$:

$$4.5 = \frac{1}{2}k(0.15)^2 = \frac{1}{2}k(0.0225)$$

$$k = \frac{4.5 \times 2}{0.0225} = \frac{9}{0.0225} = 400 \text{ N/m}$$

(b) Energy at 0.05 m displacement:

$$PE = \frac{1}{2}(400)(0.05)^2 = 200(0.0025) = 0.5 \text{ J}$$

Is this one-third of 4.5 J? No! $\frac{4.5}{3} = 1.5$ J, but we got 0.5 J.

Why? Energy depends on $x^2$. When displacement is multiplied by 3 (from 0.05 to 0.15), energy is multiplied by $3^2 = 9$. Indeed: $0.5 \times 9 = 4.5$ J ✓

(c) Work for second interval:

First interval (0 to 0.05 m): $$W_1 = \frac{1}{2}k(0.05)^2 = 0.5 \text{ J}$$

Second interval (0.05 to 0.10 m): $$W_2 = \frac{k}{2}[(0.10)^2 - (0.05)^2] = 200[0.01 - 0.0025] = 200(0.0075) = 1.5 \text{ J}$$

Ratio: $W_2/W_1 = 1.5/0.5 = 3$

The second 0.05 m requires 3 times as much work as the first—even though both intervals are the same length! This is because the spring is already stretched, so the force is larger throughout the second interval.

(d) General derivation:

For the $n$th interval, from $(n-1)\Delta x$ to $n\Delta x$:

$$W_n = \frac{k}{2}[(n\Delta x)^2 - ((n-1)\Delta x)^2]$$

Factor out $(\Delta x)^2$:

$$W_n = \frac{k(\Delta x)^2}{2}[n^2 - (n-1)^2]$$

Expand $(n-1)^2 = n^2 - 2n + 1$:

$$W_n = \frac{k(\Delta x)^2}{2}[n^2 - n^2 + 2n - 1] = \frac{k(\Delta x)^2}{2}(2n - 1)$$

Since $W_1 = \frac{k(\Delta x)^2}{2}$:

$$\boxed{W_n = (2n-1)W_1}$$

Verification: $W_1 = 1 \cdot W_1$ ✓, $W_2 = 3W_1$ ✓, $W_3 = 5W_1$ ✓

(e) Sum verification:

Total work for $n$ intervals:

$$W_1 + W_2 + \cdots + W_n = W_1(1 + 3 + 5 + \cdots + (2n-1))$$

The sum of the first $n$ odd numbers equals $n^2$:

$$= W_1 \cdot n^2 = \frac{k(\Delta x)^2}{2} \cdot n^2 = \frac{1}{2}k(n\Delta x)^2$$

This equals the potential energy at displacement $n\Delta x$, confirming our formula! ✓

Physical insight: The odd-number pattern $(1, 3, 5, 7, \ldots)$ for successive work increments is a direct consequence of $PE \propto x^2$. Each new "layer" of stretch requires more work because you're fighting a stronger spring force.

Conceptual Questions (CCI-Style)

Conceptual Doubling the Stretch

Spring A is stretched 10 cm from its natural length, requiring work $W$. Spring B (identical to A) is stretched 20 cm from its natural length. How much work is required for Spring B?

  1. $W$
  2. $2W$
  3. $4W$
  4. $8W$
Thought Process

Work to stretch from natural length is $W = \frac{1}{2}kb^2$. If you double $b$, what happens to $W$?

Show Answer

Answer: (C)

Work from natural length: $W = \frac{1}{2}kb^2$

  • Spring A: $W_A = \frac{1}{2}k(0.1)^2 = 0.005k$
  • Spring B: $W_B = \frac{1}{2}k(0.2)^2 = 0.02k$

Ratio: $\frac{W_B}{W_A} = \frac{0.02k}{0.005k} = 4$

Since work depends on $b^2$, doubling the stretch quadruples the work.

Mastery Checklist

Mental Model

The Stacking Bricks Analogy:

Imagine stacking bricks. The first brick is easy—you lift it from ground level. The second brick is harder because you're lifting against gravity AND you have to lift it higher. Each successive brick requires more effort because:

  1. You're lifting to a greater height
  2. The stack itself resists being made taller

A spring works the same way. Each bit of additional stretch requires more force than the last. The work "accumulates faster" as you go—that's why it's proportional to $x^2$, not just $x$.

Fun Fact: The World's Stiffest Springs

The springs in a car's suspension have spring constants around 25,000–50,000 N/m. But the "springs" in atomic force microscopes—used to image individual atoms—have constants of about 0.01–1 N/m. At the other extreme, the "effective spring constant" of a neutron star's crust is estimated at $10^{30}$ N/m. Hooke's Law applies across 32 orders of magnitude!

Connections

Looking back:

Looking ahead:

Real-world connections:

Previous Up Next
Work with Integrals Ch5 Sec4 Skills Pumping Work

Last updated: 2026-01-23