Hooke's Law: Comprehensive Guide with Solved Examples

Hooke's Law is a fundamental principle in physics that describes the behavior of elastic materials when subjected to external forces. Understanding this law is essential for studying mechanical systems, structural engineering, and material science.

Hooke's Law Fundamentals

In the diagram below, we examine a block attached to a spring in three different configurations:

Spring systems demonstrating Hooke's Law

In configuration (A), the spring is at its equilibrium position with no external force applied. In position (B), an applied force \( F \) compresses the spring by displacement \( \Delta x \). In position (C), the same magnitude of force \( F \) stretches the spring by displacement \( \Delta x \).

Hooke's Law mathematically expresses the relationship between the applied force and spring displacement:

\[ |F| = k |\Delta x| \]

Where:

According to Newton's Third Law, the spring exerts an equal and opposite restoring force \( -F \) when stretched or compressed.

Problems with Detailed Solutions

Problem 1: Calculating Applied Force

What magnitude of force is required to stretch a 20 cm-long spring (spring constant \( k = 100 \, \text{N/m} \)) to a length of 21 cm?

Solution

The spring elongation is calculated as:

\[ \Delta x = 21 \, \text{cm} - 20 \, \text{cm} = 1 \, \text{cm} = 0.01 \, \text{m} \]

Applying Hooke's Law:

\[ |F| = k |\Delta x| = 100 \, \text{N/m} \times 0.01 \, \text{m} = 1 \, \text{N} \]

A force of 1 N is required to produce this elongation.

Problem 2: Determining Spring Constant

A spring requires 3 N of force to compress it from 40 cm to 35 cm. What is its spring constant?

Solution

The compression distance is:

\[ \Delta x = 40 \, \text{cm} - 35 \, \text{cm} = 5 \, \text{cm} = 0.05 \, \text{m} \]

Using Hooke's Law:

\[ k = \frac{|F|}{|\Delta x|} = \frac{3 \, \text{N}}{0.05 \, \text{m}} = 60 \, \text{N/m} \]

The spring constant is 60 N/m, indicating a relatively stiff spring.

Springs in Parallel Configuration

Springs arranged in parallel configuration

When springs are arranged in parallel, they share the applied load. The combined spring constant for parallel springs is the sum of individual constants:

\[ k_{\text{parallel}} = k_1 + k_2 + \cdots + k_n \]

This arrangement results in a stiffer composite spring system than any individual spring.

Problem 3: Parallel Spring System

Two springs with constants \( k_1 = 100 \, \text{N/m} \) and \( k_2 = 200 \, \text{N/m} \) are connected in parallel. What force is required to stretch this system by 6 cm?

Solution

The equivalent spring constant for parallel arrangement:

\[ k_{\text{eq}} = k_1 + k_2 = 100 \, \text{N/m} + 200 \, \text{N/m} = 300 \, \text{N/m} \]

Applying Hooke's Law with \( \Delta x = 0.06 \, \text{m} \):

\[ |F| = k_{\text{eq}} |\Delta x| = 300 \, \text{N/m} \times 0.06 \, \text{m} = 18 \, \text{N} \]

The parallel configuration requires 18 N, significantly more than either spring alone would require.

Springs in Series Configuration

Springs arranged in series configuration

In series arrangements, springs experience the same force but different displacements. The reciprocal of the combined spring constant equals the sum of reciprocals of individual constants:

\[ \frac{1}{k_{\text{series}}} = \frac{1}{k_1} + \frac{1}{k_2} + \cdots + \frac{1}{k_n} \]

Series combinations produce a less stiff composite spring than any individual component.

Problem 4: Series Spring System

The same springs (\( k_1 = 100 \, \text{N/m} \), \( k_2 = 200 \, \text{N/m} \)) are now connected in series. What force is needed to stretch this combination by 6 cm?

Solution

Calculate the equivalent spring constant:

\[ \frac{1}{k_{\text{eq}}} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200} \] \[ k_{\text{eq}} = \frac{200}{3} \approx 66.67 \, \text{N/m} \]

Applying Hooke's Law:

\[ |F| = k_{\text{eq}} |\Delta x| \approx 66.67 \, \text{N/m} \times 0.06 \, \text{m} \approx 4.0 \, \text{N} \]

The series arrangement requires only 4.0 N, making it more easily stretched than the parallel configuration.

Elastic Potential Energy in Springs

When a spring is deformed, work is done against the restoring force. This work is stored as elastic potential energy, given by:

\[ U = \frac{1}{2} k (\Delta x)^2 \]

Where \( U \) represents the potential energy in joules (J). This energy is recoverable when the spring returns to its equilibrium position.

Problem 5: Energy Storage in Springs

How much work is required to compress a spring ( \( k = 150 \, \text{N/m} \) ) from 15 cm to 10 cm?

Solution

The compression distance:

\[ \Delta x = 10 \, \text{cm} - 15 \, \text{cm} = -5 \, \text{cm} = -0.05 \, \text{m} \]

The work done equals the stored potential energy:

\[ W = U = \frac{1}{2} k (\Delta x)^2 = \frac{1}{2} \times 150 \, \text{N/m} \times (-0.05 \, \text{m})^2 = 0.1875 \, \text{J} \]

Thus, 0.1875 joules of energy are stored in the compressed spring.

Key Takeaways