Problem 7 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The work done is 0.3125 lb-ft.

Step 1: Understand Hooke's Law

First, recall Hooke's Law, which states that the force required to stretch or compress a spring by some distance \( x \) from its natural length is given by \( F(x) = kx \), where \( k \) is the spring constant. We need to find this constant \( k \) first.

Step 2: Determine the Spring Constant

Given that a force of 10 lb is required to stretch the spring 4 in. beyond its natural length, we use Hooke's Law: \( 10 = k \times 4 \). Solve for \( k \) to find \( k = \frac{10}{4} = 2.5 \) lb/in.

Step 3: Set Up the Work Integral

Work done in stretching the spring from its natural length involves calculating the integral: \( W = \int_{0}^{x} F(x) \, dx = \int_{0}^{x} kx \, dx \), where \( x \) is the final displacement.

Step 4: Compute the Work Done from 0 to 6 inches

Plug in \( k = 2.5 \) and the displacement from 0 to 6 inches (0.5 feet since 6 inches is 0.5 feet) into the integral: \[W = \int_{0}^{0.5} 2.5 \cdot x \, dx\]Substitute the displacement limit for proper integration in inches, since we're given the force in lb.

Step 5: Integrate the Expression

Calculate \( \int_{0}^{0.5} 2.5x \, dx = 2.5 \left[ \frac{x^2}{2} \right]_{0}^{0.5} = 2.5 \times \frac{(0.5)^2}{2} \).

Step 6: Evaluate the Integral

The evaluation yields: \[2.5 \times \frac{0.25}{2} = 2.5 \times 0.125 = 0.3125 \] lb-ft.

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. In simple terms, it tells you how much force is needed to stretch or compress the spring by a unit length, like an inch or a meter. A large value of \( k \) indicates a stiffer spring that requires more force to stretch, while a small \( k \) indicates a more flexible spring.
Hooke's Law is the formula used to calculate this constant, expressed as \( F(x) = kx \), where \( F(x) \) is the force applied, and \( x \) is the displacement from the natural length.
In our exercise example, 10 pounds of force stretched the spring 4 inches. By rearranging Hooke's Law, we solve for \( k \):

• \( k = \frac{F(x)}{x} = \frac{10}{4} = 2.5 \) lb/in

This tells us that 2.5 pounds of force are needed to stretch this particular spring by 1 inch.

Work Done

Work done refers to the energy expended when a force moves an object over a distance. In the context of springs, work is the energy used to stretch or compress the spring. The general formula for work is \( W = F imes d \), where \( F \) is the force applied and \( d \) is the distance moved in the direction of the force.
However, since the force needed to stretch a spring changes with the deformation, we can't use this formula directly for springs. Instead, we must consider the changing force over the displacement. This is where calculus comes into play to provide a more accurate measurement.
The work done on springs is determined by evaluating the integral of the force over the displacement, leading us to accurately determine the actual work used in stretching or compressing a spring by a specified amount.

Calculus Integral

Integrals in calculus are a powerful tool for understanding areas under curves, among other concepts. In this context, they help in calculating the work done on a spring as it involves a force that changes with distance.
For our spring problem, the integral \( W = \int_{0}^{x} kx \, dx \) handles the varying force as the spring stretches. This lets us calculate the total work done across a continuous range of displacement.
The calculated integral from 0 to 0.5 foot with \( k = 2.5 \) lb/in was:

• \( W = \int_{0}^{0.5} 2.5x \, dx = 2.5 \left[ \frac{x^2}{2} \right]_{0}^{0.5} \)
• Result: \( 2.5 \times 0.125 = 0.3125 \) lb-ft

This method ensures we account for every tiny increment of force over the spring's extension, providing the exact work done.

Force and Displacement

Understanding force and displacement is central to solving our spring-related exercise. Force is any interaction that changes the motion of an object, and displacement refers to the distance moved in a specific direction. When considering springs, these two factors are intricately linked together.
In Hooke’s Law \( F(x) = kx \), the force \( F \) applied to stretch the spring is directly proportional to the displacement \( x \). This linear relationship means as you stretch the spring further, the force must increase proportionally.
For the exercise, we see that it took 10 pounds of force to extend the spring by 4 inches, proving this linear concept. Each additional inch of stretch beyond its natural length demands additional force, exactly regulated by the spring constant \( k = 2.5 \) lb/in.
Knowing this relationship simplifies calculating how much work is involved across a range of displacements.