Problem 1 - The Calculus with Analytic Geometry

Question

A spring has a natural length of 8 in. If a force of \(20 \mathrm{lb}\) stretches the spring \(\frac{1}{2}\) in., find the work done in stretching the spring from \(8 \mathrm{in}\). to \(11 \mathrm{in}\).

Step-by-Step Solution

Answer

The work done is 180 in·lb.

Step 1 - Understand Hooke's Law

Hooke's Law states that the force required to stretch or compress a spring by some distance is proportional to that distance: \[ F = kx \] where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the distance stretched from the natural length.

Step 2 - Calculate the Spring Constant

Given a force of \( 20 \mathrm{lb} \) stretches the spring \( \frac{1}{2} \) inch, we can find the spring constant \( k \): \[ 20 = k \cdot \frac{1}{2} \] Solving for \( k \): \[ k = \frac{20}{1/2} = 40 \]\mathrm{lb}/\mathrm{in}.

Step 3 - Set Up the Work Integral

The work done to stretch a spring from \( x_1 \) to \( x_2 \) is calculated by the integral: \[ W = \int_{x_1}^{x_2} kx \, dx \] Here, the spring stretches from its natural length 8 in. (0 in. stretched) to a length of 11 in. (3 in. stretched), so the limits of integration are from 0 to 3.

Step 4 - Evaluate the Integral

Substitute the spring constant \( k = 40 \): \[ W = \int_{0}^{3} 40x \, dx \] Find the antiderivative of \( 40x \): \[ W = 40 \int_{0}^{3} x \, dx = 40 \left[ \frac{x^2}{2} \right]_{0}^{3} \] Evaluate this antiderivative: \[ W = 40 \left( \frac{3^2}{2} - \frac{0^2}{2} \right) = 40 \times \frac{9}{2} = 180 \mathrm{in}\cdot\mathrm{lb} \]

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics and engineering. It relates the force required to extend or compress a spring to the distance it is stretched or compressed. This law is usually stated mathematically as: \( F = kx \). Here,

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant, a measure of the stiffness of the spring.
• \( x \) is the distance the spring is stretched or compressed from its natural length.

Understanding Hooke's Law helps us analyze how materials and springs behave under different loads. It is especially crucial in problems involving spring mechanics and is the starting point for calculating other quantities like work done by the spring.

Spring Constant

The spring constant \( k \), also known as the stiffness constant, is an essential factor in Hooke's Law. It defines how stiff or flexible a spring is. A higher spring constant indicates a stiffer spring. To find the spring constant, you can rearrange Hooke's Law: \[ k = \frac{F}{x} \] For example, if a force of \( 20 \mathrm{lb} \) stretches the spring by \( \frac{1}{2} \mathrm{in} \), the spring constant would be: \( k = \frac{20}{1/2} = 40 \mathrm{lb}/\mathrm{in} \). This means it takes 40 pounds of force to stretch the spring by one inch. Knowing \( k \) is crucial for calculating work or potential energy stored in the spring.

Work Integral

The work done by a spring is a measure of energy needed to stretch or compress it. Using Hooke's Law, we can find this work by integrating the force over the distance stretched. Mathematically, the work \( W \) to stretch a spring from \( x_1 \) to \( x_2 \) is given by: \[ W = \int_{x_1}^{x_2} kx \ dx \] In our problem, the spring is stretched from its natural length by 3 inches. This integral calculates the area under the force-distance curve, giving the total work done. It's like adding up all the small bits of work done over each tiny distance to get the total.

Integration in Calculus

Integration is an essential concept in calculus. It helps us calculate quantities that accumulate over an interval, such as area under a curve, total distance, and in our case, work done by a spring. In this problem, we evaluated the integral: \[ \int_{0}^{3} 40x \ dx \] We found the antiderivative of \( 40x \) to be \( 40 \frac{x^2}{2} \). Next, we evaluated it at the bounds 3 and 0 to get: \[ 40 \left( \frac{3^2}{2} - \frac{0^2}{2} \right) = 40 \times \frac{9}{2} = 180 \mathrm{in} \cdot \mathrm{lb} \] This final value represents the total work done to stretch the spring.