Problem 3 - The Calculus with Analytic Geometry

Question

A spring has a natural length of 6 in. A \(12,000-\mathrm{lb}\) force compresses the spring to \(5 \frac{1}{2}\) in. Find the work done in compressing it from 6 in. to 5 in. Hooke's law holds for compression as well as for extension.

Step-by-Step Solution

Answer

The work done is 12,000 in-lb.

Step 1: Understand Hooke's Law

Hooke's law states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula is given by \[ F = k \times x \] where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the natural length.

Step 2: Determine the Spring Constant

Given that a 12,000-lb force compresses the spring from its natural length of 6 in. to 5.5 in., we can find the spring constant (\( k \)). The displacement, \( x \), is 0.5 in. Setup the equation:\[ 12,000 = k \times 0.5 \]Solve for \( k \):\[ k = \frac{12,000}{0.5} = 24,000 \] lb/in.

Step 3: Setup the Work Integral

To find the work done in compressing the spring from 6 in. to 5 in., use the work formula for a spring:\[ W = \frac{1}{2} k x^2 \] Integrate this expression from the initial displacement \( x_i = 0 \) (6 in.) to the final displacement \( x_f = 1 \) (5 in.). The integral becomes:\[ W = \frac{1}{2} \times 24,000 \times \bigg[ \frac{1^2}{2} \bigg|_0^1 \bigg] \]

Step 4: Evaluate the Integral

Evaluate the integral:\[ W = \frac{1}{2} \times 24,000 \times (1^2 - 0^2) = \frac{1}{2} \times 24,000 \times 1 = 12,000 \] The work done in compressing the spring from 6 in. to 5 in. is 12,000 in-lb.

Key Concepts

work integration

Work integration is a key concept in physics, especially when studying springs and Hooke's law. To understand it better, let's break it down. When you compress or extend a spring, you apply a force over a distance, and this is where work comes into play. We use integration because the force is not constant; it changes as the spring compresses or extends.

Specifically, the work done on a spring is calculated using the integral of the force function over the distance it's applied. The general formula for work done in compressing or extending a spring from position \(x_i\) to \(x_f\) is given by:

\ W = \int_{x_i}^{x_f} kx \ dx \

Here, \(k\) is the spring constant, and \(x\) is the displacement. By integrating the force function \(kx\) over the specified limits, we get the total work done on the spring.

spring constant

The spring constant, represented as \(k\), is a measure of the stiffness of a spring. Strength and flexibility of springs depend heavily on this constant.

To find the spring constant, we use Hooke's law, which states that the force required to compress or extend a spring is directly proportional to the distance the spring is stretched or compressed. Mathematically, it's expressed as:
\ F = kx \

For instance, if a spring with a natural length of 6 inches is compressed to 5.5 inches by a 12,000-lb force, the displacement \(x\) is 0.5 inches. Using this information, we can find the spring constant by rearranging the formula to solve for \(k\):

\ k = \frac{F}{x} = \frac{12,000}{0.5} = 24,000 \ \text{lb/in} \

This value is essential for further calculations, like finding the work done in compressing the spring.

displacement

Displacement in the context of springs refers to how much the spring is stretched or compressed from its natural length. It's a crucial variable when applying Hooke's law.

In our example, the natural length of the spring is 6 inches. When compressing the spring to 5 inches, the displacement is the difference between these two lengths. Here, the displacement \(x\) is:
\ x = 6 - 5 = 1 \ \text{inch} \

This displacement, along with the spring constant, allows us to calculate the work done using the work integration formula. By identifying and understanding displacement, we can find how much force is needed and how much work is done in compressing or extending the spring.