Problem 43 - Beginning and Intermediate Algebra

Question

Hooke's law states that the force required to stretch a spring is proportional to the distance that the spring is stretched from its original length. A force of 200 lb is required to stretch a spring 5 in. from its natural length. How much force is needed to stretch the spring 8 in. beyond its natural length? (GRAPH CANT COPY)

Step-by-Step Solution

Answer

The force required to stretch the spring 8 inches beyond its natural length is 320 lb.

Step 1: Understand Hooke's Law formula

Hooke's Law can be expressed as F = k * x, where F is the force required, k is the proportionality constant, and x is the distance stretched beyond the natural length.

Step 2: Find the proportionality constant (k)

Given that a 200 lb force is required to stretch the spring 5 inches, we can find the proportionality constant k by substituting the values in the formula: 200 = k * 5 Now we can solve for k: k = 200 / 5 k = 40

Step 3: Calculate the force required to stretch the spring 8 inches

Now that we have the proportionality constant (k = 40), we can use Hooke's Law formula again to calculate the force required to stretch the spring 8 inches: F = k * x F = 40 * 8 F = 320 So the force required to stretch the spring 8 inches beyond its natural length is 320 lb.

Key Concepts

Proportionality Constant

In Hooke's Law, the proportionality constant, often denoted as \( k \), is a crucial factor. It represents the spring's stiffness, essentially indicating how resistant the spring is to being stretched. The greater the value of \( k \), the stiffer the spring, necessitating more force for the same amount of stretch.
The formula \( F = k \cdot x \) demonstrates that the force exerted (\( F \)) is directly proportional to the extension or compression of the spring (\( x \)), with \( k \) acting as the constant of proportionality.
In our exercise, to determine \( k \), we use the given information: a 200 lb force stretches the spring by 5 inches. Plugging these values into the formula results in:

• \( 200 = k \cdot 5 \)
• Simplifying gives \( k = \frac{200}{5} = 40 \)

This value of 40 lb/inch remains constant for any calculations involving this particular spring.

Force Calculation

Calculating the force needed to stretch a spring further involves using the newly determined proportionality constant \( k \). Here, the spring is stretched further than before, and you need to find out what force that requires.
The formula once again comes to our aid: \( F = k \cdot x \). Previously, we found that \( k = 40 \) lb/inch, and the new extension \( x = 8 \) inches beyond the spring's relaxed length.

• Substitute these values: \( F = 40 \cdot 8 \)
• This results in \( F = 320 \) pounds

This calculation shows that a force of 320 lb is essential to stretch this spring to 8 inches beyond its natural length. The procedure highlights the direct use of proportionality constant in computing force based on the stretch length.

Spring Mechanics

Spring mechanics delve into how springs react to external forces based on their inherent properties. Hooke's Law provides a simplified linear relationship for this behavior.
The primary attributes in spring mechanics include:

• Natural Length: This is the state of the spring without any force applied.
• Stiffness: Depicted by \( k \), it determines how easily a spring can be deformed, affecting both force needed and energy stored.
• Elastic Limit: Beyond which permanent deformation occurs and Hooke's law no longer holds.

When a spring is used within these elastic limits, it functions as described by Hooke’s Law: its response to applied forces remains predictable and reproducible. This makes springs invaluable in practical applications like vehicle suspension systems, mattresses, and engineering tasks requiring precise force regulation.