Problem 114 - College Algebra with Modeling and Visualization

Question

If an 80 -pound force compresses a spring 3 inches, how much force must be applied to compress the spring 7 inches?

Step-by-Step Solution

Answer

Approximately 186.67 pounds.

Step 1: Understanding Hooke's Law

First, we need to recall Hooke's Law, which states that the force required to compress or extend a spring is directly proportional to the displacement of the spring from its original position. Mathematically, this relationship is expressed as: \[ F = kx \]where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

Step 2: Finding the Spring Constant

We are given that an 80-pound force compresses the spring by 3 inches. We can use Hooke's Law to find the spring constant \( k \). Substituting the given values into the formula, we have:\[ 80 = k \times 3 \]Solving for \( k \), we divide both sides by 3:\[ k = \frac{80}{3} \] pounds per inch.

Step 3: Calculating Force for New Compression

Now, we need to find the force required to compress the spring by 7 inches. Using the same formula, \( F = kx \), and substituting for \( k = \frac{80}{3} \) and \( x = 7 \), we have:\[ F = \left( \frac{80}{3} \right) \times 7 \]Simplifying this, we perform the multiplication:\[ F = \frac{560}{3} \]Thus, the force required is approximately 186.67 pounds.

Key Concepts

Spring Constant

The spring constant, denoted by the symbol \( k \), is a measure of a spring's stiffness. It is a key component in Hooke's Law, which describes how springs behave when forces are applied to them. The spring constant tells us how much force is needed to compress or extend a spring by a specific distance.

• A larger \( k \) value means the spring is stiffer and requires more force to compress or extend.
• A smaller \( k \) value indicates a more flexible spring, requiring less force.

Knowing the spring constant allows us to calculate the force needed for different displacements. In this exercise, we determined the spring constant by using the given force (80 pounds) and displacement (3 inches) in the relation \( F = kx \). By solving \( k = \frac{80}{3} \), we identified the stiffness of the spring, which can then be used to predict how it will respond to other forces.

Force and Compression

Force and compression are closely linked when dealing with springs. According to Hooke's Law, the force required to compress a spring is directly related to the displacement from its equilibrium position. This means the further you compress the spring, the more force you must apply.

• The equation \( F = kx \) captures this relationship, where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the compression.
• To compress a spring more, a larger force is needed, if the spring constant remains the same.

In our exercise example, we first calculated the spring constant, then used it to find the force necessary to compress the spring 7 inches. Substituting into the formula \( F = 560/3 \) revealed that approximately 186.67 pounds of force is needed for this greater compression, demonstrating the direct relationship between force and displacement.

Proportionality in Physics

Proportionality in the context of physics refers to the concept where two quantities increase or decrease in a consistent, predictable manner relative to each other. Hooke's Law exemplifies this as it shows a linear relationship between force applied to a spring and its resulting compression or extension.

• Direct proportionality means doubling the displacement doubles the force required, provided the spring constant remains unchanged.
• This kind of relationship makes it easy to predict behaviors under varying forces.

In our scenario, knowing that 80 pounds compressed the spring 3 inches, we can apply the proportionality principle to find the force needed for different displacements (like 7 inches). By relying on the formula and understanding the constant \( k \), we maintain accuracy in predictions, which is fundamental in engineering and physics disciplines.