Problem 78 - Algebra and Trigonometry with Analytic Geometry

Question

According to Hooke's law, the force \(F\) (in pounds) required to stretch a certain spring \(x\) inches beyond its natural length is given by \(F=(4.5) x\) (see the figure). If \(10 \leq F \leq 18\), what are the corresponding values for \(x\) ?

Step-by-Step Solution

Answer

The values for x are \( 2.22 \leq x \leq 4 \).

Step 1: Understand Hooke's Law Equation

The force required to stretch a spring is given by Hooke's law, which states \( F = kx \), where \( F \) is the force in pounds, \( k \) is the spring constant, and \( x \) is the distance in inches beyond the spring's natural length. In this case, \( k = 4.5 \), so the equation becomes \( F = 4.5x \).

Step 2: Set Up the Inequality

We are given that the force \( F \) is between 10 and 18 pounds, inclusive. This translates into the inequality: \( 10 \leq 4.5x \leq 18 \).

Step 3: Solve for x (Minimum Bound)

To find the smallest value of \( x \), solve the inequality \( 10 \leq 4.5x \). Divide both sides by 4.5: \( x \geq \frac{10}{4.5} \approx 2.22 \).

Step 4: Solve for x (Maximum Bound)

To find the greatest value of \( x \), solve the inequality \( 4.5x \leq 18 \). Divide both sides by 4.5: \( x \leq \frac{18}{4.5} = 4 \).

Step 5: Combine the Solutions

The corresponding values for \( x \) are those that satisfy both inequalities from Steps 3 and 4. Thus, \( 2.22 \leq x \leq 4 \).

Key Concepts

Understanding the Spring Constant

When dealing with springs, the spring constant, denoted as \( k \), is an important factor. It describes how stiff the spring is. The value of the spring constant tells us how much force is needed to stretch or compress the spring by a certain amount.

- A larger \( k \) means a stiffer spring which requires more force to stretch.- A smaller \( k \) indicates a looser spring that requires less force to stretch.

Using Hooke's law, you can find the force \( F \) required to stretch the spring a distance \( x \). The formula is \( F = kx \).
For example, if \( k = 4.5 \) as in our exercise, this means for every inch you want to stretch the spring, you need 4.5 pounds of force.

Recognize that \( k \) remains constant for a specific spring unless the spring is damaged or permanently deformed.

Exploring Force and Displacement

The force \( F \) and displacement \( x \) have a linear relationship according to Hooke's law. This means as you increase the force applied to the spring, the displacement increases proportionally.

- Imagine the spring at rest. Its natural length is your starting point.- As you apply force, the spring will stretch and \( x \) measures how much beyond the natural length the spring has been stretched.

In our exercise, we need to find the displacement corresponding to specific force values \( (10 \leq F \leq 18) \). Using \( F = 4.5x \), we set up an equation for each boundary of force.

Replacing \( F \) with 10, we solve \( 10 = 4.5x \) to get \( x \approx 2.22 \) inches.
Replacing \( F \) with 18, we solve \( 18 = 4.5x \) leading to \( x = 4 \) inches. So, the displacement, \( x \), can vary between these two values.

Solving Inequalities in Hooke's Law

Understanding inequalities is essential to solve problems like our exercise. Inequalities show us a range of possible values instead of a single solution.

Given the equation \( 10 \leq 4.5x \leq 18 \), it represents a range of forces and thus a range of values for the displacement \( x \).

To find the displacement values:1. Start with solving the lower inequality, \( 10 \leq 4.5x \). Divide by 4.5 to isolate \( x \): \( x \geq \frac{10}{4.5} \approx 2.22 \).2. For the upper bound, solve \( 4.5x \leq 18 \). Again, divide by 4.5: \( x \leq \frac{18}{4.5} = 4 \).

Combine both solutions to obtain the complete range for \( x \): \( 2.22 \leq x \leq 4 \). This means any displacement value between 2.22 and 4 inches satisfies the conditions given by the force range 10 to 18 pounds, ensuring the spring behaves as described.