Problem 14 - Algebra and Trigonometry with Analytic Geometry

Question

Hooke's law states that the force \(F\) required to stretch a spring \(x\) units beyond its natural length is directly proportional to \(x\). (a) Express \(F\) as a function of \(x\) by means of a formula that involves a constant of proportionality \(k\). (b) A weight of 4 pounds stretches a certain spring from its natural length of 10 inches to a length of \(10.3\) inches. Find the value of \(k\) in part (a). (c) What weight will stretch the spring in part (b) to a length of \(11.5\) inches? (d) Sketch a graph of the relationship between \(F\) and \(x\) for \(x \geq 0\).

Step-by-Step Solution

Answer

(a) \( F = kx \); (b) \( k \approx 13.33 \); (c) approximately 20 pounds. (d) A linear graph with slope 13.33.

Step 1: Understanding Hooke's Law

According to Hooke's Law, the force \( F \) exerted on a spring is directly proportional to the displacement \( x \), which is the difference between the stretched length and the natural length. The formula for Hooke's Law is given by \( F = kx \), where \( k \) is a constant known as the spring constant that depends on the spring's characteristics.

Step 2: Calculating the Spring Constant \( k \)

In this step, we use the information given to solve for \( k \). A force of 4 pounds stretches the spring from 10 inches to 10.3 inches, which means the displacement \( x \) is: \[ x = 10.3 - 10 = 0.3 \text{ inches}. \] Using the formula \( F = kx \): \[ 4 = k \times 0.3. \] Solving for \( k \), we divide both sides by 0.3: \[ k = \frac{4}{0.3} = \frac{40}{3} \approx 13.33. \]

Step 3: Determining the Force for Additional Displacement

To find the weight required to stretch the spring to 11.5 inches, first calculate the new displacement: \[ x = 11.5 - 10 = 1.5 \text{ inches}. \] Using the previously found constant \( k \approx 13.33 \), substitute into the formula to find \( F \): \[ F = 13.33 \times 1.5 = 19.995. \] Thus, approximately 20 pounds are required to stretch the spring to 11.5 inches.

Step 4: Sketching the Graph of \( F \) versus \( x \)

In this graph, \( F \) is plotted on the y-axis and \( x \) on the x-axis. As \( F = kx \), the graph is a straight line passing through the origin with a slope of \( k \). For this spring, the slope is \( k \approx 13.33 \). This means for every unit (inch) increase in \( x \), \( F \) increases by \( k \) units (pounds). The line remains linear and positive for all \( x \geq 0 \).

Key Concepts

Direct Proportionality

When we say direct proportionality, we mean that two quantities increase or decrease together at a constant rate. In the case of Hooke's Law, the force \( F \) exerted on a spring is directly proportional to the displacement \( x \). This means if you double the amount you stretch the spring, the force required also doubles. The relationship is expressed mathematically by the equation \( F = kx \), where \( k \) is the spring constant. Understanding direct proportionality allows us to predict how changes in one quantity affect another.

Spring Constant

The spring constant \( k \) is a crucial factor in Hooke's Law and tells us how stiff or soft a spring is. A larger \( k \) indicates a stiffer spring, which requires more force to stretch it farther. Conversely, a smaller \( k \) corresponds to a softer spring. In the exercise, the spring constant was calculated by rearranging the formula \( F = kx \) and using the given force and displacement. We found that \( k \approx 13.33 \), meaning for this specific spring, each inch stretched requires about 13.33 pounds of force.

Force-Displacement Relationship

The core of Hooke's Law is the force-displacement relationship, which binds the force applied to a spring and how far it is stretched (displacement) together. Through the equation \( F = kx \), this relationship becomes apparent. In practical terms, if a spring is stretched beyond its natural length, the displacement \( x \) is the amount the spring is stretched. Given the spring constant \( k \), you can easily determine the force needed to achieve any desired displacement by plugging the values into the equation.

Linear Graph Interpretation

Mapping the relationship between force and displacement on a graph provides a visual understanding of Hooke's Law. The graph is a straight line starting from the origin, with displacement \( x \) plotted on the x-axis and force \( F \) on the y-axis. The slope of the line is the spring constant \( k \). Because the graph is linear, it reinforces the concept of direct proportionality—highlighting that any increase in displacement leads to a proportional increase in force. For the spring in the problem, the slope is approximately 13.33. This means that the graph confirms our calculations: a linear increase in force with increasing displacement.