Problem 14 - Precalculus: Functions and Graphs

Question

Hooke's law 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) \( k = \frac{40}{3} \); (c) 20 pounds; (d) Line through origin with slope \( \frac{40}{3} \).

Step 1: Understanding Hooke's Law

Hooke's Law states that the force \( F \) required to stretch or compress a spring is directly proportional to the displacement \( x \) from its natural length. Formally, this is represented by the equation \( F = kx \), where \( k \) is the constant of proportionality.

Step 2: Using Given Information to Find k

A weight of 4 pounds stretches the spring by 0.3 inches beyond its natural length (from 10 to 10.3 inches). Thus, \( F = 4 \) and \( x = 0.3 \). Substituting into the formula, we get \( 4 = k \times 0.3 \). Solving for \( k \) gives \( k = \frac{4}{0.3} = \frac{40}{3} \) or approximately \( 13.33 \).

Step 3: Calculating the Force for a New Displacement

To find the new force that stretches the spring to 11.5 inches, calculate the new displacement: 11.5 inches - 10 inches = 1.5 inches. Use \( F = kx \) with \( k = \frac{40}{3} \) and \( x = 1.5 \). Thus, \( F = \frac{40}{3} \times 1.5 = 20 \). The weight required is 20 pounds.

Step 4: Sketching the Graph of F versus x

The relationship between \( F \) and \( x \) is linear with a slope of \( k = \frac{40}{3} \). The graph passes through the origin (0,0) because there is no force when there is no displacement. The line slopes upwards, starting at the origin and continuing for \( x \geq 0 \).

Key Concepts

Direct Proportionality

Direct proportionality is a fundamental concept in physics and mathematics, and it's crucial for understanding Hooke's law. When we say that two quantities are directly proportional, we mean that as one quantity increases, the other increases at a constant rate, and vice versa. This relationship can be vividly seen in Hooke's law, which describes how a spring behaves under force.

In Hooke's law, the force required to stretch or compress a spring is directly proportional to the displacement of the spring from its natural length. Mathematically, this can be expressed as:

• \( F = kx \)

where \( F \) is the force applied, \( x \) is the displacement, and \( k \) is the constant of proportionality. This equation indicates that the change in force produces a corresponding change in displacement, maintaining a consistent ratio.

Understanding direct proportionality helps in visualizing how changes in one variable relate to changes in another. In practical terms, if you double the force applied to a spring, the displacement doubles, as long as the spring follows Hooke's law.

Constant of Proportionality

The constant of proportionality, often denoted by \( k \), is a crucial factor in equations that describe direct proportionality. In the context of Hooke's Law, \( k \) represents the stiffness or rigidity of a spring. It determines how much force is needed per unit of displacement.

Consider the problem where a weight of 4 pounds stretches a spring by 0.3 inches. By substituting these values into the formula \( F = kx \), we find:

• Given: \( F = 4 \, ext{pounds} \) and \( x = 0.3 \, ext{inches}\)
• Rearranging gives us \( k = \frac{F}{x} = \frac{4}{0.3} \)
• Hence, \( k = \frac{40}{3} \approx 13.33 \)

This value of \( k \) tells us that for every inch the spring is stretched, a force of approximately 13.33 pounds is required. Changing the constant of proportionality affects the slope of the line when graphing force versus displacement. A higher \( k \) means a steeper slope, indicating a stiffer spring.

Being aware of the constant of proportionality helps us predict how different springs react to forces, assisting in applications ranging from engineering to everyday solutions.

Linear Relationship

A linear relationship in a graph shows a straight line, illustrating a consistent rate of change between two variables. In the case of Hooke's Law, the relationship between force \( F \) and displacement \( x \) creates a linear graph, with the constant of proportionality \( k \) as the slope of the line.

This means no matter how large or small the displacement, the force required maintains a constant ratio, reinforcing the linearity of the relationship. To visualize this, we plot the equation \( F = kx \) on a graph:

• The y-axis represents force \( F \), while the x-axis represents displacement \( x \).
• The line will pass through the origin (0,0), indicating no force results in no displacement.
• As \( x \) increases, \( F \) increases linearly, forming an upward-sloping line.

This graph clearly shows how changes in one variable proportionally affect the other. For Hooke’s law, the distinction of linearity means clear predictability and simplicity in calculating outcomes.

Grasping the concept of a linear relationship makes it easier to solve problems involving Hooke's law and similar real-world applications.