Question

According to Hooke's Law, a linear relationship exists between the distance that a spring stretches and the force stretching it. Suppose a weight of 0.5 kilograms causes a spring to stretch 2.75 centimeters and a weight of 1.2 kilograms causes the same spring to stretch 6.6 centimeters. (a) Find a linear model that relates the distance \(d\) of the stretch and the weight \(w\). (b) What stretch is caused by a weight of 2.4 kilograms? (c) What weight causes a stretch of 19.8 centimeters?

Short answer

The linear model is \( d \approx 5.51w \). A 2.4 kg weight causes a stretch of 13.2 cm, and a 19.8 cm stretch is caused by a 3.6 kg weight.

Step-by-step solution

Understand Hooke's Law

Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula can be written as: \[ F = k \times d \]where \(F\) is the force applied, \(k\) is the spring constant, and \(d\) is the displacement (distance stretched) of the spring.

Convert Weights to Forces

Using the formula \( F = m \times g \), where \(m\) is the mass in kilograms and \(g\) is the acceleration due to gravity (approximately 9.8 m/s\textsuperscript{2}), convert the given weights to forces.\[ F_1 = 0.5 \times 9.8 \approx 4.9 \text{ N} \]\[ F_2 = 1.2 \times 9.8 \approx 11.76 \text{ N} \]

Create a System of Equations

Use the distances and the corresponding forces to create a system of equations:\[ 4.9 = k \times 2.75 \]\[ 11.76 = k \times 6.6 \]

Solve for the Spring Constant

Solve the equations from Step 3 to find the spring constant \(k\).\[ k = \frac{4.9}{2.75} \approx 1.78 \]Verify with the second equation:\[ k = \frac{11.76}{6.6} \approx 1.78 \]

Write the Linear Model

The linear model relating the distance \(d\) and the weight \(w\) is given by Hooke's Law rearranged:\[ d = \frac{F}{k} = \frac{w \times g}{k} \] Assuming \(g = 9.8\) and \(k = 1.78\), we get:\[ d = \frac{w \times 9.8}{1.78} \approx 5.51w \]

Find the Stretch for 2.4 kg Weight

Using the linear model, calculate the stretch for a weight of 2.4 kg:\[ d = 5.51 \times 2.4 \approx 13.2 \text{ cm} \]

Find the Weight for a 19.8 cm Stretch

Using the linear model, calculate the weight that causes a 19.8 cm stretch:\[ 19.8 = 5.51w \]\[ w = \frac{19.8}{5.51} \approx 3.6 \text{ kg} \]

Key concepts

Linear Relationships

Understanding linear relationships is key in grasping Hooke's Law. A linear relationship means that if one quantity changes, another related quantity changes in proportion. In the context of Hooke’s Law, the force applied to a spring and the distance it stretches have a direct proportionality.

For instance, if you double the weight applied to a spring, the distance the spring stretches will also double. This makes it easy to predict how a spring behaves under different weights.

Mathematically, this relationship is described with a straight line equation, which in Hooke's Law can be expressed as: \( F = k \times d \) where:

• \(F\) is the force applied.
• \(k\) is the spring constant.
• \(d\) is the displacement (stretch or compression) of the spring.

Understanding this linear relationship helps when you need to model the behavior of springs under varying forces.

Spring Constant

The spring constant \(k\) is a crucial parameter in Hooke’s Law. It tells us how stiff or flexible a spring is. A higher \(k\) value means a stiffer spring, while a lower \(k\) value indicates a more flexible spring.

To find the spring constant, you can use the formula from Hooke's Law. In our example, we had two sets of data:

• A 0.5 kg weight causing a 2.75 cm stretch, which translates to a force (\(F\)) of 4.9 N.
• A 1.2 kg weight causing a 6.6 cm stretch, which translates to a force (\(F\)) of 11.76 N.

Using these, we set up the equations: \( 4.9 = k \times 2.75 \) and \( 11.76 = k \times 6.6 \).

Solving for \(k\), we find: \( k \approx 1.78 \)

This value means that for every newton of force applied, the spring will stretch approximately 1.78 cm. This consistent ratio helps us predict the spring's behavior under any amount of force.

Force and Displacement

In Hooke’s Law, force and displacement are directly related. The force (\(F\)) refers to the weight or push/pull applied to the spring, while the displacement (\(d\)) refers to how much the spring stretches or compresses.

Here’s how we can calculate these in our example:

• We converted weights to forces using \(F = m \times g\), where \(m\) is mass in kilograms and \(g\) is the acceleration due to gravity (9.8 m/s\textsuperscript{2}).
• This gave us forces of 4.9 N and 11.76 N for the given weights of 0.5 kg and 1.2 kg, respectively.

Using these forces and the spring constant \(k\), we predicted other scenarios:

To find the stretch for a new weight, we used our linear model: \( d = \frac{F}{k} \)

\( d = \frac{2.4 \times 9.8}{1.78} \approx 13.2 \text{ cm} \)
Similarly, to find the weight for a given stretch, we used: \( w = \frac{d}{5.51} \)

\( w \approx 3.6 \text{ kg} \)

This relationship allows us to solve many practical problems involving springs and elastic materials.