Problem 33 - University Physics with Modern Physics

Question

A surgeon is using material from a donated heart to repair a patient's damaged aorta and needs to know the elastic characteristics of this aortal material. Tests performed on a 16.0-cm strip of the donated aorta reveal that it stretches 3.75 cm when a 1.50-N pull is exerted on it. (a) What is the force constant of this strip of aortal material? (b) If the maximum distance it will be able to stretch when it replaces the aorta in the damaged heart is 1.14 cm, what is the greatest force it will be able to exert there?

Step-by-Step Solution

Answer

(a) 40 N/m; (b) 0.456 N.

Step 1: Understanding the Hooke's Law

The problem involves elasticity and can be solved using Hooke's Law, which states that the force exerted by a material is directly proportional to the distance it is stretched. The formula is given by \( F = k \cdot x \), where \( F \) is the force applied, \( k \) is the force constant, and \( x \) is the displacement from its original length.

Step 2: Calculating Force Constant (Part a)

From the information provided, we know the strip is stretched by 3.75 cm (0.0375 m) with a force of 1.50 N. Plug these values into Hooke’s Law: \( k = \frac{F}{x} = \frac{1.50\, \text{N}}{0.0375\, \text{m}} \). This calculation gives the value of the force constant \( k \).

Step 3: Performing the Calculation

Calculate \( k = \frac{1.50}{0.0375} = 40 \) N/m. So, the force constant of this strip of aortal material is 40 N/m.

Step 4: Calculating Greatest Force (Part b)

For the greatest force, the strip can stretch by a maximum distance of 1.14 cm (0.0114 m). Again, use Hooke’s Law: \( F = k \cdot x = 40\, \text{N/m} \times 0.0114\, \text{m} \). Substitute the values to find the maximum force \( F \) that can be exerted.

Step 5: Performing the Calculation

Calculate \( F = 40 \times 0.0114 = 0.456 \) N. Thus, the greatest force it will be able to exert when replacing the aorta is 0.456 N.

Key Concepts

Hooke's Law

One of the fundamental principles in elasticity in physics is Hooke's Law. This is a simple yet powerful concept that helps us understand how elastic materials respond to forces. Hooke's Law explains the relationship between the force applied to a material and the amount it stretches or compresses. In mathematical terms, this is expressed as the equation \( F = k \cdot x \). Here, \( F \) is the force applied to the material, \( k \) is the force constant (also known as the spring constant), and \( x \) is the displacement or change in length of the material.A great way to visualize this is by imagining a spring. When you pull on a spring, it stretches, and the harder you pull (increasing force \( F \)), the more it stretches (increasing displacement \( x \)). Hooke’s Law tells us that this relationship is linear, meaning if the force doubles, the stretch doubles too, as long as we stay within the material’s elastic limit.

Force Constant Calculation

Calculating the force constant \( k \) is crucial for understanding a material's elasticity. In the context of the exercise, we're looking at a strip of aortal material. To find \( k \), rearrange Hooke’s Law: \( k = \frac{F}{x} \).

• Measure how much the material stretches (displacement \( x \)). In our exercise, the strip is stretched by 3.75 cm, which converts to 0.0375 m.
• Measure the force applied \( F \). Here, it's given as 1.50 N.
• Simply divide the force by the displacement to get the force constant: \( k = \frac{1.50\, \text{N}}{0.0375\, \text{m}} \). This calculation results in a force constant of 40 N/m.

The force constant is specific to the material and the conditions under which you're applying the force. It tells you how stiff or flexible the material is, with a higher \( k \) indicating a stiffer material.

Elastic Deformation

Elastic deformation describes the way materials change shape when forces are applied, but importantly, they return to their original shape when the forces are removed. It’s a reversible process, distinguishing it from plastic deformation, where the material does not return to its original form. In our exercise, the aortal material undergoes elastic deformation when stretched. When a force is applied within the elastic limits, the material deforms and stores potential energy. Once the force is removed, this energy is released as the material returns to its original length. To ensure materials like the aortal tissue are used safely and effectively, we need to understand their elastic limits and how much they can be deformed before they reach a point of no return. This is vital in medical applications, where the right amount of flexibility and strength is crucial. Knowing the maximum stretch and corresponding force, as calculated in this exercise, helps ensure that the material can perform as needed without permanent damage.