Problem 84 - Physics

Question

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{~N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{~cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See 6.9 for a discussion of the work done by a variable force.)

Step-by-Step Solution

Answer

(a) The spring constant \( k \) is 322.12 N/m. (b) The work done is 1.39 \times 10^{-3} J.

Step 1: Convert Units

First, we need to convert the given dimensions from centimeters to meters. The radius \( r \) is given as \( 0.091 \ cm \). Converting this to meters, \( r = 0.091 \ cm = 0.00091 \ m \). The length \( L \) is given as \( 2.5 \ cm \). Converting to meters gives \( L = 0.025 \ m \).

Step 2: Calculate Cross-Sectional Area

The cross-sectional area \( A \) of the collagen cylinder is calculated using the formula for the area of a circle, \( A = \pi r^2 \). Substituting the value of \( r \), we have: \[ A = \pi (0.00091 \ m)^2 = 2.601 \times 10^{-6} \ m^2 \]

Step 3: Calculate Spring Constant

To find the spring constant \( k \), we use Young's modulus formula, \( E = \frac{F/A}{\Delta L/L} \), where \( F \) is the force and \( \Delta L \) is the extension. Rearranging for \( k = \frac{E \cdot A}{L} \), substitute \( E = 3.1 \times 10^6 \ N/m^2 \), \( A = 2.601 \times 10^{-6} \ m^2 \), and \( L = 0.025 \ m \): \[ k = \frac{3.1 \times 10^6 \ N/m^2 \cdot 2.601 \times 10^{-6} \ m^2}{0.025 \ m} = 322.12 \ N/m \]

Step 4: Calculate Work Done

The work done by a variable force is calculated as \( W = \frac{1}{2} k (\Delta x)^2 \), where \( \Delta x \) is the total extension. From Hooke's Law, \( F = k \Delta x \), rearrange to find \( \Delta x = \frac{F}{k} \). Here, \( F = 3.0 \times 10^{-2} \ N \) and \( k = 322.12 \ N/m \): \[ \Delta x = \frac{3.0 \times 10^{-2} \ N}{322.12 \ N/m} = 9.31 \times 10^{-5} \ m \] Substituting \( \Delta x \) and \( k \) into the work formula gives: \[ W = \frac{1}{2} \times 322.12 \ N/m \times (9.31 \times 10^{-5} \ m)^2 = 1.39 \times 10^{-3} \ J \]

Key Concepts

Spring Constant

The spring constant, often denoted by \( k \), is a fundamental parameter in understanding the stiffness of a spring or any elastic material. It quantifies the extent to which an object resists deformation when subjected to a force. Specifically, the spring constant describes the relationship between the force applied to a spring and the resulting elongation or compression.

It's calculated using Young's modulus for materials, which is the ratio of stress to strain. In exercises like stretching a collagen cylinder, to find \( k \), we use the formula:

• \( E = \frac{F/A}{\Delta L/L} \)

Where \( E \) is Young's modulus, \( F \) is the force, \( A \) is the cross-sectional area, \( L \) is the original length, and \( \Delta L \) is the change in length. Rearranging gives:

• \( k = \frac{E \cdot A}{L} \)

This equation shows how the spring constant depends on the material's inherent properties (Young's modulus) and its dimensions (area and length). A higher \( k \) value indicates a stiffer material, which means more force is needed to achieve the same amount of deformation compared to a less stiff material.

Hooke's Law

Hooke's Law is a principle describing the extension or compression of an elastic material. Formally, it states that the force \( F \) needed to extend or compress a spring by some distance \( x \) is proportional to that distance:

• \( F = k \cdot \Delta x \)

Where \( k \) is the spring constant, and \( \Delta x \) is the displacement from the equilibrium position.

This law is vital in understanding the behavior of springs and elastic objects because it defines a linear relationship between the force applied and the change in length, provided the material does not exceed the elastic limit. It implies that as long as the material remains within its elastic range, doubling the force will double the displacement.

In practical terms, Hooke's Law can be used to measure weight, absorb energy, and design materials that require predictable stretching behavior—such as bridges or load-bearing structures.

Work Done by a Variable Force

When a variable force acts on an object, the work done is not simply force times distance, because the force changes as the object moves. For a spring or elastic material governed by Hooke's Law, the work done when stretching the material from its natural length to a final length can be calculated using Hooke's equation.

The work done \( W \) in stretching a spring is given by:

• \( W = \frac{1}{2} k (\Delta x)^2 \)

Where \( \Delta x \) is the extension or compression.

This formula calculates the work as the area under a force versus displacement graph, producing a triangular area for linear springs. This calculation is integral in engineering and physics to determine the energy stored in elastic objects and is also essential in designing systems that involve springs, such as vehicle suspensions, trampolines, or shock absorbers in various mechanical devices. Understanding the work done by variable forces allows engineers to predict system behavior under different loads efficiently.