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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.)

Short Answer

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

Step by step solution

01

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 \( L = 0.025 \ m \).
02

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: \[ A = \pi (0.00091 \ m)^2 = 2.601 \times 10^{-6} \ m^2 \]
03

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 \( L = 0.025 \ m \): \[ k = \frac{3.1 \times 10^6 \ N/m^2 \cdot 2.601 \times 10^{-6} \ m^2}{0.025 \ m} = 322.12 \ N/m \]
04

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: \[ W = \frac{1}{2} \times 322.12 \ N/m \times (9.31 \times 10^{-5} \ m)^2 = 1.39 \times 10^{-3} \ J \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Interactive Solution \(\underline{10.21}\) at presents a model for solving this problem. A spring (spring constant \(=112 \mathrm{~N} / \mathrm{m}\) ) is mounted on the floor and is oriented vertically. A 0.400 kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in \(\mathrm{Hz}\) ) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{~N} / \mathrm{m}\). Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of \(0.370 \mathrm{~s}\). If the mass of the empty car is \(1560 \mathrm{~kg}\), determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

A \(15.0-\mathrm{kg}\) block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) in \(0.500 \mathrm{~s}\). In the process, the spring is stretched by \(0.200 \mathrm{~m}\). The block is then pulled at a constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\), during which time the spring is stretched by only \(0.0500 \mathrm{~m}\). Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table

A \(0.70-\mathrm{kg}\) block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{~m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{~m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{~Pa}\). What force \(\overrightarrow{\mathbf{F}}\) must be applied to the die?

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