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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 Section 6.9 for a discussion of the work done by a variable force.)

Short Answer

Expert verified
(a) The spring constant \(k\) is approximately 322.4 N/m. (b) The work done is 1.4 \times 10^{-3} J.

Step by step solution

01

Converting Radius to Meters

First, convert the radius from centimeters to meters for consistency in units. The given radius is 0.091 cm, which is equivalent to \(0.091 \times 10^{-2} \text{ m} = 9.1 \times 10^{-4} \text{ m}\).
02

Calculating the Cross-Sectional Area

Calculate the cross-sectional area \(A\) of the cylinder using the formula for the area of a circle, \(A = \pi r^2\). \[A = \pi (9.1 \times 10^{-4})^2 = \pi (8.281 \times 10^{-7}) \approx 2.60 \times 10^{-6} \text{ m}^2\].
03

Finding the Spring Constant (k) Using Young’s Modulus

Young's modulus \(E\) relates the stress and strain in the material. Use the formula \(E = \frac{F/L_0}{ ext{strain}}\), then rearrange to find \(k\), where \(k = \frac{E imes A}{L_0}\). \[k = \frac{3.1 \times 10^6 \text{ N/m}^2 \times 2.60 \times 10^{-6} \text{ m}^2}{0.025 \text{ m}} = \frac{8.06 \text{ N}}{0.025 \text{ m}} \approx 322.4 \text{ N/m}\].
04

Writing the Expression for Work Done

The work done \(W\) by a variable force is given by the integral of force with respect to displacement: \(W = \int F \, dx\). For a spring that obeys Hooke’s law, \(F = kx\), so \(W = \int kx \, dx = \frac{1}{2}kx^2\).
05

Calculating Maximum Displacement

Use the maximum force \(F_{max} = 3.0 \times 10^{-2} \text{ N}\) and the expression \(F = kx\) to find maximum displacement \(x_{max}\). \[x_{max} = \frac{F_{max}}{k} = \frac{3.0 \times 10^{-2} \text{ N}}{322.4 \text{ N/m}} = 9.3 \times 10^{-5} \text{ m}\].
06

Calculating the Work Done

Substitute \(k\) and \(x_{max}\) into the work formula: \(W = \frac{1}{2} kx_{max}^2\). \[W = \frac{1}{2} \times 322.4 \times (9.3 \times 10^{-5})^2 = 1.4 \times 10^{-3} \text{ J}\].

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Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental concept in physics that describes how materials deform when subjected to a force. This law states that the force ( F ) needed to extend or compress a spring by some distance ( x ) is proportional to that distance. The mathematical expression of Hooke's Law is F = kx , where F is the force applied, x is the extension or compression from the natural length, and k is the spring constant. Hooke's Law is applicable as long as the material does not reach its elastic limit. Beyond this limit, the material may deform permanently and Hooke's Law would no longer be valid. In the context of the collagen exercise, Hooke's Law assists in finding the spring constant when the material is stretched. The spring constant ( k ) characterizes the stiffness of the material. A larger k indicates a stiffer material. By using Young's Modulus, which measures the ability of a material to withstand length changes under tension or compression, alongside the cylindrical dimensions of the collagen, we can calculate k . This constant allows us to understand the mechanical properties of collagen as an elastic material.
Spring Constant
The spring constant (k) is a crucial parameter in describing the elastic properties of a material. It helps determine how much force is needed to stretch or compress a material by a unit length. In mathematical terms, the spring constant is expressed in Newtons per meter ( ext{N/m}) and is indicative of the stiffness of a spring or any elastic object.To derive the spring constant from physical properties, we utilize Young's Modulus (E), which is a measure of the elasticity of the material. For the collagen rod in the exercise, E helps calculate k using the formula: \[k = \frac{E \times A}{L_0}\]where A is the cross-sectional area and L_0 is the original length of the sample. After performing these calculations, the spring constant of the collagen is found to be approximately 322.4 ext{N/m}. This means for each meter the collagen is stretched, roughly 322.4 Newtons of force is required, implying a moderate level of stiffness.
Work Done by Variable Force
The concept of work done by a variable force is essential for understanding how energy is transferred when a force is applied over a distance. This is especially relevant when dealing with forces that change in magnitude, such as in elastic materials like springs. The work done (W) on an object by a force is the product of the force and the displacement in the direction of the force.For springs that follow Hooke's Law, the work done by a variable force can be calculated using the formula:\[W = \int F \, dx = \frac{1}{2}kx^2\]Here, F = kx is the force at displacement x, and k is the spring constant. In the collagen example, we calculate the work done by first finding the maximum displacement (x_{max}) using the maximum applied force. The formula for work considers this displacement, showing that the work done by stretching collagen to its maximum extension is approximately 1.4 × 10^{-3} Joules. This value provides insight into the energy required to stretch the collagen, illustrating the energy transformation from potential energy to kinetic energy as the force is applied.

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Most popular questions from this chapter

A copper rod (length \(=2.0 \mathrm{m},\) radius \(\left.=3.0 \times 10^{-3} \mathrm{m}\right)\) hangs down from the ceiling.A 9.0 -kg object is attached to the lower end of the rod. The rod acts as a "spring," and the object oscillates vertically with a small amplitude. Ignoring the rod's mass, find the frequency \(f\) of the simple harmonic motion.

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighhone can endure without hreaking is \(7.0 \times 10^{4} \mathrm{N}\). The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2},\) its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) The mass of the man is 65 kg. He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

The length of a simple pendulum is \(0.79 \mathrm{m}\) and the mass of the particle (the "bob") at the end of the cable is \(0.24 \mathrm{kg}\). The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of \(12.0 \mathrm{Hz} .\) When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is \(4.00 \mathrm{Hz}\). Find the ratio \(m_{2} / m_{1}\) of the masses.

In preparation for shooting a ball in a pinball machine, a spring \((k=675 \mathrm{N} / \mathrm{m})\) is compressed by \(0.0650 \mathrm{m}\) relative to its unstrained length. The ball \((m=0.0585 \mathrm{kg})\) is at rest against the spring at point A. When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point \(\mathrm{B},\) which is \(0.300 \mathrm{m}\) higher than point A. Ignore friction, and find the ball's speed at point B.

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