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Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a \(250 \mathrm{~g}\) object was hung from it, the tendon stretched \(1.23 \mathrm{~cm}\). (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\). (b) Because of its thickness, the maximum tension this tendon can support without rupturing is \(138 \mathrm{~N}\). By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?

Short Answer

Expert verified
(a) \( k \approx 1994 \ N/m \); (b) Maximum stretch \( \approx 0.0692 \ m \), energy stored \( \approx 4.77 \ J \).

Step by step solution

01

Understanding 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 is: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension or compression.
02

Calculate the Force Applied to the Tendon

The force due to the weight of a mass in kilograms is given by \( F = mg \), where \( m \) is the mass in kilograms and \( g \) is the acceleration due to gravity (approximately \( 9.81 \ m/s^2 \) on Earth). For a \( 250 \ g \) object, convert the mass to kilograms (0.25 kg) and calculate the force.
03

Convert the Stretch from Centimeters to Meters

The stretch of the tendon is given as \( 1.23 \ cm \). Convert this length into meters by dividing by \( 100 \), resulting in \( 0.0123 \ m \).
04

Calculate the Force Constant \( k \) of the Tendon

Using the equation \( F = kx \), rearrange it to find \( k = \frac{F}{x} \). Substitute the values calculated in the previous steps: \( F = 0.25 \times 9.81 \) N and \( x = 0.0123 \ m \), to find \( k \).
05

Calculate Maximum Stretch Without Rupture

The maximum force before the tendon ruptures is given as \( 138 \ N \). Using Hooke’s Law, rearrange to find maximum extension \( x = \frac{F}{k} \) using \( k \) from Step 4.
06

Calculate the Energy Stored in the Tendon

The energy stored in a spring (or tendon in this case) is given by \( U = \frac{1}{2}kx^2 \). Use the maximum stretch calculated in Step 5 and \( k \) to calculate the energy stored.

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Key Concepts

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

Understanding the Force Constant
The force constant, also known as the spring constant, is a measure of the stiffness of a spring or, in this case, a tendon that behaves like a spring. Hooke's Law helps us to understand this concept by using the formula: \( F = kx \). Here, \( F \) represents the force applied, \( x \) is the distance the material stretches, and \( k \) is the force constant we are looking to find. When a force is applied, the tendon deforms proportionally according to Hooke's Law, stressing its elasticity. The force constant \( k \) is crucial as it determines how much force is needed for a certain amount of stretch. For this tendon, the problem involved hanging a 250 g object, which applied a force due to gravity, and resulted in the tendon stretching 1.23 cm. To find \( k \), we first convert all measurements to match standard units:
  • Mass in kilograms (since 250 g = 0.25 kg)
  • Stretch in meters (1.23 cm = 0.0123 m)
The force due to gravity is calculated as \( F = mg \), resulting in \( F = 0.25 \times 9.81 \). With \( F \) and \( x \), we use \( k = \frac{F}{x} \) to find \( k \). This gives us the force constant that essentially describes the tendon’s stiffness.
Determining Maximum Tension
The maximum tension is the greatest force a tendon can handle without failing. In this context, it is given as 138 N. This means that if the applied force exceeds this value, the tendon may rupture or suffer irreversible damage.Tendons have a certain limit of elasticity and strength. Beyond this limit, the material can either deform permanently or break. To find the maximum stretch the tendon can undergo without breaking, we return to Hooke’s Law, using the formula: \[ x = \frac{F}{k} \]Here, we substitute the maximum tension for \( F \) and use the value of \( k \) determined in the previous section. This calculation tells us the maximum length the tendon can safely stretch. Understanding these limits helps in designing safety protocols for activities and workloads involving tendons.
Calculating Energy in Springs
Energy stored in a spring, or tendon, is determined using the formula for elastic potential energy: \[ U = \frac{1}{2}kx^2 \]This formula showcases how the energy is dependent on both the stiffness of the material (force constant \( k \)) and the extent to which it is stretched or compressed (\( x \)). For the tendon being analyzed, reaching its maximum stretch calculated earlier gives insight into how much energy is stored at that point.This energy storage is crucial in scenarios such as dynamic movements in sports or biological mechanisms, where tendons act like springs storing and releasing energy efficiently. Moreover, understanding this energy storage helps us learn how manipulation of tendons and materials can lead to more efficient energy use and conservation in both biological and mechanical systems.
Exploring Material Elasticity
Material elasticity refers to a material's ability to return to its original shape after being stretched or compressed. Tendons, like springs, demonstrate this property vividly under Hooke’s Law, where their elongation due to an applied force can be reversed when the force is removed—provided the elastic limit is not exceeded. The capacity for elasticity doesn't mean infinite stretchability. Every material has its elastic limit determined by factors like:
  • Composition of the material
  • Environmental factors (temperature, humidity)
  • Amount and type of applied force
Recognizing this, for tendons or any elastic material, gives insight into potential applications and limitations. It ensures appropriate engineering of wearable technology or developing athletic training programs that increase performance without the risk of injury from crossing the elastic threshold. Understanding elasticity helps in optimizing the use of tendons and similar materials in practical, real-world applications.

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Most popular questions from this chapter

A \(61 \mathrm{~kg}\) skier on level snow coasts \(184 \mathrm{~m}\) to a stop from a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) (a) Use the work-energy theorem to find the coefficient of kinetic friction between the skis and the snow. (b) Suppose a \(75 \mathrm{~kg}\) skier with twice the starting speed coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier's skis and the snow.

The aircraft carrier USS George Washington has mass \(1.0 \times 10^{8} \mathrm{~kg}\). When its engines are developing their full power of 260,000 hp, the George Washington travels at its top speed of 35 knots \((65 \mathrm{~km} / \mathrm{h})\). If \(70 \%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

A \(250 \mathrm{~g}\) object on a frictionless, horizontal lab table is pushed against a spring of force constant \(35 \mathrm{~N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is compressed \(12.0 \mathrm{~cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

A constant horizontal pull of \(8.50 \mathrm{~N}\) drags a box along a horizontal floor through a distance of \(17.4 \mathrm{~m}\). (a) How much work does the pull do on the box? (b) Suppose that the same pull is exerted at an angle above the horizontal. If this pull now does \(65.0 \mathrm{~J}\) of work on the box while pulling it through the same distance, what angle does the force make with the horizontal?

II A \(60 \mathrm{~kg}\) woman steps onto an up-going escalator, which has an incline of \(32^{\circ}\) with respect to the horizontal and is moving at \(0.5 \mathrm{~m} / \mathrm{s}\). The top of the escalator is \(20 \mathrm{~m}\) above the ground level. Calculate how much work is done by (a) the friction force between the woman's feet and the escalator step, (b) gravity, and (c) the normal force on the woman's feet, as she moves from the bottom to the top of the escalator. What is the total work done on the woman as she moves from the bottom to the top?

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