Problem 35 - College Physics

Question

\(\bullet\) Tendons. Tendons are strong elastic fibers that attach mus- cles 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 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?

Step-by-Step Solution

Answer

(a) 199.19 N/m; (b) 0.692 m, 47.7 J.

Step 1: Convert mass to force

First, find the force exerted by the 250 g object. Convert the mass to kilograms by dividing by 1000; thus, 250 g = 0.250 kg. Then calculate the force using the formula for weight: \[ F = m imes g \]where \( m = 0.250 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). So, \[ F = 0.250 \times 9.8 = 2.45 \, \text{N} \].

Step 2: Use Hooke's Law to find the force constant

Hooke's Law is given by \( F = k \times x \), where \( F \) is the force exerted, \( k \) is the force constant, and \( x \) is the extension of the tendon. Rearrange it to solve for \( k \):\[ k = \frac{F}{x} \].Here, \( x = 1.23 \, \text{cm} = 0.0123 \, \text{m} \).Substitute both \( F = 2.45 \, \text{N} \) and \( x = 0.0123 \, \text{m} \) into the equation to find \( k \):\[ k = \frac{2.45}{0.0123} = 199.19 \, \text{N/m} \].

Step 3: Calculate maximum stretch without rupturing

Using the maximum tension the tendon can support (138 N), find the maximum stretch \( x_{\text{max}} \) using Hooke's Law again:\[ 138 = 199.19 \times x_{\text{max}} \].Solve for \( x_{\text{max}} \):\[ x_{\text{max}} = \frac{138}{199.19} \approx 0.692 \, \text{m} \].

Step 4: Calculate the energy stored in the tendon

The energy stored (elastic potential energy) in the stretched tendon can be found using the formula:\[ U = \frac{1}{2} k x^2 \].Substitute \( k = 199.19 \, \text{N/m} \) and \( x_{\text{max}} = 0.692 \, \text{m} \) into the formula:\[ U = \frac{1}{2} \times 199.19 \times (0.692)^2 \approx 47.7 \, \text{J} \].

Key Concepts

Force Constant and Hooke's Law

Hooke's Law is a fundamental principle used to understand how elastic materials behave under stress. According to Hooke's Law, the force required to stretch or compress a spring by some distance is directly proportional to that distance. This principle is mathematically expressed as \( F = k \times x \), where \( F \) represents the force applied to the object, \( k \) is the force constant (also known as the spring constant), and \( x \) is the extension or compression of the object. In this scenario, we are dealing with a tendon, which behaves similarly to a spring in its elastic properties.

To find the force constant of a tendon, which demonstrates this elastic behavior, we first need to calculate the force exerted on it. By hanging a 250 g object from the tendon, the force due to gravity acting on this mass can be calculated using the formula \( F = m \times g \), with \( g \) being the acceleration due to gravity (9.8 m/s²). Then, by measuring how much the tendon stretches under this force, we are able to determine the value of \( k \) using rearranged Hooke's Law: \( k = \frac{F}{x} \). For instance, in this exercise, the tendon stretched by 1.23 cm, or 0.0123 m, resulting in a force constant of approximately 199.19 N/m.

This constant tells us how stiff the tendon is; the larger the \( k \), the stiffer the tendon, and the more force it takes to stretch it by a certain amount.

Tendon Elasticity

Tendons are unique structures in the human body that connect muscles to bones. They are designed to be both strong and elastic, which means they can stretch a bit under load but will return to their original shape. The elasticity of a tendon can be analyzed using Hooke's Law, giving us insight into how much a tendon can stretch when a force is applied.

In this example, the tendon had a force constant of 199.19 N/m. This elasticity allowed it to stretch by approximately 0.0123 meters (1.23 cm) when a small force was applied. However, tendons also have a limit to how much they can stretch before they rupture. The maximum tension this tendon can withstand before rupturing was given as 138 N.

To find out how much the tendon can actually stretch without exceeding its limit, we use the maximum tension value and apply Hooke’s Law once more. By solving \( 138 = 199.19 \times x_{\text{max}} \), we discover that the maximum stretch before rupture is approximately 0.692 meters. Understanding a tendon's elasticity is crucial, especially in medical, sports, and biological engineering fields, where tendon injuries are common and must be carefully managed.

Potential Energy Calculation

When a tendon is stretched, it stores energy in the form of elastic potential energy. This energy is similar to that stored in a stretched or compressed spring and can be calculated using the formula \( U = \frac{1}{2} k x^2 \), where \( U \) is the potential energy, \( k \) is the force constant, and \( x \) is the amount of stretch or compression.

Elastic potential energy represents the energy stored in an object due to its deformation. For instance, in our tendon exercise, when the tendon is stretched to its limit of 0.692 meters, it stores a significant amount of energy. By substituting \( k = 199.19 \, \text{N/m} \) and \( x_{\text{max}} = 0.692 \, \text{m} \) into the potential energy formula, we find that about 47.7 Joules of energy is stored in the tendon at maximum stretch.

This stored energy is important as it can potentially be released when the tendon returns to its original length. Understanding this concept is valuable, especially in biomechanics, where the efficiency and safety of movements in sports and physical activities are considered.