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Chapter 7: Problem 16

BIO 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 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
The force constant of the tendon is 199.19 N/m. The tendon can stretch by 0.69 meters without rupturing and will store 47.14 J of energy at this point.

Step by step solution

01

- Calculate the force constant

Since the first part of the question is to calculate the force constant of the tendon, this can be done by rearranging Hooke's law to \(k = F / x\). Firstly, we need to convert the mass to a force using the equation \(F = mg\), where m is mass and g is acceleration due to gravity, which this is approximately \(9.8 \, m/s^2\). Thus, the force exerted by the 250 g object is \(F = 0.250 \, kg * 9.8 \, m/s^2 = 2.45 \, N\). The extension of the tendon is given as \(1.23 \, cm\), which needs to be converted to meters, \(0.0123 \, m\). So, \(k = F / x = 2.45 \, N / 0.0123 \, m = 199.19 \, N/m\).
02

- Maximum extension of the tendon

The second part of the problem is to determine the maximum extension of the tendon before rupture. We have been given the maximum tension that the tendon can withstand as 138 N. Applying Hooke's law again, \(F = kx\), rearranged, we get \(x = F / k\). Substituting the values in, \(x = 138 \, N / 199.19 \, N/m = 0.69 \, m\). Thus, it can stretch by 0.69 meters without rupturing.
03

- Energy stored in tendon

Lastly, we need to determine the energy stored in the tendon right before rupture. The energy stored in a spring, which can be applied to a tendon, is given by the formula \(E = 0.5kx^2\), where x is the extension. Substituting the appropriate values, \(E = 0.5 * 199.19 \, N/m * (0.69 \, m)^2 = 47.14 \, J\). Therefore, and the tendon stores 47.14 J of energy at the point of rupture.

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Key Concepts

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

Force Constant Calculation
Understanding the force constant is fundamental when studying the behavior of elastic materials under the force. In the context of Hooke's Law, the force constant is also known as the spring constant, denoted by the letter 'k'. It quantifies the stiffness of an elastic material. The basic form of Hooke's Law is given by the equation:
\[ F = kx \]
where 'F' is the force applied to the material, 'k' is the force constant, and 'x' is the displacement caused by the force. Hence, calculating the force constant involves rearranging the equation to
\[ k = \frac{F}{x} \]
In the exercise, once the force exerted on the tendon is calculated, and the displacement is converted to meters, the force constant is found with ease. The higher the force constant, the stiffer the material. Remember, different materials have different force constants, which means the same force can result in different displacements for different materials.
Tendon Elasticity
Tendons, much like springs, exhibit elastic properties that allow them to stretch and then return to their original length. The term 'elasticity' refers to this ability of a material to deform under stress and then recover its shape upon the removal of stress. This characteristic is crucial for tendons as they need to withstand and transmit the forces generated by muscles to move the bones.

In biomechanics, understanding tendon elasticity is vital, as it relates to both the efficiency of movement and the potential for injury. A highly elastic tendon can store and release more energy, leading to more powerful movements. Conversely, if a tendon is overstretched, as when the force exceeds its ultimate tensile strength, it can lead to rupture, which is why the measurement of maximum tension, as seen in the exercise, is clinically significant.
Energy Stored in Elastic Materials
Elastic materials have the capacity to store energy when they are deformed. This is known as elastic potential energy, and for materials obeying Hooke's Law, the energy stored can be calculated using the formula:
\[ E = \frac{1}{2}kx^2 \]
, where 'E' represents the energy, 'k' is the force constant, and 'x' is the amount of displacement or stretch from its equilibrium position. The energy storage capability is a crucial factor in applications ranging from engineering to biomechanics.

In the given exercise, calculating the energy stored in the tendon at the point of maximum stretch before rupture is vital for understanding both the potential energy that can be harnessed during physiological activities and the risk of overstretching. This knowledge helps in sports science and orthopedics to enhance performance and prevent injuries.

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Most popular questions from this chapter

In one day, a 75 kg mountain climber ascends from the 1500 m level on a vertical cliff to the top at 2400 m. The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 m. What is her change in gravitational potential energy (a) on the first day and (b) on the second day?

A small box with mass \(0.600 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline that slopes upward at \(37.0^{\circ}\) above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is \(\mu_{k}=0.400\) The spring is released and the box travels up the incline, leaving the spring behind. What minimum elastic potential energy must be stored initially in the spring if the box is to travel \(2.00 \mathrm{~m}\) from its initial position to the top of the incline?

C Pendulum. A small rock with mass \(0.12 \mathrm{~kg}\) is fastened to a massless string with length \(0.80 \mathrm{~m}\) to form a pendulum. The pendulum is swinging so as to make a maximum angle of \(45^{\circ}\) with the vertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes through the vertical position? What is the tension in the string (b) when it makes an angle of \(45^{\circ}\) with the vertical, (c) as it passes through the vertical?

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A 25.0 kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0掳 from the vertical and releases her from rest. (a) What is her potential energy just as she is released, compared with the potential energy at the bottom of the swing鈥檚 motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?
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