Human feet and legs store elastic energy when walking or running. They are not nearly as efficient at doing so as kangaroo legs, but the effect is significant nonetheless. If not for the storage of elastic energy, a \(70-\mathrm{kg}\) man running at \(4 \mathrm{m} / \mathrm{s}\) would lose about \(100 \mathrm{J}\) of mechanical energy each time he sets down a foot. Some of this energy is stored as elastic energy in the Achilles tendon and in the arch of the foot; the elastic energy is then converted back into the kinetic and gravitational potential energy of the leg, reducing the expenditure of metabolic energy. If the maximum tension in the Achilles tendon when the foot is set down is \(4.7 \mathrm{kN}\) and the tendon's spring constant is $350 \mathrm{kN} / \mathrm{m},$ calculate how far the tendon stretches and how much elastic energy is stored in it.

Step by step solution

01

Identify the given information

We are provided with the following information: - Maximum tension in the Achilles tendon, \(T = 4.7 \, \mathrm{kN} = 4700 \, \mathrm{N}\) - Spring constant of the tendon, \(k = 350 \, \mathrm{kN/m} = 350000 \, \mathrm{N/m}\)

02

Calculate the stretching in the tendon

We will utilize Hooke's Law to calculate the amount of stretching in the tendon. Hooke's Law states that the force exerted by a stretched spring is proportional to the displacement from its equilibrium position: \(F = -kx\) where: - \(F\) is the force exerted by the spring (in this case, the tension in the Achilles tendon) - \(k\) is the spring constant - \(x\) is the displacement (the stretching of the tendon) We will rearrange the formula to solve for \(x\): \(x = \frac{-F}{k}\) Now, substitute the given values: \(x = \frac{-(-4700 \, \mathrm{N})}{350000 \, \mathrm{N/m}}\) \(x = \frac{4700}{350000} \, \mathrm{m}\) \(x \approx 0.0134 \, \mathrm{m}\) So, the Achilles tendon stretches by about 0.0134 meters or 1.34 centimeters.

03

Calculate the elastic potential energy stored in the tendon

The elastic potential energy stored in a spring is given by the formula: \(U = \frac{1}{2}kx^2\) where: - \(U\) is the elastic potential energy - \(k\) is the spring constant - \(x\) is the displacement (the stretching of the tendon) We have calculated the stretching in the tendon (\(x \approx 0.0134 \, \mathrm{m}\)), and we know the spring constant (\(k = 350000 \, \mathrm{N/m}\)). So, we can now calculate the elastic potential energy: \(U = \frac{1}{2}(350000 \, \mathrm{N/m})(0.0134 \, \mathrm{m})^2\) \(U \approx 31.9 \, \mathrm{J}\) Thus, the elastic potential energy stored in the Achilles tendon is approximately 31.9 Joules.

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