written Question #2: As a person steps onto their \"tiptoes\" as shown in the fi

Question

written Question #2: As a person steps onto their "tiptoes" as shown in the figure, the force exerted by their Achilles tendon on their ankle, Ft, increases greatly in magnitude. (We will learn in chapter 8 that this increase in tendon force occurs to prevent the ankle's rotation.) The Achilles tendon to good approximation 200 mm obeys Hooke's law. (a.) A person starts from rest with their feet flat on the floor (where their Achilles tendons are unstretched). They tilt both feet upward so that they are on their tiptoes. When each ankle is 200 mm above the ground, they stop and stay there. Explain why this process cannot conserve mechanical energy and therefore requires the person to burn calories while doing it. For ful points you must correctly identify every type of energy involved in this process and explain why it is increasing or decreasing. (b.) Using a realistic value for your mass, estimate the nmber of times you would have to stand on your tiptoes to burn one calorie. Assume your entire body rises 200mm. Hint: When standing on your tiptoes, the force F in each of your Achilles tendons is typically 75% of your weight. Also, a typical spring constant for a human's Achilles tendon is 35000 N/m

Explanation / Answer

a) To understand the tiliting process we have to remember the mechanical energy conservation theory.

According to the theory net mechanical energy is the sum of potential and kinetic energy. When a body is not moving the overall energy is stored in the body as the potential energy and if the body starts to move then the potential energy converts into the kinetic energy considering the rate of conversion. Now the conservation theory states the net mechanical energy is constant and with transformation to each other it completes any mechanical process. So while I am tilting my legs I am acting against the gravitationaal force while remaing in a static position. I have not done any movement but has merely changed my position. Therefore, no conversion between potential and kinetic energy. But the position is acting against gravity so I have to appy force in the tiptoes to keep the position. Therefore, the force on the tiptoe is increasng while acting against gravity. The force elevates the feet 200 mm from the surface therefore if we consider the formula of work done then x=200mm and F= the force on the tiptoe.

So, W= F.x which is the amount of energy due to applied force which means burning the calorie using acceleration due to gravity.

b) I consider myself as an average person with weight in the range 65-75kg.

Suppose the weight is = (65 +75) / 2 = 70kg

So the displacement x= 200mm = .2m

The force Ft = 75% of the weight = 75% of mg = 75 % of 70 * 9.8m/s^2 = 514.5 N [ m= mass and g= acceleration due to gravity]

Now spring constant K = 35000N/m

Using Hooks law Ft=-Kx then overall work done or energy burning for .2m displacement = 1/2 kx^2 Joules= .5 * 35000 *.2^2 joules= 700joules = 167.304 calorie

Due to the force  514.5 N the work done or energy burning = 514.5 * .2 = 102.9 joules = 24.59 calorie [ the force is against the gravity]

The net energy = 167.304 - 24.59 = 142.714 cal

Then to burn 1 calorie I have to stand 1/142.714 = .007 times.

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