Physics: energy absorbed

In summary, a 60 kg person lands on their 2 feet on a floor with a speed of 3.5 m/s and the energy is dispatched in various ways. The calf and thigh muscles play a major role, along with the elastic stretching of the Achilles tendon. In part a), assuming the extension of the tendon is 0.010m and the maximum force is 1800 N, the energy absorbed elastically is determined by calculating the spring constant of the tendon and the energy stored when stretched to 0.010m. In part b), the fraction of energy absorbed in this way can be determined by comparing the energy of the person just before landing to the energy stored in the tendon springs.
  • #1
debzrie
4
0

Homework Statement


A 60 kg person lands on their 2 feet on a floor with a speed of 3.5 m/s.?
Energy dispatched in a variety of ways, including compression in the cartilage of your knees, in the soles of your shoes etc. calf and thigh muscles have a major role. elastic stretching of your Achilles tendon also plays a part.
a) if the extension of the tendon is 0.010m and is the maximum force (in one tendon) is 1800 N, how much energy is absorbed elastically? state any key assumption made.
b) what fraction of the energy just prior to impact is absorbed in this way?

Homework Equations


I'm so confused I have no idea :(


The Attempt at a Solution

 
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  • #2
debzrie said:

Homework Statement


A 60 kg person lands on their 2 feet on a floor with a speed of 3.5 m/s.?
Energy dispatched in a variety of ways, including compression in the cartilage of your knees, in the soles of your shoes etc. calf and thigh muscles have a major role. elastic stretching of your Achilles tendon also plays a part.
a) if the extension of the tendon is 0.010m and is the maximum force (in one tendon) is 1800 N, how much energy is absorbed elastically? state any key assumption made.
b) what fraction of the energy just prior to impact is absorbed in this way?
Think of the tendon as a spring that is stretched from 0 to .01 m in stopping the fall. What is the spring constant k of that spring? How much energy is stored in that spring when stretched to .01 m? How much energy does the person have just before landing? How much of that energy can be stored in those tendon springs?

AM
 
  • #3
still a little confused :S
 
  • #4
debzrie said:
still a little confused :S
What are you confused about?

Can you determine the energy of the person just before landing? Can you determine the spring constant of the tendons?

AM
 
Last edited:
  • #5

I would first identify the key components of this problem: a 60 kg person, landing on their feet with a speed of 3.5 m/s, and the various ways energy is absorbed in the body. From this, I can begin to formulate a solution by using the concept of kinetic energy and understanding how energy is transferred and absorbed in different forms.

a) To calculate the amount of energy absorbed elastically, we can use the equation for elastic potential energy: Eel = 1/2kx^2, where k is the spring constant and x is the extension of the spring. In this case, the Achilles tendon is acting as a spring, so we can use this equation. The key assumption made here is that the force applied to the tendon is constant throughout the extension, which may not be entirely accurate, but will provide a reasonable estimate.

First, we need to calculate the spring constant, k. We can use Hooke's Law, F = kx, where F is the force applied and x is the extension. Rearranging for k, we get k = F/x. Plugging in the values given, we get k = 1800 N / 0.010 m = 180,000 N/m.

Now, we can plug this value into the equation for elastic potential energy: Eel = 1/2(180000)(0.010)^2 = 9 J. Therefore, the amount of energy absorbed elastically is approximately 9 Joules.

b) To find the fraction of energy absorbed in this way, we need to compare the elastic potential energy to the initial kinetic energy of the person. Using the equation for kinetic energy, Ek = 1/2mv^2, we get Ek = 1/2(60 kg)(3.5 m/s)^2 = 367.5 J. Therefore, the fraction of energy absorbed elastically is 9 J / 367.5 J = 0.024, or approximately 2.4%.

In conclusion, the amount of energy absorbed elastically in this scenario is approximately 9 Joules, with the key assumption being that the force applied to the tendon is constant. This accounts for only a small fraction (2.4%) of the initial kinetic energy of the person. This calculation also highlights the importance of understanding how energy is transferred and absorbed in different forms in the human body during physical activities.
 

1. What is the definition of energy absorbed in physics?

Energy absorbed, also known as energy input, is the amount of energy that is taken in by a system or object. It can be in the form of heat, light, or any other type of energy.

2. How is energy absorbed calculated in physics?

The formula for calculating energy absorbed is E = m x c x ∆T, where E is the energy absorbed, m is the mass of the object, c is the specific heat capacity of the material, and ∆T is the change in temperature.

3. What are some examples of energy absorbed in everyday life?

Some examples of energy absorbed in everyday life include the sun's energy being absorbed by plants during photosynthesis, a person's body absorbing energy from food, and a light bulb absorbing electrical energy and converting it into light and heat.

4. How does energy absorbed relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. This means that the total amount of energy absorbed by a system must be equal to the total amount of energy released or used by that system.

5. How does energy absorbed affect the behavior of matter?

The amount of energy absorbed by a substance can affect its behavior in various ways. For example, if enough energy is absorbed, a solid can melt into a liquid or a liquid can boil into a gas. Energy absorbed can also cause changes in an object's temperature, shape, or state of matter.

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