How Much Energy is Absorbed Elastically When a Person Lands on Their Feet?

  • Thread starter Thread starter debzrie
  • Start date Start date
  • Tags Tags
    Energy Physics
AI Thread Summary
A 60 kg person landing at 3.5 m/s experiences energy absorption through various mechanisms, including cartilage compression and muscle involvement. The Achilles tendon acts like a spring, with an extension of 0.010 m and a maximum force of 1800 N, which leads to elastic energy absorption. To calculate the energy absorbed elastically, one must determine the spring constant and the energy stored in the tendon when stretched. Additionally, the fraction of energy absorbed compared to the total energy just before impact needs to be evaluated. Understanding these calculations clarifies how much energy is managed during landing.
debzrie
Messages
4
Reaction score
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

 
Physics news on Phys.org
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
 
still a little confused :S
 
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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top