The Achilles tendon, which connects the calf muscles to the heel, is the thickes

Question

The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2mm2 , its average length is 26cm , and its average Young's modulus is 1474MPa . Part A How much tensile stress is required to stretch this muscle by 5.2%  of its length? Express your answer using two significant figures. ? =
Pa SubmitMy AnswersGive Up Try Again Part B If we model the tendon as a spring, what is its force constant? Express your answer using two significant figures. k =
N/m SubmitMy AnswersGive Up Try Again Part C If a 75kg  sprinter exerts a force of 13.0 times his weight on his Achilles tendon, by how much will it stretch? Express your answer using two significant figures. x =
cm SubmitMy AnswersGive Up
The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2mm2 , its average length is 26cm , and its average Young's modulus is 1474MPa .
The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2mm2 , its average length is 26cm , and its average Young's modulus is 1474MPa . The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2mm2 , its average length is 26cm , and its average Young's modulus is 1474MPa . The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2mm2 , its average length is 26cm , and its average Young's modulus is 1474MPa . Part A How much tensile stress is required to stretch this muscle by 5.2%  of its length? Express your answer using two significant figures. ? =
Pa SubmitMy AnswersGive Up Try Again Part A How much tensile stress is required to stretch this muscle by 5.2%  of its length? Express your answer using two significant figures. ? =
Pa SubmitMy AnswersGive Up Try Again ? =
Pa ? =
Pa SubmitMy AnswersGive Up Try Again Try Again Try Again Part B If we model the tendon as a spring, what is its force constant? Express your answer using two significant figures. k =
N/m SubmitMy AnswersGive Up Try Again Part B If we model the tendon as a spring, what is its force constant? Express your answer using two significant figures. k =
N/m SubmitMy AnswersGive Up Try Again k =
N/m k =
N/m SubmitMy AnswersGive Up Try Again Try Again Try Again Part C If a 75kg  sprinter exerts a force of 13.0 times his weight on his Achilles tendon, by how much will it stretch? Express your answer using two significant figures. x =
cm x =
cm SubmitMy AnswersGive Up ? =
Pa

Explanation / Answer

part A:
let length of the tendon be l, and change in length due to streching be l'

strain (E) is defined as l'/l.

now, as given, l' = 5.2/100 x l

it implies that E= l'/l = 0.052.

also youngs modulud (Y) is defined as stress/strain, it implies that stress (s) = strain x Y

so stress(s) = 1474000000 x 0.052 = 76648000 Pa

in two significant figures it will be: 77000000 Pa

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part B:

in spring model, force (F) = Kl'

wher k is the spring constant and l' is change in length.

force can be written as stress(s) x area(a), and dividing both sides by original length (l), we get:

(s*a)/l = K* (l'/l)

but, (l'/l) is strain(E) and as discussed above it can be written as (stress/Y)

s*a/l = K*s/Y

it implies that K = a*Y/l

or K = 78.2*10^-6 x 1474*10^6/26*10^-2

K= 443333.846 N/m

in two sig. fig. it will be 440000 N/m

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part C:

lets us assume acceleration due to gravity(g) is 9.8 ms^-2

weight of sprinter = mass*g = 75*9.8= 735 N

force on tendon= 13*weight= 9555 N

now we can use the spring model derived in part B,

F= Kl' or l' = F/K

or l' = 9555/443333.846 = 0.0215526 m or 2.15526 cm

in two sig. fig, l' = 2.1 cm

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hope it helped :)

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