Question

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alternately stretches, as you bring your weight down onto your forward foot, and contracts to push you off the ground. A 70 kg runner has an Achilles tendon that is 15 cm long with a typical 1.1×10−4m2 area.

A. By how much will the runner's Achilles tendon stretch if theforce on it is 8.0 times his weight?
B. What fraction of the tendon's length does this correspond?

Answer 1

Step-by-step explanation:

Given:

Mass, M = 70 kg

Area, A = 1.1 × 10^-4 m^2

Achilles tendon length, l = 15 cm

= 0.15 m

Young modulus for tendon, E = 0.15 × 10^10 N/m^2

Young modulus, E = tensile stress/tensile strain

But,

Stress = force/area

Strain = extension/length

A.

Force = 8 × N

N = m × g

= 70 × 9.8 × 8

= 5488 N

Extension = (force × length)/(area × yound modulus)

= (5488 × 0.15)/(1.1 × 10^-4 × 0.15 × 10^10)

= 4.989 mm

= 0.005 m

B.

Fraction of length, x = extension/original length

= 0.005/0.15

= 0.033

= 3.3 %

Answer 2

a)5mm

b)3.3%

Step-by-step explanation:

first note down the what's given

`Y_(tendon) = 0.15x10^(10) N/m^(2)`

Area=
`A= 1.1x10^(-4) m^(2)`

Mass of the runner=
`m_(r)`=70kg

Fmax= 8 F (since tendon can stretch 8 times of runner's weight)

= 8
`m_(r)`g=8 x 70 x 9.8=> 5488N

L=15cm=> 0.15m

a)
`We know that, Y={Fmax.L}`/AΔL

YAΔL=Fmax. L

ΔL=
`(Fmax L)/(YA) =(5488*0.15)/(0.15x10^(10)x1.1x10^(-4) )`

ΔL=
`4.99x10^(-3) m` => 5mm

b) L=15cm= 150mm

Fractional change can be determined by,

(ΔL/L) *100

=
`(4.99)/(150) *100`

=3.33%