Answer:
(a) [tex]\triangle l=5 mm[/tex]
(b) 0.033
Explanation:
(a)
Force F=mg where m is mass and g is acceleration due to gravity whose value is taken as [tex]9.81 m/s^{2}[/tex]
However, for this case, the maximum force is 8 times the weight of runner hence F=8mg
Assume Young's modulus for tendon is [tex]0.15*10^{10} N/m^{2}[/tex]
Young's modulus is given by
[tex]E=\frac {Fl}{A\triangle l}[/tex] and [tex]\triangle l=\frac {Fl}{EA}[/tex] and substituting F with 8mg we obtain [tex]\triangle l=\frac {8mgl}{EA}[/tex]
Where E is young's modulus, l is stretched length and \triangle l is change in length
Substituting m as 70 kg, g as [tex]9.81 m/s^{2}[/tex], l as 15cm=0.15 m, E as [tex]0.15*10^{10} N/m^{2}[/tex] and A as [tex]110 m^{2}=0.000110 m^{2}[/tex]
[tex]\triangle l=\frac {8*70 Kg*9.81 m/s^{2}*0.15m}{0.15*10^{10} N/m^{2} *0.00011 m^{2}}=0.004994182 m[/tex]
[tex]\triangle l=5 mm[/tex]
(b)
Strain, [tex]\epsilon=\frac {\triangle l}{l}[/tex] and the fraction of tendon’s length is the ratio of change in length to the stretched length
The fraction of tendon, f is given by
[tex]f=\frac {\triangle l}{l}[/tex]. Substituting [tex]\triangle l[/tex] with 0.005m and l with 0.15m we obtain
[tex]\epsilon=f=\frac {0.005}{0.15}=\frac {1}{30}=0.033[/tex]
Therefore, fraction of the tendon’s length is 0.033