Answer:
[tex]\Delta U=9.176\, J[/tex]
Explanation:
In such a case where the energy stored is in the physical dimension of the material due to the elastic deformation is called the elastic potential energy.
It is mathematically given as:
[tex]U=\frac{1}{2} k.\Delta x^2[/tex].....................................[1]
where :
U= elastic potential energy
k= constant of elasticity or the spring constant
[tex]\Delta x[/tex]= change in the length
Given:
stretch in tendons for the sprinters, [tex]\Delta x_s[/tex]= 41 mm
stretch in tendons for the nonsprinters, [tex]\Delta x_n[/tex]= 33 mm
spring constant for the tendons of both group, k= 31 [tex]N.mm^{-1}[/tex]
So, the difference in the stored energy:
[tex]\Delta U=[\frac{1}{2} k. x_s^2]-[\frac{1}{2} k. x_n^2][/tex]
[tex]\Delta U=\frac{1}{2} k[ x_s^2-x_n^2][/tex]
putting the respective values
[tex]\Delta U=\frac{1}{2}\times 31 \times [ 41 ^2-33^2][/tex]
[tex]\Delta U=9176\, N.mm[/tex]
[tex]\Delta U=9.176\, J[/tex]