Answer:7.92 J
Explanation:
Given
[tex]x_1=40 mm[/tex]
[tex]x_2=33 mm[/tex]
spring constant of tendons [tex]k=31 N/mm\approx 31\times 10^3 N[/tex]
Energy stored in athletes tendon[tex]=\frac{kx_1^2}{2}=\frac{31\times 10^3\times (40\times 10^{-3})^2}{2}[/tex]
Energy stored in athletes tendon[tex]=\frac{kx_2^2}{2}=\frac{31\times 10^3\times (33\times 10^{-3})^2}{2}[/tex]
Difference in maximum stored energy[tex]=\frac{k}{2}(x_1^2-x_2^2)[/tex]
[tex]=\frac{31\times 10^3}{2}(1600-1089)\times 10^{-6}[/tex]
[tex]=7.92 J[/tex]
