Answer:
[tex]fraction = 0.11[/tex]
Explanation:
Linear kinetic energy of the bicycle is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]K_1 = \frac{1}{2}(13) v^2[/tex]
[tex]K_1 = 6.5 v^2[/tex]
Now rotational kinetic energy of the wheels
[tex]K_2 = 2(\frac{1}{2}(I)(\omega^2))[/tex]
[tex]K_2 = (mR^2)(\frac{v^2}{R^2})[/tex]
[tex]K_2 = mv^2[/tex]
[tex]K_2 = 3.1 v^2[/tex]
now kinetic energy of the rider is given as
[tex]K_3 = \frac{1}{2}Mv^2[/tex]
[tex]K_3 = \frac{1}{2}(38) v^2 [/tex]
[tex]K_3 = 19 v^2[/tex]
So we have
[tex]fraction = \frac{K_2}{K_1 + K_2 + K_3}[/tex]
[tex]fraction = \frac{3.1 v^2}{6.5 v^2 + 3.1 v^2 + 19 v^2}[/tex]
[tex]fraction = 0.11[/tex]
