Answer:43.311 J
Explanation:
Given
mass of frame[tex]\left ( m_f\right )=6.71 kg[/tex]
mass of wheel [tex]\left ( m_w\right )=0.820 kg[/tex]
radius of wheel[tex]\left ( r\right )=0.343 m[/tex]
v=3.22 m/s
Moment of inertia of each wheel\left ( I\right )=\frac{1}{2}mr^2[/tex]
[tex]I=0.0482 kg-m^2[/tex]
kinetic Energy of whole cycle=Kinetic energy of wheels and frame+rotational energy of Wheels
[tex]K.E.=\frac{1}{2}\left ( m_f+m_w\right )v^2+2\times \frac{1}{2}I\omega ^2[/tex]
[tex]K.E.=\frac{1}{2}\left ( 6.71+0.820\right )3.22^2+2\times \frac{1}{2}0.0482\times \left ( \frac{3.22}{0.343}\right )^2[/tex]
K.E.=39.037+4.274=43.311J