Answer:
Starting point, in the highest part of the hill
gravitational potential energy
Final point , in the lower part of the hill,
translational kinetic energy , rotational kinetic energy
there is not work of non-conservative forces (friction),
Explanation:
This is a problem that can be treated using the principle of the conservation of mechanical energy, where the energy must be written in two instants.
Starting point.
This point is in the highest part of the hill, therefore it has gravitational potential energy, as it is stopped there is no other energy in the system.
Em₀ = U = m g h
Final point
This point is located in the lower part of the hill, in this case the height is zero so there is no gravitational potential energy, the cyclist has a speed therefore there is a translational kinetic energy and the wheels are turning with a speed therefore each wheel has rotational kinetic energy
[tex]Em_{f} = K = \frac{1}{2} m v^{2} + 2 ( \frac{1}{2} I w^{2} )[/tex]
Where is the speed of the cyclist, I the moment of inertia of each wheel and w the angular velocity, the number two comes from the fact that there are two wheels.
If we approximate the wheels like a hoop, the moment of inertia is
I = m r²
angular and linear velocity are related
v = w r
w = v/r
let's replace
Em_{f} = K = \frac{1}{2} m v^{2} + 2 ( \frac{1}{2} m r^{2} v^{2}/ r^{2} )
Emf = ½ M v² + m v²
Emf = [tex]\frac{1}{2}[/tex] v² (M + 2 m)
As they indicate that friction is neglected, there is not work of non-conservative forces (friction), also the point of contact of the wheel with the pavement is at rest, therefore it does not generate work,