A 75kg bicyclist (including the bicycle), initially at rest at the top of a hill, coasts down the hill, reaching a speed of 14.6m/s at the bottom of the hill. The distance and height of the hill are shown. Neglect any friction impeding the motion and the rotational energy of the wheels. List the energy types at the initial and final time and whether work and loss (due to non-conservative forces) occur as well as the corresponding amounts of energy.

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

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 =  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,

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach