Work done by friction when a bicycle comes to a skidding stop My textbook contains this question as a worked out example.
A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle ? (b) How much work does the cycle do on the road ?
And it gives the answers as:
(a) 200 × 10 × cos π = -2000J
(b) 0 (Because the road doesn't move)
I understand the answers.
Here are my questions:
(1) Would it be right to say that the system (bicycle + bicyclist) had more than 2000 J of kinetic energy before before the brakes were applied? My reasoning is this: -2000J of work was needed to stop it, and there would've been some heat generated, so work done by the road served to stop the bicycle and also generate some heat. Hence the bicycle should've had more than 2000J to begin with.
(2) Is it kinetic frictional force that's acting on the bicycle? (Because the question says "skidding").
(3) Let's for a moment assume that the earth isn't spinning. So before the bicyclist applied brakes, the earth would have been moving under him/her in the opposite direction (albeit very little). So what happens to the earth when the brakes are applied? Will the earth first stop moving under him/her, and then start moving in the same direction as the bicycle till there's no relative motion between earth and bicycle? (I am not able to picture this).
Thank you!
 A: Let's take System = ground + bicycle, in which case there are no external forces acting, and hence the net work done on the system is zero.  Thus, the total energy of the system must remain the same (i.e., be conserved).
In the rest frame of the ground, the bicycle is moving at the beginning of the process and stationary at the end. The total energy is
$$
E_{\textrm{system}} = E_{\textrm{th}} + T_{\textrm{bicycle}} + T_{\textrm{ground}} + \cdots\,,
$$
where the $T$'s are kinetic energies, $E_{\textrm{th}}$ is the thermal energy of the system (split between the bicycle and the ground), and in $\cdots$ there are energies like the gravitational potential energy $U_{\textrm{grav}}$ which are irrelevant in this problem (so let's just pretend they're zero, because we're going to be computing changes in system energy anyway).
At the beginning of the process, $T_{\textrm{bicycle}} = T$ (where $T$ is some positive number) and $T_{\textrm{ground}}=0$. At the end of the process, both kinetic energies are zero. Hence, the total change in system energy is
$$
0=\Delta E_{\textrm{system}} = \Delta E_{\textrm{th}} + \Delta T_{\textrm{bicycle}} + \Delta T_{\textrm{ground}}
=\Delta E_{\textrm{th}} + (0-T) + (0-0)\Longrightarrow
T = \Delta E_{\textrm{th}}\,.
$$
In other words, the initial kinetic energy of the bicycle was all converted into thermal energy, split in some unknown way between the bicycle and the ground.
Now, to compute the $\Delta E_{\textrm{th}}$, we have to use the kinematic argument made by the OP and compute the pseudo-work done on the bicycle by the frictional force exerted by the ground.  But to understand where the energy is going to and coming from, a systems view of energy is required.  Frictional forces (and dissipative forces in general) require careful reasoning about the energy along the lines that I've outlined above. Generally speaking, it is usually best to put all objects interacting via dissipative forces inside the system in order to carefully track where the energy goes.
Therefore:

Would it be right to say that the system (bicycle + bicyclist) had more than 2000 J of kinetic energy before before the brakes were applied?

No, as pointed out, that 2000 J of energy is equal to the initial kinetic energy of the bicycle, all of which gets converted into thermal energy.

(3) Let's for a moment assume that the earth isn't spinning...

In the rest frame of the bicycle, the ground is moving backwards, and we can make the same argument: the frictional force exerted by the bicycle on the ground is (by Newton's third law) in the opposite direction as the frictional force exerted by the ground on the bicycle.  This means that the force exerted on the ground is in the opposite direction as the ground's "motion", and hence the ground slows down relative to the bicycle.
(I feel like the energy argument is more difficult in this frame, partly because the bicycle lose kinetic energy in this case, too, i.e., the wheels are initially rotating but eventually stop.  The basic idea is the same, though.)
A: 

*

*Would it be right to say that the system (bicycle + bicyclist) had
more than 2000 J of kinetic energy before before the brakes were
applied?


No. Per the work energy theorem the net work done on an object equals its change in kinetic energy. Since the only work done on the bicycle + bicyclist is the kinetic friction work, that work must equal the change in kinetic energy, i.e., -2000 J.

My reasoning is this: -2000J of work was needed to stop it, and there
would've been some heat generated, so work done by the road served to
stop the bicycle and also generate some heat. Hence the bicycle
should've had more than 2000J to begin with.

The loss of macroscopic kinetic energy of the bicycle + bicyclist equals the gain in the internal (microscopic) molecular kinetic energy of the bicycle materials, which in turn increases the temperature of the materials relative to the environment. That, in turn, results in heat transfer to the environment equal to the loss of kinetic energy.

(2) Is it kinetic frictional force that's acting on the bicycle?
(Because the question says "skidding").

Yes.
Regarding (3) and the textbook answer to (b), I think the the book answer can be misleading. Theoretically, the bicycle/bicyclist does do work on the road, but it is infinitesimal because its change in kinetic energy is infinitesimal.
Taking the bicycle/bicyclist plus the earth as an isolated system, momentum has to be conserved. The loss of momentum of the bicycle/bicyclist must equal the increase in momentum (and thus an increase in velocity) of the earth.  It's just that the resulting increase in velocity (and thus increase in kinetic energy) of the earth is infinitesimal because its mass is so large.
Hope this helps.
