Does the kinetic friction does work?
My book says that
"In fact, the displacement of the point of application of the friction force is not calculable and so neither is the work done by the friction force.".
However, in many other sources, the work done by friction can be calculated. How is that?
When analyzing friction I find it easier to focus on power instead of work. Mechanical power is given by $P= \vec F \cdot \vec v$ where $\vec v$ is the velocity of the material at the point of application of the force, $\vec F$.
For example, suppose that we have an automobile which is skidding to a stop with the wheels not turning. In this case the tires in contact with the road are moving at $\vec v$ in the forward direction and $\vec F$ is in the backward direction. So $P=\vec F \cdot \vec v=-Fv$, meaning that mechanical power is leaving the car.
At the same time the road is not moving, so for the road $P=-\vec F \cdot 0=0$. This means that mechanical power is not entering the road. The difference between the mechanical power leaving the car and the mechanical power entering the road is the thermal power generated by the skid.
If you must obtain work and not just power then $W=\int P \ dt$
Kinetic friction usually does negative work, and is often calculable.
If you take a flat object sliding on a flat surface, with known speed and mass. Measure the distance till it stops, and the work done is the distance times the friction force which can be calculated easily here.
You will have to provide context about where it was said to be uncalculable. It does sound like like you are referring to rolling body , which in ideal setup is static and not kinetic... But in practice there is kinetic friction (i.e. skidding)
From my research, the friction force doesn't do any work, because its point of application is not moving with the body, in other words, it's stationary.
Let's take an example of a sliding cubical object on the surface of the ground with an initial velocity v0. The only force applied to it is the friction force in the opposite direction of movement. So the object will slow down until it stops. Applying the energy balance on the system (object) we have the following equation: dU+dEp+dEk=W+Q. With, dU: variation of internal energy, dEp: variation of potential energy, dEk: variation of kinetic energy, W: work done on the system and Q is the heat absorbed by the system. In our case, dEp=0 (no elevation) and W=Q=0 (no work done on the system and no heat absorbed from outside).
So our equation becomes dU=-dEk=0.5m(v0)². Hence, all the kinetic energy is transformed into internal energy or into heat.
