Coefficient of Kinetic Friction Given Coefficient of Static Friction

Is there a way to calculate the coefficient of kinetic friction given the coefficient of static friction? Is there any direct relationship between the two or is it completely different between materials?

The kinetic friction coefficient is, in general, smaller that the static one. As far as I know, there is no way to calculate one from the other. But they usually do not differ by order of magnitude. These are purely empirical quantities.

• The entire concept of "static" and "kinetic" friction coefficients, and the assumption that the two coefficients are constant, is an assumption - or more formally a "model" of how friction works. It's a fairly good model for simple situations like lab experiments sliding blocks down inclined planes, and it's sufficiently simple that you can solve problems by hand rather than computer simulations - but don't get the wrong idea that it is a "physical law" with the same level of "universal truth" as Newton's laws of motion, etc. – alephzero Oct 2 '16 at 3:00
• A bit of logical thinking should convince you that it's impossible for the kinetic friction coefficient to be bigger than the static coefficient - just think about what the friction force must be when the two objects start moving relative to each other. – alephzero Oct 2 '16 at 3:03
• @ alephzero - A bit of search should reveal to you that empirically the static friction coefficient is larger than the kinetic friction coefficient. See e.g. en.wikipedia.org/wiki/Friction#Kinetic_friction – freecharly Oct 2 '16 at 3:58

I do have a simple setup which you can use to find coefficient of kinetic friction.

Take a block having mass $M$ on a rough horizontal surface. Apply force on the block in horizontal direction till the block just starts to move. This force would be $$F=\mu_sMg$$

Once the block starts moving kinetic friction will act on the block. Since coefficient of kinetic friction is numerically less than that of static there will be unbalancing of forces thus causing an acceleration, say $a$. If by any means you can calculate the acceleration then by force equation we will have-

$$F-\mu_kMg =Ma$$ $$\Rightarrow \mu_k = \mu_s - \frac {a}{g}$$

I believe this might just exist in paper and practically it seems to be a bit difficult to comprehend this. Hence @freecharly answer suits.