Question: A block sliding across a table top comes to rest. Why?
"Why?" is a philosophical question that science is ill equipped to
answer. "How?" is a question that has a better shot at being profitably
addressed by scientific considerations. So how does the sliding block
come to rest?
The way the block comes to "rest" w.r.t. the table is by the operation
of the 2nd law of thermodynamics as the macroscopic kinetic energy
tied up in the block's center of mass degrees of freedom is thermalized
via dissipative processes and shared with the other internal degrees of
freedom of the block and table as the total entropy of the block/table
system increases in concert with the evermore randomized distribution
of the total internal energy among all the joint system's degrees of
freedom. This energy sharing is very Marxist in character. Once a new
equilibrium for the block-table system is established the energy is quite
equitably shared *on average* between all the various degrees of freedom.
In fact, the classical equipartition theorem guarantees that each of the
quadratic degrees of freedom *including* each of those 3 associated with
the kinetic energy of the block's center of mass (and the 3 DOF of
table's center of mass, too, for that matter) are each allocated a ration
of (1/2)*k*T of the block/table's total available internal thermal energy
to have as their average equilibrium energy. This means that after the
block fully equilibrates with the table its center of mass is quivering
with a Brownian motion whose average kinetic energy is (3/2)*k*T where T
is the final uniform absolute temperature of the block/table system.
Since a final temperature near room temperature allows an allocation of
(3/2)*k*T= 6.1 x 10^(-21) J as the average kinetic energy for the block's
center of mass, we see that we tend to not notice its motion at the
macroscopic level when the block is just sitting on the table in
equilibrium with it.