Question

A block of mass m1 = 4.80 kg sits on top of a second block of mass m2 = 15.6 kg, which in turn is on a horizontal table. The coefficients of friction between the two blocks are µs = 0.300 and µk = 0.100. The coefficients of friction between the lower block and the rough table are µs = 0.500 and µk = 0.400. You apply a constant horizontal force to the lower block, just large enough to make this block start sliding out from between the upper block and the table.


Determine the magnitude force needed to initiate this motion.


Determine the acceleration you measure for each block in this situation

Answer 1

First, find the force between the blocks.
m1*g*mu_s
That force will produce an accelearation on the upper block of a=F/m=g*mu_s
but to get that acceleartion, on has to apply a force to the lower block..
F=totalmass*a= (m1+m2)*g*mu_s
that answers the first question.

Now, the acceleration for each block
block 1: a=m1*g*mu<k>/m1
for block2: net force actin on lower block is pushing force- force on block1, or
(m1+m2)*g*mu_s - m1*g*mu_k
and acceleration of the lower block is that divided by m2.

check my thinking.