A 40 kg slab rests on frictionless floor a 10 kg block rests on the top of the slab (see figure). The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force F. What is the acceleration of the slab if F = 50 N

(i) The limiting friction between block and slab

f_lim = $\mathrm{\mu }$_sN = 0.60 x 10g = 58.8N
As applied force (50 N) is less than the limiting friction between the block and slab, so block
will not slide over the slab. They will move together as one unit of mass 40 + 10 = 50 kg by a
net force of 50 N over the frictionless floor.
$\therefore {\mathrm{a}}_{\text{block }}=\frac{50}{40+10}=1\mathrm{m}/{\mathrm{s}}^2={\mathrm{a}}_{\text{slab }}$