If 100 N force is applied to 10kg block as shown in diagram, then acceleration produced for slab :

Let friction force between block and slab be f and both block move together with acceleration a.a=10040+10=f40⇒f=80NAs fmax=μs(10g)=0.6×(10×9.8)=58.8N which is less than 80 N, so there is slipping between block and slab. Therefore acceleration produced in slab=μk(10g)40=0.98m/s2