A 40 kg slab rests on a frictionless floor. A 10 kg block rests on the top of the slab as shown in figure.The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 ms-2, the resultingacceleration of the slab will be

Force of limiting friction for block =μsmg=0.60×10×9.8=58.8 NIf the applied force is greater than 58.8 N, then the block will move over the slab.Kinetic friction acting on the block towards right=μkmg=0.40×10×9.8=39.2 NThis is also equal to the force of friction acting on slab towards left. This is the only force acting on slab.So, acceleration of the sIab,  a=Fm=39.240=0.98 ms-2