In the figure, a slab of mass m1 = 40 kg rests on a frictionless floor, and a block of mass m2 = 9 kg rests on top of the slab. between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. a horizontal force of magnitude 103 n begins to pull directly on the block, as shown. in unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?

Draw a free body diagram of on the mass m2 = 9 kg, as shown in the picture attached with the question.
 Weight = mg = 9 x 10 = 90N downwards 
Normal reaction R = 90N upwards 
F = 103N  
and friction on the right.

Then the static friction = μs.R = 0.60 x 90 = 54N (right direction)
Now, the force applied is 103 N and the max static friction is 54 N, therefore the smaller box will slip on the larger box and in this case, kinetic friction will act. 
Fr = μk.R = 0.4 x 90 = 36 N (right direction)

The resultant horizontal force on m2 = 103 N - 36 N = 67 N (to the left)

F = ma,
⇒ a = F/m = 67/9 = 7.44m/s² 

In unit vector notation, if rigth direction is considered as +i, then a = -5.1î m/s² 

Now, on bigger box, the net horizontal force is 36N (KINETIC FRICTIONAL FORCE)

a = F/m = 36/40 = 0.9m/s² 

In unit vector notation this is -0.9î m/s² 

Divide 108.2 x 10^6 into 1.5 x 10 and the result will be in au (astronomical units)

The answer is c hope this