Question

A 40kg slab rests on a frictionless floor . A 10kg block rests on top of slab . coefficient of kintic friction between the block and slab is 0.4 . A horizontal force of 100 N applied on 10 kg block .Find resulting acceleration of slab

Answer 1

Normal reaction on 10 kg block will be equal to weight of the 10 kg block.
 which is N =mg = 10 x 9.8 = 98 N.
When 100 N force is applied on the block, say to the right side, the block tends to move tom the right. Friction on  the block will act to the left to oppose the relative motion.
 Force of friction, f = μN = 0.4 x 98 = 39.2 N.
As every force come as a pair, the pair of this friction will act on the slab to the right as shown in the diagram. 
The floor is smooth, so no other horizontal force will act on the slab.

The net horizontal force on the slab is thus 39.2 N which will produce an acceleration of 39.2/40 = 0.98 $ms^{-2)$.
Thus the slab will accelerate to the right( in the direction of the applied force).

Note:
Net horizontal force on the 10 kg block will be 100 -39.2 = 60.8 N.
This will produce acceleration of $a = \frac{60.8}{10}=6.08 ms^{-2}.$.
The slab will also accelerate to the right( in the direction of the applied force).

Answer 2

Hey there!

Normal reaction from 40 kg slab on 10 kg block = 10 * 9.81 = 98.1 N 

Static frictional force = 98.1 * 0.6 N is less than 100 N applied 

=> 10 kg blck will slide on 40 kg slab and net force on it 

= 100 N - kinetic friction 

= 100 - 98.1 * 0.4 = 61 N 

=> 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s 

Frictional force on 40 kg slab by 10 kg block = 98.1 * 0.4 = 39 N 

=> 40 kg slab will move with 

39/40 m/s 

= 0.98 m/s.

Hope It Helps you!