A 1.0-kg block of wood sits on top of an identical block of wood, which sits on top of a flat level table made of plastic. The coefficient of static friction between the wood surfaces is μ1 , and the coefficient of static friction between the wood and plastic is i-t2’(a) A horizontal force F is applied to the top block only, and this force is increased until the top block starts to slide. The bottom block will slide with the top block if and only if(b) Instead a horizontal force F is applied to the bottom block only, and this force is increased until the bottom block just starts to slide. Under what conditions will the top block slide with the bottom block?(A) If μ1 > 0 the top block will slide regardless of i-t2’

The correct option is:
(a)


It can be seen from the figure above that when a force F acts on block 1, the frictional force acts on the block in the opposite direction. When the block just begins to slide over the one beneath it, the frictional force experienced by it is the static force of frictions.
Therefore, when block 1 just begins to move, F = ƒ_s
Here, ƒ_s is the static force of friction between block 1 and block 2.
The static force of friction exerted by block 2 on block 1 is reacted by a force of same magnitude, but in opposite direction. Therefore, block 2 experiences a force ƒ_s in the forward direction and experiences a static force of friction ƒ’_s in the backward direction due to the contact forces between the plastic and its surface.

When block 2 just begins to move with block 1, the force ƒ_s should be greater than ƒ’_s, that is,

Here, N₁ is the normal force acting on block 1 and is equal to whereas N₂ is the normal force acting on block 2, and is equal to 2mg (m being the mass of block and g the free fall acceleration).
Substitute the same in equation

Thus, the condition that should exist for block 2 to move with block 1 is 2μ₂ < μ₁.. Therefore (D) is the correct option and the rest are ruled out.
(b)
When the force F acts on the bottom block, the reaction force to the static force of friction experienced by block 2, acts on block 1. Therefore, the net force acting on block 1 is ƒ_s whereas the net force acting on block 2 is F – ƒ_s – ƒ’_s.
Therefore, the block 1 slides if ƒ_s > 0
Substitute ƒs = μ₁N₁,
μ₁N₁ > 0
μ₁ > 0
Therefore, if μ₁ > 0, the top block will slide regardless of μ₂.
Thus (A) is the correct option, while the rest are ruled out.