A 3 kg block is placed over a 10 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.2. If a horizontal force of 20 N is applied to 3 kg block, accelerations of the two blocks in ms^–2 are (g = 10 m/s²)

Force of friction between the blocks f =μN=0.2×3×10

=6 N

This force acts in the backward direction on the lighter block and in the forward direction on the heavier block.

Thus ma = F−f

⟹3a = 20−6 = 14

⟹a = 14/3​ m/s²
And for the second body,
 ma = f

⟹10a = 6

⟹a = 0.6 m/s2