In the show arrangement mass of A=1 kg, mass of B= 2kg. Coefficient of friction between A and B =0.2

There is no friction between B and ground. The frictional force exerted by A on B equals.

Options

A. 2 N
B. 3 N
C. 4 N
D. 5 N

Answer

A

Explanation

(a) Block A moves due to friction. Maximum acceleration of A can be `(f_("max"))/(m) " or " (mu mg)/(m)` or `mu g=0.2xx10=2 ms^(-2)`. If both the blocks move together, then combined acceleration of A and B can be `(10)/(3)=3.33 ms^(-2)`. Since, this is more than the maximum acceleration of A . Slipping between them will take place and force of friction will be maximum of `mu m_(A)g=2N`.