question_answer

A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied to B for the blocks to move together will be:

A) 30 N

B) 25 N

C) 27 N

D) 48 N

Correct Answer: C

Solution :

[c] Minimum force on A =frictional force between the surfaces = 12 N Therefore maximum acceleration \[{{a}_{\max }}=\frac{12\,N}{4kg}=3\,m/{{s}^{2}}\] Hence maximum force, \[{{\operatorname{F}}_{max}}=total\,\,mass\times {{a}_{max}}=9\times 3=27\,\,N\]