question_answer

A block A of mass 4 kg is placed on another block B of mass5 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:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

A) 30 N                      

B) 25 N

C) 27 N      

D) 48 N

Correct Answer: C

Solution :

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