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:   [JEE ONLINE 09-04-2014]

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

\[{{a}_{\max }}=\frac{12N}{4kg}=3m\text{/}{{s}^{2}}\]

Hence maximum force,

\[{{F}_{\max }}=\]total mass \[\times {{a}_{\max }}\]

\[=9\times 3=27N\]