A 4 kg block A is placed on the top of 8 kg block B which rests on a smooth table.

A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force F applied on B to make both A and B move together, is

Options

A. 12 N
B. 24 Ns
C. 36 N
D. 48 N

Answer

B

Explanation

(b) When force is applied on A, acceleration produced will be
`(F_(A))/(m_(A)+m_(B))=(mu m_(A)g)/(m_(B))`…(i)
When force is applied on B, acceleration produced will be
`(F_(B))/(m_(A)+m_(B))=(mu m_(A)g)/(m_(A))`….(ii)
Dividing these two equations, we get ,brgt `(F_(B))/(F_(A))=(m_(B))/(m_(A))`
`therefore" " F_(B)=(m_(B))/(m_(A)).F_(A)=(8)/(4)xx12=24 N`