Question

A $4Kg$ block $A$ is placed on the top of a $8Kg$ block $B$ which rests on a smooth table. A just slips on $B$ when a force of $12$ Newton is applied on $A$. The maximum horizontal force $F$ on $A$ required to make both $A$ and $B$ move together is

• A. $18N$
• B. $24N$
• C. $36N$
• D. $48N$

Answer 1

The correct option is A $18N$

When they both move together
$a=\frac{F}{TotalMass}=\frac{F}{12}$
For $A$
$F-12=4\times \frac{F}{12}⟶$Newtons 2nd Law
$\therefore F-\frac{F}{3}=12$
$F=18N$