Q125 / 170 · Manipal (MET) 2021 · Physics

A 4 kg block $A$ is placed on the top of a block $B$ of mass 8 kg , 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 required to make both $A$ and $B$ move together is :

  • A
    12 N
  • B
    24 N
  • Correct answer
    C
    36 N
  • D
    48 N

Solution

The mass of block $A=4 \mathrm{~kg}$ Force of friction between $A$ and $B=12 \mathrm{~N}$ Mass of block $A$ and $B=4+8=12 \mathrm{~kg}$ $i e$, mass is increased 3 times. The force of friction also increases 3 times $=12 \times 3=36 N$ Hence, maximum horizontal force required $=36 \mathrm{~N}$

Hence, the correct answer is option: 'C'.

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