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AIIMS2019PhysicsLaws of Motion

Block A of mass 2 ~kg is placed over block B of mass 8 ~kg . The combination is placed over a rough horizontal surface. Coefficient of friction between B and the floor is 0.5 . Coefficient of friction between A and B is 0.4 . A horizontal force of 10 ~N is applied on block B . The force of friction between A and B is (g=10 ~m ~s ⁻² )

Options

  1. A100 ~N
  2. B40 ~N
  3. C50 ~N
  4. Dzero

Correct answer

D. zero

Step-by-step solution

Here, m_A=2 ~kg , m_B=8 ~kg , ₁=0.4, ₂=0.5 , F=10 ~N The frictional force between block B and surface is f= ₂ N= ₂ (m_A+m_B ) g=0.5 (2+8) 10=50 ~N As applied force F(=10 ~N ) < f(=50 ~N ) , the system will not move. Hence, the force of friction between A and B is zero.

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