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KCET2011PhysicsLaws of Motion

Block A of mass of 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 blocks A and B is 0.4 . A horizontal force of 10 ~N is applied on block B . The force of friction between blocks A and B is (g=10 ~ms ⁻² )

Options

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

Correct answer

D. Zero

Step-by-step solution

Total mass of blocks A and B=2+8=10 ~kg Friction between surface and combination of A and B aligned F &= R &=0.5 10 10=50 ~N aligned Here applied force on box B is 10 ~N that is less than 50 ~N . So the system will be in rest because of this there is no friction between blocks A and B .

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