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JEE MainPhysicsLaws of Motion

A block of mass 5 kg rests on a rough horizontal surface. A pushing force F is applied to the block, directed downwards at an angle of 30^ below the horizontal. If the coefficient of static friction between the block and the surface is 1 2 3 , what is the minimum value of F required to just start moving the block? [Take g = 10 m/s ^2 ]

Options

  1. A16.67 N
  2. B14.28 N
  3. C57.7 N
  4. D20 N

Correct answer

D. 20 N

Step-by-step solution

The applied force F can be resolved into two components: a horizontal component F 30^ and a vertical component F 30^ acting downwards. From the equilibrium of forces in the vertical direction, the normal reaction N is given by: N = mg + F 30^ N = 5 10 + F ( 1 2 ) = 50 + F 2 For the block to just start moving, the horizontal driving force must equal the maximum static friction: F 30^ = _s N F ( 3 2 ) = 1 2 3 (50 + F 2 ) Multiplying both sides by 2 3 : F ( 3 3 ) = 50 + F 2 3F = 50 + 0.5F 2.5F = 50 F = 50 2.5 = 20 N A

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