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JEE MainPhysicsLaws of Motion

A block of mass 5 kg is sliding at a speed of 5 m/s on a rough horizontal floor. The coefficient of kinetic friction between the block and the floor is 0.2 . A constant horizontal force is applied on the block in the direction of its motion. If the block covers a distance of 44 m in the next 4 s, the magnitude of the applied force is (Take g = 10 m/s ^2 ):

Options

  1. A15 N
  2. B25 N
  3. C5 N
  4. D37.5 N

Correct answer

B. 25 N

Step-by-step solution

From the second equation of motion, s = ut + 1 2 at^2 44 = 5(4) + 1 2 a(4)^2 44 = 20 + 8a a = 3 m/s ^2 The kinetic friction acting on the block is, f_k = _k mg = 0.2 5 10 = 10 N Applying Newton's second law in the horizontal direction, F - f_k = ma F - 10 = 5 3 F = 25 N Answer: 25 N

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