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JEE Main202431 Jan 2024Evening ShiftPhysicsLaws of MotionActual

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F → 1 is the force required to just move the block up the inclined plane and F → 2 is the force required to just prevent the block from sliding down, then the value of F → 1 − F → 2 is: [Use g = 10 m s - 2 ]

Options

  1. A25 3 N
  2. B5 3 N
  3. C5 3 2 N
  4. D10 N

Correct answer

B. 5 3 N

Step-by-step solution

Limiting friction force is, f m a x = μ m g cos θ = 0 .1 × 50 × 3 2 = 2 .5 3 N For first case: We can write, F 1 = m g sin θ + f m a x = 25 + 2 .5 3 For second case: F 2 = m g sin θ − f m a x = 25 − 2 .5 3 ∴ F 1 − F 2 = 5 3 N

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