Laws of Motion · NEET Preparation Physics
A 5 kg block is on a rough inclined plane of angle 37° to the horizontal. The coefficient of static friction is 0.5. A force of 50 N acts on the bloc…
Practice question from Laws of Motion (NEET Preparation Physics), with the correct answer and a step-by-step solution.
hard · MCQ · Q-CC6767
A 5 kg block is on a rough inclined plane of angle 37° to the horizontal. The coefficient of static friction is 0.5. A force of 50 N acts on the block parallel to the plane in the upward direction. Given g = 10 m/s², sin 37° = 0.6 and cos 37° = 0.8, the frictional force on the block is
- A.0 N
- B.10 N downward along the plane
- C.20 N downward along the planeCorrect answer
- D.50 N downward along the plane
Solution
The component of weight down the plane is mg sin 37° = 5 × 10 × 0.6 = 30 N. The applied force is 50 N upward, so a 20 N downward force is required for equilibrium. The maximum static friction is μsN = 0.5 × (5 × 10 × 0.8) = 20 N, so friction is 20 N downward along the plane.
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