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Question

# A block of weight 200 N is pulled along a rough horizontal surface at a constant speed by a force of 100 N acting at an angle of 30∘ above the horizontal. The coefficient of friction between the block and the surface is

A

0.43

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B

0.58

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C

0.75

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D

0.85

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Solution

## The correct option is B 0.58 Since the block moves with a constant velocity, no net force acts on it. Therefore, the horizontal component F cos θ of force F must balance the frictional force force, i.e., fr=F cos θ. Now, N+F sinθ=mg⇒N=mg−F sinθ Also, fr=μN=μ(mg−F sin θ)⇒F cosθ=μ(mg−F sin θ)⇒μ=F cosθmg−F sin θ=100×√32200−100×12=1√3=0.58 Hence, the correct choice is (b).

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