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in Laws of motion by (65.8k points)
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A Block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in figure. The coefficient of friction between the block and the plane is μ = 0.5 . What minimum force F should be applied perpendicular to the plane of block so that block does not slip on the plane ? (g = 10 ms–2

(A) zero 

(B) 6.24 N 

(C) 2.68 N 

(D) 4.3 N

1 Answer

+1 vote
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Best answer

Correct option: (C) 2.68 N

Explanation:

m = 2kg 

Q = 30° 

μ2 = 0.5 

N = F + mg cos 30 

mg sin = μ N 

(2)(10)(1 / 2) = 0.5 N 

∴ N = 20 

Now N = F + mg cos 30 

20 – {2 × 10 × (√3 / 2)} = F 

∴ F = 2.68 N

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