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For upper half of inclined plane v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ. For lower half of inclined plane 0 = u² + 2 g (sin θ − μ cos θ) S/2 ⇒ gS sin θ = gS (sin θ − μ cos θ) ⇒ 2 sin θ = 2 tan θ.

1. μ = 2 sin θ / cos θ
2. μ = 2 sin θ
3. μ = 2 tan θ
4. μ = 2 cos θ

Correct answer: μ = 2 tan θ

Solution

The given equations simplify to show that the coefficient of friction μ is equal to 2 times the tangent of the angle θ. This matches the option μ = 2 tan θ.

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