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A block weighing 15 N slides on a horizontal table whose coefficient of kinetic friction is 0.4. If the area of contact between block and table is 0.05 m², the shearing stress on the contact surface is:

1. 120 N/m²
2. 140 N/m²
3. 160 N/m²
4. 180 N/m²

Correct answer: 120 N/m²

Solution

Friction force = 0.4 * 15 = 6 N acting tangentially, so shearing stress = 6 / 0.05 = 120 N/m².

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