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A block of mass m lies at rest on a frictionless horizontal floor. A uniform rope of mass m/3 has one end attached to the block, and the free end is pulled horizontally by a force F. What is the tension at the midpoint of the rope?

1. 8F/7
2. F/7
3. F/2
4. 7F/8

Correct answer: 7F/8

Solution

Total mass = m + m/3 = 4m/3, so a = F/(4m/3) = 3F/(4m). Mass behind the rope's midpoint = block + half the rope = m + m/6 = 7m/6. Tension there = (7m/6)*a = (7m/6)(3F/4m) = 7F/8.

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