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Two identical blocks, each of mass m, are joined by a relaxed (un-stretched) spring and rest on a smooth horizontal floor. A steady horizontal force F is applied to one block, pulling it directly away from the other. If the spring stretch at time t is x0, what are the ground-frame displacements of the two blocks at that moment?
- Pulled block: (F*t²)/(4m) + x0/2; other block: (F*t²)/(4m) - x0/2
- Pulled block: (F*t²)/(2m) + x0; other block: (F*t²)/(2m)
- Pulled block: (F*t²)/(4m); other block: (F*t²)/(4m) - x0
- Pulled block: (F*t²)/(2m); other block: (F*t²)/(2m) - x0
Correct answer: Pulled block: (F*t²)/(4m) + x0/2; other block: (F*t²)/(4m) - x0/2
Solution
The CM accelerates at F/(2m), giving CM displacement (F*t²)/(4m). Relative to the CM the blocks are split by the extension x0, so each is offset by x0/2.
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