E None of the above How can the block get hotter if no work was done on it

E none of the above how can the block get hotter if

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E. None of the above. How can the block get hotter if no work was done on it? Reading Question - Sec. 8.6
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Point Particle System: All external forces are applied to the center of mass; correctly describes the motion of the center of mass. ! K translational = ( F " f ) d Since v=constant, no change in K, so: Can this tell us anything about internal motions of the system? r F = r f Analysis of Point Particle System
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! K trans + ! E thermal = Fd + W friction For simplicity, assume Q=0. ! E thermal = Fd + W friction Since Δ K=0. Since this is the real system, not the point particle system, forces do not all act at the same point, and they do not have to act over the same distance: ! E thermal = Fd " fd effective # 0 We must have F=-f, but not d=d eff ! Analysis of the Real System
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Actual work done by friction forces: KEY ISSUE - The point of contact where the friction force is applied moves a shorter distance than the distance the block moves. This means friction does less work than expected. A Model for Friction
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Instead of teeth stuck on each other, let them drag on opposite surfaces. Each then applies force F/2. When the top block moves by d, one tooth moves by d, while the other one just bends, meaning no displacement at the contact. Only one tooth does work across the distance d, with a force F/2, so only half of the expected amount of work is done. A More Realistic Model
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Consider 2 identical blocks as shown. Lower block is fixed. Total work done on the system (2 blocks) is just Fd (top block displacement d). Divide the increased thermal energy between the two blocks: ! E th ,1 + ! E th ,2 = Fd ! E th ,1 = Fd 2 = Fd " Fd eff d eff = d / 2 Hence, In this simple model, friction only does half the expected work. The remaining work done by the external force goes into thermal energy. 2 Identical Blocks
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