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The conservation of mechanical energy between points Ii 0 and P requires Ko + Uo = K + U, or mgR = ~mv2 + :I mgy (since the bug starts from rest). Thus, v2 2gx (R - y) = 2gd. The bug leaves the surface when the normal force has decreased to zero. From the radial component of Newton's second law, mgcosO - N = . mv2 / R, so this occurs when N = mg cos () - mv2j R 0, or v2 = gRcos(} gy = g(R- d). Combining these results, we find v2 = 2gd = g(R - d), or d ~R. lover, the center of the brick must lie to the $fthe vertical through its 8-cm edge, as shown. citential energy of the brick would have to be dbyUb - Ua mg(Yb - Ya)' If we assume that is conserved between a and b, this is equal to iurmum kinetic energy sought (Ka - Kb = Ub if Kb - 0). The numerical value is just the = = K;:,in. = = = 142 CHAPTER 8 Solution
The condition for the rope to be taut going around the rock (at 01 in Fig. 8-13) is 0 ~ T mgcos (h + mvUa (this follows from the radial component of Newton's second iaw), or v~ ~ -gacosOl' Th...
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