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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 8cm 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. 813) 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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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.
 Spring '08
 WORMER
 Physics, Conservation Of Energy, Energy, Work

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