Unformatted text preview: is 47 cm/s, find the x coordinates of its turning points. Solution
FIGURE 845 Problem 69 Solution. This situation is exactly similar to that in Problem 39, ,.where the energy principle led to ~mv~ax = mghmax = 140 CHAPTER 8 the tluning
2 mg(0.18/m)x~ax.Thus,
Xmax pai~ts accur at mg(YA 1IB), or VA  YB = (v~j2g)
(6 m/s)2/(19.6 2 m/s ) = xJ(0.47 m/s)2/2(9,8 x25.0 em. m/s )(O.18Ym) = + J1.kX = + (O.32)(20 m) = 8.24 m. Problem
72. A Ikg particle slides an a frictianless track whose height is given by y = ax4  bx2, where. 3 G= 1 m , and b= 4 illI. TheparticIe's tatal energy is 20 J, where the zero. afpatentialenergy is at y = O. Find the turning paints af its matian. Supplemeritary Problems Problem
75. A uranium nucleus has a radius of 1.43xlOIO m. An alpha particle (mass 6.7xIO27 kg) leaves the surface of the nucleus with negligible speed, subject to. a repulsive force whose magnitude is F = Ajx2, where A= 4.1xlO26 N.m2, and where x is the distance from the alpha particle to the center of the nucleus. What is the speed of the alpha par...
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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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