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Unformatted text preview: energy, satisfies Equation 87 with total energy E = K + U = ~in(dx/dt? + U{x). The turning points are the solutions of this equation whendx/dt = OJ that is, E = U(x) = 3.5 = 7.0 8.0x + 1.7x2 (where energy is in joules and displacement in meters). The quadratic formula gives x = (1.7)1(4.0:t:. v'lO.l)m= 0.488 m and 4.22 m. Solution
The potential energy is symmetric about x = Ox (U( x) = U(x, so we need only plot it for positive x. The zeros of U are at x= :t:.y'b7a = :f:l.83 m. The maxima and minimum can be found from the zeros of the derivative, dUjdx = (2x/e2)(b+ ac2  ax2)ex2/e2, which areXmin = 0 and Xmax = :1:Je2 + b/a = :1:3.51m. The corresponding energies are Umin= b == 5.0 J, and Umax = ac2e(I+blae2) = 3.43 J. We finally note that U(8 m) = 91 e7.1l J = 7.43xlO2 J. One can see that the particles with energies less than Umaxare confined to the well, with turning points on either side of :1:1.83ill, respectively. The speed ofa particle is v = J2(E  U)/m. Since U t 0 for x t 00, the speed of the particle with E = 4 J approaches J2(4 J)/(1 kg) == 2.83 mls for la...
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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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