# Problem 81 ad electron with kinetic energy 085 fj

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Unformatted text preview: ectron penetrate? (bJ At what position does the electron have its maximum speed? (c) what is this maximum speed? em. = Solution h FIGURE 8-48 Problem 79. Solution Let us neglect any possible losses in energy. and assume that the kinetic energy: associated with the car's horizontal motion is unchanged by the drop. Then the energy ptincipleimplies that,Ugrav{A) Ugrav(B) + Uspr(B}, where Aisa point befote the drop in the road, and B is the point after the drop where the springs are maximally compressed (the kinetic energy associated with the car's vertical mqtion on its springs is instantaneously zero at B). Thus,!i Ugrav(A) Ugrav(B) = mgh = Uspr(B) = ~kx2, or h = (110,000 N/m)(O.4 m)2/2(1200x9.8 N) :::':: 74.8 cm. = (a) The electron's total energy; E 0.85 fJ, equals its initial kinetic energy upon entering the region at x = O. At any other point in the region x ~O, E = K + U, so the point of maXimum penetration (the turning point) can be found from the equation K{xm) E U{xm) == o. Therefore, ax;' - bXm - 0.85 fJ == 0, or Xm [4.2 + J(4.2}2 + 4(2.7)(...
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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.

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