Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 11 - Work - Exercises and Problems: 22

Answer

(a) $KE = 2.5\times 10^{-21}~J$ (b) The spring constant of the carbon-carbon bond is 2.0 N/m.

Work Step by Step

(a) We can find the kinetic energy of the carbon atom. $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(2.0\times 10^{-26}~kg)(500~m/s)^2$ $KE = 2.5\times 10^{-21}~J$ (b) We can find the spring constant of the carbon-carbon bond. $\frac{1}{2}kx^2 = U_s$ $k = \frac{2~U_s}{x^2}$ $k = \frac{(2)(2.5\times 10^{-21}~J)}{(0.050\times 10^{-9}~m)^2}$ $k = 2.0~N/m$ The spring constant of the carbon-carbon bond is 2.0 N/m.
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