Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 212: 121a

Answer

$m = 2.1\times 10^6~kg$

Work Step by Step

We can find the energy produced by the locomotive in 6.0 minutes: $E = P~t$ $E = (1.5\times 10^6~W)(360~s)$ $E = 5.4\times 10^8~J$ We can find the mass of the train: $\Delta K = 5.4\times 10^8~J$ $K_2-K_1 = 5.4\times 10^8~J$ $\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 = 5.4\times 10^8~J$ $m = \frac{(2)(5.4\times 10^8~J)}{v_2^2-v_1^2}$ $m = \frac{(2)(5.4\times 10^8~J)}{(25~m/s)^2-(10~m/s)^2}$ $m = 2.1\times 10^6~kg$
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