College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1012: 51


$K = 2.7\times 10^{15}~J$

Work Step by Step

We can find the kinetic energy of the object: $K = (\gamma-1)mc^2$ $K = (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)mc^2$ $K = (\frac{1}{\sqrt{1-\frac{(1.80\times 10^8~m/s)^2}{(3.0\times 10^8~m/s)^2}}}-1)(0.12~kg)(3.0\times 10^8~m/s)^2$ $K = (1.25-1)(0.12~kg)(3.0\times 10^8~m/s)^2$ $K = 2.7\times 10^{15}~J$
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