College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1011: 36


(a) $F = 1.8\times 10^7~N$ (b) $a = 8200~m/s^2$

Work Step by Step

We can find the magnitude of the force: $F~t = \gamma~m~v$ $F~t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~v$ $F = \frac{m~v}{t~\sqrt{1-\frac{v^2}{c^2}}}$ $F = \frac{(2200~kg)~(0.70~c)}{(3.6\times 10^4~s)~\sqrt{1-\frac{(0.70~c)^2}{c^2}}}$ $F = \frac{(2200~kg)~(0.70)(3.0\times 10^8~m/s)}{(3.6\times 10^4~s)~\sqrt{1-(0.70)^2}}$ $F = 1.8\times 10^7~N$ (b) We can find the initial acceleration: $F = m~a$ $a = \frac{F}{m}$ $a = \frac{1.8\times 10^7~N}{2200~kg}$ $a = 8200~m/s^2$
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