College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 140: 87


a). $513N$ b). $0.151m/s$

Work Step by Step

a). Change in velocity $=47+122=169mi/hr=75.5m/s$ Therefore, $F=ma=m\frac{\Delta v}{\Delta t}=0.17\frac{75.5}{0.025}=513N$ b). $v=a_{g}t=\frac{F}{m_{g}}t=\frac{513}{85}0.025=0.151m/s$
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