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 135: 22

Answer

$7.0\,m/s^{2}$

Work Step by Step

According to Newton's second law, $F_{net}=ma$ $\implies a=\frac{F_{net}}{m}=\frac{15\,N-8.0\,N}{1.0\,kg}=7.0\,m/s^{2}$
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