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 137: 49



Work Step by Step

$Tsin15^{\circ}=ma$ $Tcos15^{\circ}=mg$ Therefore, from above 2 equations, $tan15^{\circ}=\frac{a}{g}$ $a=gtan15^{\circ}=2.626m/s^{2}$
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