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 139: 70

Answer

$0.212$

Work Step by Step

$a=\frac{v^{2}-u^{2}}{2S}$ F=coefficient of friction$ \times mg=ma$ Coefficient of friction $= \frac{a}{g}$ Therefore, coefficient of kinetic friction$=\frac{v^{2}-u^{2}}{2gS}=\frac{2.5^{2}}{2\times1.5\times9.8}=0.212$
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