Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 249: 84

Answer

$1.8m$

Work Step by Step

We can find the required distance as follows: $d=\frac{v_i^2}{2g(\mu_kcos\theta-sin\theta)}$ We plug in the known values to obtain: $d=\frac{(1.56)^2}{2(9.81)((0.62)cos 28.4^{\circ}-sin 28.4^{\circ})}$ $d=1.8m$
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