University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 168: 5.97

Answer

$\mu_k = 0.450$
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Work Step by Step

Since the block moves with constant speed, the force directed down the ramp must be equal in magnitude to the sum of the friction forces directed up the ramp. $mg~sin(\theta) = F_{N_1}~\mu_k + F_{N_2}~\mu_k$ $3w~sin(\theta) = 4w~cos(\theta)~\mu_k + w~cos(\theta)~\mu_k$ $3~sin(\theta) = 5~cos(\theta)~\mu_k$ $\mu_k = 0.6~tan(\theta)$ $\mu_k = 0.6~tan(36.9^{\circ})$ $\mu_k = 0.450$
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