Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Section 5.4 - Friction - Example - Page 85: 5.9

Answer

Dry road: 25 m/s Wet road: 12 m/s

Work Step by Step

We know that the necessary centripetal force must equal the force of friction. Thus: $\frac{mv^2}{r} = \mu_s mg \\ v = \sqrt{\mu_s gr}$ Thus, we find: a) $v_{max} = \sqrt{\mu_s gr} = \sqrt{.21\times9.81\times73} = 12 \ m/s$ b) $v_{max} = \sqrt{\mu_s gr} = \sqrt{.88\times9.81\times73} = 25 \ m/s$
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