Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Exercises and Problems - Page 88: 62


$\mu = \frac{v^2}{gr} - tan(tan^{-1}(\frac{v_0^2}{rg}))$

Work Step by Step

We find the angle of the road based on the maximum posted speed. Thus, we obtain: $\theta = tan^{-1}(\frac{v_0^2}{rg})$ We also know: $ mgcos\theta \mu+mgsin\theta=\frac{mv^2}{r}$ Thus, we find an expression for $\mu$: $\mu=\frac{\frac{v^2}{r}-gsin\theta}{gcos\theta}$ $\mu = \frac{v^2}{gr} - tan\theta$ Plugging in the value of theta gives: $\mu = \frac{v^2}{gr} - tan(tan^{-1}(\frac{v_0^2}{rg}))$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.