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 - Exercises and Problems - Page 92: 63

Answer

You could have been going $27.94 \ m/s$, which is greater than the posted speed of the road.

Work Step by Step

We find the angle of the road based on the maximum posted speed. Thus, we obtain: $\theta = tan^{-1}(\frac{v^2}{rg})=tan^{-1}(\frac{22.22^2}{(210)(9.81)})=13.48^{\circ}$ Now that we have found this, we find the maximum force of friction possible: $F_f=F_n \mu = mgcos\theta \mu = m(9.81)cos(13.48)\times .15 = 1.43m$ We know the force of friction and part of the force of gravity are the centripetal forces acting on you, so we find the maximum velocity you could have been going: $ 1.43m+mgsin\theta=\frac{mv^2}{r}\\ v=\sqrt{1.43r+grsin\theta}=\sqrt{(1.43 \times 210)+(9.81\times 210\times sin13.48)}=27.94 \ m/s$
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