Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 107: 86

Answer

$\mu_s = 0.36$

Work Step by Step

$v_0 = (45 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 12.5~m/s$ We can use this velocity to find the acceleration. $a = \frac{v-v_0}{t} = \frac{0-12.5 ~m/s}{3.5 ~} = -3.57 ~m/s^2$ We can use this magnitude of deceleration to find the coefficient of static friction. We can assume that the force of static friction provides the force to decelerate the cup at this rate of deceleration: $F_f = ma$ $mg ~\mu_s = ma$ $\mu_s = \frac{a}{g} = \frac{3.57 ~m/s^2}{9.80 ~m/s^2} = 0.36$ The coefficient of static friction $\mu_s$ is 0.36
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.