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 87: 33

Answer

.56

Work Step by Step

We know that the centripetal force is given by $\frac{mv^2}{r}$. We know that the force of friction is the centripetal force, so, using a reasonable coefficient of friction of .25, we find: $\mu =.25 \times \frac{60^2}{40^2}=.56$
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.