College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Conceptual Questions - Page 134: 23

Answer

Distance measured is S. Speed at the final point is v=0. Initial velocity is $v_{0}$. $v^{2}=v_{0}^{2}+2aS$ Now, $a=\frac{F_{f}}{m}=\frac{coefficient \,of \,friction \times mg}{m}$ $a=coefficient\, of\, friction \times g$ Thus, from $v^{2}=v_{0}^{2}+2aS$, Coefficient of friction $\times g=a=-\frac{v_{0}^{2}}{2S}$ Coefficient of friction = $-\frac{v_{0}^{2}}{2gS}$ The minus sign indicates that the direction of the friction force is opposite to that of the velocity.

Work Step by Step

Distance measured is S. Speed at the final point is v=0. Initial velocity is $v_{0}$. $v^{2}=v_{0}^{2}+2aS$ Now, $a=\frac{F_{f}}{m}=\frac{coefficient \,of \,friction \times mg}{m}$ $a=coefficient\, of\, friction \times g$ Thus, from $v^{2}=v_{0}^{2}+2aS$, Coefficient of friction $\times g=a=-\frac{v_{0}^{2}}{2S}$ Coefficient of friction = $-\frac{v_{0}^{2}}{2gS}$ The minus sign indicates that the direction of the friction force is opposite to that of the velocity.
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.