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: 47


$25.33 \ seconds$

Work Step by Step

We know that the overall mass of the system is: $=73+940=1013 \ kg$ We know that the acceleration is equal to the force of gravity minus the force of friction over the overall mass. Thus, we find: $a= \frac{(73)(9.81)-(.057)(940)(9.81)}{1013}=.159\ m/s^2$ We now can find the change in time, using the same equation that the book uses in example 5.4. Doing this, we find: $t =\sqrt{\frac{2x}{a}}=\sqrt{\frac{2(51)}{.159}}=25.33 \ seconds$
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