## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 5 - Exercises and Problems: 32

#### Answer

Please see the work below.

#### Work Step by Step

We know that if $\mu_k=0$ then $a=gsin(\theta)=(9.8)sin28=4.60\frac{m}{s^2}$ Now $t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{(2)(100)}{4.60}}=6.593s$ If $\mu_k=0.17$ then $a=gsin(\theta)-\mu gsin(\theta)=(9.8)sin(28)-(0.17)(9.8)cos (28)=3.13\frac{m}{s^2}$ $t=\sqrt{\frac{2d}{a}}$ $t=\sqrt{\frac{2(100)}{3.13}}=7.994s$ Now we can find the difference as $7.994-6.593=1.4s$

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