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

Published by Pearson

# Chapter 7 - Exercises and Problems - Page 126: 31

#### Answer

a) 1.07 J b) 1.12 J

#### Work Step by Step

We have to find the height of the 0 potential energy, which is in the middle of the brick. Thus, we find the middle of the brick's height originally: $h = \frac{.055}{2}=.0275 \ m$ For the next parts of the problem, we find how much the center of the brick is above the point of 0 potential energy: a) $U=mgh=(1.5)(9.81)(\frac{.2}{2}-.0275)=1.07 \ J$ b) $U=mgh=(1.5)(9.81)(.104-.0275)=1.12 \ J$

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