Answer
$W_{s}=1.4kJ$
Work Step by Step
Knowing that:
$W_{s}=\frac{k}{2}*\Delta x^2$
Where $k$ is the spring constant
$W_{s}=(\frac{70\frac{kN}{m}}{2})*[(0.2m)^2-(0m)^2]=1.4kJ$
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 99: 2-33
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