Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems: 31a

Answer

$39.2J$

Work Step by Step

Elastic potential energy stored in spring can be defined as: $E_{p}=\dfrac {kx^{2}}{2}$ Given the numbers, $k=19.6\dfrac {N}{cm}=19.6\dfrac {N}{10^{-2}m}=19.6\times 10^{2}\dfrac {N}{m};x=20cm=0.2m$ We get: $E_{p}=\dfrac {kx^{2}}{2}=\dfrac {19.6\times 10^{2}\dfrac {N}{m}\times \left( 0.2m\right) ^{2}}{2}=39.2J$
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.