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 - Page 204: 24



Work Step by Step

As we know that $P.E=\frac{1}{2}kx^2$ but $P.E=mgh$ $mgh=\frac{1}{2}kx^2$ here in our case $h=0.4+x$ Thus putting the values, we get $2(9.8)(0.4+x)=\frac{1}{2}(1960)x^2$ $7.84+19.6x=980x^2$ $980x^2-19.6x-7.84=0$ after solving it, we get $x=0.1$m
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.