College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 229: 35


The speed at the bottom of the swing is $8.42~m/s$

Work Step by Step

We can find the difference in height between the starting point and the bottom of the swing: $h = (20.0~m)-(20.0~m)~cos~35.0^{\circ} = 3.62~m$ The kinetic energy at the bottom will be equal to the magnitude of the change in the potential energy: $\frac{1}{2}mv^2 = mgh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(3.62~m)}$ $v = 8.42~m/s$ The speed at the bottom of the swing is $8.42~m/s$.
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.