Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems - Page 272: 11


The child's maximum speed 4.2 m/s

Work Step by Step

Let $L$ be the length of the chains. We can find the height $h$ (above the swing's lowest point) when the chains are at a $45^{\circ}$ angle. $\frac{L-h}{L} = cos(\theta)$ $h = L~[1-cos(\theta)]$ $h = (3.0~m)~[1-cos(45^{\circ})]$ $h = 0.88~m$ The maximum speed will occur at the lowest point when kinetic energy is at a maximum. We can use conservation of energy to find the child's speed at the lowest point. The kinetic energy at the lowest point will be equal to the potential energy at height $h$. $KE = PE$ $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(0.88~m)}$ $v = 4.2~m/s$ The child's maximum speed 4.2 m/s
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