College Physics (4th Edition)

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

Chapter 6 - Problems - Page 227: 13

Answer

The kinetic energy of Charles Murphy and his bicycle is $27,200~J$ The kinetic energy of John Howard and his bicycle is $1.63\times 10^5~J$

Work Step by Step

We can find the kinetic energy of Charles Murphy and his bicycle: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(70.5~kg)(27.8~m/s)^2$ $KE = 27,200~J$ The kinetic energy of Charles Murphy and his bicycle is $27,200~J$ We can find the kinetic energy of John Howard and his bicycle: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(70.5~kg)(68.04~m/s)^2$ $KE = 1.63\times 10^5~J$ The kinetic energy of John Howard and his bicycle is $1.63\times 10^5~J$.
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.