College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 138: 52



Work Step by Step

Using $v^{2}=u^{2}+2aS$ $0=25^{2}-2a\times100$ $a=3.125m/s$ So, $F=ma=2\times10^{5}\times3.125=625kN$
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.