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 135: 20

Answer

$3.0\,m/s^{2}$

Work Step by Step

According to Newton's second law, $F=ma$ $\implies a=\frac{F}{m}=\frac{F}{m_{1}+m_{2}}=\frac{F}{\frac{F}{a_{1}}+\frac{F}{a_{2}}}$ $=\frac{1}{\frac{1}{a_{1}}+\frac{1}{a_{2}}}=\frac{a_{1}\times a_{2}}{a_{1}+a_{2}}=\frac{4.0\,m/s^{2}\times12\,m/s^{2}}{4.0\,m/s^{2}+12\,m/s^{2}}=3.0\,m/s^{2}$
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.