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 136: 29

Answer

$(-2.6\,N)\hat{x}+(1.5\,N)\hat{y}$

Work Step by Step

$\vec{a}=-\hat{x}(a\cos\theta)+\hat{y}(a\sin\theta)$$=(-1.5\,m/s^{2}\times\cos30^{\circ})\hat{x}+(1.5\,m/s^{2}\times\sin30^{\circ})\hat{y}$$=(-1.3\,m/s^{2})\hat{x}+(0.75\,m/s^{2})\hat{y}$ $\vec{F}=m\vec{a}=(2.0\,kg)[(-1.3\,m/s^{2})\hat{x}+(0.75\,m/s^{2})\hat{y}]$ $=(-2.6\,N)\hat{x}+(1.5\,N)\hat{y}$
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.