Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems: 21a

Answer

11.7 N

Work Step by Step

Looking at the graphs, in graph 1 we see that the Vx goes from 2m/s at t = 0 to 5 m/s at t = 1. This means that the acceleration is 3 $m/s^2$ in the positive a direction. In the second graph, the Vy goes from 2 m/s at t = 0 to -3 m/s at t = 1. Therefore, the acceleration is 5 $m/s^2$ in the negative y direction. The package weighs 2kg and F = ma so writing a in vector notation we have a = ($3 m/s^2$)i + ($-5m/s^2$)j and then F = ($6 N$)i + ($-10 N$)j To find the magnitude we take $\sqrt (x^2 + y^2) = \sqrt (6^2 + (-10)^2) = \sqrt(136) $ This is about 11.7 N
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.