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 - Page 119: 38b

Answer

$F_x = 28N$

Work Step by Step

$Fg = mg$ $Fg = 40kg \times 9.8m/s^2$ $a=1m/s^2$ Applying the 2nd law of motion: $F_{net} = Fgsin\theta-F_x$ $F_x = Fgsin\theta-F_{net}$ $F_x = Fgsin\theta-ma$ $F_x = (392\times sin(10^{\circ}))- (40\times 1)$ $F_x = 68 -40$ $F_x = 28N$
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.