Fundamentals of Physics Extended (10th Edition)

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

Chapter 7 - Kinetic Energy and Work - Problems - Page 171: 11a


Angle between force and displacement $=62.3^{\circ}$

Work Step by Step

The magnitude of displacement is: $=\sqrt {2^{2}+4^{2}+3^{2}}=\sqrt 29 m$ We know, work done is equal to change in Kinetic Energy, so: $Work$ $Done$ $=30$ $J$ We also know that: $Work$ $Done = F dcos\theta$ Putting the values and solving: $30= 12\times \sqrt 29 \times cos\theta$ $cos \theta= \frac{30}{64.62} = 0.46$ $\theta =cos^{-1}(0.46) $ $\theta =62.3^{\circ} $
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.