University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 297: 9.42


P = 14.7 N

Work Step by Step

The work done by the applied force will be equal to the kinetic energy of the cylinder. $P~d = KE$ $P~d = \frac{1}{2}I\omega^2$ $P~d = \frac{1}{2}(mr^2)(\frac{v}{r})^2$ $P~d = \frac{1}{2}mv^2$ $P = \frac{mv^2}{2d}$ $P = \frac{(\frac{40.0~N}{9.80~m/s^2})(6.00~m/s)^2}{(2)(5.00~m)}$ $P = 14.7~N$
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.