Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 4 - Exercises and Problems: 75

Answer

Please see the work below.

Work Step by Step

We know that $a=\frac{F}{M}$ $a=\frac{F_{\circ}sin(\omega t)}{M}=\frac{F_{\circ}}{M}sin(\omega t)$ We also know that the jerk is the rate of change of acceleration $J=\frac{da}{dt}$ $J=\frac{-F\omega{\circ}}{M}cos(\omega t)$ The maximum jerk is given as $J_{max}=\frac{F_{\circ}\omega}{M}$
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