Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 11 - Exercises and Problems: 33

Answer

Please see the work below.

Work Step by Step

We know that $a=\frac{v^2}{2d}$ We plug in the known values to obtain: $a=\frac{(42)^2}{(2)(0.05)}=17640\frac{m}{s^2}$ As $F=ma$ $F=(0.145)(17640)=2558N$ Now we can find the torque as $\tau=rF$ We plug in the known values to obtain: $\tau=(0.63)(2558)=1600N.m$
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