Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 9 - Exercises and Problems - Page 164: 42


a) .00369 Ns b) 20 meters

Work Step by Step

a) We use the equation for impulse: $J=F\Delta t = (.041)(.09)=\fbox{.00369 Ns}$ b) We first find acceleration: $a = \frac{.041}{8.3\times 10^{-6}}=-4939.759 \ m/s^2$ This means: $v_0=(4939.759 \ m/s^2)(.09s)=444.57 \ m/s$ This means that the distance traveled is: $\Delta x = (444.57 \ m/s)(.09)+\frac{1}{2}(-4939.759 \ m/s^2)(.09s)^2=\fbox{20 meters}$
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