Fundamentals of Physics Extended (10th Edition)

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

Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 322: 30a

Answer

$a = (3.00~m/s^2)~\hat{i}-(4.00~m/s^2)~\hat{j}+(2.00~m/s^2)~\hat{k}$

Work Step by Step

We can express the force in unit-vector notation: $F = (6.00~N)~\hat{i}-(8.00~N)~\hat{j}+(4.00~N)~\hat{k}$ We can find the acceleration: $F = ma$ $a = \frac{F}{m}$ $a = (\frac{1}{2.00~kg})[(6.00~N)~\hat{i}-(8.00~N)~\hat{j}+(4.00~N)~\hat{k}]$ $a = (3.00~m/s^2)~\hat{i}-(4.00~m/s^2)~\hat{j}+(2.00~m/s^2)~\hat{k}$
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