Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 8

Answer

At t = 1 s: $F = 8.0~N$ At t = 4 s: $F = 0$ At t = 7 s: $F = -12~N$

Work Step by Step

The acceleration $a_x$ is equal to the slope of the $v_x$ versus time graph. At t = 1 s: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{12~m/s-0}{3.0~s-0}$ $a_x = 4.0~m/s^2$ We can use $a_x$ to find the net force: $F = ma_x$ $F = (2.0~kg)(4.0~m/s^2)$ $F = 8.0~N$ At t = 4 s: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{12~m/s-12~m/s}{6.0~s-3.0~s}$ $a_x = 0$ We can use $a_x$ to find the net force; $F = ma_x$ $F = (2.0~kg)(0)$ $F = 0$ At t = 7 s: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{0-12~m/s}{8.0~s-6.0~s}$ $a_x = -6.0~m/s^2$ We can use $a_x$ to find the net force; $F = ma_x$ $F = (2.0~kg)(-6.0~m/s^2)$ $F = -12~N$
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