## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 10

#### Answer

At t = 1 s: $F = 6.0~N$ At t = 4 s: $F = 3.0~N$ At t = 7 s: $F = 0$

#### Work Step by Step

The acceleration $a_x$ is equal to the slope of the $v_x$ versus time graph. At t = 1 s: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{6~m/s-0}{2~s-0}$ $a_x = 3~m/s^2$ We can use $a_x$ to find the net force. $F = ma_x$ $F = (2.0~kg)(3~m/s^2)$ $F = 6.0~N$ At t = 4 s: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{12~m/s-6~m/s}{6~s-2~s}$ $a_x = 1.5~m/s^2$ We can use $a_x$ to find the net force. $F = ma_x$ $F = (2.0~kg)(1.5~m/s^2)$ $F = 3.0~N$ At t = 7 s: $a_x = \frac{\Delta v_x}{\Delta t}$ $a_x = \frac{12~m/s-12~m/s}{8~s-6~s}$ $a_x = 0$ We can use $a_x$ to find the net force. $F = ma_x$ $F = (2.0~kg)(0)$ $F = 0$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.