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 2 - Kinematics in One Dimension - Exercises and Problems: 67

Answer

The cart rolled 6.9 meters.

Work Step by Step

The cart's acceleration is $g~sin(3^{\circ})$. We can compare the distance covered by the cart and Julie to find the time $t$. $\frac{1}{2}at^2 = \frac{1}{2}[g~sin(3^{\circ})]~t^2+20~m$ $\frac{1}{2}at^2 - \frac{1}{2}[g~sin(3^{\circ})]~t^2=20~m$ $t^2 = \frac{40~m}{a-g~sin(3^{\circ})}$ $t = \sqrt{\frac{40~m}{2.0~m/s^2-(9.80~m/s^2)~sin(3^{\circ})}}$ $t = 5.186~s$ We can find the distance the cart has rolled. $x = \frac{1}{2}(g~sin(3^{\circ}))~t^2$ $x = \frac{1}{2}(9.80~m/s^2)~sin(3^{\circ})~(5.186~s)^2$ $x = 6.9~m$ The cart rolled 6.9 meters.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.