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 9 - Impulse and Momentum - Exercises and Problems - Page 240: 13

Answer

The glider is in contact with the spring for 0.20 seconds.

Work Step by Step

We can find the impulse exerted on the glider by the spring. $p_0+J = p_f$ $J = p_f-p_0$ $J = m~(v_f-v_0)$ $J = (0.60~kg)[3.0~m/s-(-3.0~m/s)]$ $J = 3.6~N~s$ The impulse is equal to the area under the force versus time graph. We can find the time $t$ that the glider was in contact with the spring. $\frac{1}{2}F_{max}~t = J$ $t = \frac{2J}{F_{max}}$ $t = \frac{(2)(3.6~N~s)}{36~N}$ $t = 0.20~s$ The glider is in contact with the spring for 0.20 seconds.
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.