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 241: 33

Answer

$804\;\rm N$

Work Step by Step

We know, from Newton's second law, that $$\sum F_y=\dfrac{dp_y}{dt}$$ where we know that the momentum is given by $p_y=mv_y$. Thus, $$\sum F_y=\dfrac{d}{dt}(mv_y)$$ where both changes with time, the rocket's mass and velocity. Thus, $$\sum F_y=\dfrac{dm}{dt}(v_y)+\dfrac{dv_y}{dt}(m)$$ where $dv_y/dt=a_y$; $$\sum F_y=\dfrac{dm}{dt}(v_y)+ ma_y $$ Plugging the given; $$\sum F_y=(-0.5\times120)+ (48\times18)$$ The negative sign in front of the mass is due to the mass losing during its vertical trip. $$\sum F_y=\color{red}{\bf804}\;\rm N$$
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.