College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 228: 29

Answer

(a) There is no change in gravitational potential energy during the trajectory. (b) The change in gravitational potential energy during this flight is $-2.94~J$

Work Step by Step

(a) Since there is no height difference between the start and end of the trajectory, there is no change in gravitational potential energy during the trajectory. (b) The change in height from the start to the end of the trajectory is $-1.0~m$. We can find the change in gravitational potential energy: $\Delta U_g = mg~\Delta h = (0.30~kg)(9.80~m/s^2)(-1.0~m) = -2.94~J$ The change in gravitational potential energy during this flight is $-2.94~J$.
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