Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 7 - Kinetic Energy and Work - Problems - Page 171: 3a


$\Delta K=-5\times 10^{14}J$

Work Step by Step

Kinetic energy is defined as $$K=\frac{1}{2}mv^2$$ Therefore, the change in kinetic energy can be calculated as $$\Delta K=\frac{1}{2}(v_f^2-v_o^2)$$ The meteorite would be at rest once it collides, therefore, $v_f=0.0m/s$. Substitute this value and other known values of $v_o=15km/s=15000m/s$, $m=4\times 10^6kg$ yields a change in kinetic energy of $$\Delta K=\frac{1}{2}(4\times 10^6m/s)((0m/s)^2-(15000m/s)^2)$$ $$\Delta K=-5\times 10^{14}J$$
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