Answer
$12744000J$
Work Step by Step
We can find the required kinetic energy as
$K.E=\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K.E=\frac{1}{2}(1770)(120)^2$
$K.E=12744000J$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 7 - Work and Kinetic Energy - Problems and Conceptual Exercises - Page 211: 20
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