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: 22a

Answer

Please see the work below.

Work Step by Step

We know that: $W_1-mgd=\Delta K_1=\frac{1}{2}mv_1^2$ This can be rearranged as: $W_1=mgd+\frac{1}{2}mv_1^2$ We plug in the known values to obtain: $W_1=80.0(9.8)(10.0)+\frac{1}{2}(80.0)(5.00)=8.84\times 10^3J$
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