Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 246: 24

Answer

$6.73\frac{m}{s}$

Work Step by Step

According to the law of conservation of energy: $K.E_{bottom}+U_{bottom}=K.E_{top}+U_{top}$ $\implies \frac{1}{2}mv_{bottom}^2+0=0+mgy_{top}$ This can be rearranged as: $v_{bottom}=\sqrt{2gy_{top}}$ We plug in the known values to obtain: $v_{bottom}=\sqrt{2(9.81)(2.3.1)}$ $v_{bottom}=6.73\frac{m}{s}$
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