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: 25

Answer

$6.78\frac{m}{s}$

Work Step by Step

According to law of conservation of energy $K.E_{bottom}+U_{bottom}=K.E_{top}+U_{top}$ $\implies \frac{1}{2}mv_{bottom}^2+0=\frac{1}{2}mv_{top}^2+mgy_{top}$ This can be rearranged as: $v_{bottom}=\sqrt{v_{top}^2+2gy_{top}}$ We plug in the known values to obtain: $v_{bottom}=\sqrt{(0.840)^2+2(9.81)(2.31)}$ $v_{bottom}=6.78\frac{m}{s}$
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