Fundamentals of Physics Extended (10th Edition)

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

Chapter 8 - Potential Energy and Conservation of Energy - Problems: 33b

Answer

$4.19\dfrac {m}{s}$

Work Step by Step

İnitial potential Energy relative to point B is: $U_{0}=\dfrac {k\Delta x^{2}}{2}+mg\left( D\sin \theta +\Delta h\right)$ $ =\dfrac {170\times 0.2 ^{2}}{2}+2\times 9.8\times \left( 1\times \sin 37+0.2\times \sin 37\right)$ $ \approx 17.56J$ All this potential energy will be converted to kinetic energy. So: $\dfrac {mv^{2}_{B}}{2}=v_{0}\Rightarrow v_{B}=\sqrt {\dfrac {2U_{0}}{m}}=\sqrt {\dfrac {2\times 17.56}{2}}\approx 4.19\dfrac {m}{s}$
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