Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - General Problems - Page 196: 73

Answer

$110\frac{km}{h}>90\frac{km}{h}$ See solution below for complete answer

Work Step by Step

$W_A=(18m)(1500kg)(9.8\frac{m}{s^2})(0.60)=158,760J$ $W_B=(30m)(1100kg)(9.8\frac{m}{s^2})(0.60)=194,040J$ $v'_A=\sqrt{\frac{2\times158,760J}{1500kg}}=14.5\frac{m}{s}$ $v'_B=\sqrt{\frac{2\times194,040J}{1100kg}}=18.8\frac{m}{s}$ $m_Av_A+m_Bv_B=m_Av{'}_A+m_Bv{'}_B$ $(1500kg)v_A=(1500kg)(14.5\frac{m}{s})+(1100kg)(18.8\frac{m}{s})$ $v_A=\frac{42,430kg \frac{m}{s}}{1500kg}=28.3\frac{m}{s}$ $E_{KA}=\frac{(1500kg)(28.3\frac{m}{s})^2}{2}=600667J$ The amount of work done by the first 15m of skidding can be calculated by $W_{iA}=dF_f=(15m)(1500kg)(9.8\frac{m}{s^2})(0.60)=132300J$ $E_{iKA}=600667J+132300J=732967J$ $v_{iA}=\sqrt{\frac{2\times732967J}{1500kg}}=31.3\frac{m}{s}\times\frac{1km}{1000m}\times\frac{3600s}{1h}=110\frac{km}{h}>90\frac{km}{h}$
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