Physics: Principles with Applications (7th Edition)

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

Chapter 6 - Work and Energy - General Problems - Page 168: 88

Answer

$F=4.48\times10^4N$

Work Step by Step

$F=-ky$ $k=\frac{F}{-y}=\frac{760N}{6\times10^-5m}=1.267\times10^6\frac{N}{m}$ $mgy_i=mgy_f+\frac{ky_f^2}{2}$ $y_f^2+\frac{2mgy_f}{k}-\frac{2mgy_i}{k}=0$ $y_f^2+\frac{2(760N)y_f}{1.267\times10^6\frac{N}{m}}-\frac{2(760N)(1.0m)}{1.267\times10^6\frac{N}{m}}=0$ $y_f^2+1.2\times10^{-3}y_f-1.2\times10^{-3}=0$ Solving the quadratic equation, we get the two solutions: $y_f=100.8m$ and $y_f=0.037m$ $F=ky_f=(1.267\times10^6\frac{N}{m})(0.037m)=4.48\times10^4N$
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