Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 274: 6

Answer

$6.3\times10^{-2}\space m$

Work Step by Step

Let's apply Newton's second for the object. $F=ma$ ' Let's plug known values into this equation. $F-mg=ma$ $F=m(g+a)$ According to equation 10.2 we can write, $x=-\frac{F}{k}-(2)$ (1)=>(2), $x=-\frac{m(g+a)}{k}=-\frac{(5\space kg)(9.8\space m/s^{2}+0.6\space m/s^{2})}{830\space N/m}=-6.3\times 10^{-2}m$ The amount that the spring stretches = $6.3\times10^{-2}m$
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