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 - Problems - Page 193: 24

Answer

a) $J=407kg\frac{m}{s}$ b) $=1.5\times10^5N $ upward c) $=3.6\times10^3N$ upward

Work Step by Step

a) $J=\Delta p$ $v_f=\sqrt{2gh}=\sqrt{2(9.8\frac{m}{s^2})(2.8m)}=7.4\frac{m}{s}$ $J=m(v_f-v_i)=(55kg)(7.4\frac{m}{s})=407kg\frac{m}{s}$ b) $\sum F=F_g-mg=ma$ $F_G=m(g+a)=m(g+\frac{v_f^2-v_i^2}{2\Delta x})$ $=(55kg)\Big((9.8\frac{m}{s^2})+\frac{(7.4\frac{m}{s})^2-(0\frac{m}{s})^2}{2\times 2.81m}\Big)$ $=1.5\times10^5N $upward c) $F_G=(55kg)\Big((9.8\frac{m}{s^2})+\frac{(7.4\frac{m}{s})^2-(0\frac{m}{s})^2}{2\times 3.30m}\Big)$ $=3.6\times10^3N $upward This is why it is less painful and safer when you bend your knees when you land after jumping or falling.
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