University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 4 - Newton's Laws of Motion - Problems - Exercises: 4.52

Answer

$F = [(1.7\times 10^4)~\hat{i}+0\hat{j} - (3.4\times 10^3)~\hat{k}]~N$

Work Step by Step

$\hat{r} = (0.020~m/s^3)~t^3~\hat{i}+(2.2~m/s)~t~\hat{j} - (0.060~m/s^2)~t^2~\hat{k}$ $v(t) = \frac{dr}{dt} = (0.060~m/s^3)~t^2~\hat{i}+(2.2~m/s)~\hat{j} - (0.120~m/s^2)~t~\hat{k}$ $a(t) = \frac{dv}{dt} = (0.120~m/s^3)~t~\hat{i}+0\hat{j} - (0.120~m/s^2)~\hat{k}$ At t = 5.0 s: $a = (0.120~m/s^3)(5.0~s)~\hat{i}+0\hat{j} - (0.120~m/s^2)~\hat{k}$ $a = (0.600~m/s^2)~\hat{i}+0\hat{j} - (0.120~m/s^2)~\hat{k}$ We can find the mass of the helicopter. $weight = mg$ $m = \frac{weight}{g} = \frac{2.75\times 10^5~N}{9.80~m/s^2}$ $m = 2.81\times 10^4~kg$ We can find the net force on the helicopter at t = 5.0 s. $F = ma$ $F = (2.81\times 10^4~kg)[(0.600~m/s^2)~\hat{i}+0\hat{j} - (0.120~m/s^2)~\hat{k}]$ $F = [(1.7\times 10^4)~\hat{i}+0\hat{j} - (3.4\times 10^3)~\hat{k}]~N$
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