Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 6 - Exercises and Problems - Page 111: 72


Please see the work below.

Work Step by Step

We know that $Power=\frac{Energy}{time}$ $\implies P=\frac{mgh+F_{air}d}{t}$ $P=(mgsin(\theta)+F_{air})(v)$ This simplifies to: $sin(\theta)=\frac{(\frac{P}{v})-F_{air}}{mg}$ We plug in the known values to obtain: $sin(\theta)=\frac{(3800)(\frac{60}{3.6})-450}{(1400)(9.8)}$ $sin(\theta)=0.1334$ Now the angle is: $\theta=sin^{-1}(0.1334)$ $\theta=7.7degrees$
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