Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 63

Answer

The tractor's power output is 1420 J

Work Step by Step

We can convert the speed to units of m/s; $v = (5.0~km/h)(1~hr/3600~s)(1000~m/1~km)$ $v = 1.39~m/s$ If the hay moves at a steady pace, then the applied force from the tractor is equal in magnitude to the sum of the forces exerted on the hay which are directed down the incline. We can find the magnitude of the tractor's force $F$ as; $F = mg~sin(\theta)+F_f$ $F = mg~sin(\theta)+mg~cos(\theta)~\mu_k$ $F = (150~kg)(9.80~m/s^2)~sin(15^{\circ})+(150~kg)(9.80~m/s^2)~cos(15^{\circ})(0.45)$ $F = 1020~N$ We then find the tractor's power output: $P = F~v$ $P = (1020~N)(1.39~m/s)$ $P = 1420~J$ The tractor's power output is 1420 J.
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