Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
6
Work and Energy
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6-1: Work, Constant Force
6-3: Kinetic Energy; Work-Energy Principle
6-4 and 6-5: Potential Energy
6-6 and 6-7: Conservatino of Mechanical Energy
6-8 and 6-9: Law of Conservation of Energy
6-10: Power

Problem 59
A
F=5.5×102NF=5.5 \times 10^2N
Giancoli 6th Edition, Chapter 6, Problem 59 solution video poster
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VIDEO TRANSCRIPT

We know that the average power output of something, is going to equal force times the average velocity. Now in this case the velocity is steady, so the average velocity is just whatever that velocity is. There's no change in it, so we'll solve this for f, by dividing both sides by v, and we have the average force is, the power output divided by the speed. So that's 18 Horse power which we'll have to convert into Mks units, 746 Watts per Horse power. And that gets divided by the speed, which we'll have to change into meters per second, so we have 88 kilometers per hour, times one hour for every 3600 seconds, times 1000 meters for every Kilometer. So that's the whole denominator, and that works out to 5.5 times 10 to the 2 Newtons. In the 5th edition, the car is going 90 Kilometers per hour instead of 88. And so for the 5th edition, the answer is 5.4 times 10 to the 2 Newtons.

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