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$P=4.8 \times 10^2 W$

Giancoli 6th Edition, Chapter 6, Problem 65

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Chapter 6, Problem 65 is solved.

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Transcript for this Giancoli solution

Power equals force times average velocity, so we'll have to use the information from the question in order to find those two things, force and average velocity. We know that the force is going to be 'MA' is only one force on it assuming there's no friction just the driving force of the shock putter and that's going to be mass times the change in velocity over time. That's mass times 'VF' minus 'VI' over time. But the initial speed is zero because it starts from rest and so we can say just 'MVF' over 'T' is the force. The average velocity is the, is 'VI' plus 'VF' over 2 this is true when you have constant acceleration, initial speed again is zero so we can just say this is 'VF' over 2, so we’ll make a substitution for this and this and this formula. So we have power equals the force formula which is 'MVF' over 'T' times the average speed formula which is 'VF' over 2 and that combines to make 'MVF' squared over '2T' and we know all of those variables and so we’ll substitute it in our numbers, we have…7.3 kilograms times 14 metres per second squared divided by 2 and 1.5 seconds and this gives a power of 4.8 times 10 to the 2 watts. In the fifth edition the time is different and this occurs over 2 seconds instead of 1.5 so in the fifth edition less power is needed because the same work gets done over a longer period of time the fifth edition answer is 3.6 times 10 to the 2 watts.