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a) $PE_g=45.3J$
b) $PE_h=12J$
c) $\Delta PE = 45.3J$

Giancoli 6th Edition, Chapter 6, Problem 30

(2:48)

Chapter 6, Problem 30 is solved.

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Transcript for this Giancoli solution

So this man lifts a book up from the ground and he lifts it up to a height of 2.2 meters. So the book ends up being here, 2.20 meters. And we need to know the height of this guy because we also want to know this distance, the height above his head. Because that'll be part B, the potential energy of the book with respect to his head. So that’s going to be 2.2 meters minus his height.

In the sixth edition its 1.6 meters and we'll talk about the fifth edition at the end. So we have 2.2 minus 1.6 which is 0.60 meters. Okay, so in part A, we want to know the potential energy of the book with respect to the ground. So that’s going to be 'mg' which is the force of gravity times the height of the book with respect to the ground. So that’s 2.1 kilograms times 9.8 newtons per kilogram times 2.2 meters. And that gives ANSWER joules.

In part B we'll ask, what is the potential energy of the book with respect to the person's head? And that’s 'mg' the height above their head. Just 2.1 kilograms times 9.8 newtons per kilogram times 0.6 meters above the persons head. And that equals 12 joules. For part C, the work done by the person is going to equal the force that they exert which is going to be equal to gravity. It's going to be directed upwards so we'll say positive 'mg' times the height of the person who lifts it. And that is going to be same as we had in part A, you can see that this formula is the same as that. So we can say that this is ANSWER joules, okay. And the work done by the person is the same as the change of potential energy of the book from part A. So the work done by the person has nothing to do with part B, okay.