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a) $\Delta PE = 9.2 \times 10^5J$
b) $W_{min}=9.2 \times 10^5J$
c) Yes, the actual work will be more due to inefficiencies in muscles, joints, and energy absorbed by the inelastic compression of the ground...

Giancoli 6th Edition, Chapter 6, Problem 31

(2:20)

Chapter 6, Problem 31 is solved.

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Transcript for this Giancoli solution

The hiker in this question changes elevation from 1600 meters up to 3300 meters, in the fifth edition this becomes 3100 meters. Will work with sixth edition numbers throughout then do the fifth edition at the end. So that change in height is what's going to be important we'll call this 'h' and that equals 3300 meters minus the original starting height of 1600 meters. And we get 1700 meter change in height. So for part 'a' the question is what's the change in potential energy and that's going to be 'mgh'. And that is 55 kilograms times 9.8 Newton’s per kilogram times 1700 meters times 9.2 times ten to the five joules.

For part 'b' the minimum amount of work that this person would have to do is equal to that change in potential energy. So again we have ANSWER joules. And in part 'c' we have to think practically in real world this person would in fact do more work than this change of potential because there is inefficiency in muscles and joints , so inefficiency in muscles and joints would cause the actual work to be greater than the minimum work.

For the fifth edition let's just plug in the proper numbers for the fifth edition. We would have 3100 as the top altitude here so this difference is actually 1500 meters and we substitute that number here, and we get an answer for the fifth edition of ANSWER joules.