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a) $21.0kg$
b)$F_{g_E}=206N$
$F_{g_p}=252N$

Giancoli 6th Edition, Chapter 5, Problem 29

(2:12)

Chapter 5, Problem 29 is solved.

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Transcript for this Giancoli solution

We have this brass ball which is twenty one point zero point zero kilograms moved to a different planet with a ‘g’ of twelve. Part of a of this question is a trick question because the mass doesn’t change with changes in ‘g’. So the answer is twenty one point zero kilograms. The weight is what changes, not the mass so the mass stay constant but the weight of this brass ball will change and that’s what we calculate in part b. So we have the weight on earth ‘m’ times ‘gE’ of the earth so we’re going to use nine point eight so twenty one point zero kilograms times nine point eight newtons per kilogram or meters per second squared which is two hundred and six newton which is the weight on the earth. The force of gravity on the planet, wherever that is, is going to be mass time the different ‘g’ of the planet, so we have twenty one point zero kilograms times twelve point zero meters per second squared. So we have two hundred and fifty two newtons is the weight on the other planet. In the 5th Edition the mass is different, it’s a two point one kilogram brass ball so this becomes two point one and the 5th Edition answer is twenty point six newtons, and the 5th Edition answer for the force of gravity on the other planet is twenty five point two newtons.