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Physics 11 Online
OpenStudy (anonymous):

how much work would it take for a 75 kg person to climb the 9000 meters from sea level to the peak of mount everest? how much potential energy wold this person have when he or she reached the summit? show yoour calculations can somebody help me

OpenStudy (anonymous):

have you read the formula U = mgh ?

OpenStudy (anonymous):

huh !

OpenStudy (anonymous):

I'm not sure if can assume that g is constant. OK:\[Work=- \Delta U\]U is potential energy. \[U=-\frac{MmG}{r}\] \[\Delta U=-\frac{MmG}{6353000+9000}-(-\frac{MmG}{6353000})\] http://www.wolframalpha.com/input/?i=-%5Cfrac%7B5.97219%C3%9710%5E24+75+%286.67300+%C3%97+10%5E-11%29%7D%7B6353000%2B9000%7D-%28-%5Cfrac%7B5.97219%C3%9710%5E24+*+75+%286.67300+%C3%97+10%5E-11%29%7D%7B6353000%7D%29 So the work the gravitational field does is -6.65558131072543162062365316101183770926987440628 × 10^6, and the work you do is 6.65558131072543162062365316101183770926987440628 × 10^6 Using just U=mgh W=mg(h(2)-h(1))=75*9.8*(9000)= 6615000 The error is 0.6%, so g is fine being constant.

OpenStudy (anonymous):

no it says if not told then assume g is 10 m/s^2

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