Work Problem: Climbing 3000m Mountain with 50kg Woman

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To calculate the work done by a 50kg woman climbing a 3000m mountain, the potential energy equation U=mgh is applied, resulting in significant energy expenditure against gravitational forces. The total work done translates to a specific amount of potential energy stored. With fat providing 3.8 x 10^7 J of energy and the woman's efficiency at converting fat to mechanical energy being 20%, the amount of fat consumed can be determined by calculating 20% of the total energy required for the climb. This approach allows for a straightforward calculation of both the work done and the fat consumption during the ascent. The discussion emphasizes the relationship between gravitational work, energy conversion efficiency, and fat consumption in physical exertion.
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Homework Statement



A 50kg woman climbs a 3000m mountain high. a) how much work does she do against the gravitational forces? b) a kilogram of fat supplies 3.8 x 10^7 J of energy. If she converts fat into mechanical energy with a 20% efficiency rate, how much fat will she consume in the climb?

Homework Equations



E= Einitial + Wa or K + U = Kinitial + Uinitial + Wa

K= 1/2mv^2 and U=mgh

thanks.. I really appreciate it.
 
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Work done against the gravitational force is stored as the potential energy. From the relevant equation given by you can find this energy.
Find 20% of that energy. From that find the fat consumption.
 
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