# Physics

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What is the efficiency of an athlete who consumes 3000kcal of food and does 2.5 x 10*6 J of useful work?

• Physics -

Divide his "useful work" by the energy value of the food eaten, but converted to the same units as the work (Joules)

3000 kcal of food = 1.254*10^7 J

I get about 20% efficiency

• Physics -

Ty I was converting the J to kcal no wonder it wasn't working! Ty for the help.

Tezara

• Physics -

You could convert the J to kCal also, as long as numerator and denominator units are the same. Either both kcal or both Joules, for example.

The efficiency ratio you get should be the same.

• Physics -

1,791,686.574?

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