EC Mast gPh Hwo7 ork and Energy Google Chrome https:// mastering physics.com 586

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EC Mast gPh Hwo7 ork and Energy Google Chrome https:// mastering physics.com 58678760&offset; next Close ASU PHY 111 Spring 2016 10577 HW07 Work and Energy Problem 7.39: Food calories. Resources previous l 6 of 11 l next Problem 7.39: Food calories. Part A The food calorie, equal to 4186 J, is a measure of how much If a 75.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off the calories, assuming that all the food energy goes only energy is released when food is metabolized by the body. A into increasing gravitational potential energy certain brand of fruit-and-cereal bar contains 145 food calories per bar. Submit My Answers Give Up Part B lf, as is typical, only 21.0 of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that 100 of the food calories that are eaten are absorbed and used by the body. This is actually no true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used: the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.) Submit My Answers Give U Provide Feedback Continue

Explanation / Answer

145 Calories
1 Cal = 4186 J
Total Energy from 1 bar = 4186 * 145 J
Total Energy from 1 bar = 606970 J

Increase in Gravitational Potential Energy = 606970
m*g*h = 606970
75.0 * 9.8 * h = 606970
h = 825.8 m

Height he can climb, h = 825.8 m

If only 21% of it goes into mechanical energy then,
0.21 * 606970 = m*g*h
h = (0.21 * 606970 ) / (75.0*9.8)
h = 173.4 m

Or simply we could do, 0.21 * 825.8 = 173.4 m

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