- #1
Jacob87411
- 171
- 1
I worked this out but I must of gone wrong somewhere
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kcal = 4186 J. Metabolizing 1 g of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 100 steps, each 0.150 m high, in 61.5 s. For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is 20.0%. Therefore when your body converts 100 J from metabolizing fat, 20 J goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume that the student's mass is 56.0 kg.
(a) How many times must she run the flight of stairs to lose 1 lb of fat?
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First off, we have to convert. When 1 lb of fat is burned that is 453.592 g. Each gram of fat burned requires 9.00 kcal as the problem states which is equivalent to 37674 J. So to burn 1 g of fat you need to work off 37674 J so to burn 453.592 g of fat one must burn (453.592)(37674)=17088625 J.
Ok so to the physics:
W=Fd
The distance is the vertical distance going up the stairs which is 100(.15m)=15m.
I got confused on what the force should be, is it the force of gravity which would be 56(9.8)?
If so W=fd = (560)(15) = 8400 but only 1/5 of that work goes towards metabolizing fat so 1680 J per time up the stairs. Then you take the total amount of joules needed to be burned / 1680 and get a huge, unreasonable number..any help is appreciated
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kcal = 4186 J. Metabolizing 1 g of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 100 steps, each 0.150 m high, in 61.5 s. For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is 20.0%. Therefore when your body converts 100 J from metabolizing fat, 20 J goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume that the student's mass is 56.0 kg.
(a) How many times must she run the flight of stairs to lose 1 lb of fat?
-------
First off, we have to convert. When 1 lb of fat is burned that is 453.592 g. Each gram of fat burned requires 9.00 kcal as the problem states which is equivalent to 37674 J. So to burn 1 g of fat you need to work off 37674 J so to burn 453.592 g of fat one must burn (453.592)(37674)=17088625 J.
Ok so to the physics:
W=Fd
The distance is the vertical distance going up the stairs which is 100(.15m)=15m.
I got confused on what the force should be, is it the force of gravity which would be 56(9.8)?
If so W=fd = (560)(15) = 8400 but only 1/5 of that work goes towards metabolizing fat so 1680 J per time up the stairs. Then you take the total amount of joules needed to be burned / 1680 and get a huge, unreasonable number..any help is appreciated