How Much Speed Can You Generate Using Mechanical Energy?

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Homework Help Overview

The discussion revolves around converting the energy of one food calorie into mechanical energy to determine the speed at which a 133-lbs person could be accelerated from rest. The subject area includes concepts of energy conversion, kinetic energy, and unit conversions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conversion of food calories to joules and the implications of using kinetic energy equations. There is a focus on the accuracy of unit conversions and the assumptions made regarding energy transfer to mechanical motion.

Discussion Status

Some participants have identified potential errors in the original poster's calculations, particularly regarding the conversion of food calories. There is an ongoing exploration of how much of the food energy is applicable to mechanical motion, with suggestions to consider energy losses and the correct conversion factors.

Contextual Notes

Participants note that a "food calorie" is equivalent to 1000 standard calories, which affects the calculations. The discussion includes questioning whether all food energy can be converted to mechanical energy and how to account for energy used in non-mechanical processes.

jimmyboykun
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Homework Statement


Converting the energy of one food calorie into mechanical energy, starting from rest, to what speed could you accelerate a 133-lbs person?

Homework Equations


1 calorie= 4.184J
1 lbs= 0.453592kg
K=1/2mv^2

The Attempt at a Solution


here what I did, I converted the lbs to kg which was 60.33kg. I was under the impression to use kinetic energy.
4.184J=1/2(60.33kg)v^2
4.184J/30.165kg=v^2
sqrt 4.184J/30.165kg= 0.372m/s

This is how I solved my problem, but I got 1 out of 10. Can someone explain to me what I did wrong?
 
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Well, 1cal = 4.184J, but 1 "food calorie" = 1k.cal
I don't see you dropping 9 points just for a botched conversion.

The model you used assumes that all the food energy goes to translation of the center of mass.
There are other mechanical things - motion of arms and legs for eg, that the energy could go to.
Were you supposed to account for the food energy lost to non-mechanical sources?

Check your notes for the mechanical energy of a human being JIC you were expected to do something different.
i.e. Should you have used only a percentage of the food energy to speed?

http://en.wikipedia.org/wiki/Food_energy#Energy_usage_in_the_human_body

Do you have a model answer?
 
jimmyboykun said:

Homework Statement


Converting the energy of one food calorie into mechanical energy, starting from rest, to what speed could you accelerate a 133-lbs person?

Homework Equations


1 calorie= 4.184J
1 lbs= 0.453592kg
K=1/2mv^2

The Attempt at a Solution


here what I did, I converted the lbs to kg which was 60.33kg. I was under the impression to use kinetic energy.
4.184J=1/2(60.33kg)v^2
4.184J/30.165kg=v^2
sqrt 4.184J/30.165kg= 0.372m/s

This is how I solved my problem, but I got 1 out of 10. Can someone explain to me what I did wrong?

You didn't do much wrong at all. Except that a "food calorie" is actually 1000 of the calories that you are using.
 
ok it seems that I made a mistake in my conversion with the calories. So with that 1000 calories do I convert that into joules? if so how would I do it?
 
jimmyboykun said:
ok it seems that I made a mistake in my conversion with the calories. So with that 1000 calories do I convert that into joules? if so how would I do it?

Multiply 4.184J by 1000 to get the number of Joules in a "food calorie". What else?
 

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