# Using mechanical energy?

1. Oct 20, 2013

### jimmyboykun

1. The problem statement, all variables and given/known data
Converting the energy of one food calorie into mechanical energy, starting from rest, to what speed could you accelerate a 133-lbs person?

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

3. 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?

2. Oct 20, 2013

### Simon Bridge

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?

3. Oct 20, 2013

### Dick

You didn't do much wrong at all. Except that a "food calorie" is actually 1000 of the calories that you are using.

4. Oct 20, 2013

### jimmyboykun

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?

5. Oct 20, 2013

### Dick

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