School of Physics and Astronomy, University of Minnesota
Physics 1102, Introductory Physics II, Homework 1, group problem
The process of energy input into a human body includes the oxidation of glucose
as described in your book by the chemical reaction
C6 H12 O6 + O2 → 6CO2 + 6H2 O + energy
The energy on the right hand side is carried to your muscles (and elsewhere) by a
molecule called adenine triphospphate (ATP). This process requires oxygen which
gets into your blood through your lungs. Therefore the rate at which your body
can feed energy to your muscles by this process is limited by the rate at which
your lungs can absorb oxygen. One of the main results of atheletic training is to
increase this capacity to absorb oxygen into the blood through the lungs. On the
other hand there is another way to get useful energy to your muscles which works
without directly using oxygen. Some of the glucose in your body is hooked together
in your liver into long branched chains of molecules called glycogens and stored.
Then when you have an emergency need for a lot of energy fast (running from a
lion or something) the glycogen is broken down by a process which doesn’t use
oxygen to form ATP to feed energy to your muscles. Thus you get a big spurt of
energy from using your glycogen store which doesn’t require a big lung capacity.
The downside is that the glycogen runs out pretty fast (in about a minute) and
then takes an hour to build back up, during which time you tend to feel terrible
due to the buildup of a side product called lactic acid in your blood.
Here are some approximate experimental numbers for these processes in humans.
(They also take place in other animals.)
maximum power from oxidation of glucose by oxygen ≈ 220 calories per minute
per kilogram of body weight
maximum power from anaerobic breakdown of glycogen ≈ 260 calories per minute
per kilogram of body weight.
Assume an efficiency of .25 for the human body in answering these questions:
Consider racing upstairs. If the risers on the stairs are .2 meters in height,
at what rate (steps per minute) can you run upstairs without dipping into your
glycogen supply (and feeling sick)?
N×.2 meters× g /1 minute= 0.25*220 calories/min/kg x 4.19 joules/cal
so
N/minute =117
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Note that the mass cancels out.
If you run ’all out’ , ie as fast as you can go , how many steps will you get up
before the glycogen runs out and you start to feel so bad that you have to stop.
What was the rate (steps per minute) in this second case?
N×g×0.2=0.25 ×(260+220)×4.19× 1 minute
N= 256
which is the number of steps per minute.
To run a marathon, what strategy do these considerations suggest? For example
a steady pace throughout, a short burst at the beginning, a short burst at the end
or some other sequence of efforts?
A short burst at the end is suggested, because then the athlete can take advantage of the glycogen pathway while only dealing with the disabling recovery period
after the race is over.
Here is some data on world track records (from
http://trackandfield.about.com/od/worldrecords/tp/Men-s-World-Records.htm):
distance(meters)
100
200
400
800
100
1500
2000
3000
5000
time(seconds)
9.58
19.19
43.18
100.91
131.96
206
284.79
440.67
757.35
Calculate the average speeds for each of these runs and graph the results (speed
versus time). Qualitatively explain the form of the graph you get in terms of the
biochemical considerations above. Then, to check, suppose that the long distance
runners were depending only on the glucose oxidation process and using the value
200 cal/kgmin for that process find the number of calories per kilogram per meter
required for those long runs. Then, supposing that the number of calories per
kilogram meter is the same for all the runs, find the number of calories per kilogram
per minute consumed in the shortest runs. Compare the last answer to the value 260
calories per kilogram per minute given for the glycogen process above to see if it’s
the same. Comment on the origin of any differences. Finally convert the numbers
of calories per kilogram per minute to watts for a 68kg male and compare with the
numbers in Table 4 of your book. What are the possible origins of differences?
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The average speed drops at about 1 minute to a relatively constant value, reflecting the shift from the glycogen to the glucose source for long runs.
The metabolic energy used is 220 cal/kgmin. They are running at about 7m/s
in the long runs so the metabolic energy used per meter is
220cal/kgmin/(7m/sec × 60 sec/min)= .52 cal/kgmeter
The shortest run is 100meters so 52 cal/kg are consumed. The race took 9.58
sec so the consumption rate was
52 cal/kg/(9.58 sec/60sec/min)= 326 cal/kgmin
It is bigger that 260 cal/kgmin because the glucose pathway is also available during the dash giving a total availibility of approximately (260+200)=460 cal/kgmin.
The numbers are reasonably close given the approximate nature of metabolic rates,
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which vary from person to person.
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