EK FL 1: C/P Q17

[To be MD]

"The Cooper test determines the maximum aerobic capacity of an individual by having the subj. run for 12 minutes on a treadmill. If a man ran 2 km in this time, how much work did he do? (g = 10 m/s^2)"

a) 0 J
b) 20,000 J
c) -20,000 J
d) 20 J

_________

Me? I didn't even calculate anything. I knew someone who has run 2 km is going to use up a lot more than 0 or 20 J, so I picked choice b.

Examkrackers is trying to say that since W = F * d, there is 0 displacement and the work is 0.

The problem with this is that it's so freaking wrong. Just straight up wrong. The displacement is measured relative to the treadmill and to the work your muscles do on your bones; it has nothing to do with the displacement of your physical body relative to the stationary world. https://www.reddit.com/r/askscience/comments/3peet3/work_force_x_distance_i_just_ran_5k_on_a/


Would this kind of a question show up on an AAMC exam? I don't know whether or not to go with the stupid answer (i.e. not actually thinking about the work done and just trying to turn it into a 'displacement is 0.. hehehe' problem) or actually calculate something.

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[theonlytycrane]

displacement is 0.. hehehe' problem

Interested in this question as well. ^ How I feel about a lot of MCAT problems

 

[spicynuggetfiend]

This question is poorly worded.

While there are internal forces in your body (bonds breaking releasing chemical energy to do useful work such as muscle contraction), your body itself is the system which work is being done by or on. Thus, those forces don't really contribute to useful work because they aren't changing your inertia if you aren't moving -- aka there is no observable change in mechanical energy. Imagine if you were motionless in outer space: flailing your arms or kicking your legs will not propel you in any direction, even if you are still using energy to make those motions.

The treadmill in some ways is artificially mimicking that scenario. While the amount of work you have exerted on the environment is non-zero, the amount of useful work that you have done is 0 J because you haven't moved. I say this question is poorly worded because it is assuming the work done is mechanical, i.e. the work is useful. The one thing that gives this away is the " (g = 10 m/s^2) ", but it would probably have been better to be explicit in the wording.

Here's another example: a car's engine can be running, but if the tires are busted, the car won't move. Thus, all the work being done by the engine is dissipated as heat as opposed to contributing to movement.

 

[To be MD]

This question is poorly worded.

While there are internal forces in your body (bonds breaking releasing chemical energy to do useful work such as muscle contraction), your body itself is the system which work is being done by or on. Thus, those forces don't really contribute to useful work because they aren't changing your inertia if you aren't moving -- aka there is no observable change in mechanical energy. Imagine if you were motionless in outer space: flailing your arms or kicking your legs will not propel you in any direction, even if you are still using energy to make those motions.

The treadmill in some ways is artificially mimicking that scenario. While the amount of work you have exerted on the environment is non-zero, the amount of useful work that you have done is 0 J because you haven't moved. I say this question is poorly worded because it is assuming the work done is mechanical, i.e. the work is useful. The one thing that gives this away is the " (g = 10 m/s^2) ", but it would probably have been better to be explicit in the wording.

Here's another example: a car's engine can be running, but if the tires are busted, the car won't move. Thus, all the work being done by the engine is dissipated as heat as opposed to contributing to movement.

Thank you for my explanation. I see what you're saying.

I'm frustrated because the question starts off with this 'test' that is used to measure maximum aerobic capacity. One assumes in order to figure this out, they don't plug in a value of 0 for Joules expended to find out the aerobic capacity of the user.

I hope something like this don't show up on the real deal.

 

[To be MD]

Haha! I found the answer to my own question on the AAMC Sample Test C/P #56.

The work done by a person on a stationary bicycle in the question is non-zero. Zero isn't even an answer choice.

Fellow MCAT aficionados, if you see work on the MCAT, it's going to be non-zero on an exercise machine even if the person is stationary. Don't select "0 work" unless the question explicitly states having 0 displacement.

 

[theonlytycrane]

great follow-up. Calculate work as normal? @To be MD

 

[DesitnationMD]

Haha! I found the answer to my own question on the AAMC Sample Test C/P #56.

The work done by a person on a stationary bicycle in the question is non-zero. Zero isn't even an answer choice.

Fellow MCAT aficionados, if you see work on the MCAT, it's going to be non-zero on an exercise machine even if the person is stationary. Don't select "0 work" unless the question explicitly states having 0 displacement.

I am guessing the EK people confused their own question. If they had asked what the net work done was, sicne the guy didnt move it would be 0 J work. That doesnt mean he didn't do work while on the bike. When adapting this question from the AAMC, EK likely was not paying close attention and did not word the question correctly.

It's like rolling a rock up a hill. The net work done is 0 J. This is because any work I do on the rock while pushing it up the hill is cancelled out by the work done by gravity. I did do work on the rock, but the new work is 0 J

 

[To be MD]

great follow-up. Calculate work as normal? @To be MD

Yes!

I am guessing the EK people confused their own question. If they had asked what the net work done was, sicne the guy didnt move it would be 0 J work. That doesnt mean he didn't do work while on the bike. When adapting this question from the AAMC, EK likely was not paying close attention and did not word the question correctly.

It's like rolling a rock up a hill. The net work done is 0 J. This is because any work I do on the rock while pushing it up the hill is cancelled out by the work done by gravity. I did do work on the rock, but the new work is 0 J

Exactly.