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Semantics regarding Work problems

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    (I've summarized) A man applies to a sled 1000N parallel to the ground and pushes the sled a total displacement of 30m. The mass of the sled is 250N and [itex]\mu[/itex]k = 0.55, how much work does the man perform? If done over 12.8s, how much power does he exhibit?


    2. Relevant equations
    W=Fdcos[itex]\theta[/itex] and P=W/t


    3. The attempt at a solution
    I calculated the work that the man did and only the man and got:
    W=1,000N(30m) = 30,000J
    Thus, P=30,000/12.8s = 2,343.75J/s

    If net work was asked, I realize that you would add the work done by the man and the work done by friction to arrive at lower values, i.e., W=25,375J and J=2,021.48J/s.

    However, my question to you all regards semantics. The instructor marked my question wrong whereas I believe it to be correct because she did not ask for net/total work done on the sled/in the system, thus man+friction, only "how much work does the [man] perform?" The only answer she accepted was the net work and I want to know if I have a legitimate reason to get credit back based on what was asked in the given statement.
     
  2. jcsd
  3. Oct 20, 2011 #2

    PeterO

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    Your reasoning all sound great - but it is based on your own summary of what was being asked, so of course it correlates.
    You would need to post the actual wording of the problem if you want us to check if you are calculating the correct answer.
     
  4. Oct 20, 2011 #3
    Here you go, word for word.

    "A football player slams into a tackle sled with a force of 1000N acting parallel to the playing surface. He pushes the sled a total displacement of 30 meters. If the mass of the sled is 250N and the μk = 0.55, how much work does the football player perform? If he completes this task over 12.8 seconds, how much power does he exhibit?"
     
  5. Oct 20, 2011 #4

    PeterO

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    I like your answer.
    Your instructor is answering: "How much of the work done by the footballer is transformed into Kinetic energy of the cart?"
     
  6. Oct 20, 2011 #5

    There are so many things wrong with this problem it's tough to figure out where to start.

    Let's hope that the sled's weight is 250N, not it's mass. (If its mass is 250 kg, the football player won't be able to move it!) That makes its mass 250/9.8 = 25.5kg -- pretty wimpy sled.

    Now, you are absolutely correct in calculating that the football player's power is 2343.75w. Uh, that's over 3 horsepower. No way, no how a person can do that, unless it's actually a horse they suited up.

    Anyway, that wimpy sled has 25375J kinetic energy. Let's see, that means that it's traveling at sqrt(2*25375/25.5) = 44.6m/s at the end of the the run. Dang! Hope he doesn't get a speeding ticked!

    Wonder what his acceleration was? Let's see, that would be F/m = (1000 - 250*.55)/25.5 = 33.8 m/s^2. Hope he doesn't get whiplash!

    But wait a minute, the problem states that he took 12.8s to get to 44.6m/s, and that means his acceleration must have been 44.6/12.8 m/s^2. Something's fishy here.

    In fact, the 30m doesn't make a lot of sense either. This must have taken place in some alternate universe where the laws of physics are different!
     
  7. Oct 20, 2011 #6
    Ugh, thanks for the conformation Peter.
    And yes, this problem and the course is frustrating. It's a biomechanics course and on every assignment and every exam there are horribly designed questions, e.g., what's the most important aspect in projectile motion: velocity, release angle, release height, or landing height. I feel like arguing with her on every grade I get b/c it does eventually add up.

    Again, thanks for the help.
     
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