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2 Kinetic Questions

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data
    While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)


    2. Relevant equations
    W = F * s

    3. The attempt at a solution
    m = w/(g*s)
    m = 70J/(9.8m/s^2 * .40m
    m = 17.86 kg

    Then I take this and divided by total mass 70kg to get the percentage but I got the wrong answer. I don't know why....what am I doing wrong here?

    1. The problem statement, all variables and given/known data
    Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar?

    2. Relevant equations
    KE = 1/2m (vf^2 - vo^2)

    3. The attempt at a solution
    I didn't know how to get this started but I gave it a shot. First, I took the 5 revolution per 3 seconds and converted them to velocity.

    5rev/3s (2*pie*r/rev) = 20.9 m/s

    Then I take that answer and plug it into the kinetic equation but as expected, I was wrong. This kinetic stuff is really confusing or is it just me? :confused: Pls help?
     
    Last edited: Feb 12, 2007
  2. jcsd
  3. Feb 12, 2007 #2

    radou

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    Homework Helper

    What is the problem question here?
     
  4. Feb 12, 2007 #3
    Oops. :redface: I totally forgot.

    Question: What is the kinetic energy of this bar?
     
  5. Feb 12, 2007 #4
    Check your math. Your equation seems right, but I get a different answer when I plug in the values.
     
  6. Feb 12, 2007 #5

    radou

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    Homework Helper

    Excellent. Now, how can you calculate the angular velocity of the bar from the given information? The units should be [rad/s].

    Further on, what is the definition of rotational kinetic energy?
     
  7. Feb 12, 2007 #6
    First of all, your radius of the bar is 1m. Don't get confused.

    Second of all, look into Rotational Kinetic Energy. You're using the linear kinetic energy equation
     
    Last edited: Feb 12, 2007
  8. Feb 12, 2007 #7
    1) I checked my math already and I still get 17.86kg. It should be 70/3.92 right? Are you sure I'm not missing anything?

    2) I haven't learned about rotational kinetic energy yet. That's on chapter 9, I'm only on chapter 6.

    Edit: Is the angular speed (5rev/3s)(2pie/rev) = (10pie*rad/3s)?
     
    Last edited: Feb 12, 2007
  9. Feb 13, 2007 #8
    Can someone pls pls pls help me with the first problem? I can't figure it out at all, I don't know where I went wrong. I've been punching numbers for the last hour and I still can't seem to get the right answer and this, I know, is a really easy problem and yet I'm going crazy over it. Sigh.

     
  10. Feb 13, 2007 #9

    Doc Al

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    Staff: Mentor

    First problem

    Not quite sure what you are doing here. Try this. Call the man's total mass M and his muscle mass m. Figure out: The amount of work required to do one chin up; the amount of work his muscle mass can generate. Set them equal and solve for m/M.
     
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