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Some Applications of Newton's Laws of Motion

  1. Dec 4, 2005 #1

    KD

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    This is probably really easy, but I can't figure out what to do.

    A man of mass 80kg (W= 176 lb) jumps down to a patio from a window ledge .50m (1.6ft) from ground. He doesn't bend his knees on landing so his motion is arrested in a distance 2.0cm (.79in) What is the acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest? With what force does this jump jar his bone structure?

    If someone could just lead me in the right direction, that would be helpful.
     
  2. jcsd
  3. Dec 4, 2005 #2
    Find his acceleration using kinematics, and use [itex] \Sigma F = ma [/itex]

    edit: thats a big sigma...
     
  4. Dec 4, 2005 #3

    KD

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    I forgot to include what I have already tried.

    I separated the kinematics into two. From the ledge to the ground and then from the ground to rest. I'm jusing 9.8 as the acceleration for the first one and I'm getting 3.13 for final velocity. Then I'm using that as my initial velocity for the next set. That must be wrong because then I'm getting 245 m/s2 as my acceleration...

    I also changed the 80kg into 5.5slugs, but I doubt that is helpful.
     
  5. Dec 4, 2005 #4
    well the acceleration is going to be very high for the second part, so what you have is correct. Now use that with newton's second law to find the net force exerted on the man.
     
  6. Dec 4, 2005 #5

    KD

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    Okay, so what I need to do now is change all of that work to get it into lbs then subtract the orignal weight of the man to get the force acting on him? Or do I just keep the answer how it is?

    Thanks for your help.
     
  7. Dec 4, 2005 #6
    don't use lbs, use SI units.
    in fact, you dont even need work.

    if you know his acceleration and his mass, how would you find the force?
     
  8. Dec 4, 2005 #7

    KD

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    Okay, I know sigmaF=ma, and I know I would use that to find the force. The only reason I am converting to pounds is because I need to subtract the weight because that is a "negative" force so I need to add that part for the sigma...at least, I'm pretty sure. That is how we have been doing it in class...but maybe that's not what I do here.

    Wait, I don't need to convert all of that work to pounds, I can just use 4.448N/lb to convert the weight...
     
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