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Air Resistance of a Diver using Ek and Eg

  1. Apr 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A 57.0 kg diver dives from a height of 15.0 m. She reaches a speed of 14.0 m/s just before entering the water. What was the average force of air resistance (e.g., friction) acting on the diver?
    What is the force of friction underwater if she reaches a depth of 2.5 m before stopping? Do not neglect the buoyant force of 500 N acting on the diver once underwater

    2. Relevant equations
    Eg=mgh
    Ek=(1/2)mv^2
    Et=Ek+Eg
    W=Ef-Ei
    W=change in Ek
    W=fd

    3. The attempt at a solution
    Ek=(1/2)(57kg)(14m/s)^2
    =5586J

    W=Ek
    =5586J-0J

    W=fd
    5586J=F(15m)
    5586/15=F
    F=372.4N

    Fnet=ma
    =(57kg)(9.8N/kg)
    =558.6N

    Fnet=558.6N
    Fapp-Ff=558.6N
    372.4N-Ff=558.6N
    Ff=186.2N

    I'm not sure if I solved for friction properly. The way that I solved here doesn't work for the next step in the water, so I think initially started wrong.
     
  2. jcsd
  3. Apr 24, 2014 #2
    How does the actual velocity of the diver about to enter the water compare to what would be expected in a vacuum? What would that imply about the actual acceleration?
     
  4. Apr 24, 2014 #3
    The only acceleration would be that of gravity 9.8.
     
  5. Apr 24, 2014 #4
    What I am trying to say is that the velocity of the diver ( as given in the problem) is somewhat less than what would result from g when falling from the height (as given in the problem). Clearly the diver is accelerating less than g and that is due to the force of air resistance. You can use Newton's Second law to determine what this force is.
     
  6. Apr 24, 2014 #5
    It just occured to me that the problem or the instructor wants the problem solved by COE. In this case, unless I made a math mistake, I get an answer close to yours. I took the sum of forces approach because it was a more straightforward way of addressing the buoyant force once the diver enters the water.
     
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