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Is average force dependent on the exertion time of force?

  1. Jan 9, 2016 #1
    1. The problem statement, all variables and given/known data
    A 65 kg man jumps at a velocity of 1.8 m/s. The contact time is 0.45 seconds. Find the average force exerted by the man.
    2. Relevant equations
    F=ma+mg
    3. The attempt at a solution
    F=65*(1.8/.45)+65*9.81

    I'm not sure if the equation has to be divided by the contact time, .45 seconds, to obtain the average force acting on the man.
     
  2. jcsd
  3. Jan 9, 2016 #2

    Doc Al

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    Hint: Think in terms of momentum changes.
     
  4. Jan 9, 2016 #3

    haruspex

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    You did divide by the time. Are you just asking whether that was the right thing to do? Yes, it is, and I agree with your answer.
    You don't explain how you came up with that equation. I'm guessing that you divided the velocity change by the duration to obtain the average acceleration. If so, you are quite right that mass times average acceleration gives average force.
    Doc Al's hint gives you another route to the same answer.
     
  5. Jan 9, 2016 #4

    Doc Al

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    Yeah, I should have just said "yes". :smile:
     
  6. Jan 9, 2016 #5
    Sorry for the ambiguity in my question. I'm well aware that we have to divide the velocity by contact time to obtain the average acceleration. What really confuses me is that due to the fact that the man also has to provide mg in 0.45 seconds, alongside ma which was solved above, do i also have to divide mg by 0.45 seconds to find out the AVERAGE force provided by the man, which unit is 1 second? Please correct me if there is any misconception.
     
  7. Jan 9, 2016 #6

    haruspex

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    The mg is constant over the time period, so it must equal its own average. If you divide it by time it will no longer be a force, but something rather strange.... mass times jerk, I guess.
     
  8. Jan 9, 2016 #7
    Thanks.
     
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