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Average Net force and high jumpers

  1. May 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A high-jumper with a mass of 65kg just clears a height of 2.13m and drops onto a 0.30m thick landing mat.
    a) What average net force does the landing mat exert on the high jumper if it is compressed by 0.18m during landing?
    b) What would be the effect on the force if the high-jumper were to land in sand that compresses only 5cm

    I managed to figure out the potential energy to be= 1356.81J
    and the Velocity = 6.46 ms/-1

    Answers:
    a) 7100N
    b) 27,000J

    2. Relevant equations
    Ek=1/2mv2
    Ep=mgh
    W=Fs=Change in energy
    F=ma
    F=mg

    3. The attempt at a solution
    Cannot figure out average net force.
    I've tried a lot of things and even tried working backwards from the answer and cannot get anywhere. I just would like to know how to figure the average net force and the second question with the given information.
     
    Last edited: May 22, 2014
  2. jcsd
  3. May 22, 2014 #2

    tms

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    What forces act on the jumper?
     
  4. May 22, 2014 #3

    haruspex

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    Welcome to the forum... but perhaps you have not read the guidelines.
    You are supposed to post your attempt at a solution, as far as you could get. If you couldn't even get started, at least some insight into what is blocking you.

    However, I do sympathise with you on this problem, since in my view the question is flawed.
    Unless specifically stated otherwise, "average force" ought to mean an average over time: ∫F.dt/∫dt, which can also be written as Δp/Δt, i.e. the change in momentum divided by the elapsed time. The reason for this is that average acceleration is definitely defined as Δv/Δt, and average force ought to equal mass * average acceleration.

    The posted question gives you no way to determine the duration of the landing. You could guess, perhaps, that the force increases linearly with degree of compression, as with a spring. But in order to get the given answer, you have to interpret it as the average over distance. This you would calculate as ∫F.ds/∫ds = ΔE/Δs, the change in energy divided by the displacement. This will in general give a different answer from average over time.
     
  5. May 22, 2014 #4

    tms

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    I think the problem wants the solver to pretend the average force is a constant force, and to use the kinematics equations to find the acceleration.
     
  6. May 22, 2014 #5
    I managed to find the potential energy and the velocity and I can't get any further than that. The thing is, this is included in the exercise in my book under mechanical energy. The chapter is teaching kinetic and gravitational potential energy so somehow it has to relate to this?
    It says in the summary that potential energy is stored energy with the potential to allow work to be done. It may take many forms, including chemical and spring.
    I don't know if any of this helps.
    Also I can't really show my attempt at the question since I don't know how to attempt it or what equations to really use with the information provided.
     
  7. May 22, 2014 #6

    tms

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    If you have calculated the potential energy, then you already know how that relates to the problem. Look at the jumper as a particle dropped from a height. What is it's speed when it hits the mat?
     
  8. May 22, 2014 #7
    yeah but it asks for the mat and not the jumper and somehow i have to use the thickness and the compression of the mat to figure out the answer. I still am stuck on it and really all I need to help and for someone to explain how to do it
     
  9. May 22, 2014 #8

    haruspex

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    You should not assume all the information given is necessary.
    I gave quite a strong hint how to relate the energy to the force at the end of the penultimate sentence in post #3.
     
  10. May 22, 2014 #9
    Yeah and I tried that equation and it seemed to not work and did not give the answer.
     
  11. May 22, 2014 #10

    haruspex

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    It worked for me. Please post your working.
     
  12. May 22, 2014 #11

    HallsofIvy

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    You say you have calculated potential energy. (where? At the top of the jump?) That potential energy has become kinetic energy at the end and must be dissipated: work (energy) equals force times distance. You know the energy that must be dissipated and you know the distance so you can solve for force.
     
  13. May 22, 2014 #12
    I calculated the potential energy which is at the top of the jump. I know it becomes kinetic energy at the bottom (that's how I figured out the velocity) so I tried solving for force:

    W=Fs
    1356.81=Fx2.13m
    F=1356.81/2.13
    F= 637N
    ^ Wrong Answer
     
  14. May 22, 2014 #13
    As my previous post has, I pretty much did the same thing and got the same answer.
    I'm just getting more and more confused.
    You told me change in energy/displacement
    I did
    1356.81/2.13 and got 637N, still the wrong answer.
     
  15. May 22, 2014 #14

    tms

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    The displacement over which the force of the mat works is the amount of compression, not the height of the bar.
     
  16. May 22, 2014 #15
    Still did not work.

    1356.81/0.18
    = 7537.83 - close but still not the answer.
     
  17. May 22, 2014 #16

    tms

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    Show your equations. I don't know what those numbers are, or how you got them.
     
  18. May 22, 2014 #17
    If you saw above on the 3rd post the guy gave me a hint to what equation to use to find the force:
    "But in order to get the given answer, you have to interpret it as the average over distance. This you would calculate as ∫F.ds/∫ds = ΔE/Δs, the change in energy divided by the displacement. This will in general give a different answer from average over time."
    SO I did change in energy which I thin is 1356.81/0.18 (the compression of the mat) which did not give the right answer and I cannot really find the energy of the mat if I don't have the mass.
     
  19. May 22, 2014 #18

    haruspex

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    The height of the bar from the ground is 2.13m. The mat lies on the ground, is 0.3m thick, and compresses 0.18m. What is the vertical distance between the bar and where the jumper comes to rest.
     
  20. May 22, 2014 #19

    tms

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    How did you get that number for the energy? What equations did you use? Show the actual equations, not just the numerical results.
     
  21. May 23, 2014 #20

    haruspex

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    It comes from using 2.13m as the vertical displacement (which is not quite right).
     
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