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Bungee Jump (spring) question

  1. Apr 20, 2006 #1
    You go bungee jumping from a bridge that is 100m above a river, the bungee cord has an unstretched length of 50m and a spring constant of k=700 N/m
    You have a mass of 45kg.
    1. How fast are you falling when you just start to stretch the cord and how long does it take to reach this point?
    2. What types of mechanical energy does the system (you and the bungee cord) have just before you jump. What types of energy does the system have at the instant you come to rest at your lowest point?
    3. What is your height above the river at your lowest point.
    4. What would be the period of your oscillation if it is assumed that the bungee cord acts like a perfect spring.
    whew, thats alot.

    Ok so for 1 it is asking for velocity and time, but I have no idea on where to start or what equation to use
    For 2, just before you jump you dont have any kinetic energy but you do have gravitational PE and elastic PE correct? Am I missing any? When you come to rest at the lowest point, once again no KE, but elastic PE?
    For 3, I know that enrgy stored in the bungee cord is 1/2 k (x)^2 where x is the amount of stretch, but how do I get height from this?
    And for 4, T = 2pi square root of (m/k) is that it?

    Please help!!!
     
  2. jcsd
  3. Apr 20, 2006 #2

    andrevdh

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    1. For the first 50m you will free fall from the bridge - the cord is falling with you. Actually, this problem is more complicated if the mass of the cord is taken into account. Which it seems you do not need to consider.
    2. Correct, if you choose that point to be where gravitational PE to be zero.
    3. Energy conservation of the system since both gravity and the elastic restoring force is conservative.
    4. Yes.
     
  4. Apr 20, 2006 #3
    OK, Im almost embarassed to ask, but how do i caluclate the freefall velocity?
    And for three, so PEg = PEs = mgh=1/2k x^2 ??
    and x is the amount of stretch but how do i find this? would it be Force / K =
    (441N / 700N/m) = .63 m ?
     
  5. Apr 20, 2006 #4

    Hootenanny

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    You donot need to consider the free fall velocity, you just need to know how fast you are travelling when the rope begins to strech, you can do this by conservation of energy.


    [tex]mgh_{initial} = mgh_{50m} + \frac{1}{2}mv^2[/tex]

    Solve for v.

    Yep, that's correct.

    As andrevdh said, use conservation of energy in the equation you specified above;

    [tex]mgh = \frac{1}{2}kx^2[/tex]

    Solve for x.

    ~H
     
  6. Apr 20, 2006 #5

    andrevdh

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    Free fall is a constant acceleration problem, so use the derived equations. You take the acceleration as [itex]9.8\ m/s^2[/itex]. This will enable you to calculate your speed after you have dropped from rest through a distance of 50 meters. Also calculate the time it will take to drop through this distance out of rest.

    For the energy problems it is probably easiest to take the zero gravitational potential energy level at the point where the cord starts to stretch as the question suggest by asking you to calculate your speed at this point of the motion. This means your gravitational potential energy is negative as you drop below the 50 meter mark. If you decide to go that way the total energy at all stages of the motion will be equal to your kinetic energy at this point (50 meter mark) since both the elastic potential energy and gravitational potential energy is zero at this stage.
     
    Last edited: Apr 20, 2006
  7. Apr 20, 2006 #6
    Using this equation, I solved for v and got .298 m/s which doesnt seem right?


    Knowing that a=9.8 and the distance is 50m, I dont know what equation to use to find velocity.

    It seems like I would be solving for 2 unknown variables? I need to find h and x using the same equation?

    I know I seem hopeless but I just wanted to thank you for your help, I truly appreciate it.
     
  8. Apr 20, 2006 #7

    Hootenanny

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    I got 31.3m/s

    [tex]mgh_{initial} = mgh_{50m} + \frac{1}{2}mv^2[/tex]

    The masses cancel leaving

    [tex]100g = 50g + \frac{1}{2}v^2[/tex]

    Can you go from here?

    You are given h at the start. "from a bridge that is 100m above a river"

    Your not hopeless, stick at it and we'll help you work through it :smile:

    ~H
     
  9. Apr 20, 2006 #8

    andrevdh

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    The constant acceleration question to use is
    [tex]v^2=u^2 + 2as[/tex]
    if you take downwards as positive both the acceleration and displacement, s, the height, is positive as you fall downwards. [itex]u[/itex] is the initial velocity which is zero. Once you have solved for the velocity, [itex]v[/itex], after dropping 50 meters you can use
    [tex]v=u+at[/tex]
    to solve for how long does it take to drop out of rest, [itex]u=0[/itex], up to this speed [itex]v[/itex]
     
  10. Apr 20, 2006 #9

    Hootenanny

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    To nophun6:
    (Just to avoid any confusion)

    This question can solved one of two ways. My solution used conservation of energy priciples. Andrevdh's uses kinematic equations, either is a valid method and both will lead to the same outcome.

    Regards,
    ~H
     
  11. Apr 20, 2006 #10
    Ah, I see now, thank you both so much. The velocity I calculated was wrong because I divided something instead of subtracting, duh!
    Ok, so now I got v=31.3 m/s
    So 31.3 m/s = 0 + (9.8m/s^2)t and t= 3.19 s
    Is this correct?

    Then using PEg=PEs I calculated:
    (45kg)(9.8)(100m) = 1/2 (700 N/m)x^2 and got x = 11.22m
    So x is the amount of stretch, so the height above the river at the jumpers lowest point would be 50m - 11.22m = 38.78 m
    Is the correct?

    And finally, for the period: T= 2pi (square root of m/k)
    T= 2pi (square root of 45kg/700N/m)
    T= 1.59s
     
    Last edited: Apr 20, 2006
  12. Apr 20, 2006 #11

    Hootenanny

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    I apologise nophun6, I mislead you above, the bunjee jumper does not fall the full 100m so, there will be some potential energy left at the bottom if you like, so the equation really should be;

    [tex]mg[50+x] = \frac{1}{2}kx^{2}[/tex]

    (50+x) is the change in height when the bunjee jumper is at the bottom.

    I apologise, I forgot that the bunjee jumper doesn't go all the way down :blushing: . Sorry for wasting your time. Anyway, re-arranging the above equation leaves you with quadratic which you can solve with the quadratic equation. I Apologise for making a stupid mistake.

    ~H
     
    Last edited: Apr 20, 2006
  13. Apr 20, 2006 #12
    Thanks for the correction, ok so plugging in numbers and solving the equation, I got x = 8.5922 m
    So the change in height would be 58.5922 and since the jumper started at height 100m, the height above the river at the lowest point in the jump would be 100 - 58.5922 = 41.4078 m
    Is this correct?
     
  14. Apr 21, 2006 #13

    andrevdh

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    Yes your time and velocity is correct.

    No the stretch is wrong since the cord only start to stretch when you you reach the 50 meter mark. That is why I suggest to start the energy consideration at the 50 meter mark. At this point you can calculate your kinetic energy since you know your speed. Then by choosing this point where your gravitational potential energy is zero your total energy available for the rest of the motion is just the kinetic energy.
     
  15. Apr 21, 2006 #14

    Hootenanny

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    I got [itex]x = 8.60 m[/itex] from the quadratic;

    [tex]350x^2 - mg x - 50mg = 0[/tex]

    ~H
     
  16. Jun 27, 2007 #15
    Help please...

    In basketball you often need to perform a ‘jump shot’. TO do this, you jump vertically upwards as high as possible and make the shot. To maximize the height of this jump you initially bend your legs and then quickly ‘spring’ up.
    a)What provides the force to allow you to jump and how is this produced?
    b)If, by bending her legs, a 55.0 kg basketball player initially lowers her center of gravity 20.0cm, what is the maximum height she can jump if the maximum force her legs can supply is 2.50 * 10^3 N? It takes her 0.100s for her to leave the ground from this position.

    I have no idea where to start, please help.
     
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