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Gravity Car Distance

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Hey guys, a gravity car is driven by a weight connected to a string which in turn is connected to the axle, which the weight drives when dropped. I believe the solution is very simple, however I just cannot grasp it. My task is to calculate the distance of a gravity car given the weight of 840g (plus the falling weight, x) and the distance it falls. Initial velocity is 0m/s.The distance will be with respect to x, being the falling weight powering the car.
    Mass of car = 840g
    Mass of weight = x
    Distance Weight Falls = 250mm
    Distance = ???
    2. The attempt at a solution
    Well, the work that the falling weight will do is =mgh =.250*9.81*x = 2.4525x
    The mass of the car =.840+x
    Friction =u*Fn =.01*840+x*9.81

    Distance =???

    Thanks guys.
     
  2. jcsd
  3. Aug 23, 2011 #2

    tiny-tim

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    Welcome to PF!

    Hi PhyzWizKid! Welcome to PF! :smile:

    (have a mu: µ :wink:)
    (you mean .01*(.840+x)*9.81 :wink:)

    ok so far :smile:

    now use the work energy theorem: https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in energy …

    what is the relation between https://www.physicsforums.com/library.php?do=view_item&itemid=39" and work done? :wink:
     
    Last edited by a moderator: Apr 26, 2017
  4. Aug 23, 2011 #3
    But with the work energy theorem, change in energy requires a velocity doesnt it? (EK=1/2mV^2)
    Thanks
     
  5. Aug 24, 2011 #4

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  6. Aug 24, 2011 #5
    Ahh of course thanks :smile: .
    Now, the only problem is, when I integrate friction into the formula

    Work Done - Friction = Change in Energy

    And re-arrange for distance with respect to x, I end up with a recipricol. This is no help, as there will be a point that the mass is optimum for greatest distance (a peak in the data, but a recipricol does not have this)?:confused::confused:

    Thanks.
     
  7. Aug 24, 2011 #6

    tiny-tim

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    I don't understand. :confused:

    What is your equation?
     
  8. Aug 24, 2011 #7
    Work Done - Friction = Change In Energy
    mgh - [itex]\mu[/itex]Fn = mgh
    x*9.81*.250 - .01?(.840+x)*9.81 = (.840+x)*9.81*D
    2.4525x - .01?(.840+x)*9.81 / ( (.840+x)*9.81 ) = D

    That is a recipricol, and theoretically there should be a point where there is a optimum weight, where the car will travel farthest (quadratic), beacuse as the pushing force increases, so does its resistance (total system weight)?

    Cheers
     
  9. Aug 24, 2011 #8

    tiny-tim

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    eugh! :yuck:

    you have no distance in your https://www.physicsforums.com/library.php?do=view_item&itemid=75"

    and why do you have mgh on both sides? :confused:
     
    Last edited by a moderator: Apr 26, 2017
  10. Aug 24, 2011 #9
    Oh boy I think I have really messed up now :redface:
    The way I think of it:

    Energy Expanded by the weight - Friction opposing motion = Final Energy exerted on the Car

    a.k.a

    Energy Expanded by the weight = Final Energy exerted on the Car + Friction opposing motion

    Thanks for your time :smile:
     
  11. Aug 24, 2011 #10

    tiny-tim

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    times distance :wink:
    This I don't understand. :confused:

    Work done = ∆KE + ∆PE.

    ∆KE = 0, so there are only two non-zero terms in the equation …

    what is your "Final Energy exerted on the Car" ?​
     
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