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Potential energy -- child sleds down a slope and comes to a stop...

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A child starts from rest and slides on a sled first slope
    down over a distance of 25 m . Then, the child slides further over a
    horizontal plane. The angle of inclination is 35 ° and the kinetic
    coefficient of friction between the sled and the snow is 0.100. The mass
    of the child and the slide together 23 kg . What distance puts the child at the
    horizontal section off before coming to a halt ?(122,93) What do I wrong? thank you

    2. Relevant equations
    dE=-Fw.d Ek=(1/2m).v^2 Epot= m.g.h


    3. The attempt at a solution
    first I go from the begining to the ground then second I start from the ground to when it stops
    y=-cos35.G+N=0
    N=23*9,81N*cos(35°)
    = 184,8
    Fw= Fn*0,1= 18,48
    x=-cos55*23*9,81+18,48=m*a
    -110=m*a

    dE=-Fw*d
    m*g*hB+0-(1/2m*v^2A)-0=-Fw*d

    (hB= cos55°*25=14,34)

    23*9,81*14,34-1/2*23*v^2A=-18,48*d
    v^2A=321


    What distance puts the child at the
    horizontal section off before coming to a halt
    -1/2*m*v^2A=-Fw*d
    -1/2*23*321+18,48=d
     
    Last edited: Apr 17, 2016
  2. jcsd
  3. Apr 17, 2016 #2
    kindly recheck your calculation to see that it is correct- angle of inclination is 35 degree- but you are using 25 degrees.

    you could have used Potential energy (less workdone against friction) to calculate velocity square (final) and the calculared d straight away as K.E. lost against work done against friction.
     
  4. Apr 17, 2016 #3
    Thank you very much !! I have changed the numbers now of the degrees but I don't understand what I have to do now
     
  5. Apr 17, 2016 #4

    berkeman

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    I could be wrong, but I don't think the velocity comes into this problem. You have the initial PE of the child and sled, and that equals the work done by friction against the sled's motion along the path down the slope and on the flat. The normal force is different on the slope versus on the flat, but that is the only thing you should have to change for the two parts of the path, IMO.

    Can you try to do the problem just based on the initial PE and the work done by friction along the path?
     
  6. Apr 17, 2016 #5
    well,

    the loss in P.E. must have geenerated the kinetic energy but the friction on the incline must have dissipated part of the P.E. so till the cart comes to the horizontal end
    write the energy equation-
    P.E. - Work done against Friction= K.E. of the cart with boy
    now equate the K.E. found above with the Energy loss in going against Friction during Translation distance d on horizontal till it stops
    the mass of boy and cart may not be needed.
     
  7. Apr 17, 2016 #6
    Thank you! you mean just:
    m*g*h=-Fw*d
    23*9,81*14,3=18,48
    d=175,08
    it isn't right
    is mine hB right?
     
  8. Apr 17, 2016 #7

    berkeman

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    There are different normal forces on the slope versus the flat, so the friction force is different on the slope versus the flat. It doesn't look like you've taken that into account yet... :smile:
     
  9. Apr 17, 2016 #8
    Is there only friction on the flat?. I have put the friction of the slope =0 and changed everything in the thing
    I have post and I come still on 174,86
     
  10. Apr 17, 2016 #9

    berkeman

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    No, there is snow everywhere, so there is friction everywhere. The friction force is a bit lower on the slope, though, since the normal force is smaller. How much smaller?
     
  11. Apr 17, 2016 #10
    Thank you :) but where is the friction 0,1 then? I will try to solve it an other time because I have tried it and i can't solve it maybe another time with the help of the things you have said
     
  12. Apr 17, 2016 #11

    berkeman

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    That is the coefficient of friction between the sled and the snow. μ = 0.1 everywhere between the sled and the snow.

    When you listed the Relevant Equations, you did not list the equation that relates the friction force F to the normal force N and the coefficient of friction μ. Can you list that equation? What is N on the slope and on the flat? N is different on the slope versus the flat...
     
  13. Apr 17, 2016 #12
    Thank you :smile:
     
  14. Apr 17, 2016 #13
    Thank you I understand it now
     
  15. Apr 17, 2016 #14
    Thank you very much I understand it now :smile:
     
  16. Apr 19, 2016 #15
    I have tried it with
    Epot=-Fw.d with TWo different FN but the solution is wrong
     
  17. Apr 19, 2016 #16

    berkeman

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    Well let me try to read your mind to see where the problem is.... :smile:

    But seriously, please post your work in detail so we can try to find any errors...
     
  18. Apr 19, 2016 #17
    N1=23*9,81N*cos(35°)
    = 184,8
    Fw1= Fn*0,1= 18,48
    N2=23*9,81=225,63
    Fw2=Fn*0,1=22,563
    (hB= cos55°*25=14,34)
    Epot=-Fw.d
    23*9,81*14,34=-18,48*25-22,563*d
    3235,53=-462-22,563*d
    3235,53+462=-22,563*d
    3697,53/-22,563=d

    163,87=d
     
  19. Apr 19, 2016 #18

    haruspex

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    Be careful with signs. With that formulation, the Epot should be the change in potential energy, yes? Is that positive or negative?

    I strongly encourage you to get into the habit of working purely symbolically as far as possible. Pretend no numeric values are given, so make up variables to represent them. It has many benefits. In the present case, all the mass terms and g terms would have cancelled, simplifying the arithmetic at the end. You would also get more precise answers numerically since you would avoid accumulation of rounding errors. See https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/ for more on the subject.
     
  20. Apr 20, 2016 #19
    But how do i come to the Right solution ? What did i do Wrong in the latest answer i gave, and not only the positive of negative sign
     
  21. Apr 20, 2016 #20

    haruspex

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    As far as I can see, the signs are your only errors.
     
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