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Highest velocity reached on a hill

  1. Aug 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Alexander sled down a hill and on a snow-covered horizontal leveled ground. He wanted to calculate the highest velocity that the sled can reach. The hill is 3 meters high and 6 meters long. When Alexander sled down the hill he traveled 12,5 meters on the snow-covered ground. Alexander then makes the further assumption that the frictional force is 85% of frictional force on the ground.

    Calculate the highest velocity that can be reached by the sled.

    3. The attempt at a solution

    I know that the highest velocity is at the bottom of the hill. Basically when he is about to touch the ground. And I am assuming that I have to calculate the acceleration on the hill and then on the ground so that I can solve the problem. I also know that I am going to use this formula: v^2 - v0^2 = 2as
     

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  3. Aug 20, 2015 #2

    PeroK

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    It's not clear what you're assuming regarding friction. Is it that the coefficient of friction is 0.85 on the ground?
     
  4. Aug 20, 2015 #3
    no! The frictional force on the hill is 85% of the frictional force that is on the ground. it doesn't say how much the frictional constant is :/
     
  5. Aug 20, 2015 #4

    PeroK

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    Okay, so why not assume the frictional forces are ##F## and ##0.85F## and see what you get?
     
  6. Aug 20, 2015 #5
    i don't know how to solve the problem. Where do I start?
     
  7. Aug 20, 2015 #6

    PeroK

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    Start at the top of the hill! You know the height, length of hill, ##g##, ##0.85F##. That's enough to express the velocity at the bottom of the hill in terms of these variables.
     
  8. Aug 20, 2015 #7
    okay but I need the angle to calculate the gravitational force

    Cuz the sled is tilted downwards
     
  9. Aug 20, 2015 #8

    PeroK

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    If you know the height and the length, then you can work out the angle, surely?
     
  10. Aug 20, 2015 #9
    which angle do I need? the top or the bottom one? Sorry for asking so much :(
     
  11. Aug 20, 2015 #10

    PeroK

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    You don't need the angle if you think about energy. I suggest that's a better approach in any case.
     
  12. Aug 20, 2015 #11
    okay mgh but I dont have the mass of the sled
     
  13. Aug 20, 2015 #12

    PeroK

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    So what? Leave it as m.

    It stands to reason that if the problem is solvable, the mass is not relevant.
     
  14. Aug 20, 2015 #13
    (g=10)
    Energy = 30*m

    (mv^2)/2 = 30*m
    v^2 = 60
    v = sqrt (60)

    Is this right?
     
  15. Aug 20, 2015 #14

    PeroK

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    What about the friction?
     
  16. Aug 20, 2015 #15
    oh I forgot...

    the friction on the hill is not equal to the force cuz the velocity is not constant. the friction = normal force * mu
    F(friction)=mu * F(normal)

    Now I need the angle
     
  17. Aug 20, 2015 #16

    PeroK

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    You "know" the frictional force. It's 0.85F. Think energy again: work done by friction.
     
  18. Aug 20, 2015 #17
    but its not 0.85F it is 0,85 * the friction of the ground . which i dont know what to call
     
  19. Aug 20, 2015 #18

    PeroK

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    This is why you learn algebra in maths! If you always had the numbers, you'd only need a calculator. The magnitude of the force is also not relevant: only the relationship between the friction on the slope and on the ground. We'll call it ##F## on the ground, hence ##0.85F## on the slope.

    Now you have two variables: ##m## and ##F##.

    You work though the problem using these as variables (isn't algebra great?) and look for the moment when you can cancel them out. Like you did with ##m## when you looked at motion down the slope without friction. You worked out the energy, potential and kinetic, then cancelled the ##m## to get the velocity. This is the same strategy but it's a bit harder to get to the point where you can cancel out ##F## and ##m## and solve for ##v##.

    This is real physics, by the way, which putting numbers into a calculator is not. Enjoy it if you can!
     
  20. Aug 20, 2015 #19
    I don't know how to use the friction in an equation. We have a force down the hill and the friction and they are opposite of each other. So we have F=F1-Ffriction

    F = m * a

    F1 is the force that makes the sled go downwards.

    m*a=F1-0,85F

    Is this right so far?
     
  21. Aug 20, 2015 #20

    PeroK

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    Yes, but think Energy. It's much simpler when you have a force and distance. How much energy is lost to friction?
     
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