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Question that has me stumped. Need an answer to make an A

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    this straight from a textbook.
    A skier starts from rest at the top of a hill that is inclined at 10.5degrees with the horizontal. The hillside is 200.0 m long, and the coefficient of friction between the snow and the skies is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of thesnow before coming to rest?


    2. Relevant equations



    3. The attempt at a solution
    i tried different thing but i cant enderstand how the problem can be solved without the mass of the skier.
     
  2. jcsd
  3. Nov 12, 2009 #2
    I think you can find a and v of skier at the end of a hill without m. You should rewrite 2nd N's law eq. and you will see that m will be omitted automatically.
     
  4. Nov 12, 2009 #3

    Borek

    User Avatar

    Staff: Mentor

    I think you need mass to calculate both final speed and how far this will get the skier - trick is, if you combine both things, m should cancel out. Could be it cancels out even earlier, just start assuming mass m and see where it gets you.

    --
    methods
     
  5. Nov 12, 2009 #4
    This question had you stumped? its a very simple one for a undergrad student...like 2+2=4.......
    in fact we used to do these in 10th grade :P.....

    Anyway,
    Just Assume Mass as 'm'........you'll see why you don't need to worry about it :)
    Try Applying conservation of energy ;)

    involve Change in PE n work due to friction

    Remember W due to friction is frictional force*dist.......

    Come on..its not hard at all :)
     
    Last edited: Nov 12, 2009
  6. Nov 12, 2009 #5
    im a senior in highschool.
    and my teacher is a ratard
     
  7. Nov 12, 2009 #6
    can i do this
    using trig to find that the skier is 36.44m high and 196.65m away from the base of the hill?
    work=force x distande
    Force=0.075 x weight
    36.44W=0.075W x d
    36.44W/0.075W = d
    d= 485.866
    485.866m - 196.65m = 289.21m
    does that make sense at all?
     
    Last edited: Nov 12, 2009
  8. Nov 12, 2009 #7
    No, particularly the bold part. It looks like you are calling the coefficient of friction and the position work, and this is just plain nonsense (ie: the units do not match at all).
     
  9. Nov 12, 2009 #8
    i didnt think it sounded right. but i was trying.
    i redid it and i figured that the vf is 26.72m/s at the bottom of the hill

    now im on the horizontal part of the equation.
    the skier starts from 26.72m/s and the coefficient of friction is 0.075
    i still need help i feel so dumb
     
  10. Nov 12, 2009 #9
    You do always have this:

    [tex]
    \sum_i \mathbf{F}_i=m\mathbf{a}
    [/tex]

    So can you relate a velocity to an acceleration without time? Then it should just be a matter of adding (or subtracting) the correct forces to equal mass times that relation.
     
  11. Nov 12, 2009 #10
    but i dont have the mass.
     
  12. Nov 12, 2009 #11
    You don't need it:

    [tex]
    F_{friction}=\mu N=\mu mg
    [/tex]

    where [itex]\mu[/itex] is the coefficient of friction. Since you have a mass on both sides of the equation, it does drop out.
     
  13. Nov 12, 2009 #12
    is N the net force
    how do i find it?
    i cant understand it at all?
     
  14. Nov 12, 2009 #13
    N is the normal force, it is given by [itex]N=mg[/itex].
     
  15. Nov 12, 2009 #14
    wait
    i think i got but
    how do i relate velocity to accelration without time?
     
  16. Nov 13, 2009 #15
    Sorry for the late response, I had to take care of my fussy son then went off to bed last night.

    Finding the acceleration without time is one of your kinematics equations:

    [tex]
    v_f^2=v_0^2+2a\Delta x
    [/tex]

    So if you solve this for acceleration,

    [tex]
    a=\frac{1}{2\Delta x}\left(v_f^2-v_0^2\right)
    [/tex]

    But you know that the final velocity is zero (this is when the skier stops), so you should end up with a negative acceleration (that is a deceleration, he's slowing down).
     
  17. Nov 13, 2009 #16
    Well...err..i wouldn't say "N=Mg"...........

    "my teacher is a retard." i thought teachers in the US were pretty good....8|

    anyway...

    its like..one advice...always feed in numbers in the end only...i often find that people start substituting values from beginning...that can mess things up..

    write a simple equation in terms of variables first...

    Here...

    Law of conservation of energy
    Choosing base of slope as reference level...

    Initial Potential Energy of skier + Work done by friction as man moves down slope
    =Work done by friction as man moves on level ground

    Take initial height as h, length of slope traversed as l, distance covered on level ground as s, angle of slope as @

    Now frame the equations and show them to me :)

    Btw remember.......

    N on the slope will be Mg cos@ (plz try to derive this)
    while on level surface = Mg

    oh yeah..use trigo to find h....u know l

    Do tell me if you get it =)

    Replying a day before my Maths I Test II (25% wtg) >.< ..spent the whole day with Thomas' Calculus..so sorry if i made a mistake due to Math Overload :P..
     
  18. Nov 13, 2009 #17
    Well...err..i wouldn't say "N=Mg"...........

    "my teacher is a retard." i thought teachers in the US were pretty good....8|

    anyway...

    its like..one advice...always feed in numbers in the end only...i often find that people start substituting values from beginning...that can mess things up..

    write a simple equation in terms of variables first...

    Here...

    Law of conservation of energy
    Choosing base of slope as reference level...

    Initial Potential Energy of skier + Work done by friction as man moves down slope
    =Work done by friction as man moves on level ground

    Take initial height as h, length of slope traversed as l, distance covered on level ground as s, angle of slope as @

    Now frame the equations and show them to me :)

    Btw remember.......

    N on the slope will be Mg cos@ (plz try to derive this)
    while on level surface = Mg

    oh yeah..use trigo to find h....u know l

    Do tell me if you get it =)

    Replying a day before my Maths I Test II (25% wtg) >.< ..spent the whole day with Thomas' Calculus..so sorry if i made a mistake due to Math Overload :P..
     
  19. Nov 13, 2009 #18
    Since he had found the velocity of the skier after going down the hill, at this point, we can say [itex]N=mg[/itex] because the skier is on horizontal ground and there [itex]\cos[\theta]=1[/itex].
     
  20. Nov 14, 2009 #19
    Yes...But there is no need to bring velocity into the picture...

    tdcman92..dude..just try my method once..you'll get the answer.....
     
  21. Nov 14, 2009 #20
    No gossip about teachers/instructors here please....

    Masses will be cancel out if you right the right equation,

    Consider it as an inclined plane problem, where skier is an object and mountain is the inclined plane.Angle is also provided to you so you can express the force that puts the skier to motion by using trigonometric values(Sin,Cos etc)...
     
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