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Find the distance over which the skater will move

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data
    A skater of mass m standing on ice throws stone of mass M with speed v in a horizontal direction. Find the distance over which the skater will move in the opposite direction if the coeffecient of kinetic friction between the ice and the skater is mk

    2. Relevant equations
    [itex]V_{0x} + a_xt = V_x [/itex]
    [itex]\Delta x = (1/2)(V_{0x} + V_x)t[/itex]
    [itex] \Delta x = V_{0x}t + (1/2)a_xt^2[/itex]
    [itex] \Delta x = V_xt - (1/2)a_xt^2 [/itex]
    [itex] V_x^2 = V_{0x}^2 + 2a\Delta x [/itex]
    [itex] F= ma [/itex]
    [itex]P_{initial} = P_{final} [/itex]
    3. The attempt at a solution

    Ok, so using the last eq, P_i = P_f.

    The initial P = 0, so P_f = 0 as well.

    Well momentum for the stone = [itex] Mv [/itex], thus [itex] mv_{skater} = Mv [/itex]

    to make momentum = 0 again.

    So now I know
    [itex] v_{skater} = (Mv)/m [/itex]

    Using
    [itex]\Delta x = (1/2)(V_{0x} + V_x)t[/itex]

    I get

    [itex] \Delta x = (1/2)(Mv/m)t [/itex]

    using
    [itex] \Delta x = V_{0x}t + (1/2)a_xt^2[/itex]

    I get

    [itex] \Delta x = (1/2)a_xt^2 [/itex]

    Setting the two [itex] \Delta x [/itex] = to each other I get

    [itex](1/2)a_xt^2 = (1/2)(Mv/m)t [/itex]

    [itex] a_xt = (Mv/m) [/itex]
    [itex] t = (Mv/a_xm)[/itex]

    Now I plug t back into: [itex] (Mv/2m)t [/itex]

    [itex] (M^2v^2)/(2m^2a_x) = \Delta x [/itex]

    but my books answer is

    [itex] (M^2v^2)/(2m^2 (mewk) g ) [/itex]

    does anyone know where I went wrong?

    Looking at my answer a_x must = (mewk) g but I'm not sure how to figure this out?
     
  2. jcsd
  3. Apr 24, 2017 #2

    haruspex

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    Indeed it is.
    What equations do you know relating coefficient of friction to actual frictional force?
     
  4. Apr 24, 2017 #3

    Would I use f=ma?

    So I have ma going to the left, and then Fn(mewk) going to the right?
     
  5. Apr 24, 2017 #4

    haruspex

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    Yes.
    You have not said what you consider as left and right in the description. Since the acceleration is a result of Fnμk, they must surely point the same way.

    Please use the X2 and X2 buttons for subscripts and superscripts, and the Σ button for Greek letters etc.
     
  6. Apr 25, 2017 #5
    Ok so I have Fg = mg
    pointing to the bottom

    Fn = mg = Fg

    Fn pointing to the bottom

    Since he is going backwards from the force of throwing the ball, I set unknown force [itex] f = ma [/itex] going to the left, against the throw of the ball.

    Then I set force Fk = mg(mewk) = Fn(mewk)

    So force of friction opposing the movement to the left. (Trying to make it stay right)

    This is how it makes sense to me, is there something wrong with this thinking??

    Btw I think I get the subscript now but I still don't know how to do "mewk." sorry. I looked under info and I couldn't find how..
     
  7. Apr 25, 2017 #6

    haruspex

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    The ball has been thrown. That gave the initial speed (away from the ball). It takes no further part in the the skater's movement.
    (Of course, the sliding, and hence the friction, starts even before the skater has released the ball, but if the throw is quick this has little effect.)
    If the ball went to the right, the skater is moving left, so which way will friction act?
    Yes.
    There are two ways.

    If you want to use LaTeX, it's underscore for subscript (if more than one character, wrap in {}); for mu it's \mu:
    ##\mu_k##
    If you right-click on that you will get a pulldown from which you can select "View math as latex". Doesn't work on iPads though.

    Alternatively, do not use LaTeX, instead using the editing buttons. If you click on the Σ button you should see a menu of special characters appear under the typing panel.

    Do not put any of those special characters inside the LaTeX.
     
  8. Apr 25, 2017 #7
    Ohhhh I see now I think! acceleration of him from the initial push the the left doesn't exist so it would make no sense making my diagram with that, right?!

    One more question though: Why isn't it negative then since it is going in the negative direction?

    also sweet I see what you mean about the Σ button now haha cool I didnt know of this!
     
  9. Apr 25, 2017 #8

    haruspex

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    You have not defined the positive and negative directions.
    If the ball was thrown in the positive direction, the skater's velocity is negative. Friction opposes relative motion of surfaces in contact, so that force will be positive on the skater.
     
  10. Apr 25, 2017 #9
    Thanks I think I understand this problem now!
     
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