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Kinetic friction Homework Problem

  1. Nov 1, 2008 #1
    Hi there, I need help with understanding this guy's solution thanks!

    The kinetic friction force will be up the slide to oppose the motion.
    We choose the positive direction in the direction of the acceleration.
    From the force diagram for the child, we have ?F = ma:
    x-component: mg sin [tex]\theta[/tex] – Ffr = ma;
    y-component: FN – mg cos [tex]\theta[/tex] = 0.
    When we combine these, we get
    a = g sin [tex]\theta[/tex] – [tex]\mu[/tex]kg cos [tex]\theta[/tex] = g(sin [tex]\theta[/tex] – [tex]\mu[/tex]k cos [tex]\theta[/tex]).
    We can use this for the frictionless slide if we set [tex]theta[/tex]k = 0.
    For the motion of the child, we have
    v2 = v02 + 2a(x – x0) = 0 + 2ad, where d is the distance along the slide.
    If we form the ratio for the two slides, we get
    (vfriction/vnone)2 = afriction/anone = (sin [tex]\theta[/tex] – [tex]\mu[/tex]k cos [tex]\theta[/tex])/sin [tex]\theta[/tex];
    (!)2 = (sin 28° – [tex]\mu[/tex]k cos 28°)/sin 28°, which gives [tex]\mu[/tex]k = 0.40.
     
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Nov 1, 2008 #2
    Re: Homework Problem

    The question is a child slides down a slide with a 28 degree incline and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of the kinetic friction between the slide and the child.
     
  4. Nov 1, 2008 #3
    Re: Homework Problem

    anybody?
     
  5. Nov 1, 2008 #4

    djeitnstine

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    Gold Member

    Re: Homework Problem

    I analyzed this question and came up with something quite different. You should ask him why a = [tex]g sin \theta - \mu kg cos \theta [/tex] is true. Because I cannot see how the normal force has any role to play in the x direction. As far as I know [tex]F _{N}[/tex] is only related to static friction.
     
  6. Nov 1, 2008 #5
    Re: Homework Problem

    r u in gr.2 physics. your ignorant
     
  7. Nov 2, 2008 #6
    Re: Homework Problem

    [tex] a = gsin(\theta) - \mu kgcos(\theta) [/tex] is correct.

    Which parts do you need help with understanding? Or do you not understand any of it?

    First sums up all the forces acting on the child, then finds that the net force to allow the child to move is in the direction parallel down the slide.

    He re-writes his equation for acceleration by dividing the RHS by mass. (F = ma).

    recognising that the coefficient of kinetic friction is going to be a linear value, he takes a ratio of velocities of the child with friction over the velocity of child without friction (compensating for the factor of two as specified in the question) and lets that equal a ratio of acceleration with friction and without friction.

    Then solves for the coefficient
     
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