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Math Help - Friction

  1. #1
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    Friction

    A 2500 Kg car is heading east at 20 m/s. IT slams on the breaks to avoid hitting a deer. It takes the car 50 meters to stop, what is the Uk between the car and the road.
    __

    Well I've been absent for school for two days, which is coincidentally when he was teaching this stuff.

    So I'm not even exactly sure what Uk is.

    I assume I'd use V^2=V0^2+2da ?
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  2. #2
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    what is the Uk? the distance? the acceleration or deceleration?
    so you want the friction? oh you want to find out muon not U. O.o
    Last edited by BabyMilo; November 11th 2009 at 02:08 PM.
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  3. #3
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    I assume you want Friction

    yes, you use v^2=u^2+2ax

    where,

    * v=0m/s
    * u=20m/s
    * a=?
    * x=50m

    sub the numbers in

    0^2=20^2+2*a*50

    rearrange the equation to find a.

    since Newton's second law defines F=ma where F= force of overall system not friction

    using the equation

     2500g-F=-2500a where F = friction

    then rearrange the equation to find F.

    you should get  a=-4ms^{-2}
    and  F=14500N

    where g=9.8ms^{-2}

    To find out the reaction force, using the y directional force:
    -2500g+R=0

    Then F=\mu*R

    rearrange to find \mu
    Last edited by BabyMilo; November 11th 2009 at 02:18 PM.
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  4. #4
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    Quote Originally Posted by Caturdayz View Post
    A 2500 Kg car is heading east at 20 m/s. IT slams on the breaks to avoid hitting a deer. It takes the car 50 meters to stop, what is the Uk between the car and the road.
    __

    Well I've been absent for school for two days, which is coincidentally when he was teaching this stuff.

    So I'm not even exactly sure what Uk is.

    I assume I'd use V^2=V0^2+2da ?
    \mu_k is the coefficient of kinetic friction.

    \mu_k = \frac{f_k}{N} , where f_k is the force of sliding friction, and N is the normal force between the surfaces in contact.

    using the equation v_f^2 = v_0^2 + 2a(\Delta x) , you can determine the car's acceleration ... then use Newton's 2nd Law of motion,
    F_{net} = ma , substituting f_k for the net force ... then you can find \mu_k from the equation above.
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