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  1. #1
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    physics help...

    "A hockey puck on a frozen pond with an initial speed of 16.3 m/s stops after sliding a distance of 220.5 m. Calculate the average value of the coefficient of kinetic friction between the puck and ice."

    -i thought i knew what to do, it seemed simple enough. i found the deceleration, speed divided by time (time=distance divided by speed). And then it should be deceleration divided by gravity?(f=uN) i get the wrong answer so obviously not. am i doing something wrong??
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  2. #2
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    Once it starts sliding, the only force acting on the puck is friction so:
    F = F_{fr} \ \iff \ m\vec{a} = \mu F_{N} where F_{N} is the normal force

    So first find the deceleration of the puck: \vec{v}^{\ 2}_{f}= \vec{v}^{\ 2}_{i} + 2\vec{a}\vec{d} - You have the final velocity (it's stops!), the initial velocity, and its distance.

    So after solving for \vec{a}, use the equation from the first line:
    \begin{array}{rcl} m\vec{a} & = & \mu F_{N} \\ m\vec{a} & = & \mu (mg) \end{array}
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by sarabolha View Post
    "A hockey puck on a frozen pond with an initial speed of 16.3 m/s stops after sliding a distance of 220.5 m. Calculate the average value of the coefficient of kinetic friction between the puck and ice."

    -i thought i knew what to do, it seemed simple enough. i found the deceleration, speed divided by time (time=distance divided by speed). And then it should be deceleration divided by gravity?(f=uN) i get the wrong answer so obviously not. am i doing something wrong??
    You need to find \mu_k

    If you draw a free body diagram, you should get something like this:



    To find acceleration, use the kinematic equation v_f^2=v_i^2+2a(x_f-x_i)\implies 0=(16.3)^2+2a(220.5)\implies a=-.6~\tfrac{m}{s^2}

    Now let's do a FB analysis:

    \sum F_x=ma:

    Thus, -F_{fr}=ma

    \sum F_y=0:

    Thus, N-mg=0\implies N=mg

    Therefore, -F_{fr}=ma\implies -\mu_kN=ma\implies -\mu_kmg=ma\implies \mu_k=-\frac{a}{g} \implies\mu_k=-\frac{-.6~\frac{m}{s^2}}{9.8~\frac{m}{s^2}}\implies \color{red}\boxed{\mu_k\approx 0.06}

    Does this make sense?

    --Chris
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    ah that was too easy i'm embarrassed. but you explained it very very well, thanks!!
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