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Math Help - how fast is the skier going, using kinetic friction and length.

  1. #1
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    how fast is the skier going, using kinetic friction and length.

    A skier slides horizontally along the snow for a distance of 27 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is ľk = 0.050 Initially, how fast was the skier going?
    m/s

    not sure how to reach the answer with so few variables involved.

    equations i thought i might use were: vf = sqrt((2 * (KE) / m)

    and KE = 1/2*(m * v^2)

    thanks for any extra help.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    A skier slides horizontally along the snow for a distance of 27 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is ľk = 0.050 Initially, how fast was the skier going?
    m/s

    not sure how to reach the answer with so few variables involved.

    equations i thought i might use were: vf = sqrt((2 * (KE) / m)

    and KE = 1/2*(m * v^2)

    thanks for any extra help.
    Both of these questions are the same, just in a different form.

    Consider the following equation:
    W_{nc} = W_{friction} = \Delta KE

    -Dan
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    i've tried using both of these equations to reach the answer which is: 5.14 m/s, however, i can't seem to use the formulas correctly.

    what do you mean by chng in KE?

    please break this down for me.

    thankyou.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    i've tried using both of these equations to reach the answer which is: 5.14 m/s, however, i can't seem to use the formulas correctly.

    what do you mean by chng in KE?

    please break this down for me.

    thankyou.
    Quote Originally Posted by topsquark View Post
    Both of these questions are the same, just in a different form.

    Consider the following equation:
    W_{nc} = W_{friction} = \Delta KE

    -Dan
    The work done by friction is
    W_{friction} = \vec{f_k} \cdot \vec{s}

    Since the friction force is in the opposite direction to the displacement we have
    W_{friction} = \vec{f_k} \cdot \vec{s} = -f_ks

    Now f_k = \mu_k N
    So what's the normal force? Again we solve this using a Free-Body Diagram. In this case the answer is simple: N = w = mg, thus
    W_{friction} = -f_ks = -\mu_k mgs

    Now, the skier starts with a speed v_0. At the end, the skier is at rest, so v = 0 m/s. Thus
    \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2

    So finally we have:
    W_{friction} = \Delta KE

    -\mu_k mgs = -\frac{1}{2}mv_0^2
    which you can solve for v_0.

    -Dan
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