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Work, energy with kinetic friction

  1. Nov 2, 2004 #1
    Can anyone provide some assistance? I know this comes in two parts, one in locating the velocity, which I belive comes down to v=sq root of 2(9.8)(6.34), however the second part is creating havoc. Any suggestions on part II's formula?

    A box slides down a frictionless 6.34 m high hill, starting from rest. At the bottom of the hill, the box slides along a level surface where the coefficient of kinetic friction uk = 0.246. How far from the bottom of the hill does the box come to rest? The final answer will be 25.8m
     
  2. jcsd
  3. Nov 2, 2004 #2

    Pyrrhus

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    Homework Helper

    The First part

    [tex] mgh = \frac{1}{2}mv^2 [/tex]

    The Second Part:

    1)

    Use the kinetic constant acceleration formula

    [tex] v^2 = v_{o}^2 + 2a(x-x_{o}) [/tex]

    and Newton's 2nd Law on the box where there's friction to find the acceleration

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    2)

    Use

    [tex] \Delta E = W_{f} [/tex]

    [tex] \frac{1}{2}mv^2 = \mu mgd [/tex]
     
    Last edited: Nov 2, 2004
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