1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    The First part

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

    The Second Part:


    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]



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

    [tex] \frac{1}{2}mv^2 = \mu mgd [/tex]
    Last edited: Nov 2, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook