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Distance Block Slides Up Ramp

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A block of mass 2.96 kg is sliding up a ramp with an initial speed of 12.8 m/s. The ramp is inclined from the horizontal at an angle of 36.4 degrees and the coefficient of kinetic friction is [itex]\mu[/itex]=045. What is the displacement of the block along the ramp from the initial time until it stops?


    2. Relevant equations

    W=ΔE
    W=∫F*ds
    ΔE=(m/2)(vf)2-(m/2)(v0)2 where (vf)=0
    [itex]\Phi[/itex]=36.4
    m=2.96kg
    v0=12.8 m/s
    [itex]\mu[/itex]= 0.45

    3. The attempt at a solution

    [itex]\mu[/itex]*m*g*cos[itex]\Phi[/itex]x=-(m/2)(v0)2
    solve for x
    x=-(m/2)(v0)2/[itex]\mu[/itex]*m*g*cos[itex]\Phi[/itex]
     
  2. jcsd
  3. Apr 17, 2012 #2

    tiny-tim

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    hi getty102! :smile:

    correct method, but you forgot to include the gravitational PE (or gravitational work done) :wink:
     
  4. Apr 17, 2012 #3
    Cool, thank you. The potential energy from gravity is dependent on the height. Which in this problem is kind of what I'm looking for because if I can find the height I would know the total displacement of the block up the ramp.
     
  5. Apr 17, 2012 #4

    tiny-tim

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    xsinΦ ? :wink:
     
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