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Acceleration on a slant/coefficient of friction

  1. Feb 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A block is given an initial speed of 6.00 m/s as it slides up an inclined plane that makes an angle of 30 degrees with the horizontal. If the coefficient of kinetic friction is .330, how far up the incline will the block go?

    2. Relevant equations

    I messed around with the equation delta_x = v_initial + .5a*t^2 , not sure how the coefficient of friction works

    3. The attempt at a solution

    I tried setting a = - 11.3 m/s/s by using trig because of the slant on the horizon, then using the above distance formula to find the maximum distance it traveled. I tried multiplying the coefficient of friction into the acceleration part of the equation, but to no avail. Please help me figure out what I'm doing wrong. Maybe an explanation of coefficient of friction is all I really need, I don't know.
     
  2. jcsd
  3. Feb 5, 2008 #2

    mezarashi

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    Homework Helper

    The coefficient of friction defines the relation between the normal force (exerted perpendicular to the contact surface) and the frictional force (which is parallel to the surface).

    Ffrict = u x Fnormal

    On a ramp, the normal force is no longer equal to simply mg, which is where the angles come in. Give it another try using these principles.
     
  4. Feb 5, 2008 #3
    Okay, so I looked at it again, and this is what I got.

    I found the new acceleration by g(u)cos(30) . I ended up getting a = -2.80 m/s/s.

    From there, I plugged it into the distance formula above, leaving me with:

    delta_x = t(6.00) + .5(-2.80)t^2

    I took the derivative with respect to t to find the t that would give me the max distance.
    I got this.

    0 = 6.00 + (-2.80)t
    t = 2.14 s

    I plug that back in to the above equation to get delta_x.

    I ended up getting delta_x = 6.42 m, which is nowhere near any of the answer choices.

    Am I even on the right track?
     
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