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Coefficient of Kinetic Friction

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A 5.20 kg block is set into motion up an inclined plane with an initial speed of v0 = 7.70 m/s. The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal.

    (a) For this motion, determine the change in the block's kinetic energy.

    I got -154.154 (correct)

    (b) For this motion, determine the change in potential energy of the block-Earth system.

    I got 75.6 (correct)

    (c) Determine the frictional force exerted on the block (assumed to be constant).

    I got 26.18 (correct)

    (d) What is the coefficient of kinetic friction?

    However, I could not figure this out for the life of me. Please help with the kinetic friction?

    2. Relevant equations

    I think the coeff. can be found with f=μmgCosθ , however my answer keeps coming out wrong...

    3. The attempt at a solution

    after using the previous formula, I came out with a value of μ=3.327 (26.18=u(5.2)(9.81)(cos(30))
     
  2. jcsd
  3. Mar 16, 2009 #2
    is the answer .59?
     
  4. Mar 16, 2009 #3
    I dont understand how you get 3.327... coefficiant of friction isnt going to be more than 1

    if you calculate what you have

    (26.18=u(5.2)(9.81)(cos(30))

    u get

    26.18=44.13u

    divide by 44.13
    =.59
     
  5. Mar 16, 2009 #4
    oooooo haha, thank you sir, my calculator somehow reverted itself back to radians?

    Thanks again!
     
  6. Mar 16, 2009 #5
    no prob. you did all the hard work haha
     
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