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Coefficient of friction for surface

  1. Dec 22, 2008 #1
    a) A block weighing 9.1N requires a force of 2.8N to push it along at constant velocity. What is the coefficient of friction for the surface?
    b) A weight "W" is now place on the the block and 8N is needed to push them both at constant velocity. What is weight "W" on the block?

    This is what I got:
    Since it is constant velocity, there is no acceleration, therefore Fnet is 0, therefore Ff is 2.8N.
    Normal force is (9.1)(9.8) =89.18N
    Formula: Ff=uN

    Therefore u= 0.031
    , but that's not the right answer... Why? what did I do wrong?

    I take it I need to know a before I can do b?
  2. jcsd
  3. Dec 22, 2008 #2


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    Why did you multiply the weight of 9.1 N by 9.8 to get the normal force?
  4. Dec 22, 2008 #3
    Hm? Don't you need to?
    Fg=mg so Fg=(9.1)(9.8), and normal force is going the other direction and Fnet=0, so normal force is also 89.18.

    Isn't that right?
  5. Dec 22, 2008 #4


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    The weight would equal the mass times g, but you weren't given the mass, you were given the weight which is equal to the normal force. Look at your units: N is a unit of force, if you were given mass in kg and multiplied that by g which has units [m/s^2], you would end up with Newtons (again, a force). But what is 1N * 1m/s^2?
  6. Dec 22, 2008 #5
    ohhh. I see what I did wrong, thanks!! You're great help!
    I can't believe my mistake!
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