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Homework Help: What is the coefficient of kinetic friction for the incline?

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data
    A block(7 kg) is released from rest on an inclined plane of 23 degrees and moves 2.8 m during the next 4.7 s. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the acceleration of the block? What is the coefficient of kinetic friction for the incline?

    2. Relevant equations

    3. The attempt at a solution
    I already calculated the acceleration of the block which is 0.2535 m/s^2.


    Fx=ma=Ff-mgsin 23
    7(0.2535) = 63.14663295 * (coefficient) - 26.80415541
    coefficient = 0.452576077

    Can you check my calculation please?
    Last edited: Sep 22, 2011
  2. jcsd
  3. Sep 22, 2011 #2


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    I'm assuming that somewhere you were given a mass for the block of 7 kg., which is not mentioned in the problem statement above. However, you could ignore it entirely, since you only want to work with the acceleration along the incline, so you can divide out m early on. Anyway...

    You've attached the coefficient of kinetic friction to the wrong term in the second line: Ff should be [itex]\mu \cdot mg \cos 23º [/itex].

    One other thing: as you've written the equation for the net force along the incline, the left-hand side is "positive downhill", but the right-hand side is "positive uphill", since you've made the friction force positive and the "downhill" component of the weight force negative.
  4. Sep 22, 2011 #3
    ok i fixed everything. Could you check it please? My online hw keeps say it is wrong
  5. Sep 22, 2011 #4


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    Not quite: are you calling "uphill" or "downhill" the positive direction? The signs on the two sides of your force equation still aren't following the same choice. (With the block sliding downhill, which way does the friction force act?)
  6. Sep 23, 2011 #5
    uphill is the positive direction
  7. Sep 23, 2011 #6


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    Then, since the block is sliding downhill, its net acceleration is negative. You need to change the sign of the left-hand side of your equation now.
  8. Sep 23, 2011 #7
    hmmm... negative.... then acceleration is for x-component no?
  9. Sep 23, 2011 #8


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    What you have written,

    [tex]F_{x} = ma = F_{f} - mg \sin (23º) , [/tex]

    indicates that you are placing the x-axis parallel to the slope of the inclined plane already. That puts the "normal direction", which is perpendicular to the inclined plane, parallel to the y-axis, which is why you can write

    [tex]F_{n} = mg \cos (23º) . [/tex]

    So the friction force is [itex]F_{f} = \mu \cdot F_{n} = \mu \cdot mg \cos (23º) .[/itex] The order in which you wrote the forces for Fx indicates that you have already chosen "uphill" along the incline to be positive. But the block is going to slide "downhill", so ax will be negative. (ay is zero, since the block will not accelerate perpendicularly to the inclined plane; in fact, this was used to come up with Fn - mg cos (23º) = 0 , the second term being the component of the block's weight in the normal direction.

    So, yes, your equation should read

    [tex]ma_{x} = F_{f} - mg \sin (23º) , [/tex]

    with ax < 0 .

    You have to be careful, once you have chosen the directions for the coordinate axes, to be consistent with them in giving the signs to forces, accelerations, velocities, etc. Placing the wrong sign on a term will change the meaning of the force equation and lead to an incorrect solution. (The exception is if all the signs in the entire equation are reversed: this corresponds to pointing that coordinate axis in the opposite direction and will change the sign of the solution, but not the meaning. That is, if positive-x points "downhill" and the acceleration ax is positive, then the object is accelerating downhill; if positive-x points "uphill", then ax will come out negative, but that still representing downhill acceleration.)
    Last edited: Sep 23, 2011
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