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Please help, cannot find Frictional Force

  1. Nov 28, 2008 #1
    I am struggling a little bit on this problem. Any help is appreciated.

    1. The problem statement, all variables and given/known data
    A 5.60 kg block is set into motion up an inclined plane with an initial speed of v0 = 8.10 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.

    The first question was to find the change in kinetic energy, which ZI found to be -183.708
    The second question was to find the change in potential energy, which I found to be 82.404
    The third question is determine the frictional force exerted on the block(assumed to be constant).
    I can't seem to figure out the third question.
    After finding the frictional force, it asks for the coefficient of kinetic friction, which I'm sure i can find by dividing the frictional force by the normal force.


    2. Relevant equations

    Ff = μFn
    where Ff is the frictional force, μ is the coeifficient of kinetic friction, and Fn is the normal force.

    F=ma
    F is the force on the mass, m is the mass, a is the acceleration

    v2 = vo2 + 2a(Δx)
    v is final velocity, vo is initial velocity, a is acceleration, Δx is the displacement

    Acceleration I found to be -10.935
    Normal force = 47.57592
    3. The attempt at a solution

    I have found acceleration to be -10.935. So I found the total force to be -61.236. I have tried to solve for the frictional force by setting the total force equal to the frictional force + gravitational force and solved for frictional. I have also made other attempts but they were very random and I dont recall what made me come up with them. I have submitted 5 incorrect answeres so far.

    I tried to upload a picture of the slope and block, I'm not sure if it worked.
    Thanks in advance for any help.
     

    Attached Files:

  2. jcsd
  3. Nov 29, 2008 #2
    You need resolve your forces into components. It may be wise to choose the direction of the inclined plane to be the x-axis and thus the direction that is perpendicular to it to be the y-axis.

    Now you can say that [itex]\sum F_y=ma_Y[/itex] and[itex]\sum F_x=ma_x[/itex]
     
  4. Nov 29, 2008 #3
    Thank you for your prompt response. I'm am having a hard time understanding your help.

    I should set the direction of the incline to the x-axis. So wouldn't that make the [itex]\sum F_y=ma_Y[/itex] equal to zero? Then the [itex]\sum F_x=ma_X[/itex] would be the same becuase my acceleration is th same?

    I dont think im quite understanding what your telling me.
     
  5. Nov 29, 2008 #4
    If the direction of the plane is the x direction, then clearly the acceleration cannot be zero since the block is initially moving and then comes to rest. So why should the sum of the forces in that direction be zero?

    Looking at a Free body diagram (which I presume you have already drawn:smile: ) you should see that there are are two forces acting on the block in the direction of the plane.

    Can you identify those two forces?
     
  6. Nov 29, 2008 #5
    Sorry, what I meant by the x acceleration being the same was that I meant it would be the same as I have already calculated to be -10.935.

    The two forces acting parallel to the direction are the gravitational force of 9.81*5.6*Tan(30) and the Frictional force. I think.
     
  7. Nov 29, 2008 #6
    Tan? I do not think it shall be tan 30? Perhaps sin 30?
     
  8. Nov 29, 2008 #7
    The force parallel to the x axis( the incline) should be mgsin(30). When splitting vectors into components you use sin and cos. Also since you know the initial energy and the final energy you can make a comparison. Was energy conserved? Obviously not as friction is a non conservative force. how much energy was lost to friction over the 3.0m or how much work did the frictional force due? How is the coefficient related to your knowns?
     
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