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    aharmel0340's Avatar
    aharmel0340 Posts: 3, Reputation: 1
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    #1

    Oct 14, 2010, 05:38 PM
    how do you find the net force when the velocity is not constance meaning that acceler
    how do you find the net force when the velocity is not constance meaning that acceleration is not zero

    a= .3 m/s2 m=15kg
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
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    #2

    Oct 14, 2010, 06:45 PM

    Force is equal to mass times acceleration. F=ma. So if this is all that it gives you then the object has a force of .3*15 which is 4.5N. Your question is strange because you can only have a force when the velocity is not constant or in other words, when there is an acceleration.
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    aharmel0340 Posts: 3, Reputation: 1
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    #3

    Oct 14, 2010, 06:58 PM
    Comment on kpg0001's post
    15kg box is released on a 32˚ incline and accelerates down the incline at .3 m/s2 find the friction force impeding its motion. What is the coefficient of kinetic friction?
    aharmel0340's Avatar
    aharmel0340 Posts: 3, Reputation: 1
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    #4

    Oct 14, 2010, 06:59 PM
    Comment on kpg0001's post
    There is an acceleration! 3 m/s2
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    kpg0001 Posts: 88, Reputation: 12
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    #5

    Oct 14, 2010, 08:27 PM

    I knew there was more to it. Kinetic friction equals the normal force times the coefficient.(f=un). So first find the normal force. Since it is on a slope of 32, the force of gravity is not acting on it straight down, but at mgcos32=n=122.6. The force of friction is in the x direction. F(x)=mgsin32-friction. From the info given, the 15kg block is accelerating at .3m/s^2 which means the F(x) is equal to 4.5N. From this you have 4.5=81.1-friction. Friction=76.6. So then you go back to friction=normal(coefficient of friction), plug in the numbers, 76.6=122.6(u), u=.625(THESE CALCULATIONS WERE DONE IN RADIANS NOT DEGREES I'M SORRY FOR THE CONFUSION. THE THEORY IS CORRECT BUT THE NUMBERS ARE NOT. I POSTED THE STEPS TO SOLVE THIS TWO POSTS UP!)
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    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Oct 15, 2010, 06:53 AM

    I wonder you are taking the value of g as how much... I'm using 10 m/s^2 and 9.8 m/s^2 and am getting values a little different from yours... =/
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    kpg0001 Posts: 88, Reputation: 12
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    #7

    Oct 15, 2010, 07:46 AM

    Well lets look at the calculations without values. I wish I knew how to put all the correct symbols in! Fk=kinetic friction u=coefficent of kinetic friction n=normal force Fx=forces in the x direction Fy=forces in the y direction and a=angle. Fk=u(n). Fy=n=mgcos(a), Fx=mgsin(a)-Fk. It gives you the value of Fx because it tells you the objects mass and acceleration in the x direction. So solve for Fk. Fk=mgsin(a)-Fx. Then solve for u. u=Fk/n.
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    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Oct 15, 2010, 07:51 AM

    Try having a look at the mimetex tutorial for LaTeX for the forum :)

    https://www.askmehelpdesk.com/math-s...las-50415.html
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
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    #9

    Oct 15, 2010, 07:59 AM

    Oh thanks that's helpful! And thanks for checking my work, though it is a bit embarrassing to make such a simple mistake. I keep having to change it from radians to degrees for different classes.

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