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Homework Help: Finding the magnitude of the friction force.

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    An 8kg ball moves down a 12m high inclined plane with a speed of 6m/s. By the time it reaches the bottom, the speed is measured to be 12m/s. What is the frictional force opposing the motion? The inclined plane has an angle of 30 degrees with respect to the ground.

    2. Relevant equations

    Potential energy = mgh
    Kinetic energy = 1/2mv^2
    Work done by external forces = ΔE
    Work done: Fs (or Fscos(theta))

    3. The attempt at a solution

    The first step is to calculate the potential energy and kinetic energy, which is 960J and 576J respectively. Hence ΔE=960-576=384J.

    Now, since the height is 12m and the angle is 30 degrees, the length of the inclined plane is 12/sin30 or 24m.

    Finally, 24f=384 which yields an answer for f as 16N.

    I might have gotten the correct answer, but in the book it says that the answer is supposed to be 22N, so I'm assuming I'm missing something, perhaps it has something to do with the initial speed which I haven't included in my calculations.

    Thanks in advance,
  2. jcsd
  3. Sep 22, 2012 #2
    I think you forget initial KE.
  4. Sep 22, 2012 #3
    Please elaborate. I assumed it was 0.
  5. Sep 22, 2012 #4
    I'd very much like help on this one!
  6. Sep 22, 2012 #5


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    A bit unclear on the wording, but the initial speed is 6 m/s at the top of the ramp. Watch your signage.
  7. Sep 23, 2012 #6
    Not sure If I follow what you're saying. Could you elaborate?
  8. Sep 23, 2012 #7


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    l think the problem meant to say that the ball is projected down the ramp with an initial speed of 6 m/s, that is, it already had a speed of 6 m/s when it reached the top of the ramp. So re-do your calcs, and see azizlwl post 2.
  9. Sep 23, 2012 #8
    It doesn't travel up the plane, it travels down the plane. I.e. it has an initial speed of 6m/s when it is at the top (the beginning) and a speed of 12m/s at the bottom (the end). Hence, initial KE=0?

    EDIT: Oh! Now I see what you mean. It didn't go from rest to 12m/s meaning it had some kinetic energy at the top as well. In other words, initial KE=144J and hence, initial (total) energy= 144+960=1104J.

    Using that, I get an answer of 22N. Thank you for your help!
    Last edited: Sep 23, 2012
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