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Kinetic energy of a block sliding down a ramp

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data

    You observe that after sliding down a distance of 2.61 m along a 5 m long track that is inclined at 20 degrees to the horizontal, a block of wood has a velocity of 0.13 m/s. If the coefficient of kinetic friction between the wood and the surface of the track is 0.40, what was the block's initial velocity at the beginning of the 2.61 m distance. [Please use g=9.81m/s^2 and give your answer with 2 decimal places in the SI system of units.]

    2. Relevant equations

    See attached document.

    3. The attempt at a solution

    See attached document for work done so far.

    Note: I forgot to include what L (length of ramp) was in my original drawing of the diagram in word so it is towards the bottom of the diagram for when I solve for forces.

    I'm having trouble with finding mass. My professor likes students to answer by showing a step-by-step process in solving the problem, hence all the algebra on the attached document. I was unable to find a way to solve for mass from what I have done so far. I'm not sure if I'm missing an equation that should be there or a force in the diagram that I missed. Otherwise, I have to solve for the initial velocity. I also had trouble lining up what each force was for determining kinetic energy for the system. As the block slides down the ramp, I know that energy outputs are due to friction and gravity. I also know the equation for kinetic energy is:

    KE = 1/2mv^2

    I wondered if this equation could be applied twice for this problem but wasn't for sure. Any input is appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 27, 2013 #2

    CWatters

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    I've only briefly scanned your solution but this is the way I would do it..

    Write an equation for the net force down the slope. That will accelerate the block according to Newtons f=ma

    Should have something like...

    ma = (component due to weigh) - (friction)

    ma = mgsin(20)-μmgcos(20)

    m cancels leaving an equation for a.

    Then apply one of the standard equations of motion and solve for initial velocity. ..

    V^2 = U^2 +2as
     
  4. Feb 27, 2013 #3
    I think I'm confused by what you mean of using one of the standard equations of motion for this particular problem...=/

    From the way I did it, in setting it up all using algebra in steps and finding my unknown variables, I kept getting lost in what to do with N. I know that realistically, we probably don't to care about N since it's not along the same axis as the object is moving. But, I didn't understand what to do with in terms of setting it up for finding my unknowns. I didn't think it'd be correct to just ignore it. Would it just be equal to wx if it's on the same axis and equal to zero?

    Then for the axis on which the object is moving, would it be setup to ma to show that is is in motion and then apply it from there to the conservation of energy equation?
     
  5. Feb 27, 2013 #4

    haruspex

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    On the LHS of the PDF, looks like you're taking x as normal to the incline. Unusual, but ok.
    The normal force in the diagram is going in rather a strange direction.
    You wrote ΣFy: f-Wy= 0, but that is not so. There is acceleration down the plane.
    On the RHS you seem to have switched coordinates, now writing W cos θ = Wy, where previously you had Sin θ = Wy/W.
    Pls resolve these inconsistencies and post your new working.
     
  6. Feb 28, 2013 #5

    CWatters

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    Well the problem asks you to find an initial velocity given a final velocity, the distance travelled and the acceleration. I don't see why one of the standard equations for constant acceleration can't be applied..

    http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

    The most obvious one is V2 = U2 +2as


    N is the Normal force and it's used to calculate the Friction force Ff which is acting parallel to the direction of motion.

    Ff = μN ........... comes from the definition of μ

    N=mgCos(20)
    so
    Ff = μmgCos(20)
     
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