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Kinetic Energy Problem!

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    In the figure below, a block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0; the height difference is h and the coefficient of kinetic friction is μk. Find d in terms of the given variables (use g where applicable).

    2. Relevant equations
    F=ma
    U=mgh
    K=(1/2)mv^2



    IMAGE:
    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c08/pict_8_53.gif
    3. The attempt at a solution
    Idk where to start because it is all in terms of variables!
     
  2. jcsd
  3. Mar 9, 2010 #2

    PhanthomJay

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    You can solve it a number of ways. Try conservation of total energy. The letter variables are harder to work with than actual numbers, but you just have to tough it out and grind out the solution, once you have the proper equations.
     
  4. Mar 9, 2010 #3
    Ok that would make sense. Could you help me set up the right side of the equation?

    I know that the block has an initial U energy of 0 and an initial K energy of 1/2mv^2. I don't know how to set up an equation for the block at the end of the ramp to find d. The only thing I know it that the m must cancel because it is not a given variable

    (thanks.)
     
  5. Mar 9, 2010 #4
    Let me help you find a solution to your problem, maybe if some of it is done you can figure the rest yourself.

    We notice that we have a potential; you've denoted this potential as [itex]U=mgh[/itex], but i've never seen this notation so we will use the standard notation:

    [itex]GPE=mgh[/itex] where [itex]GPE[/itex] is the gravitational potential energy; and i assume you know intimately what that is. Let's focus now on the height. This can be calculate by using the equation:

    [itex]\frac{-h}{2}=\frac{v_f - v_0}{-2g}[/itex]

    This means that whatever you're object is, it is [itex]h[/itex] units off the ground. We simply calcuate the kinetic energy of systems like this by the relationships:

    [itex]KE=\frac{1}{2}m(gh)[/itex]

    Since the friction is not constant, the sum of the kinetic energy and the potential energy is not a constant. Now use your friction equation and all the proper details, and you can work out the differences. Plus, you never provided any values for the distances involved, or the strength of the friction.
     
  6. Mar 9, 2010 #5

    PhanthomJay

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    Write out the total conservation of energy equation.. at the start point and the end point....are you familiar with it....you know the one, the equation that relates initial and final kinetic and potential energies, when work is done by non conservative forces like friction......
     
  7. Mar 9, 2010 #6
    Unfortunately, no numbers are provided in this problem. The only information given is what is written out above.

    and in general, yes. I know that the potential energy + kinetic energy of a system is constant as long as there aren't any nonconservative forces acting upon the object

    SumForces= mA
    mA= -F-fk > if there IS friction

    I'm honestly still pretty lost. I understand it is probably a really simple process but I for some odd reason am just NOT seeing it. I did all of the other problems in this section fine! Of course THEY had numbers haha.
     
  8. Mar 9, 2010 #7
    Thanks for all the Help! I called one of my friends and he is going to explain it to me. Hopefully it will "click"
    We shall see. =]
     
  9. Mar 9, 2010 #8

    PhanthomJay

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    Yes, I'm a numbers man myself, but so be it. You have stated (more or less ) the conserv of energy equation W_nc =delta K + delta U. You don't have numbers, but in terms of letter variables, you know delta U and delta K (what's the speed of the object when it stops??), so thus you can solve for the work done by friction. Find the friction force and solve for d, using the def of work which you didn't list in your relevant equations.. Don't let the letters frighten you.:surprised
     
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