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Block of ice sliding down inclined plane - final speed

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A block of ice with mass 2.00kg slides 0.70m down an inclined plane that slopes downward at an angle of 30degrees below the horizontal. If the block of ice starts from rest, what is its final speed?

    2. Relevant equations
    W = 1/2(m)(v2)^2 - 1/2(m)(v1)^2

    3. The attempt at a solution
    W = 1/2(2.00kg)(v2)^2 - 1/2(2.00kg)(0)^2

    Fd = 1/2(2.00kg)(v2)^2

    wsin(theta) is the only force horizontaly:
    (wsin(theta))(0.70m) = 1/2(2.00kg)(v2)^2
    2 * ((2.00kg)(9.80m/s^2))(0.70)) = 2.00kg (v2)^2

    v2 = 3.70m/s
  2. jcsd
  3. Aug 26, 2007 #2


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    Homework Helper

    Ignore the forces and look at the energy.
    Work out the vertical distance from the length and the angle
    You can then work out the potential energy lost PE = mgh
    This is equal to the kinetic energy at the end KE=1/2 mv^2
  4. Aug 26, 2007 #3
    but this lesson hasnt reach PE yet... this is still on work & kinetic energy
    thanks btw
  5. Aug 26, 2007 #4


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    You forgot the sin(30), in your final line of calculations. It should be:

    2 * ((2.00kg)(9.80m/s^2)sin(30))(0.70)) = 2.00kg (v2)^2
    Last edited: Aug 26, 2007
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