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Work and kinetic energy

  1. Oct 31, 2005 #1
    A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of below the horizontal.

    If the block of ice starts from rest, what is its final speed? You can ignore friction.

    Completely Stuck. Please Help
  2. jcsd
  3. Oct 31, 2005 #2
    sorry. at an angle of 36.9 degrees below the horizontal
  4. Oct 31, 2005 #3
    Use simple trigonometry to find the height using the angle.

    Then use conservation of energy:

    Delta(K.E) + Delta(P.E) =0
  5. Oct 31, 2005 #4


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    There are two different ways to do this problem:
    1) The hard way. Calculate the component of gravitational acceleration along the slope: Since gravitational acceleration is 9.8 m/s2 down the slope, that would be a= 9.8sin(36.9). Now use d= (1/2)at2 to determine the time, t, it takes to go d= .75 m and put that into v= at.

    2) The easy way. At the top, the energy is entirely potential energy. At the bottom that potential energy has been converted to kinetic energy. Going down a 36.9 m slope 7.5 m mean going down a vertical distance 7.5 sin(36.9) so the potential energy has decreased by 9.8m (7.5 sin(36.9)) (m is the mass).
    At the bottom that has changed to kinetic energy so (1/2)mv2= 9.8m(7.5 sin(36.9)) solve for v.
  6. Oct 31, 2005 #5
    I don't think thats how I'm suppose to solve it. We have not studied conservation of energy. I was told to break weight into components parallel and perpendicular to the surface and then use the work energy theorem. work=delta k, or work=mv_2^2-mv_1^2. But I don't know how to do that.
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