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Acceleration of a block of ice - Dynamics

  1. Aug 5, 2006 #1
    A 10kg block of ice slides down a ramp 20m long, inclined at 10 degrees to the horizontal. If the ramp is frictionless, what is the acceleration of the block of ice?

    Struggling as to where to start, I have the formulae, just dont know how to use them.

    Most of the questions I've encountered had a force as a given, so now I'm getting a bit confused here.

    Would Ff = umg still be any good here?
     
  2. jcsd
  3. Aug 5, 2006 #2
    There is a component of weight acting down along the ramp, thats your force.
     
  4. Aug 5, 2006 #3
    Would that be gravity?

    Fg = mg = (10)(9.8) = 98N?
     
  5. Aug 5, 2006 #4

    Doc Al

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    Staff: Mentor

    Yes, that's the weight of the object, which acts down. Now do as cyrus said and find the component of that force along the ramp, since that is the only direction the block can move.
     
  6. Aug 6, 2006 #5
    So if gravity is "down", then to find the force along the ramp, it would be:

    sin10 = 98/X
    X = 564.36N (Force along the ramp)

    How would I account for the 20m in ramp length? I think I'm supposed to find time right...
     
    Last edited: Aug 6, 2006
  7. Aug 6, 2006 #6

    Doc Al

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    Rethink your use of trig. How can a component of a force be greater that the total force?

    To find the acceleration, the ramp length is irrelevant. But if you also need to find the time for the block to slide down the ramp, then you'll need the length.
     
  8. Aug 6, 2006 #7
    I don't understand...Along the ramp meaning the slanted part right? Then wouldn't the 98N be the Y component because its down?
     
  9. Aug 7, 2006 #8
    I think I got it:

    Is it 98sin10 = 17.01N

    F = ma
    17.01 = 10a
    a = 1.7m/s^2
     
  10. Aug 7, 2006 #9

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
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