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Find the mass, ramp, spring

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A block starts from rest and slides down a long slope with length d, angled at [tex]\theta^0[/tex] to the horizontal. The coefficient of kinetic friction between block and slope is [tex]\mu_k[/tex] At the end of the slope, it compresses a spring with spring constant [tex]k[/tex] to a compression [tex]x[/tex]. What is the mass of the block?


    3. The attempt at a solution

    [tex]\sum W = \Delta K[/tex]

    [tex]\frac{-kx^2}{2} + mgdsin\theta - \mu_k mgd = 0[/tex]

    [tex]m = \frac{kx^2}{2gd(sin\theta - \mu_k)}[/tex]
     
  2. jcsd
  3. Aug 3, 2011 #2

    SammyS

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    Is the spring on the slope, or is it on the horizontal ?
     
  4. Aug 3, 2011 #3
    I have no idea, it doesn't say. I am guessing it isn't? If it is you can't solve it right? Because we dont' know how long the spring is (it may be d/2)?

    I thought it was horizontal because it says "at the end of the slope"
     
  5. Aug 4, 2011 #4

    SammyS

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    It appears that it could be worked either way. I'll assume that it is as you say.

    What is the normal force exerted on the block by the ramp (or slope)? -- Looks like you have it wrong.

    Also, I hope you have drawn a free body diagram for the block, as it slides down the ramp.
     
  6. Aug 4, 2011 #5
    Ohhhhhh

    right-to

    Wait how it would it work out if the spring was on the ramp?
     
  7. Aug 4, 2011 #6
    It should be

    [tex]m = \frac{kx^2}{2gd(\mu_k cos\theta - sin\theta)}[/tex]
     
  8. Aug 4, 2011 #7

    SammyS

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    First do it correctly assuming it's not on the ramp.
     
  9. Aug 4, 2011 #8
    Okay so gravity and theta disappears.

    [tex]\sum W = \Delta K[/tex]

    [tex]\frac{-kx^2}{2} - \mu mg(d + x) = 0[/tex]
     
  10. Aug 4, 2011 #9

    SammyS

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    Actually, you're right. We don't know the uncompressed length -- like you said.

    So, the the total distance traveled down the ramp is d. The spring is compressed an amount x.
     
  11. Aug 4, 2011 #10

    SammyS

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    In your Original Post, you had: [itex]\displaystyle \frac{-kx^2}{2} + mgdsin\theta - \mu_k mgd = 0[/itex]

    The work done by the spring is correct: [itex]\displaystyle \frac{-kx^2}{2}\,.[/itex]

    The work done by gravity is correct: [itex]\displaystyle mgd\sin\theta\ .[/itex]

    The normal force you used is incorrect. What is the component of the gravitational force that's perpendicular to the ramp? That's equal in magnitude (equal but opposite) to the normal force, N .
     
  12. Aug 4, 2011 #11
    No I already have the right answer, I thought we aer working on the spring being on the ramp now.

    [tex]m = \frac{kx^2}{2gd(\mu_k cos\theta - sin\theta)}[/tex]
     
  13. Aug 4, 2011 #12

    SammyS

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    I missed your post #6.

    My meaning in post #9 is: Forget having the spring on the ramp, for the very reason you gave earlier.

    So - - - You're done. It all looks fine !
     
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