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Determine the speed of the block at the bottom of the inclin

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  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    You release a block from the top of a long, slippery inclined plane of length l that makes an angle θwith the horizontal. The magnitude of the block's acceleration is gsin(θ) .Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline.
    Express your answer in terms of g and the variables m, l, and θ.

    2. Relevant equations
    UG=−mgsinθ(x−l)

    3. The attempt at a solution
    i know that the correct answer does not depend on "m" so i assumed the equation would be v=l/gsinθ. thats inccorrect and i tried reversing it too and still wrong. i know that they won't be multiplied so i'm unsure the relation
     
  2. jcsd
  3. Oct 25, 2016 #2

    gneill

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    Staff: Mentor

    What was the "expression you derived in the previous part"? Something to do with velocity, acceleration, and distance perhaps?
     
  4. Oct 25, 2016 #3
    this was it : UG=−mgsinθ(x−l)
     
  5. Oct 25, 2016 #4

    gneill

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    Okay, well that appears to be an expression for gravitational potential energy. What are x and l in that equation?

    You still need to get to velocity somehow. Have you learned about conservation of energy and how kinetic energy and potential energy are related?
     
  6. Oct 25, 2016 #5
    l= inclined plane lenght
    x=UG(x=l)=0.
    k=-U
     
  7. Oct 25, 2016 #6

    gneill

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    I have to ask, is UG something different from U, and is k meant to represent kinetic energy?

    Do you know how to find the velocity if you have the kinetic energy?

    On a related subject, the problem statement tells you that the magnitude of the acceleration is given by gsin(θ). Do you know a standard kinematic equation that relates acceleration, distance, and velocity?
     
  8. Oct 25, 2016 #7
    Ug is potential energy, k is kinetic energy
     
  9. Oct 25, 2016 #8
    yes is i have kinetic energy, velcoity is the square root of k/m
     
  10. Oct 25, 2016 #9
    but i don't have kinetic energy, i have potential
     
  11. Oct 25, 2016 #10
    would speed be the √mgsinθ(x−l)/0.5m
     
  12. Oct 25, 2016 #11

    gneill

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    Not quite. Since ##k = \frac{1}{2} m v^2##, then solving for v gives ##v = \sqrt{\frac{2 k}{m}}##. Don't forget that "2".

    But you know how they are related. Once the block moves from the top of the ramp to the bottom, what happens to the potential energy? Where does the energy go?

    It's hard to tell because I can't be sure what the square root sign is meant to cover. And I'm still not sure what x is meant to be. I understand that l is the inclined plane length from post #5:
    But I'm not sure how to read your second equation. Surely you're not equating a length to a potential energy?
     
  13. Oct 25, 2016 #12
    The square root sign is meant to cover everything
     
  14. Oct 25, 2016 #13

    gneill

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    In that case I believe that it will yield the velocity. Note that the mass m cancels in the numerator and denominator. It would probably be more clear if you were to replace (x - l) with a single variable like "d" for the distance traveled along the slope.

    Note that there's also a kinematic equation that you should be familiar with that would get you to the result quickly if you know the acceleration and distance (which you do in this case).
     
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