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Block on frictionless ramp

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    i have been having trouble with this problem...if anyone could help me as soon as possible i would really appreciate it !!!(this is due today)...thank you!!!!!
    The problem:

    A block starts at rest and slides down a frictionless track.
    The acceleration of gravity is 9.81m/s^2.
    It leaves the track horizontally, striking the ground.


    [​IMG]


    A. at what height above the ground is the block released?
    B. What is the speed of the block when it leaves the track? i solved, and the correct answer was 5.289 m/s
    C What is the speed of the block when it hits the ground?


    2. Relevant equations



    3. The attempt at a solution
    i solved part B, and the correct answer was 5.289 m/s
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    for B: I did -2.4m = 1/2 (-9.81m/s^2)(t)^2
    and got t = .6995
    V = 3.7 / .6695
    =5.289 m/s
     
    Last edited: Feb 4, 2007
  2. jcsd
  3. Feb 4, 2007 #2

    cristo

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    What do the actual question specify? i.e. do you know the height h, or is that what you are asked to find in (A)?

    For part (B), is that the correct answer given to you by a book, or is that the answer you've calculated?
     
    Last edited: Feb 4, 2007
  4. Feb 4, 2007 #3
    It is an online submission thing.. you do the hw online and it says that i got B correct...you have so many guesses to get it correct...the question below is the actual question and the picture that i did was actually given...thank you for replying me!


    A block starts at rest and slides down a frictionless track.
    The acceleration of gravity is 9.81m/s^2.
    It leaves the track horizontally, striking the ground.


    [​IMG]


    A. at what height above the ground is the block released?
    B. What is the speed of the block when it leaves the track?
    C What is the speed of the block when it hits the ground?
     
  5. Feb 4, 2007 #4

    Doc Al

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    You solved part B correctly. Now make use of that information to solve parts A and C.
     
  6. Feb 4, 2007 #5
    I solved B but i dont exactly know where to start A or C from there or how B ties in with them.
     
  7. Feb 4, 2007 #6

    cristo

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    Ahh, ok, I didn't read the question properly!

    Yes, your answer to (B) is correct. Are you having trouble with (A) or (C) or both?

    For (A), consider the energy of the block whilst it's on the slope. At the top it's at rest, and so has only potential energy. As it leaves, it has less potential energy, but kinetic energy. Using the principle of conservation of energy, you can find the initial potential energy, and hence h.

    For (C), consider the x and y components of the final velocity and use the kinematic equations.
     
  8. Feb 4, 2007 #7

    Doc Al

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    Hint: There's no friction, so mechanical energy is conserved. Consider the block's total mechanical energy just as it leaves the track, then compare that point to its starting point and landing point. What changes?
     
  9. Feb 4, 2007 #8
    so would i do:
    KE = 1/2 mv^2
    1/2 (.534)(5.289m/s)^2 = 7.47J

    PE = mgh
    7.47J = .534 (9.8) h
    divide both sides by 5.23 to get 1.428 as h.
     
  10. Feb 4, 2007 #9

    Doc Al

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    That's the KE at the point it leaves the track. But what about the PE? You need both to have the total mechanical energy.

    No, since you only included part of the total energy. (But you are correct that at the starting point the energy is purely PE.)
     
  11. Feb 4, 2007 #10
    im sorry, but im a bit confused..how do i find the PE if i was not given h? to get the total mechanical energy
     
  12. Feb 4, 2007 #11

    Doc Al

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    Let's call the three positions of interest 1 (starting point), 2 (where it leaves the track), and 3 (where it lands). How does the PE change when going from positions 1 to 2 to 3?

    You found the KE at position 2. Now add the PE at position 2 to get the total energy (which is the same at all positions). Measure PE from the ground level. (You were given h--it's on the diagram!)
     
  13. Feb 4, 2007 #12
    gal m confuse with this problem too....
     
  14. Feb 4, 2007 #13
    aikenfan did u find PE for point 1
     
  15. Feb 4, 2007 #14
    so for PE at point 2 it would be :
    PE = mgh = .534(9.8)(2.4) =12.56 ?
     
  16. Feb 4, 2007 #15

    Doc Al

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    Yes, now you're cooking. :smile:
     
  17. Feb 4, 2007 #16
    the total energy would therefore be 20.02
    which would make:

    20.02 = 1/2 (.534)(v)^2
    divide both sides by .267 and get 74.98...take the square root to get 8.65m/s?
     
  18. Feb 4, 2007 #17

    Doc Al

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    OK... and that would be the speed of what? :wink:
     
  19. Feb 4, 2007 #18
    now, i have the speed at which it hits the ground, for part A how do i find the height above the ground that it was released? Thank you for taking the time to help me by the way!
     
  20. Feb 4, 2007 #19
    so u add PE and KE to find TE right
     
  21. Feb 4, 2007 #20

    Doc Al

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    OK, you've solved parts B and C. One more to go! Again, just set the total energy (which you know) equal to PE + KE and solve for h. (What's the KE when it was released?)
     
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