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Block of wood pushed up a wooden ramp

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.37 kg wood block is launched up a wooden ramp that is inclined at a 35° angle. The block’s initial speed is 9.32 m/s. What vertical height does the block reach above its starting point?

    2. Relevant equations
    DYNAMICS


    3. The attempt at a solution
    PHysics1.jpg
     
  2. jcsd
  3. May 15, 2009 #2
    no matter what I do i end up with a=g(ukcos(theata)+sin(theata)) is this right how am i setting it up wrong?
     
  4. May 15, 2009 #3

    Doc Al

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    Assuming there is friction (your first post doesn't mention any), that's the correct expression for the acceleration of the block along the incline. Was the problem statement in your first post complete?
     
  5. May 15, 2009 #4
    I believe so. I took fs=us*n....and replaced fs with us*n. Is that wrong?
     
  6. May 15, 2009 #5
    or should i assume there is no friction?
     
  7. May 15, 2009 #6

    Doc Al

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    Your initial problem statement doesn't mention friction or give a value for μ. What makes you think there's friction involved here?
     
  8. May 15, 2009 #7
    I think friction should be involved because its wood on wood, and when wood rubs against wood there is friction. Am i just thinking to much about this problem? If there is no friction then my acceleration would be just be gsin(theata). Correct?
     
  9. May 15, 2009 #8

    Doc Al

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    Most likely, but this is a physics problem so often things are ignored. You can only tell by context. For example, are you given a chart of coefficients of friction to use?
    I suspect so.
    Correct.
     
  10. May 15, 2009 #9
    We weren't given a table there is one in the book but there is no wood on wood coefficient given. Thank you Doc Al, I am going to take the appoarch without friction and see how it goes and post back on here with what I come up with thank you.
     
  11. May 15, 2009 #10
    Ok so i got the acceleration, which is negative because it slow down so can i just use Vf=Vi+(a*delta t) to find the time and then just use the kneimatics equation yf=yi+ (Viy*deltat)-(.5*g*delta t^2)? I tried that and it was wrong. Do i subsitute the gravity with the accleration i solved for?
     
  12. May 15, 2009 #11

    Doc Al

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    Of course! Use the acceleration you figured out.

    Be sure to convert your answer into the form they ask for: The vertical height, not the distance along the ramp.

    You can also use a different kinematic equation that combines the two you are using into one step. Save a little work.
     
  13. May 15, 2009 #12
    well isnt that why I am using Yf and Yi in the equation that will give me how high it went?
     
  14. May 15, 2009 #13

    Doc Al

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    No. The block moves along the ramp, you found its acceleration parallel to the ramp, and that equation will find the distance traveled along the ramp. But you can easily use that answer--with a bit of trig--to find the change in height.
     
  15. May 15, 2009 #14
    ahhhhh. Making since now!
     
  16. May 15, 2009 #15
    but wait.....wouldnt the equation i use to find the length it travels up the slope be.....
    Df=Di+(Vi*delta t)-(.5*a*delta t^2). The Vi is the Intial velocity not with respect to the y axis? and then the a is the calculated a?
     
  17. May 15, 2009 #16

    Doc Al

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    You could certainly use that if you knew (or figure out) the time. (You could also you a different kinematic formula; one that relates, a, v, and d together.)
    Right. The initial velocity you were given is parallel to the ramp, and the a needed is the one you calculated.
     
  18. May 15, 2009 #17
    Ok this is all my work but it still said its wrong......

    Physics4.jpg
     
  19. May 15, 2009 #18

    Doc Al

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    Looks OK to me. If this is one of those online systems, they can be picky about significant figures.

    The only other thing I can suggest is to do it over including friction. But unless they give you the coefficient, I don't see how you can do that.
     
  20. May 15, 2009 #19
    yeah that was the initial problem. and wood to wood coefficient can be between .25-.60 so i dont see how i could even pick one. Your the man Doc Al! I would be completely lost with out u. Even when I dont get rid of any significant figures til the end i still get 4.430 so... I am just going to email my professor and tell him I did all the work right but it didn't work.
     
  21. May 15, 2009 #20
    There is a second part to this problem that I am going to try using my numbers and see if I can get it right.........

    What speed does it have when it slides back down to its starting point?
     
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