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Kinetic energy and potential energy work

  1. Dec 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Look at the attachment


    Please tell me if there is anything wrong with my work because apparently my teacher thinks its wrong.


    test correction.jpg
     
  2. jcsd
  3. Dec 28, 2011 #2

    Doc Al

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    Staff: Mentor

    Can you please state the problem and what you are asked to find?
     
  4. Dec 28, 2011 #3
    1. calculate the work of W.

    2. calculate the tension or w.e its called of T.

    3. a/ write the potential energy in terms of m,g,r',and Zc

    b/ find the speed of the body (s) at the point β=60°

    sorry i'm learning this stuff in a different language so i'm just translating it by myself.
     
    Last edited: Dec 28, 2011
  5. Dec 28, 2011 #4

    Doc Al

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    I'm a bit puzzled by the problem. You are given both the distance traveled and the speed at point B? (Seems like you should be able to calculate the speed given the distance.)
     
  6. Dec 28, 2011 #5
    yea i did solve it and it came out to 2m/s and everything else is correct and i don't know how my teacher thinks its wrong.
     
  7. Dec 28, 2011 #6
    Yea i did solve the speed at the point c when it has traveled 60° on the ramp. Everything is on the paper i just want to know if there is anything wrong with it.
     
  8. Dec 28, 2011 #7

    Doc Al

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    I did not solve for the speed myself, but it seems as if you're saying that it comes out to be 2 m/s. Yet on your sheet it seems to be given as 3 m/s. Something's wrong somewhere.

    Did the statement of the problem include the speed at point B?
     
  9. Dec 28, 2011 #8
    No that speed is for when the body is at the point B. We have to find the speed of it when it's at point C.
     
  10. Dec 28, 2011 #9

    Doc Al

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    Once again I ask: Are you given both the distance AB and the speed at point B? (If so, you should check that the speed at B is correct by solving for it directly.)
     
  11. Dec 28, 2011 #10
    Yes i am given AB=1m and Vb=3m/s but the question was to find the speed at the point C.

    And i used the law of conservation of mechanical energy to solve for the speed at the point c and i got 2m/s. You are given 3m/s just so you use it to solve for c.
     
  12. Dec 28, 2011 #11
    For the speed at point B try the following:

    EpotA = EkinB

    mgh = m/2*v²

    v = sqrt(2gh) (h=0,5m)

    Check what you did differently, and that's where you went wrong.

    EDIT: Ah... Energy at point C... oops.
     
  13. Dec 28, 2011 #12
    no man i don't need the speed at the point b i need the speed at point c where i found it to be 2m/s. I don't know why you guys think i need to find the speed at b. I was given that stuff on the top right corner of the paper.
     
  14. Dec 28, 2011 #13

    Doc Al

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    Did you check that Vb = 3 m/s is correct? My point is that given AB = 1m, you should be able to solve for Vb. They shouldn't have to give it to you.
     
  15. Dec 28, 2011 #14
    Ok i'll check the speed at b right now and we given the speed at the point b so we can use it to find the speed at c and is my speed for c correct?
     
  16. Dec 28, 2011 #15

    Doc Al

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    Assuming that the speed at B is correct, then the speed you calculated for point C seems correct to me.
     
  17. Dec 28, 2011 #16
    i found the speed at b is 3.16m/s but he gave it to us as 3m/s.
     
  18. Dec 28, 2011 #17
    Yea that's all i wanted to know, he just counted it wrong for me for no reason.
     
  19. Dec 28, 2011 #18
    With

    Erad = 1/2 Jω², ω=v/b, b=2/6∏r

    EkinB - Erad - EpotC = EkinC

    => vB²-2r²(v²/((2/6)∏r)²-2gh=v² => v=1.32m/s says my calculator
     
    Last edited: Dec 28, 2011
  20. Dec 28, 2011 #19
    deleted what?
     
  21. Dec 28, 2011 #20
    I deleted something that I thought made no sense and replaced it with what I think the answer to be ;p

    h = cos 60 * r

    Because at point C, the object has kinetic, potential and rotational Energy.
     
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