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Finding Speed of a Mass in a Frictionless System at a Specific Time

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

    The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are:
    M1 = 2.00 kg
    M2 = 7.00 kg
    M3 = 4.00 kg
    Calculate the speed of M2 at a time 1.750s after the system is released from rest.

    (Question and diagram attached)


    The attempt at a solution

    I followed the way my professor did a similar question in my notes, however the answer is not correct. My process was as follows:

    For mass 1:
    ƩFnet = M1a
    T1 - M1g = M1a
    T1 = M1(a+g)
    T1 = 4(a+9.8)

    For mass 2:
    ƩFnet = M2a
    T2-T1 = M2a
    T2 = M2a - T1
    T2 = 7a + 2a - 19.6
    T2= 9a - 19.6

    For mass 3:
    ƩFnet = -M3a
    T2 - W3 = -M3a
    T2 = W3 - M3a
    T2 = 39.2 - 4a

    I then made the two T2 equations equal to find acceleration of the system:
    9a - 19.6 = 39.2 - 4a
    9a + 4a = 58.8
    a = 58.8/13
    a = 4.523 m/s^2(right)

    Then I found velocity of M3

    Vf = Vit + (1/2)at^2
    Vf = (0)(1.750) + (1/2)(4.523)(1.750^2)
    Vf = 6.926 m/s


    Anyone know where I went wrong? I would seriously appreciate any help. I've tried this a bunch of times and have gotten different answers for acceleration every time, all of them wrong. I only have a few more tries left. Any guidance would bed a great help!
     

    Attached Files:

  2. jcsd
  3. Oct 28, 2011 #2

    Doc Al

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    Staff: Mentor

    You mixed up your formulas. That's the one for distance. You want:
    Vf = Vi + at
     
  4. Oct 28, 2011 #3
    I'm still not getting the correct answer. I must have messed something up before that too. I just can't see where I messed up.
     
  5. Oct 28, 2011 #4

    Doc Al

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    M1 = 2 kg, not 4.

    I suggest you not plug in any numbers until the very last step. Solve it with symbols first, then plug in the masses to get the final answer.
     
  6. Oct 28, 2011 #5
    Okay, so I've done this question a bunch of times. Like, a lot of times. I only have 2 tries left to get the correct answer.

    1 Incorrect. (Try 1) 8.12 m/s
    3 Incorrect. (Try 2) -8.12 m/s
    4 Incorrect. (Try 3) 2.31 m/s
    5 Incorrect. (Try 4) 6.93 m/s
    6 Incorrect. (Try 5) 2.28 m/s
    7 Incorrect. (Try 6) 14.8 m/s
    8 Incorrect. (Try 7) 4.18 m/s
    9 Incorrect. (Try 8) 6.87 m/s

    I've attached a picture of another failed attempt.

    Please, does anyone know how to do this correctly?
     

    Attached Files:

  7. Oct 28, 2011 #6

    Doc Al

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    Did you correct the mistake I pointed out? Please repost your new solution. (A scan makes it hard to point out any errors.)

    FYI: Your three equations in post #1 are correct, so that's a good start.
     
  8. Oct 28, 2011 #7
    I did, but I wasn't really using the T1 system of numbers because if I let equations one and two equal each other, I ended up with two values I didn't know, and I would have to solve for T2 and equate that new equation to equation 3 to find acceleration.

    I thought if I did that, I would definitely end up making a mistake, as I was looking for the velocity of M3.

    Also, for clarity, do you want me to type out what was on the scan of the paper?
     
  9. Oct 28, 2011 #8

    Doc Al

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    The first thing to do, as you attempted in post #1, is to solve for the acceleration. You have three equations and three unknowns, so it works out OK. You can solve them any way you wish.
    Why not just cut and past your work from post #1, making any corrections needed? You were on the right track. (If you want, you can type out what was in the scan.)
     
  10. Oct 28, 2011 #9
    If I was on the right track with the first one, I'll go from there.

    Right now, I'm doing what you said earlier and waiting to put the values in until very last. I've got two tries left before I'm locked out with zero marks, so would you mind telling me if I'm on the right track?

    So I had:
    T1 = M1(a+g)
    T2 = M2a - T1
    T2 = M3(g-a)

    M2a - M1a + M1g = M3g - M3a
    M2a + M3a - M1a = M3g - M1g
    a(M2 + M3 - M1) = M3g - M1g
    a = (M3g - M1g)/(M2 + M3 - M1)

    a = [(4)(9.8) - (2)(9.8)] / (7+4-2)
    a = 2.1777777

    And then, of course, Vf = Vi + at.

    So,
    Vf = 0 + (2.177777)(1.750)
    Vf = 3.811111
     
    Last edited: Oct 28, 2011
  11. Oct 28, 2011 #10

    Doc Al

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    OK.
    Not OK. Fix this.
    OK.

    You are very very close. Fix the error above and try again.
     
  12. Oct 28, 2011 #11
    Oh, oh, okay. So would it be:

    T2 = M2a + T1

    Making it
    M2a + M1a + M1g = M3g - M3a
    M2a + M3a + M1a = M3g - M1g
    a(M2 + M3 + M1) = M3g - M1g
    a = (M3g - M1g)/(M2 + M3 + M1)

    a = (19.6)/(13)
    a = 1.5079

    Vf = (1.5079)(1.750) = 2.638

    Does that correct the mistake I was making?

    Edit: I submitted it and it was accepted! Thank you so much!
     
  13. Oct 28, 2011 #12

    Doc Al

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    Perfect.
     
  14. Oct 28, 2011 #13

    Doc Al

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    Just for fun, here's a 'trick' I use to solve this quickly. If you write the equations this way:

    T1 - m1g = m1a
    T2 - T1 = m2a
    m3g - T2 = m3a

    Just add the equations and the tensions cancel.

    (Of course you can't count on such a trick working for all problems.)
     
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