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Pulley / conservation of energy problem (should be really easy)

  1. Jun 27, 2008 #1
    This problem is bugging the crap out of me. It's the last one I need to do. I swear I got the right answer. Maybe you experts can help? :)

    1. The problem statement, all variables and given/known data
    The system shown in the figure below consists of a light, inextensible cord; light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is 1.00 m.

    2. Relevant equations
    PEi + KEi = PEf + KEf
    KE= (mv^2)/2
    PE= mgy

    3. The attempt at a solution
    1. Judging from what is presented, I assume that the block A moves twice as fast as block B and therefore, it will lower at double the rate that block B will rise.

    2. I figured that I would write out my equation to be:
    mgy[a] + mgy [initial] = mg(y-2/3)[a] + mg(y+1/3) + (mv^2)/2[a] + (m[v/2]^2)/2

    3. Simplified by dividing by m and making initial y to be 0.
    0 = g(-2/3) + g(1/3) + (v^2)/2 + ((v/2)^2)/2

    4. Input -9.8 for g and simplify more.
    -3.26699 = (v^2)/2 + ((v/2)^2)/2

    5. Simplify more.
    2.556165 = 1.5v

    Answer = 1.7 m/s

    HOWEVER, that was wrong, and I've checked my math several times. I must be doing something horribly wrong. Kudos and much thanks to you if you can find it, like I said, this has been bugging me for a while.
  2. jcsd
  3. Jun 27, 2008 #2

    This statement is a tactical blunder.
  4. Jun 27, 2008 #3
    Uhm... how so?

    I've looked at the system again, and it looks like block A goes down at 2x the rate that block B goes up.
  5. Jun 27, 2008 #4
    Your reasoning please.

    Follow these guidelines for any mechanical problem.:

    Draw reference, which is fixed. This reference can be the level of fixed pulley or the ground.
    Identify all movable elements like blocks and pulleys (excluding static ones).

    Assign variables for the positions of movable elements from the chosen reference.

    Write down constraint relations for the positions of the elements. Usually, total length of the string is the “constraint” that we need to make use for writing relation for the positions.

    Differentiate the relation for positions once to get relation of velocities and twice to get relation of accelerations.
  6. Jun 27, 2008 #5


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    Hi fvertk,

    I think you have some algebra errors:

    The left side is negative, yet the right side is a positive number. The next step also does not seem to follow this one.
  7. Jun 27, 2008 #6
    Well, when looking at the problem, it appeared that way initially. I looked up other responses to it, and that's what someone here had to say about it too:
    (read Doc Al's post)

    Now that I look at it again, I guess they actually ARE rising and falling at the same rate. (Is that what you're hinting at?) However, when I put that into the equation, all I get is 0. It can't be solved that way apparently. So I tried it for one block, doing mgy= mgy + mv^2/2 (where y(final) is -.5 and y(initial) is 0) and I got 3.13, which was still the wrong answer.

    PS: To be honest with you, if the blocks are the same mass, I don't understand why they are moving in the first place. Shouldn't it be completely stationary?
  8. Jun 27, 2008 #7


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    I think your original equation was okay; you just had the algebra errors I mentioned in my post. If you plug your answer back into what you have for equation 3 you'll find it doesn't work.
  9. Jun 27, 2008 #8

    Doc Al

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    Staff: Mentor

    Sounds right to me. (But check the errors that alphysicist pointed out.)
  10. Jun 27, 2008 #9
    Right, I did change that to positive in order to get the square root. It shouldn't matter what sign it is since I'm trying to find the speed, right? And at that point, I'm no longer adding/subtracting, but multiplying.

    What I did to get to step five was multiply 2.36699 by 2, then find the square root of that.
  11. Jun 27, 2008 #10


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    You have:

    3.26699 = (v^2)/2 + ((v/2)^2)/2

    So when you multiply by 2, you get:

    6.53398 = v^2 + v^2 /4

    What do you get when you take the square root of both sides?
  12. Jun 27, 2008 #11
    Okay, I see what you're saying alphysicist. Looks like I gave it the wrong sign in step 4 in my addition from step 3. But that should be okay, since I changed it anyway... I'm not sure that's the problem then.
  13. Jun 27, 2008 #12
    Why is the latter part of that v^2/4? How did you eliminate the fraction beneath the exponent? What I did was simplify the 1/2 to be from this:
    3.26699 = (v^2)/2 + ((v/2)^2)/2

    To this:
    3.26699 = 1/2((v^2) + ((v/2)^2))

    And then multiplied both sides by two, getting me this:

    6.53398 = v^2 + (v/2)^2

    Then after doing the square root, I get:

    2.556 = v + v/2
    Maybe I'm wrong though... perhaps there's an algebra trick I haven't learned yet? :/
  14. Jun 27, 2008 #13


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    Because in the last term,


    means squaring both numerator and denominator.

    \left(\frac{v}{2}\right)^2 = \frac{v^2}{4}
  15. Jun 27, 2008 #14
    Oh yeah, duh. *smacks head* Although, shouldn't my technique work just as well? Let me try yours though, hold on...
  16. Jun 27, 2008 #15


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    No, it doesn't work that way. For example,

    \sqrt{3^2 + 4^2} = 5 \mbox{ (not 7)}

    First combine the two terms on the right side. v^2 + (1/4) v^2 equals what?

    Then you might want to get v^2 by itself on the right side and then take the square root.
  17. Jun 27, 2008 #16
    I see... I must have really botched that square root then by doing two under one. Strange, I guess I had some misconception that that actually worked. Nice catch, that actually gave me the right answer. Thanks alphysicist.

    By the way, physixguru, I'm not sure what you were talking about above then. Seems like it did go 2x faster, but thanks for helping anyway, I'm just confused.
  18. Jun 27, 2008 #17


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    By the way, in an earlier post you mentioned that you did not understand why it was moving in the first place. Does that make sense now?
  19. Jun 27, 2008 #18
    Yes, in the case that block A has a faster velocity downward, it does. When I doubted that initial assessment of the system, thinking that each block has equal velocity, that's when I got confused.
  20. Jun 28, 2008 #19


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    While solving this problem you have taken the acceleration of the blocks as g, but it is not g. Since the single rope is passed through two pullyes, the tension T in each segment of the rope must be the same. Since block B is pulled up by 2T and block A by T, block A moves in the downward direction. The acceleration of block A is (mg - T)/m. And acceleration of block B is (2T - mg)/m. Solve for T and find the value of acceleration. Then apply the conservation law of energy for the block A or kinamatics equation v^2 = vo^2 + 2ah. As you have said the disatance moved by A is (2/3)m
  21. Jun 28, 2008 #20


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    Hi rl.bhat,

    I'm not sure what you are referring to; the acceleration was not calculated at all in the problem, so I don't think it was assumed to be g. Here an energy approach was used which gave the correct answer (after some algebraic answers were corrected).

    The effect of the ropes was handled by constraining the speed and distances of one mass to be twice that of the other.
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