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Another Laws of Motion question

  1. Sep 21, 2006 #1
    Hi this problem stumps me.

    Two bricks each having a mass of 0.2 kg tied at the end of a light flexible cord and are passing over a small frinctionless pulley. A 0.1kg block is placed on one of the blocks and removed after 2 seconds. How far will each block move in the first second the 0.1kg block is removed?

    My Work:

    for one of the blocks: [tex] 0.3(10) - T = 0.3a [/tex]
    for the other block: [tex] T - 2 = 0.2a[/tex]
    solving, [tex]a = m/s^2[/tex]
    Then I found the velocities of each block after the 0.1kg block is removed. i,e [tex] v = 2(2) = 4m/s [/tex]
    actually, one of the blocks will have 4m/s and the other -4m/s.

    for the block that has -4m/s velocity,
    the displacement in the first second:
    [tex]s = -4(1) - \frac{1}{2}(10)(1)[/tex]
    [tex] = -9m [/tex]

    and for the other block i got the displacement as 1m.

    BOOK SOLUTION: my book says the correct answer is 4m for each block.

    Please tell me where I went wrong. Thanks for your time!
    Last edited: Sep 21, 2006
  2. jcsd
  3. Sep 21, 2006 #2


    User Avatar

    Staff: Mentor

    After the one kg block is removed, there is not acceleration of the two blocks. One 2 kg block balances the other, and both contribute to the tension in the cord.

    One calculated the speed a 4 m/s for one block and -4 m/s for the other block, i.e. one moves up and the other down. In one second, what is the distance traveled at constant velocity?
  4. Sep 21, 2006 #3
    allright thanks. got it now:smile:

    by the way, can you please look at this thread I posted earlier today?
    https://www.physicsforums.com/showthread.php?t=132919 It is related to the same topic. I have been practicing laws of motion problems all day and I really want to get it straight. only that question is left..thanks
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