Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 Emeritus
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook