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Newton's Laws

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Two small blocks each of mass m, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth tabletop and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is a distance h above the floor. Block B is then released from rest at a distance h above the floor at time t=0.

    3. The attempt at a solution

    h= 2h
    g= 9.8m/s^2
    x= ???

    Fnet = ma
    mg = 2ma
    a= g/2

    x= 1/2at^2
    h= 1/2(g/2)t^2
    t^2 = 4h/g
    t= 2sqr(h/g)

    v= at
    v= (g/2)(2sqr(h/g)

    x= vt
    = (g/2)(2sqr(h/g)(2sqr(h/g)
    = 4hg/2g
    = 2h

    Is that right? I was wondering if the 2h distance that Block A has to travel before going off the table had a factor although it should be going at constant velocity...
  2. jcsd
  3. Sep 29, 2008 #2


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    Gold Member

    Since you have not posted the full question, I am not sure what you have been asked to find. However, I can tell you that your solution for whatever you have been asked to find is incorrect since you assumed that the force acting on each block is 1/2m, which is not necessarily the case. In your final section you use the equation x=vt, which assumes that the blocks are travelling with a constant velocity, whereas earlier you assume acceleration (which is correct, the blocks do accelerate).

    Start by posting the full question and we can go from there.
  4. Sep 29, 2008 #3
    Ah, I didn't realize that I didn't post the question itself...

    "Determine the distance between the landing points of the two blocks"
  5. Sep 29, 2008 #4


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    Okay, so the first thing you need to do is determine the velocity of each block at the point where the second block leaves the table. This section of the question is very similar to an ideal pulley question so you need to apply Newton's second law to each block individually, resulting in two equations.
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