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Homework Help: Tension and velocity question

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img141.imageshack.us/img141/9379/photo13ix6.jpg [Broken] (sorry its backwards)

    The graph above shows the velocity versus time for an object moving in a straight line. at what time after t = 0 does the object again pass through its initial position?

    A) Between 0 and 1 s
    B) 1s
    C) Between 1 and 2s
    D) 2s
    E) Between 2 and 3s

    3. The attempt at a solution

    i picked B. it shows a velocity of -1 for 1/2s and then a velocity of 1 for 1/2s seconds. i just thought it would travel backwards first and then forwards and end up at the initial position after 1 second.

    ____ ____
    [1kg]-----[2kg]----> F

    When the frictionless system shown above is accelerated by an applied force of magnitude the tension in the string between the blocks is

    A) 2F
    B) F
    C) 2/3F
    D) 1/2F
    E) 1/3F

    i don't know where to begin for this

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 27, 2009 #2
    First of all, the graph is a velocity v/s time graph and NOT displacement time graph. So in the first 1/2s velocity is not -1, acceleration is. The velocity is continuously increasing.

    Secondly, remember displacement is the signed area under the velocity time graph. Find such a point, where total area is zero.
  4. Jan 27, 2009 #3
    do you know how to draw a force-body diagram for such a system?
  5. Jan 27, 2009 #4
    oh crap. i interpretted the graph wrong. i didn't see that throughout the first second the velocity was zero.
  6. Jan 27, 2009 #5
    not sure
  7. Jan 27, 2009 #6
    Well, an FBD, is a diagram in which you isolate each massive body and draw all the possible forces that are acting on it. This way, when you apply newton's third law on them, it becomes much easier, as you dont have to bother with two bodies at once. Try to do that for this system.

    Isolate both the bodies, and draw all the forces acting on them.
  8. Jan 27, 2009 #7
    btw, "throughout" the first second, velocity is not zero. The velocity is zero only at time t = 1s.
  9. Jan 27, 2009 #8
    negative was what i meant
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