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Impulse and Momentum (2)

  1. Mar 21, 2004 #1
    Question: The 40 kg slider block is moving to the right with a speed of 1.5 m/s when it is acted upon by the forces F1 and F2. If these loadings vary as shown (in the attachment called graph), determine the speed of the block at t =6s. Neglect friction and the mass of the pulleys and cords.

    The graph doesn't really look like I drew it but you can get the picture (I hope).

    This is what I did so far:

    Equation: mV1 + summation (integral of) F * dt = mV2

    So I calculated the area under each part of the graph and added them together like so

    40*(1.5) + 30*(4-0) + 10*(6-4) + 10*(2-0) + 20*(4-2) + 40*(6-4) = 40V2

    340 = 40V2
    V2 = 8.5 m/s

    Is this right?
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2004 #2
    Here's the graph.
     

    Attached Files:

  4. Mar 21, 2004 #3
    Well, F1 acts in the opposite direction of F2 so one of them needs to get a minus sign in your calculations. The graph shows the forces as scalars (i.e it only tells you their magnitutde), not as vectors. Also, if the right cord is pulled by F1, then the force on the block is actually 4 times F1, because of the pulley system.
     
  5. Mar 21, 2004 #4
    Ohhhhhhhhhh! Thank-you. I get it now. The answer is 12 m/s.
     
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