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Help please-multiple choice

  1. Dec 17, 2004 #1
    Help please--multiple choice

    Two identical blocks are lifted at constant velocity by ropes that run over frictionless pulleys, as shown in the diagram. Block B is raised twice as fast as Block A. The magnitudes of the forces are respectively FA and FB, while the power supplied is respectively PA and PB. Which statement is correct?

    a. FB = FA; PB = PA.
    b. FB = FA; PB = 2 PA.
    c. FB = 2 FA; PB = PA.
    d. FB = 2 FA; PB = 2 PA.
    e. PA = FA; PB = FB.
     
  2. jcsd
  3. Dec 17, 2004 #2

    dextercioby

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    I believe it's d).Tell us what u think and bring arguments.That's the only way to check and (probably) correct your reasoning.

    Daniel.
     
  4. Dec 17, 2004 #3

    Doc Al

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    Since you did not supply the diagram, we can only guess as to the arrangement of the pulleys. Assuming that both ropes run over single pulleys...

    What does "constant velocity" tell you about the force needed? How does power depend upon speed?
     
  5. Dec 17, 2004 #4
    d?? I think it's a) - because it is at a constant velocity.
     
  6. Dec 17, 2004 #5

    dextercioby

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    At equal velocities but different times,does it mean the same force???????

    Daniel.
     
  7. Dec 18, 2004 #6

    Doc Al

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    Both d) and a) are half right. :smile: But more important than the answer is the reasoning behind it (as Daniel would say).
     
  8. Dec 18, 2004 #7
    Ooops - I meant b)... since power is F*v since b has 2v, then the power of b will be twice as much as a. Since it is at equillibrium then Fa = Fb.
     
  9. Dec 18, 2004 #8

    Doc Al

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    I agree: b) is the correct answer.
     
  10. Dec 18, 2004 #9
    Me too. Distance travelled by 1s : Distance travelled by 1s=1:s
    Hence PA: PB=1:2
     
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