Forces and power for two blocks lifted at different speeds

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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.
 
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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.
 
Phyzix said:
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?
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?
 
dextercioby said:
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.

d?? I think it's a) - because it is at a constant velocity.
 
futb0l said:
d?? I think it's a) - because it is at a constant velocity.

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

Daniel.
 
futb0l said:
d?? I think it's a) - because it is at a constant velocity.
Both d) and a) are half right. :smile: But more important than the answer is the reasoning behind it (as Daniel would say).
 
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.
 
Me too. Distance traveled by 1s : Distance traveled by 1s=1:s
Hence PA: PB=1:2