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Help Solving/Creating a Question

  1. Jun 11, 2006 #1
    I need to prove how Superman would 'be more powerful than a locomotive.'

    [​IMG]

    From my diagram I need to find out how much power Superman has to put out to be able to pull that train up the track. I know he has to pull the train 8.5m on an inclined plane of 15 degrees.

    -I know the train has a mass of 395 300kg.
    -I also know the train is applying a force in the opposite direction.
    -Coefficient of kinetic friction of a steel on steel is 0.57
    -And finally I know the train is putting out 2,387,200W (3, 200hp) to make that applied force.

    So Superman has to overcome the FApplied(train) + FFriction + Fgx to be able to pull the train.
    ------------------------------
    -FN = mgcosx = (395300kg)(9.81m/s^2)(cos15) = 3745757N

    -uk = FFriction/FN
    FFriction = (0.57)(FN) = 2135081N

    -Fgx = mgsinx = (395300kg)(9.81m/s^2)(sin15) = 1003673N
    ------------------------------

    Okay so I found FFriction and Fgx of the train but I still need the FApplied by the train to know how much force superman needs to apply in the opposite direction to move it up the hill.

    I know the train is putting out 2,387,200W of power in the opposite direction. But how can I calculate how much force that is? Maybe I misunderstand the difference between force applied and power?

    Please help me in anyway possible.
    -Jay
     
  2. jcsd
  3. Jun 11, 2006 #2

    Hootenanny

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    Assuming constant velocity, you can use the equation P=Fv, where v is velocity. Remember that F and v are vectors.
     
    Last edited: Jun 11, 2006
  4. Jun 11, 2006 #3
    So I researched that this train can get up to at most 56km/h (15.5m/s). I guess the easiest thing to do is assume it's constant because well, to do otherwise might be out of my current level of physics education :P.

    2, 387, 200W = F(15.5m/s)
    F=154013N

    So superman has to overcome Fgx (1003673N) + FFriction (2135081N) + FApplied(train) (154013N) = 3292767N?

    Then I guess I can use P = mv again once I find the average velocity a person walks with, to find how much power Superman puts out?
     
  5. Jun 11, 2006 #4

    Hootenanny

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    Yep, thats some strong biceps there :biggrin:
    That is the expression for momentum. If you mean to use P = mgv, then no you can't. The force of gravity and the velocity are perpendicular to each other and hence are independant. However, you could work out his power output using P = 3292767v, where v is his walking velocity.
     
    Last edited: Jun 11, 2006
  6. Jun 11, 2006 #5
    Oh my, sorry for that! I meant P=Fv. Thank you for your help!
     
  7. Jun 11, 2006 #6

    Hootenanny

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    My pleasure :smile:
     
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