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Calculating mechanical power

  1. Oct 4, 2007 #1
    A 1700 kg block of granite is pulled up an incline that has an angle of inclination θ = 29 ° with a constant speed of 2.40 m/s by a steam winch (see Figure). The coefficient of kinetic friction between the block and the incline is 0.25. How much power must be supplied by the winch?

    How would you calculate power without a time? Does this involve using a lot of kinematics?
     
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  3. Oct 4, 2007 #2

    robphy

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    You could do this with kinematics...
    but there is another formula for power (derived from the definition of work) that does not require an explicit time.
     
  4. Oct 4, 2007 #3
    Oh, could you lead me to something that discusses the formula, just so I can get an intuitive understanding of the formula before utilizing it.
     
  5. Oct 4, 2007 #4

    robphy

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    To get a feeling for the formula,
    write (using [tex]\Delta x[/tex]) the work done by a force F.
    Now, write the power supplied by this force as a ratio in terms of this work.

    To do this rigorously, one should use calculus.
     
  6. Oct 4, 2007 #5
    I'm still having a few problems, but I would like to run some numbers by:

    the tension in the rope pulling it up the incline would be 11719.7 N

    and the rigorous formula, would that be:

    m(1/2 Vf^2-1/2Vi^2)
     
  7. Oct 4, 2007 #6

    learningphysics

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    Tension is right. But the formula is not...

    You know the force of tension... so what is the work done by tension over a distance [tex]\Delta x[/tex]

    What is the work per unit time?
     
  8. Oct 4, 2007 #7
    I would go about calculating time? Acceleration is zero.
     
  9. Oct 4, 2007 #8

    learningphysics

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    You don't need the time... what is [tex]\frac{\Delta x}{\Delta t}[/tex] ?
     
  10. Oct 4, 2007 #9
    We aren't given any elapsed time. This is where I'm confused. The only given is a velocity, mass, and coefficient of friction.
     
  11. Oct 4, 2007 #10

    learningphysics

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    yup, you have velocity. use that.
     
  12. Oct 4, 2007 #11
    Could I break work into Force x Distance /time

    but isn't distance/time a velocity

    so could I use Force (the tension in the rope) times velocity
     
  13. Oct 4, 2007 #12

    learningphysics

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    exactly. the power is force*velocity.
     
  14. Oct 4, 2007 #13

    robphy

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    I guess my hints make more sense now.
    More generally, it's
    [tex]P=\vec F\cdot \vec v[/tex] (assuming that these quantities remain constant during the motion).
     
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