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Power calculation

  1. Feb 19, 2014 #1
    I dont feel all the way safe with my knowledge of power and work, so I am basically asking for someone to take a look at my work, and see if I am off base or not.

    Here is the problem (of which I do not know the answer to):

    A ski-elevator consists of a rope that skiers hold on to. The rope drags the skiers up a 250m long hill, with a 25 degree slope. The rope moves at a constant speed of 10 km/h. Assume no friction.

    a) How large is the work done by the rope on the skier with a mass of 80kg that is pulled to the top?

    b) The motor must generate enough power to drag 60 skiers to the top at the same time. How much power must the motor generate?


    What I have done:
    a) Since there is no friction, and no acceleration of the rope, I assume that the only work being done is going to provide potential energy. I draw a triangle, with hypotenuse of 250m, and a 25 degree angle, and say that 250*sin(25) is then equal to the elevation, which is 105.7m. Ep=mgh, where m=80, g=9.8 and h=105.7, and get that Ep=82.9 kJ= Work done.

    b) I assume that all skiers weigh the same, meaning that each skier requires 82.9 kJ, multiplied by 60, and I get 4974 kJ of work total. Since W=F*s, I assume that 4974=F*105.7, or that F=47.1 kN. The force is at a 90 degree angle, so I need the speed component that is also at 90 degrees. I convert 10km/h to 2.8 m/s, and draw a triangle with a hypotenuse of 2.8 and an angle of 25 degrees. Vertical velocity is then equal to 2.8*sin(25) which is 1.2 m/s.

    P=F*v, with F being 47.1 kN and v=1.2 m/s, so 47.1*1.2=56.52 kW.


    Have I made the correct assumptions here? Or is there a flaw in my logic/calculations? Maybe it is just me, but it just seems like 56.52 kW is a bit high...
     
  2. jcsd
  3. Feb 19, 2014 #2
    At the moment, I can't see a problem with your calculations. 56kW seems a bit low for 60 at the same time, but with no friction, it could work.
     
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