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Power for accelerating a ski lift

  1. Jun 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A ski-lift has a one-way length of 987 m and a vertical rise of 275 m. The chairs are spaced 21 m apart, and each chair can seat three people. The lift is operating at a steady speed of 14 km/hr. Neglecting friction and air drag and assuming the average mass of each loaded chair is 242 kg, determine the power required to operate this ski lift (a). Also estimate the power required to accelerate this ski lift in 6.4 s to its operating speed when it is first turned on (b).

    2. Relevant equations
    This is my most recent stab at the solution for part (b), is this the right idea? Can you poke a hole in it or two? I am not even sure that this makes sense. Please note part (a) is correct, but I am having a hard time arriving at a solution for part (b).

    3. The attempt at a solution
    (a) 987m / 21m =47 chairs * 242 kg= 11,374 kg total
    11,374 kg * 9.81 m/s2 * 14 km/hr ((1 m/s) / (3.6 km/hr)) ((1 kJ/kg) / (1000 m2/s2))= 119,237 W / 1000 W/kW= 119.237 kW which is correct per my homework solution.

    (b) This is the part I am struggling with, and I know there are a couple of ways to approach it.

    1/2 m (V22-V12)*sin(θ) ; Where m=11374 kg, sin(θ)= 275/987, V = 3.8888 m/s

    = 23963.5 J / 1000 J/kJ = 23.96 kJ

    P= above answer/ the given time to accelerate, so P= 23.96 kJ/6.4 s = 3.74 kW
     
  2. jcsd
  3. Jun 20, 2016 #2

    CWatters

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    For b...

    During the acceleration both the KE and PE changes.
     
  4. Jun 22, 2016 #3

    rude man

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    why the sin theta?
    [/QUOTE]
    Plus of course see post #2.
     
  5. Jun 22, 2016 #4
    I'm not sure how to approach post #2, but I think that would be some sort of ∂PE/∂t + ∂KE/∂t for total energy required to move the lift. The sin θ may be a wrong idea. Am I getting the gist that an integral might be used for this?
     
  6. Jun 22, 2016 #5
    So would I use an integral from 0 to 6.4s?

    06.4 ∂PE/∂t + ∂KE/∂t where PE= mgh and KE= 1/2mv2?
     
  7. Jun 23, 2016 #6

    rude man

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    Kind of, but looks rough. You could go
    Power P = d(KE)/dt + d(PE)/dt
    = m d(v2/2)/dt + mg dh/dt
    v = velocity along slope = v(t) i.e. not constant. We assume P is constant.
    h = height
    If you proceed that way you can come up with an equation expressing P in terms of v, dv/dt and the various known constants.
    This is an ODE with v as the dependent and t as the independent variable.
    You can integrate by parts to equate the time (6.4s) to an expression in vf and P plus the constants where
    vf = final velocity of 14 km/hr. You can then solve for P.
    Problem is the result is a transcendental equation in P. Might try wolfram alpha.
    Probably is a better way but right now I don't know of one.
     
  8. Jun 23, 2016 #7
    Thanks for your insight into this problem, I think I'll leave this alone for now.
     
  9. Jun 23, 2016 #8

    rude man

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    Considering the math I would too! :smile:
     
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