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Power for accelerating a ski lift

  • Thread starter Mqqses
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Homework Statement


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).

Homework 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).

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
 

Answers and Replies

  • #2
CWatters
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For b...

During the acceleration both the KE and PE changes.
 
  • #3
rude man
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(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)
why the sin theta?
[/QUOTE]
Plus of course see post #2.
 
  • #4
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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?
 
  • #5
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For b...

During the acceleration both the KE and PE changes.
So would I use an integral from 0 to 6.4s?

06.4 ∂PE/∂t + ∂KE/∂t where PE= mgh and KE= 1/2mv2?
 
  • #6
rude man
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So would I use an integral from 0 to 6.4s?

06.4 ∂PE/∂t + ∂KE/∂t where PE= mgh and KE= 1/2mv2?
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.
 
  • #7
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Thanks for your insight into this problem, I think I'll leave this alone for now.
 
  • #8
rude man
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Thanks for your insight into this problem, I think I'll leave this alone for now.
Considering the math I would too! :smile:
 

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