# Power for accelerating a ski lift

• Mqqses
In summary, the conversation discusses a ski-lift with a one-way length of 987 m and a vertical rise of 275 m. It also mentions the spacing of chairs, their seating capacity, and the lift's operating speed. The task is to determine the power required to operate the ski lift and to accelerate it in 6.4 seconds. The solution for part (a) is correct, but part (b) is more challenging and involves considering both the change in kinetic and potential energy during acceleration. Different approaches are explored, including using an integral, but the result is a transcendental equation in power.
Mqqses

## 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

For b...

During the acceleration both the KE and PE changes.

Mqqses
Mqqses said:
(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.

Mqqses
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?

CWatters said:
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?

Mqqses said:
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.

Thanks for your insight into this problem, I think I'll leave this alone for now.

Mqqses said:
Thanks for your insight into this problem, I think I'll leave this alone for now.
Considering the math I would too!

## 1. What is the power needed to accelerate a ski lift?

The power needed to accelerate a ski lift depends on various factors such as the weight of the lift, the angle of the slope, and the desired acceleration. It can range from a few kilowatts to several megawatts.

## 2. How is power calculated for a ski lift?

The power for a ski lift can be calculated using the formula P = Fv, where P is power in watts, F is the force needed to accelerate the lift, and v is the velocity at which the lift is moving. This formula assumes a constant force and velocity.

## 3. Can renewable energy be used to power a ski lift?

Yes, renewable energy sources such as solar, wind, and hydro power can be used to power a ski lift. However, the amount of power needed and the feasibility of using renewable energy will depend on the specific location and conditions of the ski lift.

## 4. How does the power source affect the speed of a ski lift?

The power source can affect the speed of a ski lift by determining the maximum amount of power that can be supplied to the lift. A more powerful power source can allow the lift to accelerate faster and reach higher speeds.

## 5. How does the terrain affect the power needed to accelerate a ski lift?

The terrain can greatly affect the power needed to accelerate a ski lift. A steeper slope will require more power to overcome the increased resistance from gravity, while a flatter slope will require less power. The type of terrain, such as trees or open space, can also impact the power needed for a ski lift due to wind resistance and other factors.

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