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Math Help - physics skier average power problem

  1. #1
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    physics skier average power problem

    In 2.0 minutes, a ski lift raises 3 skiers at constant speed to a height of 105 m. The average mass of each skier is 65 kg. What is the average power provided by the tension in the cable pulling the lift?

    please help with the equation.

    is that in watts?

    W
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  2. #2
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    I hope this is correct. Sorry, but all I've had was GeoPhysics so far so I'm a bit bad with Physics

    Well let's start by finding the force of the skiers.

    65kg \cdot 3 \cdot 9.8\frac{m}{s^{2}} = 1911N

    Now we need to find the work needed:

    1911N * 105m = 200655J

    Now to just divide by time to get power: (Remember to convert to seconds)

    \frac{200655J}{120s} = 1672.125W
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  3. #3
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    Quote Originally Posted by rcmango View Post
    In 2.0 minutes, a ski lift raises 3 skiers at constant speed to a height of 105 m. The average mass of each skier is 65 kg. What is the average power provided by the tension in the cable pulling the lift?

    please help with the equation.

    is that in watts?

    W
    Power is measured in Watts, in the MKS system of units. I don't recall what it is in the CGS system. In the English system of units it is measured in hp, "horsepower."
    P = \frac{W}{t}
    (Don't confuse this W with the symbol for Watts! This W is work.) Where P is the average power exerted over a time t. The work being done by the tension is equal to the weight of the three skiers (since the lift is operating at a constant velocity. If the skiers were accelerating we'd have to do a Newton's 2nd Law problem to find the tension in the cable.) The tension is in the same direction as the displacement, so W = \vec{T} \cdot \vec{s} = Ts \cdot cos(0^o) = Ts. The displacement is equal to the height h = 105 m. Thus:
    P = \frac{3mgh}{t} = 1672.13~W

    (So SnipedYou got it right after all. )

    -Dan
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  4. #4
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    Awesome solution, how do I understand when to use 0 degrees or 180 degrees? also, are these equations usually always measured in seconds, or is that just the standard to always use seconds for equations like these?

    thanks.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    Awesome solution, how do I understand when to use 0 degrees or 180 degrees?
    You need to look at the angle between the displacement and force vectors. (Place them "tail to tail" and look at the angle this makes.) Two vectors in the same direction have a 0 degree angle between them, two vectors in opposite directions have an angle of 180 degrees between them. The domain of possibilities is any angle from 0 to 180 degrees.

    Quote Originally Posted by rcmango View Post
    also, are these equations usually always measured in seconds, or is that just the standard to always use seconds for equations like these?
    In the MKS system of units time is measured in seconds as a standard. Then we have work in J and power in W.

    In the CGS system we have time in s, work in ergs, and I don't know what the unit for an "erg/s."

    In practically every system of units you are likely to run into for common use, time is going to be measured in seconds. (However this is not true for more advanced systems like the Heaviside-Lorentz system.) There is one exception: Most (or all) power companies in the US measure a unit of energy called a "kiloWatt-hr." Here, obviously, time is measured in hours. (Then, of course, we have mph, rpm, etc.)

    -Dan
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