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physics help please thanks?

Mar-12-2010 Posted under Ski Squaw

Squaw Valley ski area in California claims that its lifts can move 46400 people per hour. If the average lift carries people about 200 m (vertically) higher, estimate the maximum total power needed. (Assume an average mass per person of 70 kg.)

The energy to move vertically 200 m one 70 kg person is mgh = 70Kg*10m/sec^2*200m = 140000 Joules.

The minimum power to move vertically 200 m 464000 people is

=140000 Joules * 46400/sec (answer is in Watts)

Note this is the minimum power assuming 100 % efficiency. In reality, you almost never achieve 100 % efficiency. Hence, the maximum power is impossible to calculate without an assumed efficiency.

Ron

  1. bozo Said,

    The energy to move vertically 200 m one 70 kg person is mgh = 70Kg*10m/sec^2*200m = 140000 Joules.

    The minimum power to move vertically 200 m 464000 people is

    =140000 Joules * 46400/sec (answer is in Watts)

    Note this is the minimum power assuming 100 % efficiency. In reality, you almost never achieve 100 % efficiency. Hence, the maximum power is impossible to calculate without an assumed efficiency.

    Ron
    References :

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