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    albear's Avatar
    albear Posts: 1,594, Reputation: 222
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    #1

    Jun 20, 2007, 11:35 AM
    electrical energy
    A winch in a boatyard uses a 230V electric motor to raise objects from a boat onto the quay side, the winch takes 22 s to raise an object of mass 160kg through a height of 5.0m.
    the motor current during this time is 14 A . Calculate the efficiency of the winch.

    I know that efficiency=eng used/eng provided

    eng used would be Pe=mgh =9.8*160*5=7.85*10^3 kg

    I think energy provided would be electrical energy ( because that's the only other energy energy involved right)

    now I know that E=0.5QV

    which I thought would equal 0.5*14*22*230 (14*22 because Q=t*I)

    the mark scheme however dissagrees with me removing the 0.5 from E=0.5QV
    albear's Avatar
    albear Posts: 1,594, Reputation: 222
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    #2

    Jun 20, 2007, 11:52 AM
    Hello
    albear's Avatar
    albear Posts: 1,594, Reputation: 222
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    #3

    Jun 20, 2007, 12:13 PM
    Hello
    albear's Avatar
    albear Posts: 1,594, Reputation: 222
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    #4

    Jun 20, 2007, 12:34 PM
    I would like an answer asap if anybody can help me
    albear's Avatar
    albear Posts: 1,594, Reputation: 222
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    #5

    Jun 20, 2007, 01:11 PM
    Anybody there
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Jun 21, 2007, 05:55 AM
    Your "energy used" calculation is correct, excet the units isn't Kg (that's mass, not work). I would have called it "work done" and the units is Kg(meter^2)/(sec^2), or Joules.

    The energy that is put into an electrical device (measured in Watt-sec) is Volts x Amps x time. So it's 230V x 14A x 22s = 70,840 watt-sec. It turns out that 1 watt-sec = 1 Joule, so your units are now consistent and all you need to do is divide the work done by this amount.

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