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30 October, 13:59

An 120 volt electric motor that draws 5 amps of current operates a winch that lifts a 100 kg. a distance of 300 cm. in 5 seconds.

Find work done by the winch, input power, output power, and efficiency of this system

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  1. 30 October, 15:47
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    Work done = 3000J

    Input Power = 600W

    Explanation:

    Potential, V = 120 volt

    Current, I = 5A

    mass, m = 100 kg

    distance, x = 300 cm = 3m

    time, t = 5 sec

    Work = ?

    Work = Force X distance

    Force needed to lift 100 kg object is 100 X 9.8 m/s = 1000N

    So,

    Work = 1000 x 3m = 3000J

    Therefore, work done to lift 100 kg object is 3000J

    Input power = ?

    Input power = V x I

    = 120 volt X 5A

    = 600 W

    Output power = ?

    If the energy is conserved then,

    Output power = input power = 600W

    Efficiency = ?

    Efficiency = output power / input power X 100

    Efficiency = 600 X 100 / 600 = 100%

    Therefore, no loss of energy
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