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Current to an electric motor

  1. Feb 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A fully energized 12V battery is rated as "4 Ampere-hours." Suppose this battery is connected to a motor that is 90% efficient at converting electrical to mechanical power. How high could this battery-supplied motor lift a 8.1 kg mass?


    2. Relevant equations
    V = I*R
    P = V*I
    P = (I^2)*R

    3. The attempt at a solution

    An amp-hour is how many amps a battery can supply for one hour.
    Using the power equation:
    P = (12V)*(4A*hr) = 48 Watt*hr

    Since one watt-hr = 3600 Joules,
    (48 W*hr)*(3600) = 1728 Joules

    The motor is 90% efficient so,
    1728 * .9 = 1555.2 Joules

    I think the above is right (hopefully), but I'm not sure how to use the 1555.2 Joules to find the maximum height that this motor can lift the mass. Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Feb 12, 2008 #2
    Power isn't in watt-hours, a watt hour is energy

    Right idea still, 12V*4Ahr is 48 Watt*hr, but that itself equals 172800 joules

    90% of that is straight up how much work the motor can do. Work...mass...looking for distance...
     
  4. Feb 12, 2008 #3
    So, now at 90% of 172,800 J = 155,520 J

    Using conservation of energy:
    155,520 J = mgh
    155,520 J = (8.1 kg)*(9.8 m/s^2)*(h)
    h = 1959.18 m ?

    This seems like a large value but reasonable because the mass is not that great.
     
  5. Feb 14, 2008 #4
    Before I submit my answer, would you agree with using conservation of energy and the answer that I got?
     
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