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Energy to lift a mass, voltage, amps, time, distance confuse

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A student investigates the efficiency of a small electric motor. He uses a motor to lift it a mass through a constant distance of 1 m. He times how long it takes to lift the masses and makes a record of the potential difference and the current of the motor.

    In the book that I am reading there is a table of values. I'll give you values for two of the masses:

    Mass lifted = 0.01g; voltage of motor = 2.4V; current in motor = 0.20A; time to lift the mass = 22 seconds

    Mass lifted = 0.03g; voltage of motor = 2.4V; current in motor = 0.22A; time to lift the mass = 24.4 seconds

    Questions are:
    (i) What is the useful work done in Joules
    (ii) What is the electrical energy supplied?

    2. Relevant equations
    The book says useful work done = mgh

    The book does not give a formula for electrical energy supplied but I think it is E = IVt

    3. The attempt at a solution

    (i) For the 0.01g mass, useful work done= mgh = (0.01/1000) x 10 x 1 = 0.0001 J


    For the 0.03g mass, useful work done= mgh = (0.03/1000) x 10 x 1 = 0.0003 J

    (ii) For the 0.01g mass, electrical energy supplied E = IVt = 0.2 x 2.4 x 22 = 10.56 J

    For the 0.03g mass, electrical energy supplied E = IVt = 0.22 x 2.4 x 24.4 = 12.88 J


    I think there is something wrong here. How can the work done to lift the 0.01g mass be 0.0001 J but the energy supplied be 10.56 J ? Such a large difference. Does the motor just provide too much energy for the task? Is the excess just lost as heat and noise?

    The 0.01g is very low weight isn't it. May be the author meant 0.01 kg and that would give 0.1 J useful work done.

    What's going on? Are my answers incorrect or is this the way its meant to be.

    Please note that this is not homework. I'm just working through a book and there are no answers for this section. So please don't fear that you'll be "giving" me the answers for a homework. I'll be happy to accept straight explanations .... without riddles:smile::smile:

    Yours gratefully

    Barclay
     
    Last edited: Nov 2, 2015
  2. jcsd
  3. Nov 2, 2015 #2

    JBA

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    How did you know to use 1 meter for the h in your mgh calculation?
     
  4. Nov 2, 2015 #3
    The book described the experiment as lifting the mass 1m

     
    Last edited: Nov 2, 2015
  5. Nov 2, 2015 #4

    gneill

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    Staff: Mentor

    Edit your post, select "More Options". You should see a "Delete" option beside the thumbnail.
     
  6. Nov 2, 2015 #5
    Thanks for that
     
  7. Nov 2, 2015 #6

    billy_joule

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    Science Advisor

    All your working is correct.

    For small DC motor motors the rule of thumb efficiency is often stated as 50% (ie half of the electrical input energy is lost as heat).
    And that is the peak efficiency, when the motor is operating away from that point, at a different RPM, the efficiency is even lower. When the motor is stalled or free running (no load) then efficiency is 0% as no work is being done.

    With that said, I think It may be a typo.
     
  8. Nov 2, 2015 #7

    JBA

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    Sorry, I initially read right past that. (that is one problem created by having to scroll back and forth when evaluating a lower post)

    With that resolved, I don't see any errors in your calculations; so unless the two questions are meant to be totally independent of each other, I am as confused by the results as you are.
     
  9. Nov 2, 2015 #8
    That's a relief.

    I've finally met someone on the forum who is confused. I thought I was the only one.

    So the book have messed up somewhere I think
     
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