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Electrical Energy Storage, Capactor, Battery, Motor

  1. Sep 11, 2008 #1
    Here is my problem [the last one of a lot that I already got all right and have to turn in by midnight - so you can sense my frustration, lol.]

    A .39microF capacitor is charged by a 1.5 V battery. After being charged, the capacitor is connected to a small electric motor.

    a) Assuming 100% efficiency, to what height can the motor lift a 2.2 gram mass?

    b) What initial voltage must the capacitor have if it is to lift a 2.2 g through a height of 1 cm?

    What I tried doing was finding the stored energy using U = 1/2CV^2. Once I got that I plugged it into u=mgy, because for some reason I thought my 'y' would give me the height. The answers I got were wrong though, and I've been wracking my brain and reading the book trying to figure it out, but I'm at a loss.

    Thanks in advance for any help, :):).
  2. jcsd
  3. Sep 11, 2008 #2
    In principle, your method is correct (conservation of energy).

    Calculator error? Units? Answer key is wrong?
  4. Sep 11, 2008 #3
    Yes! Thank you! I totally forgot that Joules had 'kg' in it and not 'g.' I'm so happy now I can sleep in peace, :).
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