Electrical Energy Storage, Capactor, Battery, Motor

AI Thread Summary
The discussion revolves around calculating the height a motor can lift a 2.2-gram mass using a charged capacitor and a 1.5 V battery. The user initially attempted to find the stored energy using the formula U = 1/2CV^2 and then applied it to the equation u = mgy, but encountered errors in their calculations. A key point raised is the importance of using consistent units, specifically converting grams to kilograms for accurate results. The user ultimately resolved their confusion by recognizing the unit discrepancy, leading to a successful understanding of the problem. This highlights the significance of unit conversion in physics calculations.
yayirunin2car
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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?

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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, :):).
Diana
 
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In principle, your method is correct (conservation of energy).

Calculator error? Units? Answer key is wrong?
 
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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