How High Can a Motor Lift a Mass Using a Charged Capacitor?

AI Thread Summary
A 0.45 µF capacitor charged by a 1.5 V battery can be used to determine how high a small electric motor can lift a 4.7 g mass, assuming 100% efficiency. The key formulas to consider are U=1/2CV^2 and Q=CV, which help calculate the energy stored in the capacitor. To find the height the motor can lift the mass, one must equate the stored energy to the gravitational potential energy required to lift the mass. Using U=1/2CV^2 is the most efficient method for calculating the energy. This approach ensures that all energy from the capacitor is utilized in lifting the mass.
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Homework Statement



A 0.45 µF capacitor is charged by a 1.5 V battery. After being charged, the capacitor is connected to a small electric motor. Assuming 100% efficiency, to what height can the motor lift a 4.7 g mass?

Homework Equations



I really need some direction to start off this problem. I think that I probably need to use U=1/2QV=1/2CV^2=Q^2/2C and/or Q=CV but I have no idea how to go about doing this.
 
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The whole idea of this problem is that if the efficiency is 100% no energy will be lost. So all http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html" will be used to lift that mass. Now what would be the energy necessary to lift that mass to a height h?
 
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So...would I have to use Q=CV to find out Q and then use U=1/2QV to find out U ?
 
map7s said:
So...would I have to use Q=CV to find out Q and then use U=1/2QV to find out U ?

You could use any of the 3 formulas you mention, though since you know C and V the fastest would be to use U=\frac{CV^2}{2}
 
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