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

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Homework Help Overview

The problem involves a 0.45 µF capacitor charged by a 1.5 V battery, which is then connected to a small electric motor to lift a 4.7 g mass. The discussion centers around calculating the height to which the motor can lift the mass, assuming 100% efficiency.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of energy equations related to capacitors, specifically U=1/2QV and U=1/2CV^2. There are questions about how to derive charge (Q) and energy (U) from the given values of capacitance (C) and voltage (V).

Discussion Status

The discussion is ongoing, with participants exploring different equations and their relationships. Some guidance has been offered regarding the use of specific formulas, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants are operating under the assumption of 100% efficiency, which is a critical aspect of the problem. There is also a focus on ensuring that all energy from the capacitor is utilized in lifting the mass.

map7s
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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 [tex]C[/tex] and [tex]V[/tex] the fastest would be to use [tex]U=\frac{CV^2}{2}[/tex]
 

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