Physics capacitor question...

[Xiah]

Hi guys i'm pretty new here and i'm just looking for abit of help really. I'm attempting my physics assignment and i can't for the world of me figure out how to work this one out and searching online for equations is just making it more confusing for me! The question is below.

A 4700 uF capacitor is charged to a p.d. of 10v. When the capacitor is discharged through a small electric motor, the motor lifts a weight of 0.1N through a height of 0.12m.
I need to work out the energy stored in the charged capacitor - I've done this one I think! E = 1/2 CV^2 which is 1/2 * 4700 * 100
The gravitational potential energy gained by the weight
The efficiency of the energy transfer - This one i think i can do!

Any help at all would be greatly appreciated so thanks in advance!!

 

[ferdem]

You have overlooked "uF". Stored energy in the cap= 0.5*4700e-6*100 =235 mJ (millijoule)
0.1N does 0.1*0.12=12mJ much work.
e=12/235=0.05 (!)

 

[ServoAmp]

Wtransfer is the energy transferred = Work physical
Wloss = energy lost in the transfer
Total energy = Wloss + Wtransfer = Work Capacitor
Efficiency = Wtransfer/total energy

Work (physical) = .1N*.12M =.012Nm
Work Capacitor = 0.5*4700E-6*100 = 0.235 Joules = .235 Nm
Wloss = Wcapacitor=Wphysical = .235-.012 = .223
Efficiency = .012/.235 = .051 = 5%

Work (physical) = F*d
**broken link removed**)
Work Capacitor = 0.5CV^2
Capacitance - Wikipedia, the free encyclopedia
I Joule = 1N*1m
Joule - Wikipedia, the free encyclopedia

 

[Xiah]

I thought the gravitational potential energy gained by the weight would be = mgh 0.1 * 9.8 * 0.12 = 0.1176?

Thank you both so much for your help by the way!

 

[_Eduardo_]

It´s a weight of 0.1N not a mass of 0.1Kg
Already was multiplied by 9.8 ( weight = mg )

 

[sagar474]

yes _Eduardo_ you are correct

 

[Xiah]

Much appreciated thankyou