tricky physics question (for me anyway!)

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2008-10-26 12:53:11

dom117: Member; +17|6914

____[cell]____
| |
| |
| |
|___[motor]___| <--- an attempt at the circuit diagram!

basically, its a motor lifting a 500g load through a height of 75cm in 12 seconds.

take g = 9.8ms-2 and the current = 30mA

assuming the motor to be 100% efficient and the internal resistance of the cell to be negligible, calculate the EMF of the cell.

help would be much appreciated! thnx

Dom

2008-10-26 12:55:45

CanadianLoser: Meow :3 :3; +1,148|6958

could you post some relevant equations just to help remind me

2008-10-26 12:56:37

Scorpion0x17: can detect anyone's visible post count...; +691|7216|Cambridge (UK)

ooh... can't remember the exact equations, but you need to calculate the work required to lift the weight that distance in that time... then... erm... EMF I can't remember... but it probably involves ohms (or someone else's) law(s)...

Summat like that, anyway...

Or I could be way off the mark...

Last edited by Scorpion0x17 (2008-10-26 12:57:50)

2008-10-26 12:59:15

dom117: Member; +17|6914

internal resistance = (EMF-terminal voltage) / current

pd = current x resistance

power delivered to load = current ^2 x load resistance

lost my formula sheet so im afraid thats all ive got to go with..dont worry if you cant! x

2008-10-26 13:01:02

CanadianLoser: Meow :3 :3; +1,148|6958

Scorpion0x17 wrote:
ooh... can't remember the exact equations, but you need to calculate the work required to lift the weight that distance in that time... then... erm... EMF I can't remember... but it probably involves ohms (or someone else's) law(s)...

well Work = Force x Displacement
and if force = ma then

Work = (mass)(acceleration)(Displacement)
Work = 3.675 J

idk what to do next or if that is even needed.

2008-10-26 13:05:51

JoshP: Banned; +176|6139|Notts, UK

*gets out huge physics textbook

power = energy/ time

charge = current x time

p.d = power / current

resistance = p.d / current

that should be all you need.

add me on xfire if u need more help, i'm bored atm lol.

2008-10-26 13:07:39

Flaming_Maniac: prince of insufficient light; +2,490|7157|67.222.138.85

P = W/t = [(.75m)(.5kg)(9.8m/s²]/12s = I²R

I think that might help some, but I don't understand some of the terminology you are using, like what EMF is...looking into it now.

edit: Oh is EMF voltage? (guessing from the way you wrote an equation at the top, how I would write Voltage = amperage x resistance)

If so then you use what I wrote to solve for R plugging in the known value for I, then multiply R by I.

2008-10-26 13:09:01

JoshP: Banned; +176|6139|Notts, UK

EMF = electromotive force

"The electromotive force of any source of electrical energy is the energy converted into electrical energy per unit charge supplied"

2008-10-26 13:10:38

Flaming_Maniac: prince of insufficient light; +2,490|7157|67.222.138.85

JoshP wrote:
EMF = electromotive force

"The electromotive force of any source of electrical energy is the energy converted into electrical energy per unit charge supplied"

Yeah I just figured that out (see edit), it's what you silly Brits call voltage.

2008-10-26 13:27:56

JoshP: Banned; +176|6139|Notts, UK

Flaming_Maniac wrote:
JoshP wrote:
EMF = electromotive force

"The electromotive force of any source of electrical energy is the energy converted into electrical energy per unit charge supplied"
Yeah I just figured that out (see edit), it's what you silly Brits call voltage.

but it's not really, it's the voltage supplied by the battery/cell

it's do to with internal resistance, gimme a second while i read mah textbook, i'll edit the post when i work it out

right, i reckon that EMF = the force supplied by the cell/battery that makes the electrons move

Voltage/Potential difference = potential difference across the 2 terminals of the battery

more to come still, it's complicated

edit: amperage LOL WAT? u mean current lol

Last edited by JoshP (2008-10-26 13:38:51)

2008-10-26 13:32:26

Flaming_Maniac: prince of insufficient light; +2,490|7157|67.222.138.85

JoshP wrote:
Flaming_Maniac wrote:
JoshP wrote:
EMF = electromotive force

"The electromotive force of any source of electrical energy is the energy converted into electrical energy per unit charge supplied"
Yeah I just figured that out (see edit), it's what you silly Brits call voltage.
but it's not really, it's the voltage supplied by the battery/cell

it's do to with internal resistance, gimme a second while i read mah textbook, i'll edit the post when i work it out

edit: amperage LOL WAT? u mean current lol

amperage, current, same deal

edit: maybe what's closer to what we call it is voltage difference or voltage drop across a circuit.

2008-10-26 13:40:11

JoshP: Banned; +176|6139|Notts, UK

Flaming_Maniac wrote:
JoshP wrote:
Flaming_Maniac wrote:
Yeah I just figured that out (see edit), it's what you silly Brits call voltage.
but it's not really, it's the voltage supplied by the battery/cell

it's do to with internal resistance, gimme a second while i read mah textbook, i'll edit the post when i work it out

edit: amperage LOL WAT? u mean current lol
amperage, current, same deal

edit: maybe what's closer to what we call it is voltage difference or voltage drop across a circuit.

yes- E = I(R + r) for the circuit that i'm about to draw in mspaint an upload see below, it's about the difference between the voltage across the battery, and the actualy voltage "produced" by the battery which loses a bit of itself going across the battery due to internal resistance

Last edited by JoshP (2008-10-26 13:46:06)

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