# The GPe questiona dn the mass and distanceH (1 Viewer)

#### GaganDeep

##### fffff
How did u guys do it, i jst lft em. For the Distance between two rods, i jst didn't knwo the F
the net force force = attractive (RIght?) - mg
i didn't know what m and i didn't know what the net force was lol, how do u do it?

##### Member
yeah how did you find D between the rods, once i did f=mg, i did the
f/l= K.I1.I2/d

then re-arranged to get D, ehich turned out to somthing x10^-3 which makes sense to me as a distance.

#### GaganDeep

##### fffff
but that's whent eh rod is statinary,, in not current in second rod right

#### Lions_Fist

##### Member
Hmm... don't ask why, but for some reason I took g = 10 and just did the obvious conversion in my head... Stupid stupid stupid...

Ohh well, anyway I ended up with an answer about 2.7 cm I think?

#### shinji

##### Is in A State Of Trance
i got 2 x 10^-3 m

so yeah. did the same crap as shady01

##### Member
yes i know you cant get poles that close, but think about it the current is really weak. 50A is fuck all(i hope) for a 2.6 m long pole therefore i think they would have to be that close. I thought it was one of those questions where the examiners hadnt really thought about it and just made random measurements up.

I'm keen thought how did others apporach it? becuase i can see myself being rong but its all about conviencing yourself your right so you dont bog down ahaha.

#### GaganDeep

##### fffff
wat did u use if I2 ?

#### Ether

##### New Member
i think you use the inverse of teh gradient where i forgot the axes but you get the inveerse, use that F=LxKxI1/m where 1/m (inverse of gradient) = I2/d or something.

you use the previous part in the qeustion mass for F=mg

and you sub in everything and then find d

i didnt get those small numbers, i think i got something larger, cant remember

#### Mumma

##### Member
I got 0.139 meters

#### brucefeng644

##### New Member
ermm...attractive?? wasnt the rods repelling that way there was a force measured??? and thus the currents in the conductors must be going in opposite directions???

#### Ether

##### New Member
brucefeng644 said:
ermm...attractive?? wasnt the rods repelling that way there was a force measured??? and thus the currents in the conductors must be going in opposite directions???

no, if you looked at the table the weight measured on teh balance was decreasing, hence the rod was being lifted up by the attraction force between the 2 rods

#### Mumma

##### Member
brucefeng644 said:
ermm...attractive?? wasnt the rods repelling that way there was a force measured??? and thus the currents in the conductors must be going in opposite directions???
No, the weight was decreasing as the current was being increased. Thus attracted.

#### XcarvengerX

##### Chocobo
I couldn't think on the exam, so I just wrote F is constant and I2 is variable, then make d the subject. I know it is wrong, but at least I got some marks for the substitution...

#### cjh

##### New Member
What was the answer to that GPE question? with the satilitie moving from 10000km to 20000km then 80000km?

#### Ether

##### New Member
lol one of the annoying questions, so i remember the answer i got

1.7MJ, lol highly doubt it's right, but let's hope not lol

#### klaw

##### Member
For the GPE question this is what I put:
KL says:
you know how it said that 1 MJ of work had to be done to get from 10000km to 20000 km?
KL says:
I just said that GPE at 20000 km - GPE at 10000 km was equal to 1 MJ
KL says:
so then -Gmm/r-(-Gmm/2r)=1 MJ, where r = 10000 km
KL says:
so -Gmm/r=2 MJ
KL says:
and at 80000, it's -Gmm/8r
KL says:
which = -Gmm/r * 1/8 = 2 * 1/8 = .25 MJ
KL says:
and so the work required to get from 20000 km to 80000 km was 1 MJ - 0.25 MJ
KL says:
= 0.75 MJ
KL says:
anything wrong with what I've done?

and for the rods question it said in the question that the upper rod had a current of 50A running through it. So I just took a point from the line of best fit and substituted it in. Force = weight of the rod - the force experienced then just rearrange the equation.

Last edited:

#### andrewbode

##### New Member
find the gradient of the line and muktiply by 9.6 to get F/Iand make the equation f/I=KLI/d

#### Gecko888

##### Gecko88
This is what I did:

invert that then substitute in values

Last edited:

#### Gecko888

##### Gecko88
For GPE I got the same as KLAW, but I wrote the answer in joules: 750, 000 J

#### iamben

##### New Member
i made Gmm a constant and worked out that if 10000 to 20000 takes -1.0M.j then from 0 to 10000 takes -2.0M.j. Thus total to get from 0 to 20000 = -3.0 M.j

It takes -0.25M.j to get from 0 to 80000 so to get from 20000 to 80000 its -0.25 - (-3.0) = 2.75M.j

Can anyone verfiy?