working out base units

Apr 2008
5
0
uk
I am try to work out the correct base unit for a given equation.
The equation is [FONT=Thorndale, Times New Roman, serif]Es/[/FONT][FONT=Thorndale, Times New Roman, serif]X2[/FONT]

[FONT=Thorndale, Times New Roman, serif]Es = J[/FONT]
[FONT=Thorndale, Times New Roman, serif]X2 = m

givining J/m
[/FONT]
I know that
[FONT=Thorndale, Times New Roman, serif] 1 J = 1 kg m2 s-1[/FONT]
so it would be [FONT=Thorndale, Times New Roman, serif]1 kg m2 s-1 [/FONT] [FONT=Thorndale, Times New Roman, serif]/ divided by Meter[/FONT]
[FONT=Thorndale, Times New Roman, serif][/FONT]I think the answer is [FONT=Thorndale, Times New Roman, serif]kg m1 s-1[/FONT]
thanx
feenon :rolleyes:[FONT=Thorndale, Times New Roman, serif][/FONT]
[FONT=Thorndale, Times New Roman, serif][/FONT]
 
Apr 2008
11
2
Pittsburgh, PA
I am try to work out the correct base unit for a given equation.
The equation is [FONT=Thorndale, Times New Roman, serif]Es/[/FONT][FONT=Thorndale, Times New Roman, serif]X2[/FONT]

[FONT=Thorndale, Times New Roman, serif]Es = J[/FONT]
[FONT=Thorndale, Times New Roman, serif]X2 = m

givining J/m
[/FONT]
I know that
[FONT=Thorndale, Times New Roman, serif] 1 J = 1 kg m2 s-1[/FONT]
so it would be [FONT=Thorndale, Times New Roman, serif]1 kg m2 s-1 [/FONT] [FONT=Thorndale, Times New Roman, serif]/ divided by Meter[/FONT]
I think the answer is [FONT=Thorndale, Times New Roman, serif]kg m1 s-1[/FONT]
thanx
feenon :rolleyes:
Is X2 supposed to have units of \(\displaystyle m\) or \(\displaystyle m^2\)? That will obviously change your answer, but I'm not 100% sure what you mean by "x2"
 
Apr 2008
1
0
If I got feenon question he's doing a kind of dimensional analysis:

ES = J = N x m

X2 = m

So; Es/X2 = N = m x a = kg x m x s^(-2).
 
Apr 2008
5
0
uk
re base units

I am try to work out the base units for [FONT=Thorndale, Times New Roman, serif]Ks[/FONT] in the equation. [FONT=Thorndale, Times New Roman, serif][/FONT]
[FONT=Thorndale, Times New Roman, serif]Es = ½ Ks X2[/FONT]
[FONT=Thorndale, Times New Roman, serif][/FONT]
[FONT=Thorndale, Times New Roman, serif]Es = [/FONT] strain potential energy
[FONT=Thorndale, Times New Roman, serif]Ks = Spring constant.[/FONT]
[FONT=Thorndale, Times New Roman, serif]X2[/FONT] = is the extension or compression of the spring

I have made [FONT=Thorndale, Times New Roman, serif]Ks[/FONT] the subject of the equation [FONT=Thorndale, Times New Roman, serif][/FONT][FONT=Thorndale, Times New Roman, serif]Ks = [/FONT][FONT=Thorndale, Times New Roman, serif]Es / [/FONT][FONT=Thorndale, Times New Roman, serif]X2
I have worked the base unit of
[/FONT][FONT=Thorndale, Times New Roman, serif]Ks [/FONT][FONT=Thorndale, Times New Roman, serif] to be [/FONT] [FONT=Thorndale, Times New Roman, serif]kg m s-2 m-1[/FONT]
but i'm not sure if this is correct ?

feenon

ps I'm a she not a he:)
[FONT=Thorndale, Times New Roman, serif][/FONT]
 
Apr 2008
5
0
uk
base units

sorry having problems with the format
it is X squared
feenon
 

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
sorry having problems with the format
it is X squared
feenon
Okay so if I've got you right you have
\(\displaystyle E_S = \frac{1}{2}K_Sx^2\)
and you want the units for \(\displaystyle K_S\).

So
\(\displaystyle K_S = \frac{2E_S}{x^2}\)

In terms of units then
\(\displaystyle \frac{J}{m^2} = \frac{N~m}{m^2} = \frac{N}{m}\)

\(\displaystyle = \frac{kg~m/s^2}{m} = \frac{kg}{s^2}\)

-Dan
 
Apr 2008
5
0
uk
base units

Thank you for your help. I still don't understand how you get from

\(\displaystyle
\)
\(\displaystyle \frac{J}{m^2}
\)

to
\(\displaystyle
\)
\(\displaystyle \frac{N~m}{m^2}
\)


:confused:
Feenon
 
Apr 2008
5
0
uk
base units

Sorry !
Not sue how you get from J/m(squared) to N m /m(squared)
Feenon
 

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
Sorry !
Not sue how you get from J/m(squared) to N m /m(squared)
Feenon
\(\displaystyle J = N~m\)
and
\(\displaystyle N = kg~m/s^2\)

-Dan