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Old Apr 15th 2008, 02:01 PM   #1
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working out base units

I am try to work out the correct base unit for a given equation.
The equation is Es/X2

Es = J
X2 = m

givining J/m

I know that
1 J = 1 kg m2 s-1
so it would be 1 kg m2 s-1 / divided by Meter
I think the answer is kg m1 s-1
thanx
feenon
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Old Apr 15th 2008, 03:22 PM   #2
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Originally Posted by feenon View Post
I am try to work out the correct base unit for a given equation.
The equation is Es/X2

Es = J
X2 = m

givining J/m

I know that
1 J = 1 kg m2 s-1
so it would be 1 kg m2 s-1 / divided by Meter
I think the answer is kg m1 s-1
thanx
feenon
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"
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Old Apr 15th 2008, 09:36 PM   #3
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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).
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Old Apr 16th 2008, 03:17 AM   #4
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re base units

I am try to work out the base units for Ks in the equation.
Es = Ks X2

Es = strain potential energy
Ks = Spring constant.
X2 = is the extension or compression of the spring

I have made Ks the subject of the equation Ks = Es / X2
I have worked the base unit of
Ks to be kg m s-2 m-1
but i'm not sure if this is correct ?

feenon

ps I'm a she not a he
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Old Apr 16th 2008, 03:20 AM   #5
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base units

sorry having problems with the format
it is X squared
feenon
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Old Apr 16th 2008, 04:45 AM   #6
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Originally Posted by feenon View Post
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
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Old Apr 16th 2008, 06:40 AM   #7
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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}
$



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Old Apr 16th 2008, 06:43 AM   #8
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base units

Sorry !
Not sue how you get from J/m(squared) to N m /m(squared)
Feenon
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Old Apr 16th 2008, 08:13 AM   #9
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Originally Posted by feenon View Post
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
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