Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Apr 15th 2008, 02:01 PM #1 Junior Member   Join Date: Apr 2008 Location: uk Posts: 5 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    Apr 15th 2008, 03:22 PM   #2
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 Originally Posted by feenon 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"   Apr 15th 2008, 09:36 PM #3 Junior Member   Join Date: Apr 2008 Posts: 1 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 16th 2008, 03:17 AM #4 Junior Member   Join Date: Apr 2008 Location: uk Posts: 5 re base units I am try to work out the base units for Ks in the equation. Es = ½ 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    Apr 16th 2008, 03:20 AM #5 Junior Member   Join Date: Apr 2008 Location: uk Posts: 5 base units sorry having problems with the format it is X squared feenon   Apr 16th 2008, 04:45 AM   #6

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 Originally Posted by feenon 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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See the forum rules here.   Apr 16th 2008, 06:40 AM #7 Junior Member   Join Date: Apr 2008 Location: uk Posts: 5 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}$ Feenon   Apr 16th 2008, 06:43 AM #8 Junior Member   Join Date: Apr 2008 Location: uk Posts: 5 base units Sorry ! Not sue how you get from J/m(squared) to N m /m(squared) Feenon   Apr 16th 2008, 08:13 AM   #9

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 Originally Posted by feenon 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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