Physics Help Forum working out base units

 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
Junior Member

Join Date: Apr 2008
Location: Pittsburgh, PA
Posts: 11
 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

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,554
 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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

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

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,554
 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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Tags base, units, working

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post THERMO Spoken Here Kinematics and Dynamics 3 Sep 18th 2014 02:07 PM Mooki Kinematics and Dynamics 7 Aug 23rd 2010 03:37 PM labview1958 Electricity and Magnetism 0 Jul 21st 2010 12:46 AM topsquark Physics Resources 7 Jan 26th 2009 10:38 PM arbolis General Physics 2 Aug 2nd 2008 03:21 PM