# working out base units

#### feenon

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 [FONT=Thorndale, Times New Roman, serif][/FONT]
[FONT=Thorndale, Times New Roman, serif][/FONT]

#### Kalter Tod

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
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"

#### peterkentt

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).

#### feenon

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]

#### feenon

base units

sorry having problems with the format
it is X squared
feenon

#### topsquark

Forum Staff
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

#### feenon

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

#### feenon

base units

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

#### topsquark

Forum Staff
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