Joules and Kinetic Energy [units wise]

thomas49th
Messages
645
Reaction score
0
K.E is in joules

joules = 1Nm

Nm = m/s² . m
= m²/s²

therefore

k.e = 0.5mv²

m²/s² = kg . m²/s²

there is an extra kg. I don't see how

there is the same problem with GPe
GPE = m gh
m²/s² = kg . m²/s²

How can that be. The units of 0.5kgv² need to equal Nm

Thanks :)
 
Physics news on Phys.org
N = kg m s-2 ( from f=ma )

Energy = Nm = kg m2 s-2
 
i was just on my way to delete this post when i realized whay a stupid mistake i'd made.
F = ma
N = kg . m/s²
:blushing:
cheers :)
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top