Recent content by dannolul

I can get the slope of the line, but then I dont know what the slope is (units or whatever it is) then i get lost from there.

In class we havnt really talked about high k constant and only worked with small k constants. So it seemed a little skeptical as I didnt think it should be that high. Appreciate the help.

Oh Sorry, was on my phone and rushing. I expanded ET1=ET2 to get (Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom) Rearanged i got k = (mv^2+2mgh)/x^2 so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2 =479137.945N/m

I expanded ET1=ET2 to get (Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom) Rearanged i got k = (mv^2+2mgh)/x^2 so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2 =479137.945N/m