- #1
Johnson
- 29
- 0
A particle of mass m moves without friction subject to a force F(x) =
−kx + [itex]\frac{kx3}{A2}[/itex], where k and A are positive constants. It is projected
from x = 0 to the positive x direction with initial velocity v0 =
A[itex]\sqrt{\frac{k}{2m}}[/itex]. Find:
(a) the potential energy V (x),
(b) the kinetic energy T(x),
(c) the turning points of the motion.
So I don't know if I am on the right track, I feel I am missing something. For V(x) (potential energy), i got:
[itex]\frac{1}{2}[/itex] kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]
Then I found T(x) (kinetic energy) to be:
T[itex]_{o}[/itex] - [itex]\frac{1}{2}[/itex]kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]
I know I am given an initial V[itex]_{o}[/itex], but where would I plug that into find V(x) or T(x)?
Any help is very much appreciated.
Regards
−kx + [itex]\frac{kx3}{A2}[/itex], where k and A are positive constants. It is projected
from x = 0 to the positive x direction with initial velocity v0 =
A[itex]\sqrt{\frac{k}{2m}}[/itex]. Find:
(a) the potential energy V (x),
(b) the kinetic energy T(x),
(c) the turning points of the motion.
So I don't know if I am on the right track, I feel I am missing something. For V(x) (potential energy), i got:
[itex]\frac{1}{2}[/itex] kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]
Then I found T(x) (kinetic energy) to be:
T[itex]_{o}[/itex] - [itex]\frac{1}{2}[/itex]kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]
I know I am given an initial V[itex]_{o}[/itex], but where would I plug that into find V(x) or T(x)?
Any help is very much appreciated.
Regards