Rectilinear motion of a particle

  • Thread starter Johnson
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  • #1
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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 in to find V(x) or T(x)?

Any help is very much appreciated.

Regards
 

Answers and Replies

  • #2
29
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Anyone? Do I need to include other attempts?
 
  • #3
Hi,
It seems to me you've done very well so far.
You did correctly work out, that under a central force expressible as:
[itex]
\vec{F} = -\vec{\nabla}{V}
[/itex]
So that in your case,
[itex]
V(x) = -\displaystyle \int_0^x{Fdx}
[/itex]
Here, under the constraints on the given question, it is okay to take the potential energy equal to some base/initial value of your choice(at x = 0), say even zero.
But recall that m*x''t = F(x), so it may be best to solve a differential equation for x(t), in order to find out the kinetic energy, which will be, in this case K = mx'(t)^2/2;
Try that,
Daniel
 

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