3Forces

IA Dynamics and Relativity 3.1 Force and potential energy in one dimension
To define the potential, consider a particle of mass
m
moving in a straight line
with position
x
(
t
). Suppose
F
=
F
(
x
), i.e. it depends on position only. We
define the potential energy as follows:
Definition
(Potential energy)
.
Given a force field
F
=
F
(
x
), we define the
potential energy to be a function V (x) such that
F =
dV
dx
.
or
V =
Z
F dx.
V
is defined up to a constant term. We usually pick a constant of integration
such that the potential drops to 0 at infinity.
Using the potential, we can write the equation of motion as
m¨x =
dV
dx
.
There is a reason why we call this the potential energy. We usually consider
it to be an energy of some sort. In particular, we define the total energy of a
system to be
Definition (Total energy). The total energy of a system is given by
E = T + V,
where V is the potential energy and T =
1
2
m ˙x
2
is the kinetic energy.
If the force of a system is derived from a potential, we can show that energy
is conserved.
Proposition. Suppose the equation of a particle satisfies
m¨x =
dV
dx
.
Then the total energy
E = T + V =
1
2
m ˙x
2
+ V (x)
is conserved, i.e.
˙
E = 0.
Proof.
dE
dt
= m ˙x¨x +
dV
dx
˙x
= ˙x
m¨x +
dV
dx
= 0
Example. Consider the harmonic oscillator, whose potential is given by
V =
1
2
kx
2
.
Then the equation of motion is
m¨x = kx.
This is the case of, say, Hooke’s Law for a spring.
The general solution of this is
x(t) = A cos(ωt) + B sin(ωt)
with ω =
p
k/m.
A and B are arbitrary constants, and are related to the initial position and
velocity by x(0) = A, ˙x(0) = ωB.
For a general potential energy
V
(
x
), conservation of energy allows us to solve
the problem formally:
E =
1
2
m ˙x
2
+ V (x)
Since E is a constant, from this equation, we have
dx
dt
= ±
r
2
m
(E V (x))
t t
0
= ±
Z
dx
q
2
m
(E V (x))
.
To find
x
(
t
), we need to do the integral and then solve for
x
. This is usually
not possible by analytical methods, but we can approximate the solution by
numerical methods.