Download Motion and Potential Energy Graphs

Class Notes 4, Phyx 2110
Motion and Potential Energy Graphs
I.
INTRODUCTION
15
Often one is concerned with the motion of an object
that is acted on by a single, conservative force. For
example, the object might be a satellite in orbit around
the Earth where the single, conservative force is the force
of gravity on the satellite. For such a force we can define a
potential energy PE, and the body speeds up and slows
down in such a way that its total mechanical energy
ME is conserved, i.e., is constant.
In these notes we discuss how a graph of the potential
energy vs position can be used to discern the motion of
the body. For concreteness, we consider two different
conservative forces: one due to the gravitational pull of
the Earth (near its surface) and the other due to an ideal
spring.
There are two important concepts to keep in mind
when using a potential energy graph to understand an
object’s motion:
(1) The negative of the slope of the graph is equal to the
force on the object. That is, the force is the negative of
the spatial derivative of the potential-energy function.
(2) The difference between the total mechanical energy
ME of the object (which is constant) and the potential
energy PE equals the kinetic energy KE. That is KE =
M E − P E.
II.
GRAVITY
The potential energy associated with a mass m at some
vertical height y above the Earth’s surface is given by
P Eg = mg(y − y0 ).
(1)
In this expression, y0 can be any height; for example you
can choose y0 = 1 m if you wish. Above that point, the
potential energy increases and is positive; below it the
potential energy decreases and is negative. Don’t worry
about the sign of the potential energy: it’s only changes
in potential energy that are interesting – and those can
be either positive or negative.
Everywhere, the slope of the graph is mg. Thus the
gravitational force on the mass is −mg (“−” because the
force points down; remember, we defined the positive direction to be up). That is, the gravitational force is given
by
d
Fg = − [mg(y − y0 )] = −mg.
dy
(2)
Figure 1 shows the gravitational potential energy for
a 1 kg object near the surface of the Earth. Suppose we
place such an ogject 0.75 m above y0 (point A on Fig.
10
5
0
-1.5
-1
-0.5
0
0.5
1
1.5
-5
-10
-15
y-y (m)
0
Figure 1. Potential energy mg(y-y0) of a 1 kg mass vs height y-y0
2) and let go. We know that gravity will accelerate the
object downwards with acceleration g. As it does so it
will lose potential energy P Eg and gain kinetic energy
KE in such a way that its total mechanical energy ME
remains constant.
This motion can be deduced by looking at the
potential-energy graph using the two concepts outlined
above. Initially KE = 0 and from the graph we see that
P Eg = +7.5 J. Therefore ME = +7.5 J . However at
point A there is a force acting because the slope is not
zero. The object will thus accelerate in the direction
of the force, which is negative (downwards). As long
as there are no other forces, the objects’s PE will decrease as it falls and its KE will increase. At point B
(y − y0 = +0.25 m), for example, P Eg has dropped to
+2.5 J; thus KE has increased to 5.0 J [so that KE + PE
= +7.5 J (the total mechanical energy) has remained the
same]. If the object can fall below the y = y0 mark (don’t
be prejudiced: y0 doesn’t have to be at the surface of the
Earth, it can be anywhere), P Eg becomes negative. At
C (y − y0 = −0.5 m), for example, P Eg equals −5 J, so
the KE must equal 12.5 J (that is, 12.5 J + (− 5 J) =
+7.5 J). Until the Earth gets in the way (not pictured)
the object continues to lose P Eg and gain KE.
III.
THE SPRING FORCE
The potential energy associated with a mass attached
to a spring depends on how much the spring is stretched
or compressed. Let the equilibrium length of the spring
be L0 , and it’s current length be L. L can be greater
(stretch) or less (compression) than L0 . The potential
energy associated with a spring that is either stretched
or compressed is
2
15
1.2
10
1
ME = 7.5 J
0.8
5
C
B
0.6
0
-1.5
-1
-0.5
B
0
A
ME = 0.49 J
0.5
1
1.5
0.4
-5
0.2
-10
-0.15
-15
-0.1
C -0.05
0
0
0.05
A
0.1
0.15
x (m)
y-y (m)
0
Figure 4. Potential energy ½ kx2, illustrating the change in
potential energy as the mass moves from A to B to C.
Figure 2. Potential energy of 1 kg mass, illustrating the
decrease in potential energy as the mass moves from point
A to B to C.
1.2
1
0.8
0.6
0.4
0.2
0
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
x (m)
Figure 3. Potential energy 1/2 kx2 vs position x.
lower curve: k = 40 N/m; upper curve: k = 200 N/m.
P Esp =
1
2
k (L − L0 ) ,
2
(3)
where k is the spring constant – a quantity that measures the stiffness of the spring. (k is measured in N/m).
The value of k depends on what the spring is made of
and how loosely or tightly coiled the spring is. We usually replace the difference (L − L0 ) by x, so that
P Esp =
1 2
kx .
2
(4)
As discussed above, the negative of the slope (derivative) of the potential energy is the force. From this equation for P Es p we thus have
Fsp
d
=−
dx
1 2
kx
2
= −kx.
(5)
Figure 3 shows two potential energy graphs, one for
an object attached to a spring with spring constant k =
40 N/m (lower curve) and one for an object attached
to a spring with spring constant k = 200 N/m (upper
curve). In contrast to gravity, where the force is the
same at every position (height), for a spring the force
sometimes points to the right (for negative values of x
— compressions) and sometimes to the left (for positive
values of x — stretches). The magnitude of the force
is zero (equilibrium) when x = 0 (no compression or
stretch) and grows as the magnitude of x grows.
Now, let’s see how motion works for an object attached
to a spring. The potential energy for an object attached
to a spring is pictured in Fig. 4 for a value of k = 200
N/m. We start by stretching the spring by 0.07 m (point
A). If the object is released from rest it has a total mechanical energy of +0.49 J. At point A, the potential
energy has a positive slope so the object will experience a force in the negative direction. It accelerates, and
as it does so it gains KE and loses P Esp . At point B
(x = +0.05 m), the objects’s PE is +0.25 J and its KE
must be 0.24 J. As the object passes through x = 0 (the
position of no stretch) it experiences instantaneously no
force (the PE slope is zero there) but it is moving with
a KE of 0.49 J. It continues past x = 0 and begins to
compress the spring, slowing down as it does so. Finally,
at x = −0.07 m (C) the object has lost all of its KE and
P Esp is back to +0.49 J. At that instant the object is at
rest, but it feels a force pushing to the right (the PE slope
is negative, so the force points in the positive direction),
so it begins to accelerate to the right. Eventually the
object gets to x = +0.07 m again and the whole process
repeats over and over. Thus the object oscillates back
and forth, with kinetic and potential energy constantly
being exchanged as it oscillates.