Transcript Physics 430: Lecture 7

Physics 430: Lecture 7
Kinetic and Potential Energy
Dale E. Gary
NJIT Physics Department
Chapter 4—Energy
We are now going to take up the conservation of energy, and its
implications. You have all seen this before, but now we will use a powerful,
more mathematical description.
 You will see that the discussion is more complicated that the other
conservation laws for linear and angular momentum. The main reason is
that each type of momentum comes in only one flavor, whereas there are
many forms of energy (kinetic, several kinds of potential, thermal, etc.).
 Processes transform one type of energy into another, and it is only the total
energy that is conserved, hence the additional complication.
 We will be introducing new mathematical tools of vector calculus, such as
the gradient and the curl, which you may be familiar with, or not. I will
give you the needed background as they come up.

September 22, 2009
4.1 Kinetic Energy and Work
An obvious form of energy is energy of motion, or kinetic energy. We will
use the symbol T, which is perhaps strange to you but is very much
standard in Classical Mechanics. The kinetic energy of a particle of mass m
traveling at speed v is defined to be:
T  12 mv2
 Consider such a particle moving on some trajectory through space while its
kinetic energy changes, on moving from position r1 to r1 + dr. We can take
the time derivative of the kinetic energy, after writing v 2  v  v, so that r1 dr



dT 1 d
 2 m ( v  v)  12 m( v  v  v  v )  mv  v
dt
dt
But the first term on the right is the force F  p  mv. Thus, we can write
the derivative of kinetic energy as dT
 Fv
dt
Finally, multiplying both sides by dt and noting that v dt = dr, we have
dT  F  dr
Work-KE Theorem
September 22, 2009
r1+dr
Line Integrals and Work

The equation just derived is only valid for an infinitesimal displacement, but
we can extend this to macroscopic displacements by integrating, to get:
r2
T   F  dr
r1
which says that the change in kinetic energy of a particle is equal to the
sum of force (in the direction of the displacement) times the incremental
displacement.
 However, note that this is the displacement along the path of the particle.
Such an integral is called a line integral. In evaluating the integral, it is
usually possible to convert it into an ordinary integral over a single variable,
as in the following example (which we will look at in a moment).
 With the notation of the line integral
2
T  T2  T1   F  dr  W (1  2)
1
where the last is a definition, defining the work done by F in moving from
point 1 to point 2. Note that F is the net force on the particle, but we can
also add up the work done by each force separately and write:
T2  T1  Wi (1  2)
i
September 22, 2009
Example 4.1: Three Line Integrals

Evaluate the line integral for the work done by the 2-d force F = (y, 2x)
going from the origin O to the point P = (1, 1) along each of the three paths:
a) OQ then QP
b) OP along x = y
c) OP along a circle

Path a):
P
Q
P
O
Q
Wa   F  dr   F  dr   F  dr
a
1
1
1
0
0
0
  ydx   2 xdy  0  2 dy  2

Path b):
O
P
Wb   F  dr   F  dr
b

P
0
O
1
x2
 ydx  2 xdy  0 3xdx  3  1.5
2 0
1
September 22, 2009
Q
Example 4.1: Three Line Integrals

Path c): This is a tricky one. Path c can be expressed as
r  ( x, y )  (1  cos q , sin q )
so
dr  (dx, dy )  (sin q , cos q )
This is a parametric equation, using q as a parameter along the path. With
this parameter, F = (sin q, 2(1cos q)). With this substitution:
 /2
Wc   F  dr  
c
0
sin
2
q  2(1  cos q ) cos q  dq
 2   / 4  1.21

The point here is that the line integral depends on the path, in general (but
not for special kinds of forces, which we will introduce in a moment).
September 22, 2009
4.2 Potential Energy and
Conservative Forces
We must now introduce the concept of potential energy corresponding to
the forces on an object. As you know, not every force lends itself to a
corresponding potential energy. Those that do are called conservative
forces.
 There are two conditions that a force must satisfy to be considered a
conservative force.
 The first condition for a force F to be conservative is that F depends only
on the position r of the object on which it acts. It cannot depend on
velocity, time, or any other parameter.
 Although this is restrictive, there are plenty of forces that satisfy this
condition, such as gravity, the spring force, the electric force. You can often
see this directly, such as for the gravitational force:
GmM
F (r )   2 rˆ
Depends only on r.
r

September 22, 2009
Non-Conservative Forces

The second condition for a force to be conservative concerns the work done
by the force as the object on which it acts moves between two points r1
and r2 (or just points 1 and 2, for short)
2
2
W (1  2)   F  dr
1
1
Reusing our earlier figure, we saw in Example 4.1 that the force described
there was NOT conservative, because it did different amounts of work for
the three paths a, b, and c.
 Forces involving friction, obviously are not conservative, because if you
were sliding a box, say, on a surface with friction along the three paths
shown, the friction would do work Wfric (1  2)   f fric L , where L is different
for the three paths. Such forces are non-conservative.

September 22, 2009
Conservative Forces

The force of gravity, on the other hand, has the property that the work
done is independent of the path. You know that if the height of point 1 and
point 2 differ by an amount h, then you will drop in height by h no matter
what path you take. In fact
Wgrav (1  2)   mgh

independent of path.
The conditions for a force to be conservative, then, are:
Conditions for a Force to be Conservative
A force F acting on a particle is conservative if and only if it satisfies
two conditions:
1. F depends only on the particle’s position r (and not on the velocity
v, or the time t, or any other variable); that is, F = F(r).
2. For any two points 1 and 2, the work W(1  2) done by F is the
same for all paths between 1 and 2.
September 22, 2009
Potential Energy

The reason that forces meeting these conditions are called conservative is
that, if all of the forces on an object are conservative we can define a
quantity called potential energy, denoted U(r), a function only of position,
with the property that the total mechanical energy
E  T  U (r )

is constant, i.e. is conserved.
To define the potential energy, we must first choose a reference point ro, at
which U is defined to be zero. (For gravity, we typically choose the
reference point to be ground level.) Then U(r), the potential energy, at any
arbitrary point r, is defined to be
Potential Energy
r
U (r )  W (ro  r )    F(r)  dr
ro

In words, U(r) is minus the work done by F when the particle moves from
the reference point ro to the point r.
September 22, 2009
Example 4.2: Potential Energy of a
Charge in a Uniform Electric Field

Statement of the problem:


A charge q is placed in a uniform electric field pointing in the x direction with
ˆ . Show that this force is
strength Eo, so that the force on q is F  qE  qEo x
conservative and find the corresponding potential energy.
Solution:

The work done by F in going between any two points 1 and 2 along any path
(which is negative potential energy) is:
2
2
2
W (1  2)   F  dr  qEo  xˆ  dr  qEo  dx  qEo ( x2  x1 )
1

1
1
This work done is independent of the path, because the electric force depends
only on position, i.e. the force is conservative. To find the corresponding
potential energy, we must first choose a reference point at which U is zero. A
natural choice is to choose our origin (the point 1), in which case the potential
energy is
U (r )  W (0  r)  qEo x

You may recall that Eo x is the electric potential V, so that qV is the potential
energy.
September 22, 2009
Several Forces
The potential energy can be defined even when more than one force is
acting, so long as all of the forces are conservative. An important example
is when both gravity Fgrav and a spring force Fspr are acting (so long as the
spring obeys Hooke’s Law, F(r) = kr).
 The work-kinetic energy theorem says that if we move an object subject to
these two forces along some path, the forces will do work independent of
the path (depending only on the two end-points of the path) given by
T  Wgrav  Wspr  (U grav  U spr )
 Rearrangement shows that

(T  U grav  U spr )  0
hence total mechanical energy is conserved. Extended to n such forces:
Principle of Conservation of Energy for One Particle
If all of the n forces Fi (i=1…n) acting on a particle are conservative, each with its
corresponding potential energy Ui(r), the total mechanical energy defined as
E  T  U  T  U1 (r)  U n (r)
is constant in time.
September 22, 2009
Nonconservative Forces
As we have seen, not all forces are conservative, meaning we cannot define
a corresponding potential energy. As you might guess, in that case we
cannot define a conserved mechanical energy.
 Nevertheless, if there are some conservative forces acting, for which a
potential energy can be defined, then we can divide the forces into a
conservative part Fcons, and a nonconservative part Fnc, such that
T  Wcons  Wnc
 U  Wnc

which allows us to write
(T  U )  Wnc
What this says is that mechanical energy (T + U) is no longer conserved,
but any changes in mechanical energy are precisely equal to the work done
by the nonconservative forces.
 In many problems, the only nonconservative force is friction, which acts in
the direction opposite the motion so that the work f  dr is negative.

September 22, 2009
Example 4.3: Block Sliding Down
an Incline






We did this problem using forces in lecture 2. Let’s now apply these ideas
of energy to arrive at the same result.
N
f
As before, we have to identify the forces, and set a
coordinate system, but this time we write down the
h
potential and kinetic energies in the problem.
q
The kinetic energy, as always, is T = ½ mv2.
mg
The gravitational potential energy is U = mgy, where we can set y = 0 (and
hence U = 0) at the ground level.
The friction force does negative work Wfric = fd, but recall that f = mN where
N  mg cos q . Putting all of this together, (T  U )  Wfric becomes
2
2
1
1
2 mv f  2 mvi  mgy   mmgd cos q
where d is the distance along the incline, and y is the change in height.
If the block starts out with zero initial velocity at the top of the incline, and
we ask what is the speed v at the bottom, then y = h = d sin q, so
2
1
or
v  2 gd (sin q  m cosq )
2 mv  mgd sin q   mmgd cos q
September 22, 2009
Comparison with Example 1.1
When we did this problem using forces, we obtained equation of motion
x  g (sin q  m cos q )
2
1
d

g
(sin
q

m
cos
q
)
t
2
from which, after integration, we got the expression
2d
x  g (sin q  m cos q )t
t
 Comparing with our just derived expression
g (sin q  m cos q )
v  2 gd (sin q  m cosq )
they may seem quite different. What is happening is that using forces we
can get the velocity versus time, whereas with energy we are only getting
the speed at the end points. Energy considerations are very powerful if
you just want to know the result at a particular point, in which case you
can ignore the details of the motion in getting there. If you instead need
to know the path taken, or the details along the path, you have to use the
tools of Newton’s Laws.
 However, we will find in a few weeks that these energy considerations do
contain all of the information of Newton’s Laws, and we will build the
tools necessary in Lagrangian mechanics to get the equation of motion
starting from energy. This allows us to attack much more complicated
problems. For this reason, it is important to get good at energy
problems. Here is an example you probably have seen before.

September 22, 2009
Problem 4.9

Statement of the problem:


(a) The force exerted by a one-dimensional spring, fixed at one end, is F = kx,
where x is the displacement of the other end from its equilibrium position.
Assuming that this force is conservative (which it is) show that the corresponding
potential energy is U = ½ kx2, if we choose U = 0 at its equilibrium position.
Solution to (a):

We start with the definition of potential energy:
2
2
U ( x)  W (1  2)   Fdx  k  xdx  12 k ( x22  x12 )
1

1
But we choose U = 0 at x = x1, which amounts to choosing x1 = 0, so that
U ( x )  12 kx2
September 22, 2009
Problem 4.9, cont’d

Statement of the problem:


(b) Suppose this spring is hung vertically from the ceiling with a mass m suspended
from the other end, and constrained to move in the vertical direction only. Find the
extension xo of the new equilibrium position with the suspended mass. Show that
the total potential energy (spring plus gravity) has the same form ½ ky2 if we use
the coordinate y equal to the displacement measured from the new equilibrium
position at x = xo (and redefine our reference point so that U = 0 at y = 0).
Solution to (b):

The new equilibrium position is reached when the force of the stretched
spring kxo equals the force of gravity on the mass mg. Thus
xo 

m
g
k
xo
To define the potential energy at the new equilibrium position, we
have to examine the work done in displacing the mass a distance y:
y
U ( y)  W (0  y )    ( Fgrav  Fspr )dy
0
y
  (mg  k ( xo  y)dy  mgy  kxo y  12 ky2  12 ky2
0
September 22, 2009
y=0