Before
moving on to the second type of energy, let us look at conservative and
dissipative forces. Recall that we
defined the work to be
_{}
What would happen if we then
traversed a second path, only this time going from x_{2} to x_{1}? In some of the cases, we would find that the
total work done would become zero. This
is the hallmark of a conservative force.
Specifically, we say that a force is conservative if
If the force is not
conservative, it is called a dissipative
force.
Example:
The
force of gravity is a conservative force.
Let a box of mass m be moved around the path shown below and acted on
only by gravity.
Can break the work up
into 4 pieces W_{12}, W_{23}, W_{34},
W_{41}. Then
_{}
Whenever
a force is conserved, we can define a potential energy for that force as
follows:

_{} 
(12.1) 
This is interpreted
physically as the potential energy due to the force at position x_{2}
relative to x_{1}.
Example:
What
is the gravitational potential energy of an object of mass m at a
distance h above the ground?
Define
the ground to be at x_{1} = 0.
Then since the force is constant
_{}
Now let’s use (11.2) to find an important relationship
between work and energy. Let us define W_{ext}
to be the work done by all nonconservative forces. We call all of these forces the external forces. We can see from the definition of the
potential energy that the work done to change altitude is just the opposite of
the change in potential energy. So,
using (11.2) and generalizing this to all potential energies, we can write
_{}
where the sum is over the
change in potential energies for each conservative force. Moving the potential energy to the right
side of the equation,

_{} 
(12.2) 
This is the workenergy theorem, which states that the
total external work done on a system must equal the change in total mechanical
energy, where the total mechanical energy is defined to be E = K
+ U. Notice that if there are no
external forces, then the mechanical energy is constant and we can write
_{}
Example:
A
cannon shoots a shell straight up with an initial velocity of 50 m/s. How high does the shell go before falling
back to earth, neglecting air resistance?
Solve this using forces and energy conservation.
Forces:
Using
the equations for position, we have
y(t)
= y_{0} + v_{0}t  ½gt^{2}
To find the maximum,
need to find where the time when the velocity is 0
_{}
Thus,
_{}
Energy Conservation:
Since
there are no nonconservative forces acting on the shell, the total energy must
remain constant. The highest point will
be reached when the kinetic energy is zero.
Thus
_{}
Recall
that we calculated the work done by a nonuniform gravitational field and by a
spring earlier. At that time, we saw
that the work done by a gravitational field was just
_{}
Now recall the
definition of the potential energy of a force.
It is related to the work done by a force by the relation
dW
= dU
It is standard to define
the zero potential point for an inverse square field to be at infinity. Then the work supplied by gravity in
bringing an object of mass m to a distance R from a mass M is
just (setting r_{0} = and r' = R)
_{}
and the potential energy
at that point, relative to infinity,
is
_{}
Notice that the
potential energy is negative. Thus, we
say that gravity creates a potential energy well.
Similarly,
the work done by the spring as it is stretched a distance x is
W
= ½kx^{2}
where we are now taking
the unstretched position to be x = 0.
Then the potential energy with respect to the unstretched position is
U
= ½kx^{2}
Example:
A
0.1 kg ball sits on a spring gun which is aimed upward. If the spring is compressed 0.05 m and has a
spring constant of 250 N/m, how high does the ball go when the gun is fired?
Using
the workenergy theorem, we see that
_{}
^{ }
Thus,
_{}
By
combining the linear and rotational motion, we can describe fairly complex
motion in a simple manner.
Example:
A
solid 5 kg cylinder of radius 0.25 m rolls without slipping down a 3 m long frictionless
ramp inclined at an angle of 30°. What
is its speed when it reaches the bottom?
Since
there are no dissipative forces, we see that
_{}
So
far, we have only considered the amount of work that has been done, but we have
not asked how rapidly did it occur. The
rate at which work is performed (or energy is transferred) is called the power.
In a manner analogous with the definition of velocity, we can define the
average power to be
_{}
If we let the interval Dt go to
zero, the we get the instantaneous power

_{} 
(12.3) 
From dimensional
analysis, we see that the unit of power is a joule per sec. This unit has its own name, and is called a
Watt (W). If the force is constant, we
can rewrite (12.3) as

_{} 
(12.4) 
This is the power
supplied by the force F to the object. Again, if we are considering rotational motion, the rotational
power for a constant torque is given by
_{}
Example:
A
mass of 100 kg is pushed along the floor at a constant rate of 2 m/s. If the coefficient of sliding friction is
0.25, at what rate is work being done in order to keep it in motion?
Here we are interested
in the power supplied by the force F.
By Newton's third law, we see that it is equal and opposite to the force
of friction. Thus
_{}