- [Yiheng] We learned already
that even the most general
case of particle motion
can be studied using normal
and tangential components,
which means that from any
given position of this particle
we can always draw a tangential axis,
that is tangent to the path
pointing towards the direction of motion,
and a normal axis,
that is perpendicular
to the tangential axis
and points towards the
center of curvature.
And the result in the force
acting on this particle
can be completely resolved
into two components,
one along the tangential direction,
the other along the normal direction.
And then according to Newton's second law,
the resulting force along
the tangential direction
equals to m, the mass of the particle,
times at, which is the
acceleration of this particle
along the tangential direction.
And also we know this kinematic equation
that atds equals to vdv.
Combining these two
equations, we get this.
And if this particle has moved
from position 1 to position 2,
we can integrate this
equation from s1 to s2,
and we can get this equation here.
Notice the left-hand side
is simply the total work
done to this particle
during this process,
and this is known as the
principle of work and energy.
You might ask, "What happened
to the force along the normal direction?"
Well, since work is defined
as the magnitude of the force
multiplied by the displacement
along its direction,
and since the normal force
is always perpendicular to the path,
therefore there's never displacement
along the direction of the normal force,
therefore the normal
force never does work.
So this left-hand side,
the summation u1-2 this equation
is the total work done to this particle
of all the external
forces during this process
moving from position 1 to position 2.
Now, the left-hand side is
the work that we learned.
What is the right-hand side?
For any particle with a speed,
we can define its kinetic energy
to be t equals to 1/2 mv squared.
v is the speed of the particle.
Kinetic energy is the energy associated
with a motion of the object.
Just like work, it is also a scalar.
It has the same unit,
joule in the SI unit system
and foot pound in the
US customer unit system.
But unlike work,
which is a parameter
associated with a process,
kinetic energy is associated
with a status of an object.
And as you can see,
because it equals to 1/2 mv squared,
m, the mass of the
object, must be non-zero
and the velocity squared
must be non-negative.
Therefore, the kinetic energy
is always non-negative.
So at a state 1,
the kinetic energy of the
particle is 1/2 mv1 squared.
And at state 2,
the kinetic energy of the particle
is 1/2 mv2 squared.
Therefore, the principle
of work and energy
that you saw on the previous screen
becomes the total work
done to this particle
by all external forces
during the process 1-2
equals to T2, the kinetic
energy at state 2,
minus T1, the kinetic energy at a state 1.
In other words, the change in
the particle's kinetic energy
equals to the total external
work down to this particle,
or this can be rewritten in this form.
So the initial kinetic
energy of the particle
plus the external work
done to this particle
during this process
equals to the final kinetic
energy of the particle.
The principle of work and energy
can also be applied to
a system of particles.
In this equation,
this is the total kinetic energy
of all the particles in the system
at state 1 or the initial state.
This is the total work done to the system
by all the external forces
during the process from 1-2.
And this is the total kinetic
energy of all the particles
at state 2 or the final state.
When we are using this equation,
keep in mind that we
are making an assumption
that there's no energy loss
due to particle interaction.
For example, if the particles
are colliding into each other,
there will be energy loss in
the form of heat or sound,
then this equation will not apply.
When an object is sliding on the surface,
the frictional force exerted on the object
is evaluated by mu k
times the normal force N.
Mu k is the coefficient
for kinetic friction.
And when we count for the
work done to this object,
we can still use the magnitude
of the frictional force
multiplied by the
displacement of this object
to count for the work
done by frictional force.
Although it is not the true work done
by the frictional force
because if you recall,
the frictional force
is the resulting force
of numerous horizontal forces
acting on the numerous
contacting surfaces,
and the displacement of these
forces are not necessarily s.
However, it is still
reasonable to use this term
because it counts for the total
effect of the true work done
by the frictional force, mu k N s prime
as well as the heat loss during friction.
Let's look at this example.
There's a 10-kilogram crate
traveling along this smooth slope,
and if at this point shown
at x equals to 16 meter,
it has a speed of 20 meter per second,
we need to determine its speed
where it gets to the bottom of the slope
where x equals to 0.
If you try to solve this problem
using equation of motion
as well as kinematics,
it will be very, very difficult.
This is a great example to be solved
using the principle of work energy
because it involves the direct correlation
of position and speed.
If we do a quick free body
diagram of this particle,
at any given position,
it is only subjected to two forces,
its weight and the normal
force exerted by the slope.
Because the slope is smooth,
we don't need to consider friction.
Now, because the normal force
is always perpendicular to its path,
therefore the normal
force never does work.
So we're going to solve this problem
using the principle of work and energy,
but notice that in this equation,
the total work done to this
crate during this process
is only the work done by its weight force.
So to evaluate the work
done by the weight force,
we need to first specify the initial
and final positions of this particle.
And keep in mind that the
work done by the weight force
only depends on the change
in the vertical position of this particle.
Therefore, initially at
x equals to 16 meter,
the vertical position of
this particle y1 is 16 meter,
and at the second state
x equals to 0 meter,
the vertical position y2 is 0 meter.
Therefore, during this process,
the work done by weight
equals to positive
1,569.6 joule.
It is positive because this
particle is moving downwards,
therefore weight force
is doing positive work.
And as the initial state,
the kinetic energy of this particle
is evaluated from its initial speed,
20 meter per second,
to be 2,000 joule.
At state 2,
T2 equals to 5 times v2 squared,
v2 is our unknown that we need to solve.
Applying the principle of work energy,
substitute in the evaluated values,
we can solve for v2 to
be 26.7 meter per second.