- [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.