Dynamics Lecture 16: Principle of work and energy

- [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.