WEBVTT

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- [Yiheng] We learned already

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that even the most general
case of particle motion

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can be studied using normal
and tangential components,

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which means that from any
given position of this particle

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we can always draw a tangential axis,

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that is tangent to the path

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pointing towards the direction of motion,

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and a normal axis,

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that is perpendicular
to the tangential axis

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and points towards the
center of curvature.

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And the result in the force
acting on this particle

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can be completely resolved
into two components,

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one along the tangential direction,

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the other along the normal direction.

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And then according to Newton's second law,

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the resulting force along
the tangential direction

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equals to m, the mass of the particle,

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times at, which is the
acceleration of this particle

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along the tangential direction.

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And also we know this kinematic equation

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that atds equals to vdv.

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Combining these two
equations, we get this.

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And if this particle has moved

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from position 1 to position 2,

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we can integrate this
equation from s1 to s2,

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and we can get this equation here.

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Notice the left-hand side

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is simply the total work
done to this particle

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during this process,

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and this is known as the
principle of work and energy.

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You might ask, "What happened

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to the force along the normal direction?"

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Well, since work is defined

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as the magnitude of the force

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multiplied by the displacement
along its direction,

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and since the normal force

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is always perpendicular to the path,

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therefore there's never displacement

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along the direction of the normal force,

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therefore the normal
force never does work.

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So this left-hand side,

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the summation u1-2 this equation

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is the total work done to this particle

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of all the external
forces during this process

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moving from position 1 to position 2.

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Now, the left-hand side is
the work that we learned.

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What is the right-hand side?

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For any particle with a speed,

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we can define its kinetic energy

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to be t equals to 1/2 mv squared.

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v is the speed of the particle.

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Kinetic energy is the energy associated

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with a motion of the object.

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Just like work, it is also a scalar.

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It has the same unit,
joule in the SI unit system

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and foot pound in the
US customer unit system.

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But unlike work,

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which is a parameter
associated with a process,

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kinetic energy is associated
with a status of an object.

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And as you can see,

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because it equals to 1/2 mv squared,

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m, the mass of the
object, must be non-zero

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and the velocity squared
must be non-negative.

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Therefore, the kinetic energy
is always non-negative.

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So at a state 1,

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the kinetic energy of the
particle is 1/2 mv1 squared.

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And at state 2,

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the kinetic energy of the particle

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is 1/2 mv2 squared.

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Therefore, the principle
of work and energy

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that you saw on the previous screen

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becomes the total work
done to this particle

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by all external forces
during the process 1-2

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equals to T2, the kinetic
energy at state 2,

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minus T1, the kinetic energy at a state 1.

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In other words, the change in
the particle's kinetic energy

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equals to the total external
work down to this particle,

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or this can be rewritten in this form.

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So the initial kinetic
energy of the particle

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plus the external work
done to this particle

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during this process

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equals to the final kinetic
energy of the particle.

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The principle of work and energy

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can also be applied to
a system of particles.

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In this equation,

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this is the total kinetic energy

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of all the particles in the system

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at state 1 or the initial state.

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This is the total work done to the system

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by all the external forces

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during the process from 1-2.

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And this is the total kinetic
energy of all the particles

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at state 2 or the final state.

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When we are using this equation,

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keep in mind that we
are making an assumption

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that there's no energy loss
due to particle interaction.

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For example, if the particles
are colliding into each other,

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there will be energy loss in
the form of heat or sound,

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then this equation will not apply.

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When an object is sliding on the surface,

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the frictional force exerted on the object

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is evaluated by mu k
times the normal force N.

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Mu k is the coefficient
for kinetic friction.

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And when we count for the
work done to this object,

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we can still use the magnitude
of the frictional force

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multiplied by the
displacement of this object

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to count for the work
done by frictional force.

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Although it is not the true work done

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by the frictional force

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because if you recall,

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the frictional force
is the resulting force

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of numerous horizontal forces

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acting on the numerous
contacting surfaces,

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and the displacement of these
forces are not necessarily s.

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However, it is still
reasonable to use this term

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because it counts for the total
effect of the true work done

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by the frictional force, mu k N s prime

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as well as the heat loss during friction.

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Let's look at this example.

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There's a 10-kilogram crate

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traveling along this smooth slope,

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and if at this point shown
at x equals to 16 meter,

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it has a speed of 20 meter per second,

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we need to determine its speed

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where it gets to the bottom of the slope

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where x equals to 0.

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If you try to solve this problem

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using equation of motion
as well as kinematics,

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it will be very, very difficult.

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This is a great example to be solved

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using the principle of work energy

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because it involves the direct correlation

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of position and speed.

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If we do a quick free body
diagram of this particle,

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at any given position,

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it is only subjected to two forces,

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its weight and the normal
force exerted by the slope.

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Because the slope is smooth,

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we don't need to consider friction.

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Now, because the normal force

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is always perpendicular to its path,

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therefore the normal
force never does work.

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So we're going to solve this problem

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using the principle of work and energy,

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but notice that in this equation,

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the total work done to this
crate during this process

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is only the work done by its weight force.

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So to evaluate the work
done by the weight force,

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we need to first specify the initial

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and final positions of this particle.

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And keep in mind that the
work done by the weight force

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only depends on the change

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in the vertical position of this particle.

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Therefore, initially at
x equals to 16 meter,

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the vertical position of
this particle y1 is 16 meter,

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and at the second state
x equals to 0 meter,

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the vertical position y2 is 0 meter.

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Therefore, during this process,

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the work done by weight

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equals to positive

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1,569.6 joule.

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It is positive because this
particle is moving downwards,

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therefore weight force
is doing positive work.

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And as the initial state,

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the kinetic energy of this particle

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is evaluated from its initial speed,

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20 meter per second,

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to be 2,000 joule.

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At state 2,

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T2 equals to 5 times v2 squared,

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v2 is our unknown that we need to solve.

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Applying the principle of work energy,

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substitute in the evaluated values,

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we can solve for v2 to
be 26.7 meter per second.