0:00:05.250,0:00:06.570
- [Yiheng] We learned already

0:00:06.570,0:00:10.410
that even the most general[br]case of particle motion

0:00:10.410,0:00:13.470
can be studied using normal[br]and tangential components,

0:00:13.470,0:00:16.620
which means that from any[br]given position of this particle

0:00:16.620,0:00:19.650
we can always draw a tangential axis,

0:00:19.650,0:00:21.180
that is tangent to the path

0:00:21.180,0:00:23.640
pointing towards the direction of motion,

0:00:23.640,0:00:25.110
and a normal axis,

0:00:25.110,0:00:27.540
that is perpendicular[br]to the tangential axis

0:00:27.540,0:00:29.973
and points towards the[br]center of curvature.

0:00:30.990,0:00:33.630
And the result in the force[br]acting on this particle

0:00:33.630,0:00:37.170
can be completely resolved[br]into two components,

0:00:37.170,0:00:39.300
one along the tangential direction,

0:00:39.300,0:00:41.150
the other along the normal direction.

0:00:42.390,0:00:44.910
And then according to Newton's second law,

0:00:44.910,0:00:47.160
the resulting force along[br]the tangential direction

0:00:47.160,0:00:49.350
equals to m, the mass of the particle,

0:00:49.350,0:00:52.920
times at, which is the[br]acceleration of this particle

0:00:52.920,0:00:55.110
along the tangential direction.

0:00:55.110,0:00:57.660
And also we know this kinematic equation

0:00:57.660,0:01:00.390
that atds equals to vdv.

0:01:00.390,0:01:03.750
Combining these two[br]equations, we get this.

0:01:03.750,0:01:05.580
And if this particle has moved

0:01:05.580,0:01:08.790
from position 1 to position 2,

0:01:08.790,0:01:12.513
we can integrate this[br]equation from s1 to s2,

0:01:13.590,0:01:16.863
and we can get this equation here.

0:01:17.700,0:01:19.920
Notice the left-hand side

0:01:19.920,0:01:22.680
is simply the total work[br]done to this particle

0:01:22.680,0:01:24.033
during this process,

0:01:25.200,0:01:28.383
and this is known as the[br]principle of work and energy.

0:01:29.940,0:01:31.500
You might ask, "What happened

0:01:31.500,0:01:34.500
to the force along the normal direction?"

0:01:34.500,0:01:37.140
Well, since work is defined

0:01:37.140,0:01:39.180
as the magnitude of the force

0:01:39.180,0:01:43.350
multiplied by the displacement[br]along its direction,

0:01:43.350,0:01:44.700
and since the normal force

0:01:44.700,0:01:47.700
is always perpendicular to the path,

0:01:47.700,0:01:49.800
therefore there's never displacement

0:01:49.800,0:01:52.230
along the direction of the normal force,

0:01:52.230,0:01:55.410
therefore the normal[br]force never does work.

0:01:55.410,0:01:57.090
So this left-hand side,

0:01:57.090,0:02:00.010
the summation u1-2 this equation

0:02:00.960,0:02:04.230
is the total work done to this particle

0:02:04.230,0:02:07.920
of all the external[br]forces during this process

0:02:07.920,0:02:10.653
moving from position 1 to position 2.

0:02:11.880,0:02:15.840
Now, the left-hand side is[br]the work that we learned.

0:02:15.840,0:02:17.240
What is the right-hand side?

0:02:19.650,0:02:21.750
For any particle with a speed,

0:02:21.750,0:02:23.760
we can define its kinetic energy

0:02:23.760,0:02:26.880
to be t equals to 1/2 mv squared.

0:02:26.880,0:02:29.520
v is the speed of the particle.

0:02:29.520,0:02:32.220
Kinetic energy is the energy associated

0:02:32.220,0:02:34.500
with a motion of the object.

0:02:34.500,0:02:36.990
Just like work, it is also a scalar.

0:02:36.990,0:02:41.340
It has the same unit,[br]joule in the SI unit system

0:02:41.340,0:02:45.330
and foot pound in the[br]US customer unit system.

0:02:45.330,0:02:47.070
But unlike work,

0:02:47.070,0:02:51.123
which is a parameter[br]associated with a process,

0:02:52.200,0:02:55.953
kinetic energy is associated[br]with a status of an object.

0:02:57.690,0:02:59.130
And as you can see,

0:02:59.130,0:03:02.490
because it equals to 1/2 mv squared,

0:03:02.490,0:03:06.030
m, the mass of the[br]object, must be non-zero

0:03:06.030,0:03:10.500
and the velocity squared[br]must be non-negative.

0:03:10.500,0:03:14.493
Therefore, the kinetic energy[br]is always non-negative.

0:03:15.540,0:03:18.840
So at a state 1,

0:03:18.840,0:03:23.840
the kinetic energy of the[br]particle is 1/2 mv1 squared.

0:03:24.060,0:03:25.860
And at state 2,

0:03:25.860,0:03:27.390
the kinetic energy of the particle

0:03:27.390,0:03:30.780
is 1/2 mv2 squared.

0:03:30.780,0:03:33.540
Therefore, the principle[br]of work and energy

0:03:33.540,0:03:35.670
that you saw on the previous screen

0:03:35.670,0:03:39.480
becomes the total work[br]done to this particle

0:03:39.480,0:03:43.110
by all external forces[br]during the process 1-2

0:03:43.110,0:03:47.970
equals to T2, the kinetic[br]energy at state 2,

0:03:47.970,0:03:51.123
minus T1, the kinetic energy at a state 1.

0:03:52.140,0:03:56.370
In other words, the change in[br]the particle's kinetic energy

0:03:56.370,0:04:00.960
equals to the total external[br]work down to this particle,

0:04:00.960,0:04:04.353
or this can be rewritten in this form.

0:04:05.940,0:04:09.120
So the initial kinetic[br]energy of the particle

0:04:09.120,0:04:12.270
plus the external work[br]done to this particle

0:04:12.270,0:04:13.590
during this process

0:04:13.590,0:04:17.163
equals to the final kinetic[br]energy of the particle.

0:04:20.130,0:04:21.780
The principle of work and energy

0:04:21.780,0:04:24.783
can also be applied to[br]a system of particles.

0:04:26.010,0:04:27.450
In this equation,

0:04:27.450,0:04:29.310
this is the total kinetic energy

0:04:29.310,0:04:31.200
of all the particles in the system

0:04:31.200,0:04:33.603
at state 1 or the initial state.

0:04:35.580,0:04:38.220
This is the total work done to the system

0:04:38.220,0:04:40.290
by all the external forces

0:04:40.290,0:04:42.423
during the process from 1-2.

0:04:45.540,0:04:49.500
And this is the total kinetic[br]energy of all the particles

0:04:49.500,0:04:51.843
at state 2 or the final state.

0:04:53.220,0:04:54.840
When we are using this equation,

0:04:54.840,0:04:57.030
keep in mind that we[br]are making an assumption

0:04:57.030,0:05:00.510
that there's no energy loss[br]due to particle interaction.

0:05:00.510,0:05:04.290
For example, if the particles[br]are colliding into each other,

0:05:04.290,0:05:07.830
there will be energy loss in[br]the form of heat or sound,

0:05:07.830,0:05:11.043
then this equation will not apply.

0:05:13.680,0:05:16.650
When an object is sliding on the surface,

0:05:16.650,0:05:19.140
the frictional force exerted on the object

0:05:19.140,0:05:22.770
is evaluated by mu k[br]times the normal force N.

0:05:22.770,0:05:26.580
Mu k is the coefficient[br]for kinetic friction.

0:05:26.580,0:05:30.030
And when we count for the[br]work done to this object,

0:05:30.030,0:05:33.840
we can still use the magnitude[br]of the frictional force

0:05:33.840,0:05:36.510
multiplied by the[br]displacement of this object

0:05:36.510,0:05:40.290
to count for the work[br]done by frictional force.

0:05:40.290,0:05:43.170
Although it is not the true work done

0:05:43.170,0:05:44.250
by the frictional force

0:05:44.250,0:05:45.840
because if you recall,

0:05:45.840,0:05:48.540
the frictional force[br]is the resulting force

0:05:48.540,0:05:50.730
of numerous horizontal forces

0:05:50.730,0:05:54.090
acting on the numerous[br]contacting surfaces,

0:05:54.090,0:05:58.920
and the displacement of these[br]forces are not necessarily s.

0:05:58.920,0:06:02.310
However, it is still[br]reasonable to use this term

0:06:02.310,0:06:06.960
because it counts for the total[br]effect of the true work done

0:06:06.960,0:06:09.990
by the frictional force, mu k N s prime

0:06:09.990,0:06:12.543
as well as the heat loss during friction.

0:06:15.750,0:06:17.550
Let's look at this example.

0:06:17.550,0:06:19.740
There's a 10-kilogram crate

0:06:19.740,0:06:22.380
traveling along this smooth slope,

0:06:22.380,0:06:26.130
and if at this point shown[br]at x equals to 16 meter,

0:06:26.130,0:06:28.620
it has a speed of 20 meter per second,

0:06:28.620,0:06:30.180
we need to determine its speed

0:06:30.180,0:06:32.040
where it gets to the bottom of the slope

0:06:32.040,0:06:33.423
where x equals to 0.

0:06:34.710,0:06:36.360
If you try to solve this problem

0:06:36.360,0:06:39.690
using equation of motion[br]as well as kinematics,

0:06:39.690,0:06:42.030
it will be very, very difficult.

0:06:42.030,0:06:43.980
This is a great example to be solved

0:06:43.980,0:06:46.620
using the principle of work energy

0:06:46.620,0:06:49.920
because it involves the direct correlation

0:06:49.920,0:06:51.573
of position and speed.

0:06:53.820,0:06:57.210
If we do a quick free body[br]diagram of this particle,

0:06:57.210,0:06:59.040
at any given position,

0:06:59.040,0:07:01.920
it is only subjected to two forces,

0:07:01.920,0:07:05.760
its weight and the normal[br]force exerted by the slope.

0:07:05.760,0:07:07.530
Because the slope is smooth,

0:07:07.530,0:07:09.900
we don't need to consider friction.

0:07:09.900,0:07:11.820
Now, because the normal force

0:07:11.820,0:07:15.030
is always perpendicular to its path,

0:07:15.030,0:07:18.330
therefore the normal[br]force never does work.

0:07:18.330,0:07:20.760
So we're going to solve this problem

0:07:20.760,0:07:23.070
using the principle of work and energy,

0:07:23.070,0:07:25.170
but notice that in this equation,

0:07:25.170,0:07:28.800
the total work done to this[br]crate during this process

0:07:28.800,0:07:31.653
is only the work done by its weight force.

0:07:34.140,0:07:37.560
So to evaluate the work[br]done by the weight force,

0:07:37.560,0:07:39.840
we need to first specify the initial

0:07:39.840,0:07:42.390
and final positions of this particle.

0:07:42.390,0:07:45.090
And keep in mind that the[br]work done by the weight force

0:07:45.090,0:07:46.500
only depends on the change

0:07:46.500,0:07:49.410
in the vertical position of this particle.

0:07:49.410,0:07:53.250
Therefore, initially at[br]x equals to 16 meter,

0:07:53.250,0:07:58.250
the vertical position of[br]this particle y1 is 16 meter,

0:07:58.470,0:08:01.683
and at the second state[br]x equals to 0 meter,

0:08:02.670,0:08:06.150
the vertical position y2 is 0 meter.

0:08:06.150,0:08:08.970
Therefore, during this process,

0:08:08.970,0:08:11.610
the work done by weight

0:08:11.610,0:08:14.290
equals to positive

0:08:15.510,0:08:18.720
1,569.6 joule.

0:08:18.720,0:08:22.080
It is positive because this[br]particle is moving downwards,

0:08:22.080,0:08:25.263
therefore weight force[br]is doing positive work.

0:08:26.970,0:08:29.880
And as the initial state,

0:08:29.880,0:08:32.220
the kinetic energy of this particle

0:08:32.220,0:08:34.710
is evaluated from its initial speed,

0:08:34.710,0:08:36.270
20 meter per second,

0:08:36.270,0:08:38.700
to be 2,000 joule.

0:08:38.700,0:08:40.140
At state 2,

0:08:40.140,0:08:44.250
T2 equals to 5 times v2 squared,

0:08:44.250,0:08:46.593
v2 is our unknown that we need to solve.

0:08:47.460,0:08:50.760
Applying the principle of work energy,

0:08:50.760,0:08:53.910
substitute in the evaluated values,

0:08:53.910,0:08:58.383
we can solve for v2 to[br]be 26.7 meter per second.