[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:05.25,0:00:06.57,Default,,0000,0000,0000,,- [Yiheng] We learned already
Dialogue: 0,0:00:06.57,0:00:10.41,Default,,0000,0000,0000,,that even the most general\Ncase of particle motion
Dialogue: 0,0:00:10.41,0:00:13.47,Default,,0000,0000,0000,,can be studied using normal\Nand tangential components,
Dialogue: 0,0:00:13.47,0:00:16.62,Default,,0000,0000,0000,,which means that from any\Ngiven position of this particle
Dialogue: 0,0:00:16.62,0:00:19.65,Default,,0000,0000,0000,,we can always draw a tangential axis,
Dialogue: 0,0:00:19.65,0:00:21.18,Default,,0000,0000,0000,,that is tangent to the path
Dialogue: 0,0:00:21.18,0:00:23.64,Default,,0000,0000,0000,,pointing towards the direction of motion,
Dialogue: 0,0:00:23.64,0:00:25.11,Default,,0000,0000,0000,,and a normal axis,
Dialogue: 0,0:00:25.11,0:00:27.54,Default,,0000,0000,0000,,that is perpendicular\Nto the tangential axis
Dialogue: 0,0:00:27.54,0:00:29.97,Default,,0000,0000,0000,,and points towards the\Ncenter of curvature.
Dialogue: 0,0:00:30.99,0:00:33.63,Default,,0000,0000,0000,,And the result in the force\Nacting on this particle
Dialogue: 0,0:00:33.63,0:00:37.17,Default,,0000,0000,0000,,can be completely resolved\Ninto two components,
Dialogue: 0,0:00:37.17,0:00:39.30,Default,,0000,0000,0000,,one along the tangential direction,
Dialogue: 0,0:00:39.30,0:00:41.15,Default,,0000,0000,0000,,the other along the normal direction.
Dialogue: 0,0:00:42.39,0:00:44.91,Default,,0000,0000,0000,,And then according to Newton's second law,
Dialogue: 0,0:00:44.91,0:00:47.16,Default,,0000,0000,0000,,the resulting force along\Nthe tangential direction
Dialogue: 0,0:00:47.16,0:00:49.35,Default,,0000,0000,0000,,equals to m, the mass of the particle,
Dialogue: 0,0:00:49.35,0:00:52.92,Default,,0000,0000,0000,,times at, which is the\Nacceleration of this particle
Dialogue: 0,0:00:52.92,0:00:55.11,Default,,0000,0000,0000,,along the tangential direction.
Dialogue: 0,0:00:55.11,0:00:57.66,Default,,0000,0000,0000,,And also we know this kinematic equation
Dialogue: 0,0:00:57.66,0:01:00.39,Default,,0000,0000,0000,,that atds equals to vdv.
Dialogue: 0,0:01:00.39,0:01:03.75,Default,,0000,0000,0000,,Combining these two\Nequations, we get this.
Dialogue: 0,0:01:03.75,0:01:05.58,Default,,0000,0000,0000,,And if this particle has moved
Dialogue: 0,0:01:05.58,0:01:08.79,Default,,0000,0000,0000,,from position 1 to position 2,
Dialogue: 0,0:01:08.79,0:01:12.51,Default,,0000,0000,0000,,we can integrate this\Nequation from s1 to s2,
Dialogue: 0,0:01:13.59,0:01:16.86,Default,,0000,0000,0000,,and we can get this equation here.
Dialogue: 0,0:01:17.70,0:01:19.92,Default,,0000,0000,0000,,Notice the left-hand side
Dialogue: 0,0:01:19.92,0:01:22.68,Default,,0000,0000,0000,,is simply the total work\Ndone to this particle
Dialogue: 0,0:01:22.68,0:01:24.03,Default,,0000,0000,0000,,during this process,
Dialogue: 0,0:01:25.20,0:01:28.38,Default,,0000,0000,0000,,and this is known as the\Nprinciple of work and energy.
Dialogue: 0,0:01:29.94,0:01:31.50,Default,,0000,0000,0000,,You might ask, "What happened
Dialogue: 0,0:01:31.50,0:01:34.50,Default,,0000,0000,0000,,to the force along the normal direction?"
Dialogue: 0,0:01:34.50,0:01:37.14,Default,,0000,0000,0000,,Well, since work is defined
Dialogue: 0,0:01:37.14,0:01:39.18,Default,,0000,0000,0000,,as the magnitude of the force
Dialogue: 0,0:01:39.18,0:01:43.35,Default,,0000,0000,0000,,multiplied by the displacement\Nalong its direction,
Dialogue: 0,0:01:43.35,0:01:44.70,Default,,0000,0000,0000,,and since the normal force
Dialogue: 0,0:01:44.70,0:01:47.70,Default,,0000,0000,0000,,is always perpendicular to the path,
Dialogue: 0,0:01:47.70,0:01:49.80,Default,,0000,0000,0000,,therefore there's never displacement
Dialogue: 0,0:01:49.80,0:01:52.23,Default,,0000,0000,0000,,along the direction of the normal force,
Dialogue: 0,0:01:52.23,0:01:55.41,Default,,0000,0000,0000,,therefore the normal\Nforce never does work.
Dialogue: 0,0:01:55.41,0:01:57.09,Default,,0000,0000,0000,,So this left-hand side,
Dialogue: 0,0:01:57.09,0:02:00.01,Default,,0000,0000,0000,,the summation u1-2 this equation
Dialogue: 0,0:02:00.96,0:02:04.23,Default,,0000,0000,0000,,is the total work done to this particle
Dialogue: 0,0:02:04.23,0:02:07.92,Default,,0000,0000,0000,,of all the external\Nforces during this process
Dialogue: 0,0:02:07.92,0:02:10.65,Default,,0000,0000,0000,,moving from position 1 to position 2.
Dialogue: 0,0:02:11.88,0:02:15.84,Default,,0000,0000,0000,,Now, the left-hand side is\Nthe work that we learned.
Dialogue: 0,0:02:15.84,0:02:17.24,Default,,0000,0000,0000,,What is the right-hand side?
Dialogue: 0,0:02:19.65,0:02:21.75,Default,,0000,0000,0000,,For any particle with a speed,
Dialogue: 0,0:02:21.75,0:02:23.76,Default,,0000,0000,0000,,we can define its kinetic energy
Dialogue: 0,0:02:23.76,0:02:26.88,Default,,0000,0000,0000,,to be t equals to 1/2 mv squared.
Dialogue: 0,0:02:26.88,0:02:29.52,Default,,0000,0000,0000,,v is the speed of the particle.
Dialogue: 0,0:02:29.52,0:02:32.22,Default,,0000,0000,0000,,Kinetic energy is the energy associated
Dialogue: 0,0:02:32.22,0:02:34.50,Default,,0000,0000,0000,,with a motion of the object.
Dialogue: 0,0:02:34.50,0:02:36.99,Default,,0000,0000,0000,,Just like work, it is also a scalar.
Dialogue: 0,0:02:36.99,0:02:41.34,Default,,0000,0000,0000,,It has the same unit,\Njoule in the SI unit system
Dialogue: 0,0:02:41.34,0:02:45.33,Default,,0000,0000,0000,,and foot pound in the\NUS customer unit system.
Dialogue: 0,0:02:45.33,0:02:47.07,Default,,0000,0000,0000,,But unlike work,
Dialogue: 0,0:02:47.07,0:02:51.12,Default,,0000,0000,0000,,which is a parameter\Nassociated with a process,
Dialogue: 0,0:02:52.20,0:02:55.95,Default,,0000,0000,0000,,kinetic energy is associated\Nwith a status of an object.
Dialogue: 0,0:02:57.69,0:02:59.13,Default,,0000,0000,0000,,And as you can see,
Dialogue: 0,0:02:59.13,0:03:02.49,Default,,0000,0000,0000,,because it equals to 1/2 mv squared,
Dialogue: 0,0:03:02.49,0:03:06.03,Default,,0000,0000,0000,,m, the mass of the\Nobject, must be non-zero
Dialogue: 0,0:03:06.03,0:03:10.50,Default,,0000,0000,0000,,and the velocity squared\Nmust be non-negative.
Dialogue: 0,0:03:10.50,0:03:14.49,Default,,0000,0000,0000,,Therefore, the kinetic energy\Nis always non-negative.
Dialogue: 0,0:03:15.54,0:03:18.84,Default,,0000,0000,0000,,So at a state 1,
Dialogue: 0,0:03:18.84,0:03:23.84,Default,,0000,0000,0000,,the kinetic energy of the\Nparticle is 1/2 mv1 squared.
Dialogue: 0,0:03:24.06,0:03:25.86,Default,,0000,0000,0000,,And at state 2,
Dialogue: 0,0:03:25.86,0:03:27.39,Default,,0000,0000,0000,,the kinetic energy of the particle
Dialogue: 0,0:03:27.39,0:03:30.78,Default,,0000,0000,0000,,is 1/2 mv2 squared.
Dialogue: 0,0:03:30.78,0:03:33.54,Default,,0000,0000,0000,,Therefore, the principle\Nof work and energy
Dialogue: 0,0:03:33.54,0:03:35.67,Default,,0000,0000,0000,,that you saw on the previous screen
Dialogue: 0,0:03:35.67,0:03:39.48,Default,,0000,0000,0000,,becomes the total work\Ndone to this particle
Dialogue: 0,0:03:39.48,0:03:43.11,Default,,0000,0000,0000,,by all external forces\Nduring the process 1-2
Dialogue: 0,0:03:43.11,0:03:47.97,Default,,0000,0000,0000,,equals to T2, the kinetic\Nenergy at state 2,
Dialogue: 0,0:03:47.97,0:03:51.12,Default,,0000,0000,0000,,minus T1, the kinetic energy at a state 1.
Dialogue: 0,0:03:52.14,0:03:56.37,Default,,0000,0000,0000,,In other words, the change in\Nthe particle's kinetic energy
Dialogue: 0,0:03:56.37,0:04:00.96,Default,,0000,0000,0000,,equals to the total external\Nwork down to this particle,
Dialogue: 0,0:04:00.96,0:04:04.35,Default,,0000,0000,0000,,or this can be rewritten in this form.
Dialogue: 0,0:04:05.94,0:04:09.12,Default,,0000,0000,0000,,So the initial kinetic\Nenergy of the particle
Dialogue: 0,0:04:09.12,0:04:12.27,Default,,0000,0000,0000,,plus the external work\Ndone to this particle
Dialogue: 0,0:04:12.27,0:04:13.59,Default,,0000,0000,0000,,during this process
Dialogue: 0,0:04:13.59,0:04:17.16,Default,,0000,0000,0000,,equals to the final kinetic\Nenergy of the particle.
Dialogue: 0,0:04:20.13,0:04:21.78,Default,,0000,0000,0000,,The principle of work and energy
Dialogue: 0,0:04:21.78,0:04:24.78,Default,,0000,0000,0000,,can also be applied to\Na system of particles.
Dialogue: 0,0:04:26.01,0:04:27.45,Default,,0000,0000,0000,,In this equation,
Dialogue: 0,0:04:27.45,0:04:29.31,Default,,0000,0000,0000,,this is the total kinetic energy
Dialogue: 0,0:04:29.31,0:04:31.20,Default,,0000,0000,0000,,of all the particles in the system
Dialogue: 0,0:04:31.20,0:04:33.60,Default,,0000,0000,0000,,at state 1 or the initial state.
Dialogue: 0,0:04:35.58,0:04:38.22,Default,,0000,0000,0000,,This is the total work done to the system
Dialogue: 0,0:04:38.22,0:04:40.29,Default,,0000,0000,0000,,by all the external forces
Dialogue: 0,0:04:40.29,0:04:42.42,Default,,0000,0000,0000,,during the process from 1-2.
Dialogue: 0,0:04:45.54,0:04:49.50,Default,,0000,0000,0000,,And this is the total kinetic\Nenergy of all the particles
Dialogue: 0,0:04:49.50,0:04:51.84,Default,,0000,0000,0000,,at state 2 or the final state.
Dialogue: 0,0:04:53.22,0:04:54.84,Default,,0000,0000,0000,,When we are using this equation,
Dialogue: 0,0:04:54.84,0:04:57.03,Default,,0000,0000,0000,,keep in mind that we\Nare making an assumption
Dialogue: 0,0:04:57.03,0:05:00.51,Default,,0000,0000,0000,,that there's no energy loss\Ndue to particle interaction.
Dialogue: 0,0:05:00.51,0:05:04.29,Default,,0000,0000,0000,,For example, if the particles\Nare colliding into each other,
Dialogue: 0,0:05:04.29,0:05:07.83,Default,,0000,0000,0000,,there will be energy loss in\Nthe form of heat or sound,
Dialogue: 0,0:05:07.83,0:05:11.04,Default,,0000,0000,0000,,then this equation will not apply.
Dialogue: 0,0:05:13.68,0:05:16.65,Default,,0000,0000,0000,,When an object is sliding on the surface,
Dialogue: 0,0:05:16.65,0:05:19.14,Default,,0000,0000,0000,,the frictional force exerted on the object
Dialogue: 0,0:05:19.14,0:05:22.77,Default,,0000,0000,0000,,is evaluated by mu k\Ntimes the normal force N.
Dialogue: 0,0:05:22.77,0:05:26.58,Default,,0000,0000,0000,,Mu k is the coefficient\Nfor kinetic friction.
Dialogue: 0,0:05:26.58,0:05:30.03,Default,,0000,0000,0000,,And when we count for the\Nwork done to this object,
Dialogue: 0,0:05:30.03,0:05:33.84,Default,,0000,0000,0000,,we can still use the magnitude\Nof the frictional force
Dialogue: 0,0:05:33.84,0:05:36.51,Default,,0000,0000,0000,,multiplied by the\Ndisplacement of this object
Dialogue: 0,0:05:36.51,0:05:40.29,Default,,0000,0000,0000,,to count for the work\Ndone by frictional force.
Dialogue: 0,0:05:40.29,0:05:43.17,Default,,0000,0000,0000,,Although it is not the true work done
Dialogue: 0,0:05:43.17,0:05:44.25,Default,,0000,0000,0000,,by the frictional force
Dialogue: 0,0:05:44.25,0:05:45.84,Default,,0000,0000,0000,,because if you recall,
Dialogue: 0,0:05:45.84,0:05:48.54,Default,,0000,0000,0000,,the frictional force\Nis the resulting force
Dialogue: 0,0:05:48.54,0:05:50.73,Default,,0000,0000,0000,,of numerous horizontal forces
Dialogue: 0,0:05:50.73,0:05:54.09,Default,,0000,0000,0000,,acting on the numerous\Ncontacting surfaces,
Dialogue: 0,0:05:54.09,0:05:58.92,Default,,0000,0000,0000,,and the displacement of these\Nforces are not necessarily s.
Dialogue: 0,0:05:58.92,0:06:02.31,Default,,0000,0000,0000,,However, it is still\Nreasonable to use this term
Dialogue: 0,0:06:02.31,0:06:06.96,Default,,0000,0000,0000,,because it counts for the total\Neffect of the true work done
Dialogue: 0,0:06:06.96,0:06:09.99,Default,,0000,0000,0000,,by the frictional force, mu k N s prime
Dialogue: 0,0:06:09.99,0:06:12.54,Default,,0000,0000,0000,,as well as the heat loss during friction.
Dialogue: 0,0:06:15.75,0:06:17.55,Default,,0000,0000,0000,,Let's look at this example.
Dialogue: 0,0:06:17.55,0:06:19.74,Default,,0000,0000,0000,,There's a 10-kilogram crate
Dialogue: 0,0:06:19.74,0:06:22.38,Default,,0000,0000,0000,,traveling along this smooth slope,
Dialogue: 0,0:06:22.38,0:06:26.13,Default,,0000,0000,0000,,and if at this point shown\Nat x equals to 16 meter,
Dialogue: 0,0:06:26.13,0:06:28.62,Default,,0000,0000,0000,,it has a speed of 20 meter per second,
Dialogue: 0,0:06:28.62,0:06:30.18,Default,,0000,0000,0000,,we need to determine its speed
Dialogue: 0,0:06:30.18,0:06:32.04,Default,,0000,0000,0000,,where it gets to the bottom of the slope
Dialogue: 0,0:06:32.04,0:06:33.42,Default,,0000,0000,0000,,where x equals to 0.
Dialogue: 0,0:06:34.71,0:06:36.36,Default,,0000,0000,0000,,If you try to solve this problem
Dialogue: 0,0:06:36.36,0:06:39.69,Default,,0000,0000,0000,,using equation of motion\Nas well as kinematics,
Dialogue: 0,0:06:39.69,0:06:42.03,Default,,0000,0000,0000,,it will be very, very difficult.
Dialogue: 0,0:06:42.03,0:06:43.98,Default,,0000,0000,0000,,This is a great example to be solved
Dialogue: 0,0:06:43.98,0:06:46.62,Default,,0000,0000,0000,,using the principle of work energy
Dialogue: 0,0:06:46.62,0:06:49.92,Default,,0000,0000,0000,,because it involves the direct correlation
Dialogue: 0,0:06:49.92,0:06:51.57,Default,,0000,0000,0000,,of position and speed.
Dialogue: 0,0:06:53.82,0:06:57.21,Default,,0000,0000,0000,,If we do a quick free body\Ndiagram of this particle,
Dialogue: 0,0:06:57.21,0:06:59.04,Default,,0000,0000,0000,,at any given position,
Dialogue: 0,0:06:59.04,0:07:01.92,Default,,0000,0000,0000,,it is only subjected to two forces,
Dialogue: 0,0:07:01.92,0:07:05.76,Default,,0000,0000,0000,,its weight and the normal\Nforce exerted by the slope.
Dialogue: 0,0:07:05.76,0:07:07.53,Default,,0000,0000,0000,,Because the slope is smooth,
Dialogue: 0,0:07:07.53,0:07:09.90,Default,,0000,0000,0000,,we don't need to consider friction.
Dialogue: 0,0:07:09.90,0:07:11.82,Default,,0000,0000,0000,,Now, because the normal force
Dialogue: 0,0:07:11.82,0:07:15.03,Default,,0000,0000,0000,,is always perpendicular to its path,
Dialogue: 0,0:07:15.03,0:07:18.33,Default,,0000,0000,0000,,therefore the normal\Nforce never does work.
Dialogue: 0,0:07:18.33,0:07:20.76,Default,,0000,0000,0000,,So we're going to solve this problem
Dialogue: 0,0:07:20.76,0:07:23.07,Default,,0000,0000,0000,,using the principle of work and energy,
Dialogue: 0,0:07:23.07,0:07:25.17,Default,,0000,0000,0000,,but notice that in this equation,
Dialogue: 0,0:07:25.17,0:07:28.80,Default,,0000,0000,0000,,the total work done to this\Ncrate during this process
Dialogue: 0,0:07:28.80,0:07:31.65,Default,,0000,0000,0000,,is only the work done by its weight force.
Dialogue: 0,0:07:34.14,0:07:37.56,Default,,0000,0000,0000,,So to evaluate the work\Ndone by the weight force,
Dialogue: 0,0:07:37.56,0:07:39.84,Default,,0000,0000,0000,,we need to first specify the initial
Dialogue: 0,0:07:39.84,0:07:42.39,Default,,0000,0000,0000,,and final positions of this particle.
Dialogue: 0,0:07:42.39,0:07:45.09,Default,,0000,0000,0000,,And keep in mind that the\Nwork done by the weight force
Dialogue: 0,0:07:45.09,0:07:46.50,Default,,0000,0000,0000,,only depends on the change
Dialogue: 0,0:07:46.50,0:07:49.41,Default,,0000,0000,0000,,in the vertical position of this particle.
Dialogue: 0,0:07:49.41,0:07:53.25,Default,,0000,0000,0000,,Therefore, initially at\Nx equals to 16 meter,
Dialogue: 0,0:07:53.25,0:07:58.25,Default,,0000,0000,0000,,the vertical position of\Nthis particle y1 is 16 meter,
Dialogue: 0,0:07:58.47,0:08:01.68,Default,,0000,0000,0000,,and at the second state\Nx equals to 0 meter,
Dialogue: 0,0:08:02.67,0:08:06.15,Default,,0000,0000,0000,,the vertical position y2 is 0 meter.
Dialogue: 0,0:08:06.15,0:08:08.97,Default,,0000,0000,0000,,Therefore, during this process,
Dialogue: 0,0:08:08.97,0:08:11.61,Default,,0000,0000,0000,,the work done by weight
Dialogue: 0,0:08:11.61,0:08:14.29,Default,,0000,0000,0000,,equals to positive
Dialogue: 0,0:08:15.51,0:08:18.72,Default,,0000,0000,0000,,1,569.6 joule.
Dialogue: 0,0:08:18.72,0:08:22.08,Default,,0000,0000,0000,,It is positive because this\Nparticle is moving downwards,
Dialogue: 0,0:08:22.08,0:08:25.26,Default,,0000,0000,0000,,therefore weight force\Nis doing positive work.
Dialogue: 0,0:08:26.97,0:08:29.88,Default,,0000,0000,0000,,And as the initial state,
Dialogue: 0,0:08:29.88,0:08:32.22,Default,,0000,0000,0000,,the kinetic energy of this particle
Dialogue: 0,0:08:32.22,0:08:34.71,Default,,0000,0000,0000,,is evaluated from its initial speed,
Dialogue: 0,0:08:34.71,0:08:36.27,Default,,0000,0000,0000,,20 meter per second,
Dialogue: 0,0:08:36.27,0:08:38.70,Default,,0000,0000,0000,,to be 2,000 joule.
Dialogue: 0,0:08:38.70,0:08:40.14,Default,,0000,0000,0000,,At state 2,
Dialogue: 0,0:08:40.14,0:08:44.25,Default,,0000,0000,0000,,T2 equals to 5 times v2 squared,
Dialogue: 0,0:08:44.25,0:08:46.59,Default,,0000,0000,0000,,v2 is our unknown that we need to solve.
Dialogue: 0,0:08:47.46,0:08:50.76,Default,,0000,0000,0000,,Applying the principle of work energy,
Dialogue: 0,0:08:50.76,0:08:53.91,Default,,0000,0000,0000,,substitute in the evaluated values,
Dialogue: 0,0:08:53.91,0:08:58.38,Default,,0000,0000,0000,,we can solve for v2 to\Nbe 26.7 meter per second.