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Lecture Comments (24)

1 answer

Last reply by: Professor Dan Fullerton
Fri Jun 17, 2016 5:40 PM

Post by Hitendrakumar Patel on June 17 at 04:54:00 PM

A rodeo rider is riding a bucking bronco when he is thrown off. At the instant he leaves the horse he is located 1.6m above the ground and is moving straight up at 4.0m/s.

A) What maximum height above the ground does the rider reach?

B) At what speed will the rider hit the ground?

2 answers

Last reply by: Hitendrakumar Patel
Fri Jun 17, 2016 6:11 PM

Post by Hitendrakumar Patel on June 17 at 04:18:14 PM

A boy fires a pebble with his slingshot. The pebble leaves the slingshot at a speed of 35m/s. How high above the ground will the pebble rise if it is fired straight up?If the pebble is fired so that it goes in a arc(instead of straight), what is the max height, assuming it is moving side ways at 10m/sat the top of the arc? I do not get how to do the second part of the question.

0 answers

Post by Christopher Pieri on May 3, 2015

Gazuntite, brother

2 answers

Last reply by: Micheal Bingham
Sun Jan 4, 2015 11:47 AM

Post by Micheal Bingham on January 3, 2015

In example 6, right when he falls off of the mud slide, how come he only has kinetic energy and not also potential energy since he is elevated above the ground with a height?

1 answer

Last reply by: Daniel Fullerton
Fri Oct 31, 2014 6:31 AM

Post by Foaad Zaid on October 31, 2014

If the base of the cliff is the final point, why isn't it included in the PEg? I understand how it was convenient to break the problem into two different steps, but I just thought that the height for the mgh equation would be 35 meters.

3 answers

Last reply by: Daniel Fullerton
Fri Oct 31, 2014 6:27 AM

Post by Foaad Zaid on October 30, 2014

Hello Professor, In example 2, around 15:41, what happened to the toy's spring final potential energy? I don't see any zeros in the equation that would allow us to cancel it out. Thank you.

1 answer

Last reply by: Professor Dan Fullerton
Thu Oct 30, 2014 6:42 PM

Post by Foaad Zaid on October 30, 2014

For the jet fighter example, would it be possible to use the KEi +PEi = KEf + PEf approach? I don't understand why finding the total kinetic energy first was necessary. Thank you in advance.

1 answer

Last reply by: Professor Dan Fullerton
Fri Aug 29, 2014 6:32 PM

Post by Jungle Jones on August 29, 2014

In example 5, why did you say that gravitational potential energy = kinetic + work done by friction?
Where did the work done by friction get in there?

2 answers

Last reply by: Okwudili Ezeh
Wed Aug 27, 2014 8:56 AM

Post by Okwudili Ezeh on August 26, 2014

Was there any reason why you chose not to cover Mechanical Advantage?

1 answer

Last reply by: Professor Dan Fullerton
Sun Dec 8, 2013 10:05 AM

Post by Emmil Zarrugh on December 7, 2013

When would it be incorrect to find the velocity of something using the kinematics approach, but correct to find the velocity using the energy approach? For example, if an arrow is shot straight upwards from a bow whose string exerts a force on the arrow, one can only find the velocity of the arrow as it leaves the bow using the energy approach and not the kinematics approach because work must be considered, right?

Conservation of Energy

  • The energy of a system includes its kinetic energy, potential energy, and internal energy.
  • Energy is conserved in a closed system.
  • Energy can be transformed from one type into another.

Conservation of Energy

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:08
  • Law of Conservation of Energy 0:22
    • Energy Cannot Be Created or Destroyed.. It Can Only Be Changed
    • Mechanical Energy
    • Conservation Laws
    • Examples
  • Kinematics vs. Energy 4:34
    • Energy Approach
    • Kinematics Approach
  • The Pendulum 8:07
  • Example 1: Cart Compressing a Spring 13:09
  • Example 2 14:23
  • Example 3: Car Skidding to a Stop 16:15
  • Example 4: Accelerating an Object 17:27
  • Example 5: Block on Ramp 18:06
  • Example 6: Energy Transfers 19:21

Transcription: Conservation of Energy

Hi everyone! I am Dan Fullerton and I would like to welcome you back to Educator.com0000

Today we are going to talk about conservation of energy.0003

Our goals and objectives for this unit are to recognize situations in which total energy and mechanical energy are conserved and to apply conservation of energy -- to analyze energy transitions and transformations in a system.0009

So the Law of Conservation of Energy -- one of the big ideas in Physics.0021

Energy cannot be created or destroyed; it can only be changed to different forms.0026

So mechanical energy is kinetic plus gravitational potential energy, plus spring potential energy.0032

And we also have conservation laws for total energy and if there is no friction, conservation of mechanical energy.0039

Let us see how these work.0046

Let us assume as we talk about conservation of energy that we have a jet fighter with a mass of 20,000 kg, and it is coasting through the sky at an altitude of 10,000 m with a velocity of 250 m/s.0050

Let us try and figure out what its total energy is.0061

Well its total mechanical energy (Etotal) is going to be its gravitational potential energy because it is so high up, plus the kinetic energy it has due to its velocity, so that is going to be mgh + 1/2 mv2.0064

Now its mass is 20,000 kg -- (g) we are going to estimate at about 10 m/s2 with an altitude or height of 10,000 m plus 1/2 times its mass (20,000 kg) times the square of its velocity (2502).0083

Or that implies then that its total mechanical energy will be about 2.63 times 109 J.0102

Now that jet is going to dive to an altitude of 2,000 m -- it is going to go from 10,000 to 2,000 m. 0117

Find the new velocity of the jet -- and we are going to assume that we are not losing any energy to friction.0124

We are not running the engines at the moment so we are not gaining any energy or converting any other types of energy -- just a simple first pass calculation.0133

So now for our total energy, we are going to follow the same formula -- gravitational potential energy plus kinetic, which is still (mgh) + 1/2 mv2.0141

That is going to be -- then as we solve for this -- let us see if we can find out its new velocity; we will get velocity all by itself.0153

I could say that mv2 then -- if I multiply both sides by two and rearrange this a little bit -- mv2 is going to be 2 × total energy - (mgh).0161

Multiply all of this by 2 and then subtract the (mgh) from one side.0174

Now then if I want just the velocity -- velocity is going to be 2 × total energy - (mgh), all divided by that mass, and I need to take the square root of that.0180

When I substitute in my values -- that is 2 times our total energy, which was 2.63 × 109 J...0197

...We have a little bit more to put in there -- × 109 - mgh (20,000) × g (10) × our new height (2000) all divided by the mass (20,000 kg) and we need the square root of all of that.0207

Therefore, our velocity must be -- when I plug all of that into a calculator -- not doing that one in my head -- it comes out to be about 472 m/s.0230

So it was going 250 m/s -- it dove -- it converted some of that gravitational potential energy into kinetic energy and therefore increased its speed.0243

And that is why oftentimes when you are talking about aerial combat, a saying among pilots is that altitude is life; altitude is energy. It really is.0252

That is what allows them to convert very quickly that altitude into velocity, which is so important for winning dog fight scenario battles.0262

Let us take a look at how we can analyze the motion of an object from an energy approach and then from a kinematics approach -- two different ways of solving the same sort of problem -- things we have been doing.0272

Let us drop an object -- any object from a height of 10 m and see if we can find its velocity right before it hits the ground, that split second before it contacts the ground.0284

If we start with an energy approach, we know that the energy at the top of its path -- when it is at 10 m -- must equal the energy at its bottom by conservation of energy.0296

When it is at the top -- that is gravitational potential energy, and at the bottom it is all converted into kinetic energy.0307

Therefore, we could write that (mg) times the height -- when it is at its highest point -- must equal 1/2 mv2 at its lowest point.0315

If I multiply both sides by 2 here and -- this is nice -- I can divide out the (m)'s on both sides and then I will get 2gh = V2 or V = square root of 2gh.0323

And if I substitute in my values -- V = 2 × -- let us assume (g) is roughly 10, our height (10), and the square root of 200 is going to be about 14.1 m/s.0343

Let us do that from a kinematics approach, back from some of our earlier lessons.0363

If we drop an object from a height of 10 m -- well if we do that, let us look at the vertical analysis.0367

Our initial velocity -- V0 must be 0 m/s and V final is what we are looking for, so δy is going to be 10 m, which we will call down our positive direction.0373

Our acceleration is going to be about 10 m/s2 down and the time -- we do not know either.0387

For solving for final velocity, we will use one of our kinematic equations, and the one that is probably most helpful right now will be to write that Vf2 or V2 = V02 + 2A(δy).0395

Or again, this nice little trick -- our initial velocity is 0 so that term goes away -- V2 = 2A(δy) or V = the square root of 2A(δy).0412

But let us look at this for a second -- (a) is our acceleration which is the acceleration due to gravity.0428

So we could write then that (a) = (g), and δy is the change in height -- so δy = h.0436

I can then rewrite this equation as V = the square root of 2, and instead of (a) here, I am going to write (g), and instead of δy, I am going to write (h).0445

That should look mighty familiar -- V = the square root of 2(gh).0456

And of course when I plug in my values again -- V = square root of 2 × 10 × 10 -- I will get the same numeric answer as well, 14.1 m/s.0465

Two different ways of solving the same problem -- one with the conservation of energy approach, one with your traditional kinematics approach.0477

Let us take a look at a problem of a pendulum.0488

A pendulum comprised of a light string -- meaning we are going to ignore its mass, it is an ideal pendulum -- of length (L) swings mass (m) back and forth.0492

So this must be our mass (m), it is going to go back and forth and it has some length (L) -- same length here.0500

And as it does this -- at its highest point here, it has gravitational potential energy.0508

At its lowest point, it must have kinetic energy -- that is where it is going the fastest.0514

Then it is going to convert that kinetic energy back into potential energy and for a split second it is going to stop here with no kinetic energy -- all potential -- and back and forth, and back and forth.0520

So if I were to make a graph of what this would look like from an energy perspective -- if we put energy on this axis vs. (x) displacement here -- well when it is in its middle position all of its energy is kinetic.0530

So right there let us put some amount of kinetic energy, and kinetic energy we will make it green.0545

At its farthest (x) displacement, its kinetic energy must be 0 -- over here and over here -- corresponding to these points where it is no longer moving.0550

So our graph of kinetic energy is probably going to look something like this upside down U shape.0560

It should be pretty symmetric -- should be perfectly symmetric -- my art skills are a little off.0569

On the other hand, if we wanted to take a look at potential energy, we know that at these points at its maximum displacement, all of its energy is potential.0573

So then what I am going to do is at that same point I am going to put potential energy (PE) or (U) over here, which at its lowest point here, all of its energy is kinetic -- none of it is potential.0585

So we have 0, and back over here, when it is at its highest point on the left, again, all of the energy is potential and kinetic is 0 again.0600

Over here we get an alternate curve that looks kind of like this.0608

And what we are really doing is as the pendulum swings back and forth is trading off gravitational potential energy and kinetic energy.0613

The entire time though, assuming we do not have any non-conservative forces, we do not have any friction -- we are going to have a constant total energy.0621

It is going to remain the same the entire time, just different amounts, and different divisions of gravitational potential energy vs. kinetic energy.0632

We could look in a little bit more detail over here as well -- so all potential energy here, potential energy here, and kinetic here.0643

And we have already done this derivation once, but the potential energy due to that height difference (h) -- well we determine that this adjacent side was L cos(θ) here on our string...0650

...If that entire length there is (L), then (h) is equal to (L) - L cos(θ), which is (L) times the quantity -- 1 - cos(θ).0665

Its maximum potential energy is going to be (mgh) or (mg) and (h) is (L) × the quantity -- 1 - cos(θ).0678

Its kinetic energy -- max on the other hand -- is 1/2 mv2, which also must be equal to the potential energy maximum because of conservation of energy, so that also has to be equal to (mgL) × 1 - cos(θ).0695

So we could solve to find the maximum velocity of our ideal pendulum here by solving this equation for velocity.0716

And hey! We are here. Why not do it?0725

I could write then that if this is the maximum velocity -- what I am going to do is rearrange this a little bit... 0728

...I multiply both sides by 2, we get this nice little -- divide (m) out of both sides, so I can write that as 1/2 v2 = (gl) × 1 - cos(θ), multiplying both sides by 2, then v2 = 2gl × 1 - cos(θ).0735

Just remembering that this is the maximum velocity, therefore, the maximum velocity is going to be -- take the square root of both sides -- it is going to be the square root of 2gl, 1 - cos(θ).0757

So we could use conservation of energy that way to solve for the maximum velocity of our ideal pendulum, and of course that is going to occur when our pendulum is down in that position or its maximum kinetic energy is 0 potential energy.0773

All right, let us look at an example of a cart compressing a spring.0789

The diagram here shows a toy cart possessing 16 J of kinetic energy traveling on a frictionless horizontal surface toward a horizontal spring.0792

If the cart comes to rest after compressing the spring a distance of 1 m, find the spring constant of the spring.0801

Well, we can use conservation of energy to solve this.0808

The way I would do that is I would look and say that its initial kinetic energy, which is 16 J, must equal its final total energy which is all spring potential energy.0811

Therefore, 16 J must equal -- well the formula for the potential energy stored in a compressed spring is 1/2 Kx2.0823

We want to rearrange this to solve for (K), the spring constant, so I could write that as (K) is equal to...0833

...multiply both sides by 2 and we get 32/x2, which is going to be 32 over our displacement of (1 m), 12 or 32N-m for my spring constant.0838

Looking at a little bit more involved example -- here we have a pop-up toy compressing a spring.0859

The pop-up toy has a mass of 0.020 kg and a spring constant of 150N-m.0867

A force is applied to the toy to compress the spring 0.05 m. 0873

Calculate the potential energy stored in the compressed spring.0878

All right, well our first step there -- the potential energy stored in the compressed spring is 1/2 Kx2 or 1/2 × 150N-m, our spring constant, × 0.050 m2 or about 0.1875 J.0881

Let us take this one a little bit further -- that toy is then activated and all that compressed spring's potential energy is converted to gravitational potential energy.0906

The spring unleashes and the toy pops up; it starts going very quickly and it slows down, slows down, slows down until it gets to its highest point then it stops.0915

Let us find the maximum vertical height that it was propelled to, and we can do that by conservation of energy.0923

In this case, our initial potential energy, which was all stored in a spring is turned into final potential energy, which is gravitational.0929

Therefore, 0.1875 J must equal (mgh), so if we solve for the height, (h) is going to be 0.1875 J/mg, or 0.1875 J/mass (0.2) × (g), which we are going to estimate as 10, which gives us a maximum height of about 0.9375 m.0939

Let us see if we cannot take this to another type of problem. 0972

A car initially traveling at 30 m/s slows uniformly as it skids to a stop after the brakes are applied. 0978

Sketch a graph showing the relationship between the kinetic energy of the car as it is being brought to a stop and the work done by friction in stopping the car.0985

Well our car starts out and its initial velocity is 30 m/s; its final velocity -- if it is coming to a stop -- is 0 m/s, and we want a graph of the kinetic energy vs. the work done by the force of friction.0993

Now when no work is done by friction, of course it is going to have all or its maximum kinetic energy, so we should expect a nice, high point here.1017

When it has come to a stop after a lot of work has been done by friction -- it has no velocity, it must have no kinetic energy.1025

Well we have a uniform slow down, therefore that is all we need.1033

Let us take a look at accelerating an object -- the work done in accelerating an object along a frictionless horizontal surface is equal to the change in the object's momentum? No. Velocity? No. Potential energy? No. Kinetic energy -- this is the Work-Energy Theorem.1048

When you do work on an object, you change its energy. 1070

If you are doing work on a frictionless horizontal surface to it, what type of energy are you giving it?1073

It must be kinetic energy or energy of motion. You are increasing its velocity.1077

Let us take a look at a block on a ramp.1084

A 2 kg block sliding down a ramp from a height of 3 m above the ground reaches the ground with a kinetic energy of 50 J.1087

Find the total work done by friction on the block as it slides down the ramp.1093

Well let us start by making a diagram -- there is our ramp, we will put our block on it.1099

Now we know that its gravitational potential energy at the top must be equal to its kinetic energy at the bottom plus whatever work was done by friction.1106

So if we want the work done by friction then -- that is just going to be the potential energy due to gravity at the top minus its kinetic energy at the bottom, which will be (mgh) at the top minus the kinetic energy (50 J)...1116

...therefore the work done by friction will be m (2 kg) × g (10) × our height difference (3 m) - 50 -- 2 × 10 = 20 × 3 = 60 - 50 for a total of 10 J.1135

Let us take a look at a little bit more creative, fun example.1160

Andy the adventurous adventurer, while running from evil bad guys at an Amazonian rainforest, trips, falls, and slides down a frictionless mudslide of height 20 m as depicted here.1164

Once he reaches the bottom of the mudslide he has the misfortune to fly horizontally off of a 15 m cliff.1175

He has gravitational potential energy up here, but as he slides down, all of that is going to be converted into kinetic energy and he is going to be moving completely horizontally, so all of the velocity here are corresponding to his kinetic energy.1182

And at that point he becomes a projectile. How far from the base of the cliff does Andy land?1196

Well let us treat this as two problems -- a conservation of energy problem here to find his horizontal velocity and to review our projectile problems down here to figure out where he lands.1202

First step -- let us find what his potential energy due to gravity is up here.1213

That is going to be (mgh) and that is going to be equal to his kinetic energy down here -- 1/2 mv2 by conservation of energy.1221

So the height difference is 20 m, so as we do some of our cancellations -- (m)'s cancel out -- we can say again that V = square root of 2gh, or that is going to be square root of 2 × 10 × its height (20 m) from when he goes over the cliff, or 20 × 10 = 200 × 2 = 400, and the square root of 400 = 20 m/s.1231

He is going to fly off the cliff horizontally with a velocity of 20 m/s.1256

Now we have a projectile problem. 1261

Horizontally, he is going to have a constant velocity of 20 m/s. 1265

If we can find how long he is in the air, we can solve for his displacement horizontally, (d) or δx, which is going to be his velocity times the time.1271

To figure out how long he is in the air, then we have to look at vertical motion.1281

Vertically, his initial velocity is 0;, his final velocity we do not know;, and δy is going to be 15 m from the time he goes off the mudslide to hitting the ground at the bottom of the cliff.1286

So that is 15 m -- we will call down the positive y direction again; (a) our acceleration is going to be 10 m/s2 down and let us see if we can solve for time.1300

The equation that I would use to do this is I would say that δy = V initial × t + 1/2 at2.1312

V initial = 0, so that whole term goes away. 1322

We could then write that (t) must be 2 δy/a(square root) or 2 × 15 m/10 m/s2(square root).1327

So (t) equals -- when I plug all that into my calculator I come up with the time in the air of about 1.73 s, and if he is in the air 1.73 s vertically, he must be in the air 1.73 s horizontally.1343

So d or δx is just going to be velocity × time -- δx (d) will be Vt or that is going to be 20 m/s × our time of 1.73 s; it is going to give him a displacement horizontally of about 34.6 m.1357

This distance right there must be 34.6 m from the base of the cliff.1379

That is conservation of energy problem combined with our knowledge of projectile motion.1387

Hopefully that gets you a good start with conservation of energy.1392

Thank you for your time and for watching

Make it a great day.1399