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Lecture Comments (5)

1 answer

Last reply by: Yilmaz Kahraman
Thu Apr 7, 2016 2:57 PM

Post by Yilmaz Kahraman on April 7 at 02:54:57 PM

lol this dude cant do physics. the answer's 200j.

2 answers

Last reply by: Yilmaz Kahraman
Thu Apr 7, 2016 2:53 PM

Post by Zach Basel on May 11, 2014

On number 1, the vertical distance is 20 m, not 70 m.

Energy & Energy Conservation

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
  • Question 1 0:18
  • Question 2 1:27
  • Question 3 1:44
  • Question 4 2:33
  • Question 5 2:44
  • Question 6 3:33
  • Question 7 4:41
  • Question 8 5:19
  • Question 9 5:37
  • Question 10 7:12
  • Question 11 7:40

Transcription: Energy & Energy Conservation

Hi everyone and welcome back to 0000

I am Dan Fullerton and in this mini-lesson, we are going to look at energy and energy conservation by going through the first page of the APlusPhysics worksheet on energy and energy conservation. 0003

You can find the link to this worksheet down below the video. 0013

Let us get started. 0017

Number 1 -- A 1 kg rock is dropped from a cliff 90 m high. 0019

After falling 20 m, the kinetic energy of the rock is approximately...0023

Well, the way I would do that is, I would set this up to say that its fallen 70 m. 0027

If it started at 90 and it has fallen to 20 m, the distance it has fallen is 70 m. 0035

Let us call that our 0-point of potential energy, so this must be 70 m above that, so potential energy is mg(δh) and that will be 1 kg × 9.8 m/s2 and our difference in height between those two is going to be 70 m. 0043

So that is going to be right around 70 × 9.8 -- close to 10 -- around 700 J because what happened is if we had gravitational potential energy here, that difference is converted into kinetic energy by that point, so 700 J, Number 3 would be a nice, easy answer to Number 1. 0064

Number 2 -- If the speed of a car is doubled, the kinetic energy does what? 0087

Well, kinetic energy is 1/2 mv2 and (v) is squared here, so if you double that, you are going to get four times the kinetic energy, so it is quadrupled. 0092

Number 3 -- A constant force is used to keep a block sliding at constant velocity along a rough horizontal track. 0105

As the block slides, there could be an increase in its...0112

As it slides, we could have an increase in -- Well, if it is on a horizontal track, gravitational potential energy is not going to increase.0116

Its internal energy could. 0125

As it is sliding, some of the energy could be going into friction, or making it warmer, increasing its heat -- its temperature, its internal energy. 0127

Gravitational potential energy and kinetic energy -- No, that cannot be the case. 0136

Internal and kinetic energy -- Well, it cannot be kinetic energy because it is moving at constant velocity, so our best answer here is Number 2. 0141

Number 4 -- As an object falls freely, the kinetic energy of the object... 0154

Well, it is going faster and faster, so the kinetic energy must be increasing. 0158

Number 5 -- An object weighing 15 N is lifted from the ground to a height of 0.22 m. 0165

The increase in the objects gravitational potential energy is approximately...0171

Well, that is going to be either the work done in lifting it, or we could calculate potential energy directly. 0175

Either way we do it, we are going to find that potential energy is mgH or if we found the work to lift it there, that is going to be force times the distance, in which case the force would be (mg) and the distance would (H). 0181

It is going to give us the same thing. 0193

The weight, 15 N, times the height to which it was lifted, 0.22 m, is going to give us 15 × 0.22 or about 3.3 J. 0195

Number 6 -- A half kg ball is thrown vertically upward with an initial kinetic energy of 25 J. 0214

Approximately how high will the ball rise? 0221

Well, there are a few ways you could do this too. 0224

You could calculate its initial velocity knowing its kinetic energy and mass and use kinematics to figure out where it is going to stop at its highest point... 0226

...or probably more easily, if it starts with a kinetic energy of 25 J down here, that is all kinetic -- at its highest point up here, it is going to be all potential energy equal to 25 J with no kinetic energy. 0233

If its potential energy up here is 25 J, that is equal to mgH or 0.5 kg × 9.8 m/s × its height. 0247

So the height is going to be 25 J/0.5 kg divided by 9.8 m/s2 or that is going to be 25 divided by 5 and it is going to be right around 5 or 5.1 m. 0259

Number 7 -- A 45 kg boy is riding a 15 kg bike with a speed of 8 m/s. 0282

What is the combined kinetic energy of the boy and the bike? 0289

Kinetic energy is 1/2 mv2, so that will be 1/2 and their combined mass is 60 kg and their speed is 8 m/s2. 0293

So that is going to be 0.5 × 60 × 82 or about 1920 J. 0306

Number 8 -- The work done in moving a block across a rough surface and the heat energy gained by the block can both be measured in... 0320

Well work and heat or work and any type of energy are all going to be measured in joules. 0328

Number 9 -- A 50 kg child running at 6 m/s jumps on to a stationary 6 kg sled. 0338

The sled is on a level, frictionless surface. Calculate the speed of the sled with the child after she jumps onto the sled. 0345

A couple of ways we could do this too, but let us see.0353

We have a child running at 6 m/s so it has some initial momentum and jumps on to a sled that is stationary and has an initial momentum of 0. 0357

We could use conservation of momentum to find the speed of the sled after the child jumps on to the sled. 0364

Just to review that, to mix things up a little bit, why do we not do it that way. 0370

We have a child. We have a sled and we have a total, our momentum before and momentum after. 0374

Initially the 50 kg child at 6 m/s -- 50 × 6 is going to give us 300 kg-m/s as the initial momentum and the sled is stationary, giving us a total momentum is 300. 0388

Afterwards, they really become one object, so they have a combined mass now of 60 kg and some unknown velocity, so our total here is 60 v, so 300 must equal 60 v by the law of conservation of momentum, or v = 300/60, which is going to be 5 m/s. 0401

That is the speed of the sled with the child after she jumps on to the sled. 0427

In our next problem -- Calculate the kinetic energy of the sled with the child. 0432

Kinetic energy is 1/2 mv2, so that will be 1/2 × 60 kg (mass) × 5 m/s2, so that will be 1/2 × 60 × 52 or 750 J. 0436

After a short time, the moving sled with the child aboard reaches a rough level surface that exerts a constant frictional force of 54 N on the sled. 0459

How much work must be done by friction to bring the sled with the child to a stop? 0468

The sled had 750 J of energy and in order to be brought to a stop, you must be down to 0 J of kinetic energy, so how much work must be done by friction? 0472

Friction must do 750 J of work in order to bring the sled to a stop. 0485

Now notice when it does that, the friction is going to be operating in the opposite direction, so when you are talking about work by friction, you will have to think about whether we are talking about positive, negative, but really friction is going to have to do 750 J in order to take that 750 J away that it had in kinetic energy. 0492

That concludes page 1 of the energy and energy conservation worksheet. 0510

If everything went great -- Terrific -- You are probably ready for AP level questions and if it did not, now is a great time to go back and review energy and conservation of energy. 0515

Thanks so much for your time and make it a great day! 0524