WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone! I am Dan Fullerton and I would like to welcome you back to Educator.com
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Today we are going to talk about conservation of energy.
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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.
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So the Law of Conservation of Energy -- one of the big ideas in Physics.
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Energy cannot be created or destroyed; it can only be changed to different forms.
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So mechanical energy is kinetic plus gravitational potential energy, plus spring potential energy.
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And we also have conservation laws for total energy and if there is no friction, conservation of mechanical energy.
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Let us see how these work.
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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.
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Let us try and figure out what its total energy is.
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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.
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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).
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Or that implies then that its total mechanical energy will be about 2.63 times 10^9 J.
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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.
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Find the new velocity of the jet -- and we are going to assume that we are not losing any energy to friction.
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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.
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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.
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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.
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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).
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Multiply all of this by 2 and then subtract the (mgh) from one side.
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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.
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When I substitute in my values -- that is 2 times our total energy, which was 2.63 × 10^9 J...
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...We have a little bit more to put in there -- × 10^9 - 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.
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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.
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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.
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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.
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That is what allows them to convert very quickly that altitude into velocity, which is so important for winning dog fight scenario battles.
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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.
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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.
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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.
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When it is at the top -- that is gravitational potential energy, and at the bottom it is all converted into kinetic energy.
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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.
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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.
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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.
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Let us do that from a kinematics approach, back from some of our earlier lessons.
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If we drop an object from a height of 10 m -- well if we do that, let us look at the vertical analysis.
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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.
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Our acceleration is going to be about 10 m/s² down and the time -- we do not know either.
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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).
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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).
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But let us look at this for a second -- (a) is our acceleration which is the acceleration due to gravity.
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So we could write then that (a) = (g), and Δy is the change in height -- so Δy = h.
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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).
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That should look mighty familiar -- V = the square root of 2(gh).
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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.
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Two different ways of solving the same problem -- one with the conservation of energy approach, one with your traditional kinematics approach.
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Let us take a look at a problem of a pendulum.
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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.
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So this must be our mass (m), it is going to go back and forth and it has some length (L) -- same length here.
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And as it does this -- at its highest point here, it has gravitational potential energy.
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At its lowest point, it must have kinetic energy -- that is where it is going the fastest.
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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.
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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.
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So right there let us put some amount of kinetic energy, and kinetic energy we will make it green.
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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.
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So our graph of kinetic energy is probably going to look something like this upside down U shape.
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It should be pretty symmetric -- should be perfectly symmetric -- my art skills are a little off.
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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.
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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.
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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.
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Over here we get an alternate curve that looks kind of like this.
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And what we are really doing is as the pendulum swings back and forth is trading off gravitational potential energy and kinetic energy.
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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.
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It is going to remain the same the entire time, just different amounts, and different divisions of gravitational potential energy vs. kinetic energy.
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We could look in a little bit more detail over here as well -- so all potential energy here, potential energy here, and kinetic here.
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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...
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...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).
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Its maximum potential energy is going to be (mgh) or (mg) and (h) is (L) × the quantity -- 1 - cos(θ).
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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).
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So we could solve to find the maximum velocity of our ideal pendulum here by solving this equation for velocity.
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And hey! We are here. Why not do it?
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I could write then that if this is the maximum velocity -- what I am going to do is rearrange this a little bit...
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...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).
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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).
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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.
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All right, let us look at an example of a cart compressing a spring.
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The diagram here shows a toy cart possessing 16 J of kinetic energy traveling on a frictionless horizontal surface toward a horizontal spring.
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If the cart comes to rest after compressing the spring a distance of 1 m, find the spring constant of the spring.
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Well, we can use conservation of energy to solve this.
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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.
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Therefore, 16 J must equal -- well the formula for the potential energy stored in a compressed spring is 1/2 Kx².
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We want to rearrange this to solve for (K), the spring constant, so I could write that as (K) is equal to...
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...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.
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Looking at a little bit more involved example -- here we have a pop-up toy compressing a spring.
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The pop-up toy has a mass of 0.020 kg and a spring constant of 150N-m.
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A force is applied to the toy to compress the spring 0.05 m.
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Calculate the potential energy stored in the compressed spring.
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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.
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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.
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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.
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Let us find the maximum vertical height that it was propelled to, and we can do that by conservation of energy.
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In this case, our initial potential energy, which was all stored in a spring is turned into final potential energy, which is gravitational.
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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.
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Let us see if we cannot take this to another type of problem.
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A car initially traveling at 30 m/s slows uniformly as it skids to a stop after the brakes are applied.
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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.
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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.
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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.
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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.
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Well we have a uniform slow down, therefore that is all we need.
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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.
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When you do work on an object, you change its energy.
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If you are doing work on a frictionless horizontal surface to it, what type of energy are you giving it?
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It must be kinetic energy or energy of motion. You are increasing its velocity.
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Let us take a look at a block on a ramp.
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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.
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Find the total work done by friction on the block as it slides down the ramp.
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Well let us start by making a diagram -- there is our ramp, we will put our block on it.
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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.
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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)...
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...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.
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Let us take a look at a little bit more creative, fun example.
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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.
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Once he reaches the bottom of the mudslide he has the misfortune to fly horizontally off of a 15 m cliff.
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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.
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And at that point he becomes a projectile. How far from the base of the cliff does Andy land?
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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.
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First step -- let us find what his potential energy due to gravity is up here.
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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.
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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.
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He is going to fly off the cliff horizontally with a velocity of 20 m/s.
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Now we have a projectile problem.
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Horizontally, he is going to have a constant velocity of 20 m/s.
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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.
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To figure out how long he is in the air, then we have to look at vertical motion.
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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.
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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.
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The equation that I would use to do this is I would say that Δy = V initial × t + 1/2 at².
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V initial = 0, so that whole term goes away.
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We could then write that (t) must be 2 Δy/a(square root) or 2 × 15 m/10 m/s²(square root).
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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.
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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.
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This distance right there must be 34.6 m from the base of the cliff.
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That is conservation of energy problem combined with our knowledge of projectile motion.
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Hopefully that gets you a good start with conservation of energy.
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Thank you for your time and for watching Educator.com.
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Make it a great day.