Learn how you can calculate the launch velocity of an object by using the total energy of a system. Energy that is conserved can be transferred within a system from one object to another changing the characteristics of each object, like velocity. Created by Sal Khan.
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- Is this called a certain equation or not? Is it applicable to every situation? What is Δx equal to? (in most situations; in here it means spring compression)(2 votes)
- Δx or dx stands for 'delta x' which indicates displacement. This is calculated from taking the initial x position to the final x position (final x - initial x). In this case for the spring, the initial position is the uncompressed mark and the final position is the compressed stage, so delta x would measure how much the spring moved (which in this case more displacement = more compression).(2 votes)
- [Instructor] So we have a spring here that has a spring constant of four Newtons per meter. What we then do is take a 10-gram mass and we put it on top of the spring and we push down to compress the spring by 10 centimeters. We then let go. And what I'm curious about is what is going to be the magnitude of the velocity of our ball here, of our 10-gram mass, right as the spring is no longer compressed or stretched, or essentially when the ball is being launched? Pause this video and see if you can figure that out. And I'll give you a hint, the energy in this first state, the total energy has got to be equal to the total energy of this second state. We can't create or destroy energy. All right, now let's work through this together. So let's call this first scenario state one. So in state one, what is the total energy going to be? Well, it's going to be the sum of the gravitational potential energy, so that's mg times the height in state one, plus our elastic potential energy, that's 1/2 times the spring constant times how much we've compressed that spring in state one squared, plus our kinetic energy, so that's 1/2 times our mass times the magnitude of our velocity in state one squared. And that has got to be equal to, as we just talked about, the total energy in state two. Well, what's that going to be? Well, that's your gravitational potential energy in state two plus your elastic potential energy in state two plus your kinetic energy in state two. Now let's think about which of these variables we know and which ones we need to solve for. So, first of all, mass, your spring constant, the strength of your gravitational field, well, we know what these are going to be. These are going to be our mass is equal to 10 grams. The strength of the gravitational field, also the acceleration due to gravity near the surface of the earth, is 9.8 meters per second squared. Our spring constant is four Newtons per meter, and I like to remind myself what a Newton is, a Newton is kilogram meter per second squared. So this is also equal to four kilogram meter per second squared, and then we also have a meter over there. And actually, these meters will cancel out. And that's useful because it's reminding us that we want everything to be in kilograms and meters. And with that in mind, actually, let me rewrite our mass right over here as 0.01 kilograms. And then let's think about what's going on specifically in each of those states. So what is going to be our h one, our initial height? Well, I didn't give it to you, but what really matters is the difference between h one and h two. So we could just define h one right over here to be equal to zero. So let me write that down. H one is equal to zero. And if we say that, then what is h two going to be? H two would then be equal to 10 centimeters, but remember, we want everything in kilograms and meters, so 10 centimeters is the same thing is 0.1 meters. What is our spring compression in scenario one going to be? Well, that is going to be 10 centimeters. But once again, we wanna write that in terms of meters. So I'll write that as 0.1 meters. And then what is our spring compression in scenario two going to be? Well, we're completely uncompressed and unstretched, so that is going to be zero meters. And then what is going to be our velocity, or at least the magnitude of our velocity in state one? Well, we're stationary, so it's zero meters per second. And what is going to be the magnitude of our velocity in state two? Well, that's exactly what we wanna solve for. That is our launch velocity. So let's see if we can simplify this and then solve for v two. So we know that h one is equal to zero. So that simplifies that right over there, that term is zero. We know that v one is zero. So that simplifies that term there. We know that delta x in scenario two is equal to zero. So that simplifies that term right over there. And so we can now rewrite all of this, and I'll switch to one color just to speed things up a little bit, as 1/2 k times delta x one squared is equal to mgh sub two plus 1/2 mv sub two squared. And now let's just try to solve for this character. So let's subtract mgh sub two from both sides. So we're going to have 1/2 k times delta x sub one squared minus mgh sub two is equal to 1/2 mv sub two squared. Now let's see, if we multiply both sides by two over m, then that will get rid of this 1/2 m over here. So let me multiply that. I'm kind of doing two steps at the same time. One way to think about it is I'm multiplying times the reciprocal of the coefficient on the v squared. So right over here, it's m over two, so the reciprocal is two over m. I'm gonna have to make sure that I'm gonna multiply it times that whole side. And so what do we now have? This is going to be equal to, let's see, it's going to be k delta x sub one squared over m minus two gh sub two is going to be equal to v sub two squared. And then to solve for our launch velocity, we just take the principal route, the square root of both sides. So I could say v sub two, which is equal to our launch velocity, is equal to the square root of our spring constant times delta x sub one squared over our mass minus two times the gravitational field times h two. And so we just have to plug in the numbers now. This is going to be equal to, and I'll switch colors just to ease the monotony. I'm gonna use a new color right over here. This is going to be equal to the square root of, and I wanna make sure all the units work out. So I'm actually gonna write this version of my spring constant, so I can work with all the units. So it's gonna be four kilograms per second squared. And now delta x one we know is 0.1 meters. So if we square it, it's going to be times 0.01 meters squared. And then all of that is going to be over our mass, which we know is 0.01 kilograms, 0.01 kilograms, and then minus two times 9.8 meters per second squared times height in the second scenario, which we already know is 0.1 meters, so times 0.1 meters. Let me extend this radical right over here. And let's look at the units first to just to make sure we're getting the right units. So this kilogram is going to cancel with this kilogram. And then we have, over here, we're going to have something that's in terms of meters squared per second squared, and then over here, we're gonna have something in terms of meters squared per second squared. So that makes sense. We're gonna have a difference of two meters squared per second squared. And then when you take the principal route, you're going to get meters per second, which is the unit for the magnitude of velocity. So now we just have to get our calculator out and calculate this. So the .01s will cancel out. So this part right over here is going to be four. And then from that, I am going to subtract two times 9.8 times .1, close the parentheses. That's going to be equal to that. And then I needed to take the square root of all of that business, and I get this right over here. So this is approximately 1.43 meters per second, 1.43 meters per second. And we're done.