You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 10-m-high hill, then descends 15 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.0 m and that a loaded car will have a maximum mass of 400 kg. For safety reasons, the spring constant should be 10% larger than the minimum needed for the car to just make it over the top. What is the maximum speed of a 350 kg car if the spring is compressed the full amount?

Question

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 10-m-high hill, then descends 15 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.0 m and that a loaded car will have a maximum mass of 400 kg. For safety reasons, the spring constant should be 10% larger than the minimum needed for the car to just make it over the top.

What is the maximum speed of a 350 kg car if the spring is compressed the full amount? <VIDEO>

Accepted Answer

Welcome back, everyone. We are looking at an elaborate transportation system that we have established with our colleague. You see we live on opposite sides of the hill and for us on side one, we are located 8.5 m below the peak of the hill. Why our colleague which we will label as side two is m below the peak and the difference between us and them then in terms of altitude will be negative 3.5 m. Now our elaborate transportation system that we have designed is going to be a track with a carriage that uses a launching system via elastic cord. Now the elastic cord we are told is able to stretch up to three m and we are just going to say that it starts not being stretched at all with a distance of zero m. Now the maximum mass that can be carried M max uh in the carriage is going to be 500 kg. And we are tasked with finding what is the maximum speed of the carriage when launched when the carriage is only holding 420 kg. Remember we have to find that final velocity there All right, that's a lot of information to process. But let's break this down. We have all of these different variables and how can we relate all of these variables? Well, what we can do is we can use the conservation of energy equation just as follows. So we are going to have our initial kinetic energy plus our initial elastic potential energy plus our initial gravitational potential energy is equal to the final values for all of our different energies. We actually have formulas for each of these terms. And so before plugging in anything, I just want to establish our full equation that we will be using. Here, we have one half M times our initial velocity squared plus one half times our constant here. Our chord constant times our initial deformation minus our final deformation squared plus our mass times gravity times our initial altitude is equal to one half M V F squared plus one half X F K. Sorry K uh Our final deformation minus our final deformation squared plus M G Y final. I know that's a lot of terms. We'll dive into those as we go along here. We have all of these terms except for one term really, which is this chord constant right here. We will need this chord constant in order to solve for our desired final velocity. So step one is going to be finding that chord constant. And what I'm going to do is I'm going to use side one or our side as reference and use this above equation right here with the variables that we know. But before doing that, let's go over our answer choices here for the potentials of that final velocity, we have a 17.2 m per second. B 12.6 m per second. C 16.1 m per second or d 11. m per second. All right. Well, what do we know about why? One? We are going to classify here, the, the peak as 0.0, right? And I don't mean 0.0 as in the decimal place, I mean point as in a place. And our initial placement here, we will say that that is going to be zero m. Now also for side one, because we are going to be transferring the cart over, right? We are going to have a final velocity of zero because we're launching from side one to side two. So that's gonna be transferred over. And we also have an initial velocity of zero. And what else do we have? We have that our initial deformation minus our final deformation is equal to three. All right. With all of this in mind, we are ready to go ahead and find our desired chord constant here. So let's go ahead and do that. Shall we, we have using our, our extended equation here. We have zero plus one half times K times three squared plus zero is equal to zero plus zero plus our maximum. We are wanting to use the maximum mass here because we're trying to find what that maximum chord constant is uh times gravitational acceleration, times our uh al altitude right, or distance away from the peak on side one. And then that's it. OK. Wonderful. So, in order to isolate this K term right here, I'm actually going to multiply both sides of our equation by 2/3 squared, 2/3 squared. And what you'll see is that, that cancels out the constants on either side of K. And what we are left with for our minimum minimum constant here is K is equal to two times our maximum mass times our initial height on side one, sorry, our final height on side one, that should be a one there all divided by three squared. So let's go ahead and plug in the values that we know. This is two times 500 times 9. times 8.5 all divided by three squared, which gives us 9.26 times 10 to the third. But that is not our desired cod constant. Because if you remember, we have that the maximum max mass of the carriage is 500 kg and the cord has a 15% greater spring constant than the minimum constant. So what we need to do is we need to multiply the value that we found by 1.15 to get our maximum cord constant here. Let me go and scroll down a little bit. So we have a little bit more room. So let's go ahead and do that. We have our cord constant. Our final K value is equal to 1.15 times 9.26 times 10 to the third, which gives us 1.6 times 10 to the fourth newtons per meter. Wonderful. So now that we have our final chord constant, we are now going to go over to our colleague side side two. And on side two, remember we are trying to find that final velocity of the cart and once it is launched only holding 420 kg here. So let's look at all the variables we have for side two and I know they're up there. I will rewrite them down here just to make sure we have everything we have that the difference between us and our colleague in terms of vertical distance is 3.5 m. Once again, we are going to have that our initial velocity is equal to zero. We are going to have that our peak height. We just established as zero m and we still have that our deformation is equal to X I minus X F equal to three. With that, let's plug in our values to our equation. Once more we have on the left hand side of our equation one half times our final chord constant times three squared, it's equal to one half times our mass times our final velocity squared plus M G times negative 3.5. Wonderful, great. So now let's go ahead and plug in all the values that we know. This is equal to 9/2 times one point oh six times 10, 2. The fourth is equal to 2, 10 times our final velocity squared minus 4, 20 times 9.8 times 3.5. Wonderful. So I'm going to add this term on the right hand side here to both sides. And then I will be able to divide both sides by 2 10 as well. And what does this give us? Well, let me scroll down just a little bit more. What we get is that our final velocity squared is equal to 47,700 plus 14, 4 oh six divided by 2 10. Finally to isolate our final velocity term, we're gonna take the square root of both sides and you'll see that the power on the left hand side of the equation disappears. And when you plug this into your calculator, we get our final answer of 17.2 m per second. Great. So now scrolling back up to our answer choices here. What we can see is that corresponds to our final answer. Choice of a Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

Verified Solution

Key Concepts

Potential Energy

Spring Potential Energy

Conservation of Energy