As a 15,000 kg jet plane lands on an aircraft carrier, its tail h...

Question

As a 15,000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a spring with spring constant 60,000 N/m. If the spring stretches 30 m to stop the plane, what was the plane's landing speed?

Accepted Answer

Welcome back, everyone. We are making observations about a trolley that is moving on a railing system. We are told that the loaded trolley has a mass of 8500 kg and we are told as well that the braking system of the trolley uses a hook with elastic cords. Now the force constant of the breaking cord is 47 kg per meter or 47 times 10 to the third newtons per meter. And we are told that the total deformation of the elastic cord for the breaking action is eight m. Now, once it comes to a complete and final stop, we can say that the final velocity of the trolley is now zero. And we are tasked with finding what is the initial speed of the trolley before it starts breaking. So looking at all these different terms here, what are we going to do to relate them all to one another? First thing that comes to my mind is that we are going to use the conservation of energy principle. And so what we can do is we can say that the final kinetic energy plus the uh uh final elastic potential energy plus the final gravitational potential energy is equal to the initial energies of all those individual values here. And I want to point out something just, just real quick because we have a trolley on a rail system, we can assume that it is flat. We are not told that it elevates or de-elevates during its breaking process. So the gravitational potential energy is not going to matter. In this case, all we need to do is focus on these four terms right here. I'm actually we have formulas for each of these terms. So I'm going to expand this equation a little bit. What we have is one half times mass times our final velocity squared plus one half times our breaking constant, our force constant for the elastic chords, times our elastic chord deformation squared equal to one half times our mass times our initial velocity squared plus one half times K times our initial position of our chord minus the equilibrium position of the chord squared. Now, one term I want to highlight here is this term at the end assuming that the chord is not stretched at all before the breaking action. What we can assume is that X knot is zero and its equilibrium point is always classified as zero for the elastic chord. And the difference difference away from it distance away from it, we can just go ahead and cancel out this term right here because that is just going to be zero. Wonderful So outside of that, let's go ahead and plug in all of our other values here, we have one half times 8500 times our final velocity of zero squared. So this whole term will be zero plus one half times 47 times 10 to the third times our deformation of eight squared is equal to our mass over two, which is 8500 over two times our initial velocity squared plus zero. Of course. And what I'm gonna do is in order to isolate our initial velocity term, I'm going to multiply both sides by two, over eighty five hundred two, over 8500. You'll see that the twos cancel out on the left hand side, on the right hand side, the constant terms cancel out. And what we are left with is that our initial velocity? Oh Let me go ahead and label that as initial velocity squared is equal to our fourth constant of 47 times 10 to the third times eight squared all divided by 8500. Finally, to isolate our initial velocity term. I'm going to take the square root of both sides which will get rid of the power on the left hand side of the equation. So now that we have all of our terms on one side, we can go ahead and solve for our initial velocity here. And if you plug into your calculator, the final answer that we get for the trolleys initial speed is 18.8 m per second which corresponds to our final answer choice of b Thank you all so much for watching. Hope this for your help. We will see you all in the next one.

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As a 15,000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a spring with spring constant 60,000 N/m. If the spring stretches 30 m to stop the plane, what was the plane's landing speed?

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