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a) $v=3.0 \times 10^2 m/s$
b) $F_{fr}=2.1 \times 10^3N$

Giancoli 6th Edition, Chapter 6, Problem 57

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Chapter 6, Problem 57 is solved.

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Transcript for this Giancoli solution

In part a, of this question we'll assume there's no friction and we'll find the speed of the glider, based on knowing that the initial potential in kinetic, equals the final potential in kinetic energy. And we'll notice that in the end, at ground level it will have no gravitational potential because we use the ground as our reference level. And substituting for each of this and flipping the two sides around, we have that one half 'mvf' squared is 'mg', gliders initial height, plus one half 'mv' initial squared that's the gliders initial launch, velocity is 'vi'. The 'm' is a constant or a common factor that cancels everywhere. We'll multiply everything by 2 as well, to get rid of this one half. And take the square root of both sides, and we'll end up with 'vf', so 'vf' is the square root of 2 'g' initial height plus initial speed squared. Substituting numbers gives us, 2 times 9.8 meters per second squared times initial height of 3,500 meters plus 500 kilometers per hour we’ll divide that by 3.6 to turn it into meters per second and square that. This gives us 296.4 meters per second which we would write with two sig figures as 3.0 times 10 to the 2 meters per second. Just the perspective, this is about 1,067 kilometers per hour, which sounds fast, but isn't quite that fast, it's not even Mac 1, speed of sound is 340 meters per second. That's faster than a regular jet liner though.

Now part b, well things are a little bit complicated and we have to know what distance this glider covers; the total distance. And just knowing its height is not enough. Because friction is going to be acting over this full distances, the blue line here, so we need to know, over what distance is friction acting, because that will affect how much work it does. We'll have this flight path is inclined at an angle theta, and we have this distance, 'd', here. We know that sine theta is opposite over hypotenuse, so the height divided by the distance 'd', in solving for 'd' it gives us 'h' over sine theta. We know theta and h, and so that's why we want to express 'd' in terms of those. Now we'll write our conservation of energy equation, with this new scenario, knowing that friction is doing somewhere. So the initial potential in kinetic, we're going to equal the final kinetic. And that's all we said in the first part, in part a, but now we'll also include the work done by friction here. Making substitutions straight to those terms, we have mg initial height, plus one half 'm' initial speed squared, equals one half 'm' final speed squared, plus the force of friction times 'd', the distance that it acts over. But we'll just write 'h' over sine theta, because that's what 'd' is. This' an algebra to solve for friction, we'll take this term to the other side, so it's going to be minus, I'm a flip the two sides around so that we have friction on the left. And we'll also factor out m, from each of these three terms.

So if force friction times height over sine theta, equals 'm' bracket ,'g' times 'hi' plus 'vi' squared minus 'vf' squared, and that's all over 2, because each of them is multiplied by half, and then we'll multiply everything by sine theta over 'h' in order to isolate the force of friction. So it's going to cancel like that. We end up with the force of friction is equal to 'm' sine theta over 'h', times 'g' initial height plus the difference in squares of the velocities, divided by 2. Substituting the numbers, we have the force of friction is 980 kilograms in the 6th edition, times sine of 10 degrees, divided by 3500 meters, times 9.8, times 3,500 meters, plus 500 kilometers per hour, which will change into meters per second by dividing by 3.6. Square that, that's the initial velocity. Minus 200 divided by 3.6, square that as well. Divide that result by 2 and the answer will be, 2061.6 approximately 2.1 times 10 to the 3 Newtons. That's the average force of friction, over the flight path in this glider. In the 5th edition, there's only a very minor change of 1,000 kilograms here instead of 980, and you actually get the same answer of 2.1 times 10 to the 3 when you round it to two significant figures.