Work and Energy Legacy Problem #31 Guided Solution

A good problem solver reads the problem carefully, develops a mental picture of what's going on, maybe even constructs a diagram representing the physical situation, then identifies the known and the unknown quantities, recording them right there on the diagram, then uses physics ideas to plot out a strategy to get from known quantities to unknown quantities. In this problem, a very difficult problem, you're going to need to employ these habits in order to successfully get from the known information to the unknown information. I read of a sledder who's sliding along at the bottom of a sledding hill. We have a speed of 12.6 meters per second at the bottom of the sledding hill. She then begins to slide into an upward-inclined embankment and travels up the embankment until she finally comes to a resting position. The embankment, or incline, is inclined at 16 degrees above the horizontal, and we're told there's a coefficient of friction along the incline of 0.128. We wish to determine a height to which she will travel. I'm going to base my solution around the five-term equation that you see here at the bottom of the page, and in that five-term equation, there's a couple of terms that I can cancel. First of all, I can cancel the PEI, because at the bottom of the hill, I'm going to call that zero height, and as such, as she goes into the hill, she has no potential energy. And I'm also going to cancel the KEF, because when she's finally stopped and gets to this highest height possible, the KEF value is going to be zero joules. So my equation then becomes KE initial plus W and C equal MGHF, or PEF. You should write this down. I'm going to take this three-term equation, I'm going to expand it, and I'm going to expand it to one-half MVI squared plus F friction times D cosine of 180 degrees is equal to MGHF. Now, I'm going to keep doing some algebra, and I'm going to save the substitution of numbers until the very end. And as I do the algebra, you'll want to write down each step and follow along as we do it. Now the F friction value is the friction along the inclined plane. It's always a mu times F norm. What's odd about an inclined plane is that the F norm is not going to balance Fg, but rather the perpendicular component of Fg. So the F friction is mu times F norm, or mu times MG cosine of 16 degrees. So now my equation becomes the KEI term, one-half MVI squared, plus the work done by non-conservative forces term, which would be mu MG cosine 16 degrees times D is equal to the final potential energy MGHF. Now I'm going to notice that there's a mass in every term of this three-term equation. So I'm going to divide through by M and cancel all the M's. The equation becomes one-half VI squared plus mu G cosine 16 degrees times D times the cosine of the angle between F and D, which is 180, is equal to G times HF. Now what I wish to do is I wish to get rid of this D. I don't know what D is, but I know that D is related to the final height and to the angle of incline. I can look at my diagram, which I've drawn, and in my diagram I would notice that the D is the diagonal distance along the inclined plane, and the HF is a vertical height, the side opposite 16 degrees on a right triangle. So I can say that the sine of 16 degrees is equal to HF over D, and then I can say that D is equal to HF divided by the sine of 16 degrees. I'm going to take this HF divided by the sine of 16 degrees and substitute it into my equation for D, and the equation now becomes one-half VI squared plus mu times G times the cosine of 16 degrees times HF over sine of 16 times the cosine of 180 degrees is equal to G times HF. Now the next step involves getting the HF terms on the same side of the equation by themselves, so I'm going to have to take this work term that is the second term on the left side of the equation, and I'm going to have to subtract it from each side so that I get all the HF terms by themselves. Doing so yields this equation, one-half VI squared equals G HF minus, here's the long term, minus mu times G times the cosine of 16 degrees times HF divided by sine 16 degrees times the cosine of 180 degrees. Now this is up minus that term, and there's also a cosine of 180 degrees in there, and I want to kind of simplify this equation a little bit. So I'm going to take the cosine of 180 and change it to negative one, and then take the negative term in front and change it to positive. Now on the right side of the equation, I have two terms with HF in it, my unknown quantity, so I'm going to factor HF out of that side of the equation. So the right side of the equation becomes HF times the long quantity G plus, here we go again, mu times G times cosine of 16 degrees divided by the sine of 16 degrees. Now if I substitute 9.8 in for G, and I substitute in the value of mu, 9.8 for G, and cosine 16, sine 16, do that whole thing, I evaluate what's inside the parentheses to be 14.1746. So I have one half times VI squared equals HF times 14.1746, and I can substitute in my initial velocity of 12.6 for VI on the left side, and I can divide through each side of the equation by this 14.1746, and when I do, I get a final answer for HF of 5.6002.