Work and Energy Legacy Problem #32 Guided Solution

This problem is a very difficult problem that's going to demand that you do some good visualizing of the physical situation, that you're very diligent about how you organize known information, that you're using good physics understanding, and that most importantly, that you're employing the habits of an effective problem solver. Here we read about a boy who starts on top of a sledding hill. He slides down the hill, we're told the hill is inclined at 32 degrees, cross the level surface, and finally slides to a stop. What we're asked to calculate is the distance he slides along the level surface before he finally comes to a stop. We're given the mass of the boy, 27.5 kilograms, the coefficient of friction which exists along both the hill and the plateau, that's the mu in the equation F right equal mu F norm, and it's 0.128. We're given the initial height of the boy up on the hill, that's a vertical height, distance above the plateau at the bottom, and it's 8.45 meters. Finally, we're given the incline angle of 32 degrees. Now, to do this problem, we need to center it around the use of this 5-term equation you see written on this page. As we use the equation, we're going to begin by eliminating terms which are 0. For instance, the boy starts from rest, so the KE initial term is 0. And finally, the boy is stopped, so the KE final term is 0. We call the level plateau at the end of the run the 0 height point, and we can also cancel the PEI. Wow, you'd think this would be an easy problem, since now the 5-term equation is reduced to two terms. The two terms are PEI plus W and C, and that's all equal to 0. Now, for the PEI, that ought to be easy to calculate. It's simply MGHI, and we know the M is 27.5, and we know the HI is 8.45 meters. So we can go MGHI and find this value for potential energy. It's initially 2277.275 joules. Now we get to the hard part, and the hard part is in the W and C term. There's work being done along the incline plane, and work being done along the level surface. Now, I have to divide this up into two terms, because work is calculated as F-friction times the D, times the cosine of the angle between the friction and the distance of motion. Now I have to divide it into two sections, because the friction along the incline plane is different than the friction along the level surface. For the incline plane, the friction value would be mu F-norm, but F-norm is MG times the cosine of 32 degrees. For the level surface, the friction is also mu times F-norm, but F-norm along the level surface is mu MG. So as I go to interpret this W and C term, I divide it into two terms, the work along the incline, plus the work along the level surface. I'm going to discuss the work along the incline first. That's the harder of the two terms. That's going to be mu MG times the cosine of 32 degrees, multiplied by the D, multiplied by the cosine of the angle between F and D, which is going to be 180 degrees. Now for the mu MG cosine 32 degrees, you can calculate that, since you know mu is given as 0.128, and you know M is 27.5 kilograms. So substitute into this equation and find out what the mu MG cosine 32 degrees is. It comes out to be 29.2543. We're not done yet with this work term, because we also have to multiply that by the D. And you'll notice the D, over which the force works along the incline plane, is not given, so we'll have to calculate it. So we draw the incline, and we know that the vertical along the incline, the vertical side, is 8.45 meters. And we know that the angle the incline makes with the horizontal is 32 degrees. So this 8.45 vertical height is the side opposite 32 degrees. And the hypotenuse is the D that we're looking for. So we can say the sine of 32 is equal to 8.45, divided by this hypotenuse of D. Now we can rearrange that equation to say that D is going to be HF divided by the sine of 32 degrees. And when you plug in 8.45 for the H initial, you get 15.9458 meters. So the D in this FD cosine theta equation is 15.9458. I'll multiply that by the friction along the incline, and by the cosine of 180, and I get negative 466.4834 joules. Now what I have is the P initial, plus the work done along the incline, plus the work done along the level surface, is equal to zero. Now the work done along the level surface is the friction force times D times the cosine of 180. Now the friction force is mu mg, and the D, I keep in there simply the D along the level surface, and that's my unknown in this equation. Now if I go times the cosine of 180 degrees, I get a negative there. Now I'm going to take this negative term from the left side of the equation, and I'm going to add it to both sides so that it cancels from the left, and it shows up on the right side as mu mg times D along the level surface. So now what do I have? I have MGHI, which is 2277.275, plus work along the incline, which is negative 466.4834, is equal to mu times M times G times my unknown quantity D. Now I can group the two terms left side together. I get 1810.7916, and then I can divide it by mu and M and G, and that will give me my answer. It comes out to be 52.4928, and I can round that to 52.5 meters.