A child slides down a slide with a $34 ^\circ$ incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. I love solving really challenging questions like this. So, fasten your seat belt, we got lots to talk about in this question. This child is sliding down the slide with an incline of 34 degrees, we have to figure out what the coefficient of kinetic friction is, and we're told that the final speed of the child in the case where there is friction is half of what it would have been had there been no friction. And so, the case with friction we'll call case 1. So, we'll put a subscript 1 on the everything there. And in the second case where there is no friction, I'll label that 2, subscript 2. So, v1 final at the bottom of the slide is half of v2 final. And initially the child starts at rest in both cases. Case 1, where there's friction. In case 2 there is no friction. We're going to write down Newton's second law with the friction case. And so, we have 3 forces, normal force, gravity and friction. And we'll find an expression for the coefficient of kinetic friction and we will not be able to evaluate this expression because we don't know what the acceleration is. And then we'll express that acceleration in terms of v1 final because we have a connection between v1 final and v2 final. We're told that v1 final is half of v2 final, and we're trying to relate what we have in the friction case to what we know in the non friction case. So, we'll write acceleration in terms of v1 final, substitute for v2 final, and then get an expression for acceleration in the first case with friction in terms of the final velocity, in the second case where there's no friction. And I'm going to go through this slowly in a second, I'm just giving you an overview of what's going to happen so you kind of can prepare yourself. And then we can actually calculate with this v2 final is in the case with no friction because there's only two forces involved. And we'll calculate the acceleration in the case with no friction and then calculate the final speed. And then after we do that, we'll return back to our expression for a1 and substitute for the v2 final squared and that's what this is. And then... And now we have an expression for acceleration 1, substitute that back into this equation here. And blah, blah, blah, it turns into something remarkably simple, surprisingly simple looking, 3/4 tan 34 is going to give us our answer. So, we have lots to do, though, to explain that properly. What I just did is kind of like how you should read a textbook by the way, you should always glance at the captions under the images and pictures, you should look at all the headings through to the paragraphs, what are the section titles and so on, just give yourself an overview of what's coming. And it actually creates questions in your mind that help you guide yourself through when you actually dig into the actual reading after you do the overview. OK. So, you have an overview now of how the solution is going to go. And so, there's not going to be too many big surprises but what you need are the details. So, Newton's second law with friction. We have the x component of gravity pulling the child down the slide. This theta in here equals this theta which is the angle of incline, and the component of gravity is down the ramp, and so, it's in the positive direction, defining x to be positive down the ramp. And we have friction going up the ramp, and so, it's negative. And together, those make the net force in the x axis. And so, it is ma. And we'll label the a, a1, to say that this is the acceleration in the case with friction case 1. And then we can make expressions for friction and for x component of gravity. Friction is a coefficient of kinetic friction times normal force. And normal force has to be equal to the y component of gravity because there's no acceleration perpendicular to the ramp and the child's not coming off the ramp or off the slide or into the slide, so, the normal force up has to equal the gravity component, y component down. And so, that's what this says. And then the y component of gravity is gravity mg times cos theta because this is the adjacent leg of the gravity triangle. OK. So, that's friction. And then X component of gravity is mg sine theta because it's the opposite leg of that triangle. And then we'll substitute for Fgx as this, mg sine theta. And then we'll substitute for force of friction as muK mg cos theta and then we'll copy the ma1. And so this is equation 1, version b with substitutions for each of the x component of gravity and friction forces. And we'll do a bit of work on it. We can... We're gonna solve for coefficient of kinetic friction. And say we take this to the right hand side and take this left hand side and divide everything by m. And that makes a minus a1 and a positive muK g cos theta. And then switch the sides around so that we have the unknown on the left. And so we have muK g cos theta equals g sine theta minus a1, and then divide both sides by g cos theta. And that's as far as we can go with this work here until we consider the case with no friction. Well, actually I guess we're gonna go a little bit further. We're gonna express this acceleration in terms of final velocity because we have a connection between final velocity in the case with friction, to the final velocity in the case without friction. So, final velocity squared equals initial velocity squared plus 2 times acceleration times distance. And I've got a subscript 1 and everything here and then we can solve for a1 by knowing that the initial velocity is zero, the child starts from rest. And divided by 2d on both sides and switch the sides around, and we have a1 equals vif squared over 2d. Now, here's where the connection to the case without friction comes. We're going to make a substitution by saying v1 final is half of v2 final. So, that substitution just happened. And the 2 and the denominator there gets squared. So, that's 4 and it's in the denominator there, so, it combines with this to make 8. So, we have a1 equals v2f squared over 8d. Then... I don't know, I kind of feel like I should explain that and have the 8 come up, I don't know. Let's say v2f squared over 4, all divided by 2d. That's the same as going v2f squared over 4, multiplied by the reciprocal, the denominator so, that's 1 over 2d and so, then I think more clearly that shows it, it's v2f squared over 8d. OK. So, now we'll find v2f. Because in case 2 where there's no friction, we can get rid of this arrow and there's only one force in the x direction then. And so, it's Fgx, Fg x equals the net force, and so, it's m times a2, the acceleration in case 2 where there's no friction. And so, we have mg sine theta because that's the x component of gravity, equals ma2 divide both sides by m. And we have acceleration in second case is g sine theta. And from that we can figure out v2f because v2f squared equals v1f squared plus 2a2 d. It's the same distance in both cases, you know, the child falls the same, the slide is the same length in both cases, so, we don't need a subscript on the d. And... Well, there we go, v1f. That's initial there by the way, not a final. That's v2 initial squared. That's just the regular kinematics formula, and that's zero. OK. So, v2f squared equals 2 times a2 times d. And a2 is g sine theta. So, that's where this g sine theta comes from. So, we have v2 final squared equals 2g sine theta d, and we're not going to take the square root of it because a1 is v2f squared over a d. So, what is this thing squared? This thing squared is 2g sine theta d and so that's what we substitute in here. Normally I do my substitutions in red, so... Why not? Let's do that. So, we have 2g sine theta d, that was substituted in for v2f squared in this expression for a1. And the 2 over the 8 makes 1 over 4. And we have, and the d's cancel as well. And we have g sine theta over 4 is acceleration 1. That's nice. And then we go back to our equation way back here for kinetic coefficient of friction and we substitute for a1 now. And so I just copied it here, and then we'll write in a1 is g sine theta over 4,