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Aloha, we're going to be solving problem 66 in chapter 4.
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A child slides down a slide with a 34 degree incline.
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I've drawn here.
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And at the bottom, her speed is precisely half what it would be, would have been if the slide had been frictionless, calculate the coefficient of kinetic friction between the slide and the child.
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So i'll say that if it's frictionless, the final speed will be, i'll define it as capital v.
00:36
And we know that her, in this case, we're told that the final velocity is half of that.
00:54
So that's our given information.
00:56
And you want to find the coefficient of kinetic friction, muay, between the child and the slide.
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So you can start by drawing a free body diagram.
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It's going to be the force of gravity, always pointing down the structure.
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Toward the center of the earth.
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A normal force coming from the slide, which is perpendicular to the surface.
01:26
And she's sliding down, so the force of friction will be opposing that motion going up the slide in that direction.
01:37
And i think we're just considering when she's sliding down, so not when someone's pushing her down.
01:43
So that's the only forces acting on her.
01:48
And we can define a coordinate system so that the x direction is along the plane of the slide.
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And here is the angle theta.
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Same with this angle.
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And this quadrant here sums to 90 degrees.
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So this is 90 minus theta.
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But this quadrant here in blue also sums to 90.
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So that means this angle here is theta.
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So we can redraw this free body diagram.
02:34
On a coordinate system that just looks a little more intuitive with y being in the vertical direction.
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You're free to rotate our vectors around together.
02:50
And m g is at this angle of theta relative to the y direction here.
03:00
And we know that in the y direction the forces are going to sum to zero.
03:08
She's not moving in that direction at all.
03:10
She's sliding down.
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So this is newton's second law.
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And we need the component of gravity that's in the y direction, which isn't just mg.
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It's a component that's in this direction, which is the adjacent side of this triangle.
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I've just drawn here.
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Here's a right angle in this corner.
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So mg is the hypotenuse.
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So this is, this side here will be equal to the cosine theta will be this side over this hypotenuse, which is the length.
03:46
This factor magnitude mg times cosine theta fg in the y direction that's the mg over so that's how we got that and then from that equation we know we can see then the n is equal to mg cosine theta and in the x direction whatever her acceleration is we know that the only force well there's two forces.
04:32
One is going to be the component of gravity, which is pulling her down, and that's going to be mg times the sign of that angle, this component here.
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And then we're going to subtract because it's in the opposite direction, the frictional force.
04:58
So to consider what her final speed is without friction, we would just have only, i'll write this in red, without friction, we would only have the force of gravity, no frictional force, which means that the acceleration would then just be g sine theta in that case.
05:57
And then you can then use kinematics to figure out what her final speed would be.
06:16
So we have that equation as g...