A skier starts at rest before shooting down a slope that is...

A skier starts at rest before shooting down a slope that is 127 m long and inclined at an angle of 18.5°. The skier and equipment together have a mass of 45.3 kilograms. When she reaches the bottom of the hill, she slides for another 72.4 meters along level ground before she starts going up another hill, at an angle of 24.1°. For the entire trip, the coefficient of kinetic friction between the skis and the snow is 0.115. This is a long one: take it in small bites! Use forces and accelerations to answer all these questions. (3 pts) Draw a diagram showing all the forces that are acting on the skier as she descends the first hill. If any forces need to be broken up into components, show those components, and write out formulas for those components. (3 pts) What is the skier's velocity at the bottom of the first hill? Express your answer in m/s, and in miles/hour. (Hint: find the acceleration down the hill first.) (3 pts) What is the skier's velocity when she reaches the bottom of the second hill? Express your answer in m/s, and in miles/hour. (3 pts) Draw a diagram showing all the forces that are acting on the skier as she goes up the second hill. If any forces need to be broken up into components, show those components, and write out formulas for those components. (4 pts) How many meters up the second slope does she get before she starts sliding back down? (4 pts) Explain using forces and accelerations why it will take her longer to slide down the second hill than it does to go up.

Transcript

00:01 So for this problem, we have a skier who's going down a slope and then across a flat area and then back up another slope.

00:08 Her mass is 45 .3 kilograms and there is a kinetic friction term acting between her skis in the snow constantly the whole way.

00:19 And the coefficient of kinetic friction there is 0 .115.

00:23 So for this first part of the problem, we're asked to draw out the force diagrams for all of the different components.

00:30 Which i've already done here.

00:33 So there's a component of the normal force here, acting in the y direction, there's the weight, and then there's the frictional force acting back up the hill.

00:43 So i've tilted our x and y axis here a little bit so that we can say the following.

00:51 We can say that the sum of the forces in the y direction has to be equal to zero because she's not sinking into the snow and she's not flying off the surface of the, of the snow and the only forces we have acting in the y direction here are n and the y component of her weight and the sum of the forces in the x direction is equal to her mass times her acceleration in the x direction and the forces we have acting here are wx and the force due to friction so some of these terms have other things we need to plug in here we know that the weight has to be equal to mass times her acceleration, which is the acceleration due to gravity.

01:45 And in this case, if we look at the triangle breakdown of this lower triangle here, if you look at how these triangles work out, it turns out that the inner angle here between the slope and the weight actually turns out to be this angle, which i'll call theta, which is 18 and a half degrees.

02:06 So for wx, you can see that wx is actually a cross.

02:12 So this is sine.

02:17 So this is going to be mg sine theta, and the y component, wy, is going to be mg cosine theta.

02:32 And we also know that the force from friction is equal to mu times the normal force.

02:41 So we need to, in this next part, of the problem, find out how fast she's going when she reaches the bottom of the first hill.

02:50 So to do this, we need to find her acceleration in the x direction.

02:56 So let's start by writing m -a -x equals, and i'm going to pull this equation here for wx.

03:05 So this is m -g -sign -theta, and i'm going to pull this equation right here for our frictional force.

03:16 This is minus mu times n.

03:20 And now from our y component equation up here, we know that the normal force has to be equal to the y component of the weight.

03:30 So in this case, that's mg cosine theta.

03:33 So we can rewrite this as mg times sine theta minus mu times cosine theta.

03:45 All right, we can divide m on both sides here.

03:49 A x equals g times sine of theta minus mu times cosine of theta.

03:59 All right, if you plug in the numbers from this problem, you get the following.

04:11 You will find that she is accelerating at 2 .04 meters per second down the hill.

04:23 Now the next part of this is that we want to find how fast she's going at the bottom of the hill.

04:30 And for this, we're going to use the following kinematic equation.

04:36 We're going to use v squared, or v final squared, is equal to v initial squared plus 2 times the acceleration times delta x.

04:50 So in this case, her initial velocity is zero because she starts from rest at the top of the hill.

04:57 So v final is just equal to 2 times a x times delta x.

05:03 And this gives us a final velocity of 22 .8 meters per second or 51 .0 miles per hour.

05:36 So she's really going quite quick.

05:39 Now the second part of this problem, we want to find how fast the skier is going when she reaches the end of the flat segment.

05:50 Now in the flat segment, our forces change a little bit.

05:55 Her weight is acting directly down.

05:58 There's a normal force that balances that out, and there's the force due to friction acting against the direction of our velocity.

06:09 So for this, we know that n is going to be equal to mg, because she's not sinking into the snow or flying away.

06:18 And we know that the mass times the acceleration and the x, direction is going to be equal to fmue.

06:25 We have to keep in mind here that she is accelerating in the negative x direction.

06:31 So we know that this is mu times n and we know that n is m g so this is mu m g so the acceleration in the x direction while she's on the flat part of the snow is mu times gravity and we know that gravity axing in the negative direction here so our answer is actually going to be negative it's going to be negative 1 .13 meters per second squared all right so now we're going to use that same equation for finding velocities and we know that v final squared is equal to the initial squared plus two times the acceleration times delta x all right now, in this case, her initial velocity isn't zero.

07:31 Her initial velocity, when she gets to this flat part of the slope, is 22 .8 meters per second.

07:39 So v final is going to be equal to the square root of v .0 squared plus two times our acceleration in the x direction, which is this minus 1 .13 meters per second, times delta x, which in this case is, i believe, 73 meters.

08:00 Let's see, 72 .4 meters.

08:04 So if we plug all of those in here, we get, we get 18 .9 meters per second.

08:24 So she's still going pretty quick when she reaches the bottom of the second hill...

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