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Question 2 A 25 kg child slides down a slide with the following configuration. It is 4 meters long and it slants downward at 308 There is friction between the child...

Question

Question 2 A 25 kg child slides down a slide with the following configuration. It is 4 meters long and it slants downward at 308 There is friction between the child and the slide. The coefficient of kinetic friction is 0.20. If the child is assigned potential energy of zero at the bottom of the slide, what is the potential energy of the child at the top of the slide? How much work would it take to move the child from the ground up to the top of the slide with no excess kinetic energy; that is, t

Question 2 A 25 kg child slides down a slide with the following configuration. It is 4 meters long and it slants downward at 308 There is friction between the child and the slide. The coefficient of kinetic friction is 0.20. If the child is assigned potential energy of zero at the bottom of the slide, what is the potential energy of the child at the top of the slide? How much work would it take to move the child from the ground up to the top of the slide with no excess kinetic energy; that is, the child's kinetic energy is zero at the bottom and is zero at the top of the slide? How much energy does friction take out f the child when the child slides down the slide? How much kinetic energy does the child retain at the bottom of the slide? How fast is the child moving at the bottom of the slide?



Answers

A child weighing 140 $\mathrm{N}$ sits at rest at the top of a playground
slide that makes an angle of $25^{\circ}$ with the horizontal. The child keeps
from sliding by holding onto the sides of the slide. After letting go
of the sides, the child has a constant acceleration of 0.86 $\mathrm{m} / \mathrm{s}^{2}$ (down
the slide, of course).(a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum
values for the coefficient of static friction between the child and the
slide are consistent with the information given here?

Hello, Everyone in this problem, we're has to find the magnitude off the frictional force. There's acting on a child as they slide down a slide. So we're told that dizzy, that the child has a constant velocity or speed, which means that their initial and final Kandic energies are going to be the same and so there's no change into kinetic energy. However, we're also told that the height off the off the slide is 3 m and so if the boy or girl slides down the slide, then their potential energy is going to change. And this changing potential energy or two freed up quote unquote potential energy has to go somewhere. And so what happens is that what we're asked to find is this frictional force what we can think off? We can think off this dissipating, uh, potential energy as coming from the frictional force that is acting during the slide. So what happens is as the potential energy changes, there is a change that is equivalent to the mass of the person times the gravitational acceleration times to change in height, which is going to be just the height off the slide. And so this changing potential energy. Is there going to be equivalent to the energy that is dissipated away through through some kinetic friction of the slide? And so that's where the length off the of the slide is gonna come in because the work that the frictional force does is gonna be the same as the change in a potential energy. So and we knew that the work is the force Times distance. All right, so we're asked to find FF, which is gonna be the friction of force pointing along the slide so f f we can just re arrange for so using the separation over here, we find that FF is the change of potential energy divided by the length of the slide. And so that is just m times, g times, Delta H, where Delta H is the height of the slide divided by the length of the slide. And if you put in all the numbers, then you find that this works out to be 105

We have a child, it's lies down a 34 degree incline, and we want to know the cowfish of kinetic friction between the child and the slide. If the child has half the speed when there is friction compared thio when there is not frictions at the bottom of the incline. So first off we start with a free body. Diagram was a component of gravity that goes down the incline in a component of gravity that goes into the pipeline. CO sign date. An Sisk MG signed data and we can write our first equation for when there is no friction. So we have mg signed data again. No friction in this first situation peoples in May. So my acceleration is g signed data and I can get an acceleration of, um, point a 5.48 year again. This is the acceleration the child has when there is no friction. So, using a cinematic the equation, I could say the initial velocity zero So my final velocity squared equals to a deep. Then we could square root that to get a final velocity of square to 80 Now, using the same reasoning initial velocity of zero my final velocity for this second acceleration for when it does have friction is going to be equal to this value divided by two. So you have the spirit of two. Aye, We'll call this one. Do you divided by two, equals the square to a to D. Now here, if you'd like you can What? The two inside of the square root. So that means I would have to mold by this inside part by four to bring that to into the square root. So this becomes ate A to D spit right in. And then we can counsel up a square roots. We can cancel out the distances because that will be the same. And we can reduce this to an eight to a one for a two. So that tells us that my second acceleration for when there is friction is going to be 5.48 divided by four, which gives us 1.37 meters per second squared. So going back to my situation in which there is friction, I can write a new equation minus friction again. In this case, I have the coefficient of kinetic friction times mg coastline data which is my normal force. In this case, masses cancel out and I want to solve for the coefficients of kinetic friction. And then I want to get me by itself and using my second acceleration. This is for when we do have friction, I get a coefficient of friction of zero point five 05

We have a child sliding down a slide and there's friction. Uh We're modeling between the child and the slide and we want to know um what's given to what we're told is that the speed that the child reaches at the end of the slide is half of what it would be if there was no friction. So somehow, if we could um make this extremely slippery, the child would be going twice as fast as uh as the child was when there was just regular regular friction here. So we have our free by a diagram of the child going down the slide. So I'm saying acceleration is this way to the right and down in this way there's no acceleration. So that says that there um this component of the weight component in this direction, which is W. Cosign data has to be equal to the normal force. And then in this direction we have minus the friction force plus the wait times sine of theta equals the mass times acceleration. Uh substituting for our friction relationship, F F F equals U K times the normal. Um And then the normal is this and then the weight is empty. We wind up, this becomes this and wait is empty again. And so then we can see the mass cancels out. And so are acceleration is G times the whole quantity Sine of theta minus UK co sign of data. Now in the in the um ideal case or the frictionless case we have the acceleration. What I called it A prime is G times sine of theater because U K zero. Now we know from cinematics that re squared the velocity at the end is to the times A would be as the distance over which that acceleration acted or basically the length of the slide. And we know that V prime equals two. V. We also know that V prime squared is two D. A. Prime. So that means um that is to for the squared. Now let's see here, we know that what we had here. We can plug this into here. So we get to D. G. Sign of data. The square is to the A. So we get um plugging that into here. We get 48 P. A. And then A. Is here. So we get eight D. G. Times this expression and we can see these cancel out Gs cancel out and there's two cancels with this eight. And we get a four. So we're solving from UK, we get the UK is four thirds times a tangent of the angle. And that angle we were given to be all right. What did they tell us that angle was? Uh let's see here 28 degrees. Yeah. So we have an angle of 28 degrees and plugging that in. We get that that is 0.40 Um is the static cowfish or the kinetic coefficient of friction between the slide and the child.

This question The mass off child has given us a musical too 15 kg The initial itis ethical to 1.7 meters. Now we have to find the internal energy off the system the internal energy is equal to They need sell gravitational potential energy off the child. We know that gravitational potential energy is given us G. But we're G is the acceleration due to gravity. So we can write 15 kg in tow. X elation due to gravity Really, which is 9.81 meter per second. Describe into initial height. This will give us the initial gravitational potential energy as no. 50 George No, For the second part of the question, we have to fine that we're did this generated energy goes this in teller energy is used up in overcoming the friction, overcoming the friction. And when this energy is used up, then it changes. In Judah he identity This heat energy goes to the child the slide and here


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