 Hey everybody, welcome to Tutor Terrific. Today I'm gonna do the same problem I did in my last physics video, but I'm gonna do it a different way, this time involving energy conservation rather than Newton's laws. And we are still gonna have to use Newtonian kinematics here at the end. And we're gonna find the same thing. We're gonna find the time it takes this soap bar to travel down the entirety of this eight meter ramp. The ramp is inclined at eight degrees and the coefficient of kinetic friction is still 0.05, just like before. This time we're gonna make use of energy conservation. I want you to understand that the soap is released from rest at this height, which we need to calculate. What type of energy does it have? Well, it has gravitational potential energy, MGH. As the soap travels down the ramp, it gains speed because it accelerates and it eventually has some kinetic energy equal to 1.5 MV squared at the bottom. And also, friction is involved. If friction is kind of a robbing force of energy, it robs the system of some of its energy and that energy it robs is equal to the work done by friction. So the friction force times the distance over which the friction acts. We are going to set up an equation for energy conservation and what it's gonna allow us to find is the final velocity down here at the bottom. Once we have that, we can use kinematic equations to find the acceleration and then find the time required. All right, so as we said, at the top of the ramp, the total energy of the system is equal to MGH. But I have to find that H value. That can be done by taking the hypotenuse of this right triangle and multiplying it by the sine of eight degrees. So eight meters times sine of eight degrees will give us this length, and I've computed that. It is 1.11 meters. Okay, so that height's important. Gravity only cares about the height difference, not the lateral difference in location. So that's the total energy at the beginning. This, according to conservation of energy, will be equal to the kinetic energy that the soap has at the bottom. And I assume this height is zero, so there's no other potential energy down there. Now we have to add to this the energy lost to the system by friction. So there's kinetic friction that's acting over the distance D, which happens to be eight meters. So we have to find the velocity. And the only thing really stopping me is that I don't know the friction force. I need an expression for the friction force, and I'm gonna use the free body diagram for the soap to get it. So let's set that up quickly after rotating the axis so the normal force points straight up. Here's the normal force. Here's gravity at an angle, eight degrees. Here's the friction force. We know that the friction force in general, kinetic friction, is equal to mu k Fn. So I've got mu k here, 0.05, but I also need an expression for the normal force. As you can see from this diagram, if we split gravity into its components, is equal to the vertical component of gravity, Fgy. Fgy can be found like it did in the last video. This is an eight degree angle, the same as the angle of the incline. Fgy can be found by taking the cosine of this angle, saying it's equal to Fgy over Fg, and solving for Fgy. We will get that the expression for Fgy is Fg cosine 8.0 degrees, and I can replace Fg with mass times gravity. So we have Mg cosine 8.0 degrees. Now that's equal to the normal force. So what I could do right now is take this expression and put it right here. So now I have that the friction force is equal to mu k times Mg cosine 8.0 degrees, the vertical component of gravity. So now I'm gonna make some more space here. I'm gonna rewrite this expression with Fk substituted for this expression I have down here. Mgh equals 1 half MV squared plus mu k Mg cosine 8 degrees times D. Okay, now we have our full expression which allows us to solve for the velocity. Notice how I didn't give you the mass of the soap. The reason you didn't get the mass again is because it cancels, because it's in every single term of this equation so I can divide it out. So now our job is to get V squared alone. So let's subtract all of this stuff to the other side. So we'll have gh minus mu k g cosine 8 degrees times D equals 1 half V squared. Okay, so the last step is to multiply by two and then square root. So we get that V equals the square root of all of this times two. So 2gh minus 2 mu k g cosine 8 degrees D. All right, so now we're gonna get that velocity by plugging all this information in and the velocity is, according to the calculator, 3.74 meters per second. Okay, I'm gonna move all this stuff up, use that velocity to get to time. Now what I need is the acceleration. In order to get the time, I will need to figure out what A is to plug it in to one of the kinematic equations, the first or the second that has time in it. Okay, that's fine, but how am I gonna get that? Well, if I have the final velocity, I know the initial velocity is zero because I'm releasing it from rest and I have the distance, I can use the time-independent kinematic equation, the third one, to find that acceleration. So if we take this and we solve for A, we will get the following. We'll get that A equals V squared minus V naught squared over two times x minus x naught. Okay, two of these things are zero. The initial velocity is zero and the initial position we can always take to be zero when we know the total displacement. We could just set this difference equal to the displacement. So let's plug this stuff in. The final things we have are V squared, which is 3.74 meters per second. Square that, don't forget to square it, and we divide that two times the final displacement, eight meters. And we compute this in our calculator. We get the following. A equals 8., excuse me, 0.874, 225. Now the reason that's, and this is meters per second squared, the reason that's slightly different than my answer from the last video is because of me rounding the velocity a little too much, most likely, but it is very, very close to the old one within the sig figs that we are allowed to use. So now that I have the acceleration and I have the final velocity, check this out, I don't need to use the second kinematic equation to find time like I did in the last video. I can use the first one, which is excellent. So let's take this one and solve for time. So we get that time would equal the initial, final velocity minus the initial velocity over the acceleration. That's how time's solved for in that first equation. And again, the initial velocity is zero. So this just boils down to plugging in the 3.74 meters per second for V and the 0.874, 225 meters per second squared for A. When you compute this, you get the following. About 4.27 seconds. And now if you go back to your original problem, you have two sig figs for most of your measurements. And so you have to round that to 4.3 seconds, which is the exact same thing you got in the other method. But some would argue that this method is a lot easier. Make sure when you can to take advantage of energy conservation, as it could save you some time when you would use a Newtonian dynamics method with forces and sum of forces equations. All right guys, thanks for watching this video. This is Falconator, signing out.