 Hi, I'm Zor. Welcome to Unisor Education. We continue talking about mechanical work and we will solve a couple of problems. This is part of the course called Physics for Teens, presented on Unisor.com. By the way, if you found this lecture on YouTube, just searching the YouTube material, you better switch to Unisor.com because the same lecture is referenced from the website, but it's presented there as part of the course. So you have basically all the components of the course including textual material for each lecture, including problems, exams, etc. Plus the same website contains prerequisite for this course, which is called Mass for Teens. And the site is completely free. There are no advertisements, no financial strings attached. So I do recommend you to go to the whole course and take the whole course actually. First maybe mathematics as prerequisite and then the physics. Okay, now let's solve a few problems related to the work, mechanical work. Now the first problem is this following. You have an inclined slope and a certain force, which we don't know yet, which one, is pushing the object of mass m up the slope. Slope has angle phi and the height to which we move it is h. Also it is known that the object is moving up the slope with acceleration a. So these are four components which basically are describing what exactly we are doing. Now unfortunately we have the friction and there is a coefficient of friction given. Now using these components of the movement and the friction we have to calculate amount of work. Our force is performing, which is multiplication of the force. In this case force is constant obviously because there is a constant acceleration. Everything is constant here. So the force is constant so we just multiply it by the distance this force is acting. Now the distance is actually a simple thing because this is the hypotenuse of this triangle. Right triangle and there is a catatose h. So obviously distance is equal to h divided by sine of phi. Now the force is more important and it's kind of more difficult to calculate. And here is what we can do to get to the force. Let's think about this object. Now we know basically everything about this movement. So most likely we can apply the second Newton's law. Well since we know that there is an acceleration in the mass, I definitely know that m times a is equal to the total force which is acting in this particular direction. Now what kind of forces are acting on our object? Well obviously unknown force f which is pushing it up. What else? Well obviously there is a weight. Now the weight let's have it here. Weight p is equal to mg where g is free fall acceleration. And this is not actually the force which we would like to use in our calculations. We would like to represent this force as a combination of two forces. One acting parallel to the slope and one acting perpendicularly to the slope. Why? Because the one which is parallel is actually contributing or preventing the movement along the slope. And the force which is perpendicular to the slope is also very important because this is the source of friction and we have the coefficient of friction here. So let me put it this way. The final force which is equal to m times a is given. It's equal to the combination of three forces. One force is pushing it upwards which is our force f which we would like to find out the work it performs. Now there is a force which prevents this motion. This is the component of the weight. Now if this is phi then this is phi and this is phi. So this is minus because it prevents it. It goes to an opposite direction. mg times sin of phi. And also preventing the motion which means it's also going against the movement is the friction force. Which is this force times coefficient of the friction. So minus m times g times cosine phi times mu. From this we can find f. m times a plus m times g times sin phi plus m times g times cosine phi times mu. And knowing f and knowing s multiplying them we will get the work. Now in the textual material for this lecture I also ask to calculate with real numbers this work if certain numbers are given for acceleration, mass, angle and coefficient of friction. So try to calculate it yourself when you finish listening to this lecture. Solve this problem again. Derive the formula like this. Use the formula and real values which I present to get the final number. I also get the final number as an answer and I hope I didn't make any arithmetic mistake. You can check your calculations against mine. So that's it for this first problem. Let's go to the next one which is kind of similar. Okay the next is slightly different but still similar. The same inclined with a slope angle of slope phi. Now I don't have the height and I don't have acceleration. I don't know that. What I do know is that the object which is initially at the very top of this slope at rest is pushed down by force f, known force f. In this case we do know the force. Now we also know that at the very end it has a linear speed of v. So whenever it reaches the end of the slope under the actions of this force f, then the final speed is v. And I also have coefficient of friction mu. And again my problem is to find out work. Now f we know, so we don't know the s, we don't know this distance. However we do know many other things from which we will derive the distance. So first of all what we can say is we can write more or less the same type of equation based on the second Newton's law. And basically see what happens. Again m times a which is where a is acceleration of the movement along this surface. It's equal to the sum of all the forces which are acting. Now obviously our force f is acting. What else is acting? Well if the body object is here then this is its weight. Then again two components, one is parallel and another is perpendicular to the slope. Now this component of the weight actually helps us because it also directed down. And we are moving our object down the slope. So I have to add m times g times sin phi. Right? Because this is also angle phi and this is angle phi. Now the friction which is a combination of mg cos phi and mu is definitely preventing us. It directed this way. So this is an equation which is very similar to the one which was before in the first problem. The only difference is I have plus here instead of minus because we are moving down and not up. So in this case my force of gravity, well more precisely component of the force of gravity helps us. Another component of the gravity which is the friction that actually also as in the previous problem goes against our movement. Now in this case we know f which means we can find out a. So a is equal to f divided by m plus g sin phi minus mu cos phi. Something like this right? Now we know a which is good. Now we can find out the time. Why? Because a times g is equal to final speed v. So now we can find the time which is equal to v divided by a. So we know a, we know v, final speed. So that's how we find the time. And now s is equal to at square over 2. This is the standard formula of kinematics of the movement with a constant acceleration at the time. We know a, we know t. So basically we know everything. And that gives you the s and knowing f you have the work. And again in this particular problem I have given in the textual material for the lecture the real values for given variables. Try to calculate everything else here and check against the number which I provide on the website on theunisor.com for this lecture. That's my problem number two. And let's go to the third one. Okay, now let's say we are on a known planet. So we don't know anything about this planet. We do have however an object of mass m which is on certain height above the ground. And then we just let it go and it goes down and at the very end it hits the ground with a speed v. Now which force is acting on it? Only one force, the gravity, right? Now my question is what is the work which gravity force performs to basically get this result? Well we don't know the height, we don't know the time, we don't know the free fall acceleration. We don't know anything except mass and the final speed. Well let's try to, nevertheless let's try to calculate the work. Okay, now let's assume that the time is t of the falling. The free fall acceleration is a and the distance or the height it goes the distance is s. Well obviously I know from the kinematics this. Also from kinematics I know this. Now we have three different unknown variables without which we can't really do anything. Well we do have two equations, we need something else, right? Well we do know that the force of gravity f is equal to m times e, right? Okay, so what can we, what can we find from this? Well the work is equal to f times s. Well let's try to manipulate with these numbers, with these equations to get to this equation. Obviously since we have only two equations with three unknowns, well we will have to probably use one of the unknowns as a base and calculate others. Like for instance we will take a as an unknown. Well then t is equal to v divided by t, by a, sorry. Now if I will substitute it to this, I will have s is equal to a t square which is v square divided by s square and divided by 2. Which is equal to v square divided by 2 a. Now knowing f and knowing s, basically we express everything with unknown acceleration of the free fall a. However what's interesting is that the w is equal to f times s which is m a times v square divided by 2 a and a cancels and the result is m v square divided by 2. So this is basically the final result and as we see we have expressed the work performed by the force of gravity expressed only in terms of the mass and the final speed. And if you remember when we were talking about the work, mechanical work I introduced the concept, I was talking about that the work is actually as a measure of the result, not how we achieve this particular result. So that's very, very important, no matter how we go to a certain goal, it's the goal which measures amount of work we have to really exhort by going there. Of course I'm not talking about some silly loops like for instance we have to go from here to there, we go half a distance, then return back, then half again, then back and say no, I'm talking about normal way of pursuing your goal. Now in this particular case it's straightforward, we don't waste any energy, we don't waste any time or anything like this. It goes straightforward. However what is interesting is that no matter what happens, our goal is the final speed of the object of mass m and that's the goal which defines completely amount of work which is necessary to achieve that goal. Well that's it, I do recommend you to go to Unisor.com and look through the textual procedures, look through the calculations, do it yourself, I think it's a very useful exercise, I think if my answers are correct and basically that's it for today. Thank you very much.