Question 75 part of a series of questions all about a block on an inclined plane with various different attributed forces and facts to go with it on a few things that we have to work out. So we haven't no 0.2 kilogram block on inclined plane, and it's connected by a string going over a pulley to a no 0.6 kilogram hanging book. We simply have to determine the acceleration of the system. If there's no friction between the block on the surface of the implying plane eso with no friction, this becomes really quite a nice, straightforward question that's model are blocks as particles because we don't need to consider too much. That's complicated about this. And so we have a situation here where we should just model our forces on the weight of ah block. Here is no 0.6 most blood by G. Onda. Of course, our weight off this block is no point to multiplied by G. Please don't read the size of the arrows to me, anything in terms of their after scaled forces are merely drawing this arrow bigger so that I can better express it is in the diagram in a moment cause we have tension that's putting us in both directions on a normal force is acting this way. Um, this question specifies this is a 30 degree angle. So will pop that in. And we should also say that this this force that weight can be split into two components. One that's parallel one that's perpendicular to sleep, which would make this angle here also 30 degrees. She's quite nice. So you've got body one on 22. So we call this 11 and this one, too. For a start. We want to consider the forces sort of generally balancing out. Certainly this body one is not descending into the slope. And so it's normal force, especially when we don't have friction. It doesn't really matter to us. We're only concerned about motion parallel to the slope. And so these are the only equations with to take, so we'll just simply take the idea that power out. It's like we have attention. Uh, um was attract from that, uh, no points to G sign 30 because, of course, we need to take the appropriate parallel component, and that has got to be equality. North Point to okay and likewise a body to weaken. Draw a quick set of equations, which is no 0.6 g minus. T is equal to no 0.6 pay. Um, on. Of course we can do a bit of rearranging. We tell. Do in a second. I want to explain something for us, which is that I have chosen very specifically the body to is the body descending on body one is the one a sending. It's going up the slope. Uh, so there is a reason for this, and that is basically that well, we don't consider friction, which simplifies the problem incredibly, but the more massive block is going to be wanting to dissent quicker than the less massive block on. If it's on a slope, it's downwards forces its forces down the slope, parallel to the slope, or even less wanting to descend than they are. If it's just perpendicular, we going. So it's obvious in this situation, the body to is the one descending okay, s. So we can certainly rearrange these equations a little bit. Let's get both in terms of attention. So let's say the tea is equal to on rather than using theatrical points or no points. Six. Let's go for a general case because it's always nice to see what's happening in the general case. So simply say, T is equal to M one. A. Plus g son 30. We'll keep the angle is 30. We might as well, and in this one t is simply equal to em to I mean in brackets G minus a. Okay, so equating these two tensions, we can simply say that M two G minus a is equal to and one a plus g sign feta course at 30. In this question, Andi, it's probably best to expend everything out because we're about to rear end for acceleration. So we said we want, uh, something like T G minus end to a or AM, too, as I've just written it there. And that's gonna be equal to M one. A. Plus m. One g. It's sign 30. Lovely, very lovely on. So hence getting all the cheese until one side and all the days until one side we now have in expressing G into minus m one sign 30. I think I got that right. Absolutely have is equal to m one plus m to all in parentheses. A. That's a rearrange they on. This is also straightforward. It's simply the left hand side. G two minus m. One sign 30 divided by the sum of the two masses and one plus m two. Andi, these air nice numbers. Simply put, the M one is no part in the empty is no 10.6 but the same 30. But that's just 1/2. So actually, we can, from our substitution, simply say that acceleration is equal to G. Apply the numbers now, and two is 9.6 and M one is no point to. But Sign 30 is 1/2 so good 1/2 that for no port one that's good. And then on the bottom, we have m one plus Empty said no 10.6 plus door point to it is no 0.0.8, and so the acceleration is simply equal to five takes G, which we put into a calculator, and we reasonably quickly establish that our acceleration is 6.12 uh, five or 6.13 You should probably express it so few fewer significant figures 6.13 meters. A second squared on. That is all the question is asking us. It's not asking about any tensions or any other forces, so we'll leave it at that. And of course, we'll carry on what these slow questions Emily next.