In this exercise, we have the system that shown here in the figure it consists of a block that's lining on a surface, and the water has a mass and one of 0.85 kg that is connected by a string, ah to another block that has a mass in two equal to 0.42 kg. And this block, this counterweight hangs from the from the string, and the string passes through a pulley that has a mass of 0.35 kg and has us a shape of a hollow cylinder within a radius of 0.2 m and an outer radius of 0.3 m. The block and one is initially sliding with a speed V zero of 0.82 m per second and the proficient friction. The friction coefficient between the surface and the block, and one is your 10.25 and we want to know using conservation and methods in question. A what is the Speed V of the block after in ah, it travels a distance 0.7 m, So in order to use conservation, energy conservation of energy, we need to remember that the initial potential energy you I plus the initial kinetic energy ke I is equal to the final Put the show energy You f was the final puttin show Connecticut GF plus the work done by the friction on the block and one uh, no notice that initially I'm going to choose the zero the potential energy for block and 1 July exactly at the center of massive lock in one. So it doesn't the petition idiot doesn't change, but the potential energy for block and two changes And I'm gonna choose the zero the potential energy to lie exactly at the initial ah point that of the center of massive block to So the initial potential energy is just zero. The initial Connecticut energy is equal to one half of m one times V zero squared, plus one half of them too. Times e zero square noticed that they are connected by a string. So, uh, which ever speed with them have the other has as well plus one half of I times Omega square. I is the moment of inertia off the buoy and this is equal to the final. Ah, potential energy noticed that m one traveled 0.7 m. So this means that M two descended 0.7 m. So the variation in potential energy is minus m two g times d. The final kinetic energy is one half of him one V squared, plus one half of them to be squared. Plus, I Omega squared over to actually in the left hand side. I should have written these omega zero square plus the work. And the work is just the normal force times the the distance traveled and the normal forces m one g Ah, and naturally, it's normal force times mute. So it's a no normal force, which is m one g times, mu times the This is the total work done. And now I'm gonna put this equation on hold in order to calculate the moment of inertia I off the of the pulley. The moment of inertia is equal to the massive apuuli times the inner radius squared plus the outer radius squared, divided by two. Okay, so keep this in mind, I'm gonna try to organize the terms a little bit, so have also, in order to organize the terms, I'm gonna need to remember that easy uh, Omega zero is people too. V zero, divided by r two and omega is V divided by r two. So we have one half of m one plus m two times v zero square, plus one half off I. The zero squared divided by r two square is equal to minus m two. I'm gonna write it like this. G de m one mu mine is in to plus the square times over two times m one plus in two. Plus I the square over to our two squared So you have V zero square times one half of m one plus m two plus one half of I divided by r two squared is equal to g d. I am one Mu min is in to plus the square times one half of m one plus m two plus I over to are two squared so v zero squared. Actually, I'm gonna just gonna go ahead and divide both sides by ah, this term here. So everything gets a little simpler. Have zero square is equal to G d. M. One mu minus m two divided by one half of m one plus in two plus I over to our two squared plus V square. So we're left with V is equal to the square roots of V zero square plus G d times in two miners in one mu divided by one half of em. One was in two. Thus I over to our two squared and I can soups into the numbers we have. So V is equal to the square root of 0.82 squared meters per second square plus G, which isn't went 8 m per second square times d is your 0.7 m times m two. She's zero point 42 minus and one that 0.85. Actually, I should have written that kilogram here. Just get it drank. So have kilogram. Mine is your 0.85 kg times zero points, 25 divided by one half of 0.45 was 0.85 kg plus hi and I, as we saw before, ISS em capital limb, which is just 0.35. So it's, uh is your 0.35 times divided by two and then and much applied by 0.2 You should have your program have your winter 2 m square. Let me just give it a little tighter 0.2 m square plus 0.3 m square. So we actually this is I. So we should still divided by two times are it's usually two times is there 103 m swear And from this we get a V is equal to one point 59 meters. We're second in question be. We have to find what is the angular speed of the Pooley And the angular speed is just the leaner speak V divided by the outer radius are too. So this is 1.59 m per second, divided by 0.3 meters which is equal to 53 radiance per second and this concludes our exercise.