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ssm The three objects in the drawing are connected by strings that pass over massless and friction-free pulleys, The objects move, and the coefficient of kinetic friction between the middle object and the surface of the object and the surface of the table is 0.100 . (a) What is the acceleration of the three objects? (b) Find the tension in each of the two strings.

a) 0.6 $\mathrm{m}$ b) $ \mathrm{s}^{2}$

104 $\mathrm{N}$ and

230 $\mathrm{N}$

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we begin this question by using Newton's second law to calculate the acceleration off the objects. For that we can do the following. So there is a small treat that we can use in the situations which of the following. It's a system composed by three objects, but all of these objects move together, so they move with the same acceleration. Then, when we are interested in calculating the acceleration of the system, we can choose a reference frame and three to the system. As a wall like this, let me truths my reference frame as being these weird reference frame. Okay, in this reference frame, have you considered the wall system? Therefore, the Net force acting in the world system is given by the mass off the wall system, which is m one plus m Truth plus M three times the acceleration off the world system. Now, what is the Net force that is acting on the world system in the direction off my access? There are the following forces Wait Number one tension tension again. Tension, prime tension, Prime wait number tree on a fictional force that points the left. It's important to mention that I knew X pointing to the left because this block is heavier than this block, so they system. If moves is moving to the right and the tension on this point of the left. Then knowing this information, we can write the net force as follows minus W one, because it goes to the negative direction plus t because thes tension points in the positive direction minus t because that one points to the negative direction plus de prime minus t prime flows w tree and then you have the frictional force minus have f and these is the total mass off the system and one plus m two plus m tree times eight. Then I noticed that there are several cancellations because of the pairs off action and reaction off tensions. So there is a simplification here and a simplification here and we are left with the following The way you treat mine is the way you want. Minus the frictional force is equals to the full mass off the system m one plus m true plus M three at times eight. Then remember that we can write the weight force as mass times acceleration of gravity so M tree times G minus and one times G minus the frictional force. The system is moving, so the frictional force is given by the normal times the kinetic frictional coefficient. Shall we have miners? Genetic with pornography, Chef times and normal force. Bean equals two and one plus m shoot plus M tree times eight them. You can see from the problem that the normal force is clearly equal to the weight. Number two The reform M Tree Times D minus and one times G Miners mieux que times The weight number two, which is m two times g, is equals true and one plus and two plus m three times a. Then we can factor G on the left hand side to get M tree minus m one miners muche eight times, m two times d These peoples two took home us and one plus and chew plus m tree times the acceleration. Then you can stand this term to the other side to get that the acceleration is the times and three miners and one miners muche eight times m chu, divided by m one plus m chu plus m treat. Then we can plug in the values to get 9.8 times 25 miners. Stan minus 0.1 times 80 divided by 25 plus 10 plus 80. These results in an acceleration off approximately zero 0.60 meters per second squared now before showed. In the next item, I have to organize my board. Okay, the second item. We have to calculate what are detentions in each off the strings. For that, we have to apply Newton's second law to this. Enter these blocks separately, beginning my block Number one, we get the following still using the same reference frame. Newton's second law tells us that the net force acting on that block is equals to its mass times. Its acceleration. The net force acting on block number one is plus the tension that points in the positive direction, minus the weight that points the negative direction. Dan Attention minus Wait one is M. One times eight and that's it. Now it's so for attention to get em one times a plus. The weight number one which is m one times cheap they're afford attention is given by m one times a plus g plugging in the values that are given by the problem. We have 10 times the acceleration that we had calculated plus 9.8. These results in attention off 104 you times. So this is the tension in the string, the string on the left. Now for the tension on the string off the right, we have to apply Newton's second lower to block number three. By doing that, we get the following the Net force. Acting on that block is equals to its mass times its acceleration. Then notice that there are two forces acting on this block. The tension force that points in the negative direction and the weight force that points to the positive direction. Then week number tree minus. The tension is ecos to m three times a day. There is solve this equation for detention by sending this term to the other side on this term to the other side. So the tension is given by the weight number tree minus M three times a plug in the values that were given by the problem. We have the weight number three, which is 25 times 9.8 minus 25 times the acceleration off the road 0.60 then these results in attention off 230 Newton. So this is the tension own this string off the right

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