SOLVED:Two masses, m_{1}=1 \mathrm{kg} and m_{2}=3 \mathrm{kg} , are connected by a string that is draped over a pulley, as shown above. Mass 1 is positioned on a frictionless horizontal surface, while mass 2 hangs freely. The masses are released from rest. Determine the acceleration of mass 2 . (A) 2.5 \mathrm{m} / \mathrm{s}^{2} (B) 3.3 \mathrm{m} / \mathrm{s}^{2} (C) 5.0 \mathrm{m} / \mathrm{s}^{2} (D) 6.7 \mathrm{m} / \mathrm{s}^{2} (E) 7.5 \mathrm{m} / \mathrm{s}^{2}

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Video Transcript

So let's first draw the diagram showing both the masses are the objects. So here I have it fully been here. I haven't object with Mass and one on DDE here. I haven't object with Mass and two and they're connected by a string. Now let's mark all the forces. So there should be tension on Duh he had. There should be the gravitational fools off. Empt! Oh said that will be called to enter JI. And here we must have and one g on. And what else? That's it. My Let's start by. So there are two ways to solve this problem. So let me just start with the basics. Basically. So Solution one. Let's go. In fact so we're going toe right? Newton slow for these two masses. So for the first loss, So we have the Newton's Star Lord and over here study Newton's second law right on over here. I keep on telling mute and start Now this is Newton second now, now the half pension acting on mosques and once we're solving for the mark, the object m one only. So we have tension acting over here, and since the surfaces frictionless, we don't have to go into the fictional Oh here suggested, is going to be able to m oney on and solving for the extra addiction. The hides under direction City is going to be in 18 x. And I forgot to say that there is a normal force from the surface on the object as well. On those, these two are going to be equal. So n is goingto be equal to m one g. But, uh, we don't You don't need to write this our song for this because you're not going to need this. So now we got the attention. Now let's right the same thing, Farge and do and the acceleration off these two masses are going to be seen. How do you know that? That's because what the masses are connected by a string that is not stretchable. So the distance moved by this to object in a particular interval of time is going to be saying, which means that the velocity is going to be seen and hence the exploration is also going to be seeing. So that's I have used the same notation, eggs and eggs for both of them. Even though over here we're solving for the article direction. So this is actually white, so we have em to t minus t to be equal to em. Two x And the reason I'm choosing entity to be positive because the direction off the movement is downward. Therefore, the object is moving downward, and hence mpg has to be better than tea in orderto moved off. Check downward. So now I can substitute this expression off tension over here, so I'll get em to t minus and one ex to be equal. Tow them to eggs, and that way I get X to the equal do and duty over one. Bless em to on. Let me just try down the line. Do so in one is one Katie and to his creek, 80 on dhe d can be taken to be then need a post seconds guy. So you put those values over here on calculating that part, we find a the acceleration to be 7.5 meter ballistic and squish. So the kite option is going to be option. Now there is one other matter that, um you can used to solve the problem. So that is going to be really fast. That process is really fast, but it can be a little bit tricky. You need practice to be ableto solves that to be able to use this process for complicated problems. So basically, yeah, you're going to take this holy combination as one system. So here's my M one and two are both a single system by boat. So basically, I'm taking in one and endure as a single system, and I'll be solving for that system directly. Instead, off taking M one as one system into and as the other system drying the free body. Diagram off those two and solving sensor that I'll be taking in one place and tow both these objects as one single system on. Then I'll be solving that. So if I'm considered in these two as a single system, the tension forces become internally. So if I write the Newton's law now, these two forces own come into account because things to our Internal Forces and Newton's law are then rewrite Newton. Second, love your writing down the external forces so these two forces just cancel out on DDE. The next step is to see what is the external force that's acting on the whole system. So and the external force is the one that is causing the movement. For example, you have m one g and end over here on energy as well. But the only thing that is causing the two masses to move is entity. Everyone there is not playing any role because the normal force just cancels this force and one day on their foreign money has no part in moving these two objects. So the only force in the external force that got that is causing everyone and m to remove is empty. So I can't say that the net force acting on the whole system is empty. Oh gee, the next step is to see ah, the total how many masses are involved or how many objects are involved on dhe. Whether or not the acceleration off, both the objects are equal. So, as I said before, the acceleration is going to be cool because these two mothers are connected by by unsuitable string on dhe. Therefore, you can say that the total mass is goingto be m one less in tow on the acceleration off the whole system. So months off the system is in one place int o the net force acting on the whole system is in today and acceleration off the system is just going to be So Now we will use this, uh, three questions and we will drive down on second look so f night on the system. This equal toe marks off the system times acceleration off the system on the net. Force on the system is empty. Mars off. The system is in one pleasant do. Um, acceleration off the system is eight. This way, I get a Toby, Enter ji over and one less. And to on this expression, if you compare it, uh, solution one. You get the same expression on throwback metal as well. So again, if you substitute the values, they should come out. Toby. 7.5 meters per second squared. That matches on option. So those are the two matters for solving the problem? The first matter is going to be dealing. You see, that's the basic method standard one. The second and 3rd 0 you need a lot of practice to be able to use this off complicated problems, but eventually you'll catch up. So yes, the kind of bounce it is. Option. Thank you.

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Robert Jansen

Physics SAT Subject Test

Physics SAT Subject Test

Robert JansenPhysics SAT Subject Test

Robert Jansen

Physics SAT Subject Test