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Problem 91 Hard Difficulty

In the drawing, the rope and the pulleys are massless, and there is no friction. Find $(\text { a) the tension in the rope and }(b) \text { the acceleration }$ of the $10.0-\mathrm{kg}$ block. (Hint: The larger mass moves twice as far as the smaller mass.)


13.7 $\mathrm{N}$
1.37 $\mathrm{m} / \mathrm{s}^{2}$


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Sharieleen A.

October 27, 2020

Thought I needed a tutor to help with Physics 2012, but this helps a lot more.

Video Transcript

to complicate the value off the tension. For the first item, we can use Newton's second law on these block. So for the first item we have that the net force acting on these direction, which are you call the Y direction is equal to the mass off block number chew. These is a block number choo times Next iteration off block number. Truth, then the net force that acts on this block is composed by true three forces. So really, it's two times the tension minus Don't wait number true and this is the mass number two times acceleration number. Truth. I noticed that Block number two we'll move downwards. Therefore its acceleration is negative. So let us include these information in this equation. So here we have minus and two times a truth where the minus sign is because this block we'll move down Onley. One equation is insufficient because we have to discover what is attention on what is the acceleration. So we need more equations. Another equation that we can use his Newton's second law for these block on the horizontal direction which have called the X direction. So for the other book on the X direction we have that the net force acting. No need in that direction is it goes to the mass off block number one times acceleration off block number one A one them. The net force that acts on these access is equal to the tension force. So the tension is because the acceleration off block number one times its mass. But it's still not sufficient because we have 123 announced and only two equations for them. What can we do? We can use the hint. The hint tells us that the larger mass moves twice a Sfar as this mother Mass. So it's telling us that the acceleration off the larger mass, which is mass number one, is equals to two times the acceleration off this mother must, which is a truth. Now that we have three equations and three announce, we are able to solve the system before serving it. Let me organize my board to solve this system of equations. We begin by using the result from Equation three into equation truth. So using the question tree in Equation two, we get the following. The tension is because, too, and one times true times, a true and this gives us letting say equation 2.1, then by using equation 2.1 into equation number one, we get true times M one times true times A true miners wait number true is equals to minors. I m too times a truth. Then we can solve this equation for h you in order to get the acceleration number two. For that, we sent this term to the other side on this term to the other side to get work times. I am one times a true plus and true times eight you Is it close to the weight number two? Then we can factored acceleration over truth and write it as four times m one plus and chew is equals to delete number. True, then the acceleration Number two is equal to the weight number two which is m two times g, divided by four times and one plus m two bloody And the violence that the problem gave us We get three times nine point Kate divided by four times 10 plus street and these results in an acceleration off approximately zero 0.6 83 meters per second squared then using this value for a tree. We can go back to Equation 2.1 and company to detention. These results in the following detention is it close to M one which is 10 times truth times a truth which is 0.6 83 These results in attention off approximately starting 0.7 mutants. Then for the next item, we have to calculate what is a one the acceleration off the larger block on for a one. We just have to use this value for eight years into equation number three. By doing that, we get a one is equals to two times 0.683 and these results in approximately one point 37 meters per second. Squared on these is the acceleration off the larger book.