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Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is 3.0 $\mathrm{kg}$ . In Case $\mathrm{B}$ the mass of block 1 (the block behind) is 6.0 $\mathrm{kg}$ , and the mass of block 2 is 3.0 $\mathrm{kg}$ . No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 5.8 $\mathrm{N}$ does act on block 2 in both cases and opposes the motion. For both Case $A$ and Case $B$ determine (a) the magnitude of the forces with which the blocks push against each other and $(b)$ the magnitude of the acceleration of the blocks.

2.9 $\mathrm{N}$ 3.9 $\mathrm{N}$ -0.97 $\mathrm{m} / \mathrm{s}^{2} $-0.64 $\mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

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or this question. We have to use Newton's second law. So let me. I loved my reference frame as these one. This is my Y axis, and this is my sex is now. Let us apply Newton's second law to the case. Eight for the block number one. So we are engaged A and then for block number one x axis. We have the following the net force acting in block number one in the X direction. Is it close to the mask off that block times its acceleration? The net force on the X axis That's acting block number one is composed by only one force, which is this contact force exerted by block number two on block number one. Then the contact forced is because of the mass off the block number one times its acceleration, which is that No, no, we can apply neutral second law on the Y axis. By doing that, we get the following the Net forces acting on the vertical direction. Is it close to the mass off block number? World Times acceleration on the vertical direction Since it's not moving and we will not move from the vertical direction, its acceleration is close to zero, then normal number one minus wait. Number one is there close to zero. Meaning that normal number one is the cost of the weight. Number one. Now we do the same for block number two. For a block number two on the X axis, we get the net. Force acting is equal to the mass off block number. Two times he's acceleration, which is equal to the acceleration off the block number one because they're moving together, Then the net force is composed No, by true forces, the Contact Force and the Frictional force. So will you get contact force that points in the positive direction minus fictional force that points the negative direction is because of the mask off block number two times its acceleration before moving further. That I correct something that I have forgot this week. This force isn't positive because there is the contact force that Block Number two exerts on block number one and it's points to the left. So in our reference frame, it is a negative force. So miners FC is equals to m one times acceleration in the X direction now continuing for a block for block number two why direction we get the following the net force acting on that block in that direction. Is it close to the mass off the block times its acceleration on that direction, it's It will not move in that direction. It's not moving, so acceleration is it goes to zero. There are only two forces acting on the vertical direction, which are the normal force, pointing to the positive direction on the weight force point in the negative direction. Then the normal number choo is it goes to the weight number two. No, By doing that, we got four equations, one shoe tree, four off course. We have to discover what is the Contact force. And then what is the acceleration? So we must work with these two equations that includes both of these factors. Let me call this equation one. And these older equation, too. If we add equation number one and the question of virtue, we get minus F C plus F C minus F F Zico's true M one plus and true times a X before the contact forces are. Simplify it and we end up with minus F f is equals to M one plus M two times a x them the acceleration That's direction is given by minus frictional force divided by M one plus and truth. This equation will be very user food during this question. So I put it up there on the board. Now that we know the acceleration, we can go back either to equation number one or equation number two and calculate what is the contact force. It's much simpler if we use equation number one. Then you'll be using this equation. So apart into equation number one, the Contact Force is given by and one we from minus sign Times aux. But the X he's given by minus f f divided by m one plus m too. So we got em one times f f divided by m one plus and truth. So these are the equations for the Contact Force and for the acceleration on both cases. So now we just have to use these equations and calculate for Earth items A and B Let me clear the board and finish the question. The contact forced in situation A is given by SI is equal to M one times 5.8, divided by m one plus and true So this is six. Then the contact force is. It goes to 5.8 divided by True because it can simplify. Here on Deasy's 2.9 Newtons, these is the force with which one block pushes against the order for the second situation. Contact Force is given by and one which now is he goes to six times 5.8 reaches the frictional force divided by and one place and true, which is a question nine. We can again divide by treat so that we have three and true, then 11.6 divided by three. And finally this results in approximately 3.9 Newtons. Now, for the second item, we went to complete the acceleration. So the first situation acceleration he's given by minus 5.8, provided by three plus tree, which is six. These results in approximately minus 0.97 meters per second squared on situation be. The acceleration is given by minus 5.8, divided by nine and this is approximately minus 0.624 meters per second squared. And then he finished the question. Acceleration is negative because the blocks are being is going down until they stop

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