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A 4 kg block is moving at a speed of 2.71 m/s. What is the force required to bring the block to a stop in 0.0009 s?...

Question

A 4 kg block is moving at a speed of 2.71 m/s. What is the force required to bring the block to a stop in 0.0009 s?

A 4 kg block is moving at a speed of 2.71 m/s. What is the force required to bring the block to a stop in 0.0009 s?



Answers

A block of mass 0.89 slug moves with a force of 17.0 lb. Find the block's acceleration.

Here for the solution. The net force acting on blog is zero. That is sigma ethical to F. One plus two plus three which is equal to zero. And we consider it as the equation one. Hello everyone is the force applied by the lady on block. After is the force applied on the man on block by the man on block and after is the additional force now we re arrange occassion. Want to obtain an expression for F. Three. So by rearranging we get after equal to minus and break it at one plus F. Two. And we consider it as a question too. Now for the calculation the total work done by the lady is that Afghan factory equal to 5.5. And and break it cost 30° except minus signed 30 degree white cap. Or it will be equal to 4.76 and X cap plus minus 2.75 White Cap. So now the total work done by the man is that it's happened factory called to 3.5. And and bracket cost 37 degree. Xscape minus sign 37 degree. Why Cap all? It will be equal to mm hmm 2.80 and X. Caplis 2.11 N. Y. Cap. Now with the substitute 4.76 and X. Caplis minus 2.75 And white cap for F. One and 2780 and X. Caplis 2.11 And wake up for F. Two in decoration too. two. When the additional force keeps the block at rest, that is f. three. Oh sorry. So it will become after equal to minus in bracket 4.76 and Xscape blessed In Bracket -2.75 and white cap plus and back at 2.8 and X cap plus 2.11 and white cap. Or by simplifying this, we get equal to minus 7.6 and X cap plus 0.64 newton white cap. Therefore the additional force keeps the block at rest is -7.6 X capital, us 0.64 and white cap. So this is a complete solution. Please go through this. Thank you.

Solving party of this problem. So total mass M. Is equal to M. One plus M. Two here, which is equal to three kg plus two kg, which is equal to five kg according to the given data. Now, actual reason is equal to F made by total mass M. So just putting the value here so I can attend newton by five kg, Which is equal to two m per second squared. As the answer for party. Now solving part B. So I can write F. A. Is equal to M. B. A, which is equal to two kg, multiplication and two m per second squared, which is equal to four newton. Now solving party. So prompted Newton's third law. The force on A by B. Is equal to the force by be on A. So uh huh. A. Will be equal to F. B. Which is equal to four Newton according to the Newton's 3rd law.

So here we have the quick freedom free body diagram of the system we have our two masses with in contact with one another. And we can then apply Newton's second law on the first block. And we can say that some of forces on the first block Would be equal to the mass of the first block, multiplied by the acceleration. We can say that the f the force applied minus the force of block to exerted on one would be equal to them. I'm someone hey. And so we can find that. Then the force exerted would be equaling two Sub one times the acceleration plus the force exerted On Block one by Block Too. For these summer forces on the second block, this would be equal to the mass of the second block, multiplied by the acceleration. And we have the fun force of block. One on block two would be equaling M. Sub two. Okay, so with that for a part a we can solve for the force of uh force of one to the force exerted on block to buy block one. This would be equal to M sub two times a Equalling. Then 5.70 kg. Multiplied by 2.45 m/s squared. And this is giving us then 21.8 newtons. My apologies. 2.45. multiplied by 5.7. 13 point 965 Newtons. At this point, we can say that The magnitude of the force exerted on block to buy Block one would be equal to the magnitude of the force exerted by block too, on black one. And so given this, we consider the F would be equaling two M sub one times the acceleration A plus F. Sub 2 1 equaling. Then Here we have 3.20 kg. Multiplied by oh 2.45 m/s squared. And this would be plus then 13.965 Newtons. And so the force exerted here would then be equal to 21.8. Newtons rounded to three significant digits. For part B. The contact force between the blocks, we can say this would be the contact force and this would then be equal to F sub 1 2 or Fs up to one. Or again simply 13.965 Newtons. We have to round two, Three significant digits. So this will be approximately equal to 14.0 Newton's. And so for part C, The Net Force Acting on Block one I would then be equaling to the force applied minus F seven 21 Equalling then 21.805. Newtons will around at the very end -13.965 Newtons. And we find that the net force On block one would be equaling to 7.84 Newton's. That is the end of the solution. Thank you for watching

In this problem, we have a pulley at point B that's being pulled down with a force. I'm of 250 Newton's and we wanna find the speed of the block point A when it travels 1.5 m. So that's we'll call. That s 1.5 m. We wanna find its speed which will call V two and also given the mess of the block, which is 10 kilograms. Well, by the way we see a force and we see a distance. So we know that we're dealing with the work here. Since you wanna find the velocity, we know that there's a kinetic energy involved as well, since kinetic energy involves movement so we can use the work energy principle, which is just initial kinetic energy, plus the work being done eyes equal to final kinetic energy. Because this system is initially at rest on the initial kinetic energy zero because it's not moving velocity zero, then we just have the work being done is equal to the final kinetic energy, which is one half and B two squared. We'll call this equation one. So let's look at the work being done. So work is just force times distance, but there are two forces involved. We have the force from point B that's pushing down, and we also have the force due to the weight of the block. The block is going to be pushing up. Yeah, by that force. But the weight is going to be pushing in the opposite direction. So the forces are the foresters, the fully minus the force through the weight, which is just mass times gravity, I mean, multiply that times the distance travels, which were given as 1.5. Now we can just plug in our numbers. So force we have 2 50 minus 10 for the mats, 10 7.8. And of course, all of this time is 1.5. So for the work being done, we get 228. Let's put this in too hard pleasing. So we have. On the left side, 228 is equal to one half m. The mass of the block, 10 times B two squared B to the velocity is what we're looking for. So let's just all 42. So V two is equal to 2 28 and then one half times. Tennis five. So what's the viable size by five? So you have to 28 divided by five, and then to get rid of the square, take the square root. So so then the velocity be to is equal to 6.75 meters per second.


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