a-4-mathrmkg-box-is-accelerated-upwards-by-a-string-with-a-breaking-strength-of-80-mathrmn-what-is-t

Question

A 4 $\mathrm{kg}$ box is accelerated upwards by a string with a breaking strength of 80 $\mathrm{N}$ . What is the maximum upward acceleration that can be applied on the box without breaking the string?
(A) 2.5 $\mathrm{m} / \mathrm{s}^{2}$
(B) 5.0 $\mathrm{m} / \mathrm{s}^{2}$
(C) 7.5 $\mathrm{m} / \mathrm{s}^{2}$
(D) 10 $\mathrm{m} / \mathrm{s}^{2}$

Sachin Rao · Accepted Answer

the problem. We have given a four carry boxes accelerated apart by a string with a breaking strength of 18. You can the massive boxes for Katie. What is the maximum of our acceleration that can be applied in the box without breaking the stream? Suppose that exploration is a in the poor direction. This is tension T and in the downward direction M G force will be there. So we will write. According to the Newton&#x27;s second law equation, Net force should be able to mass into observation. T minus mg should be called UMA&#x27;s into exploration, so they have to find the maximum acceleration. So we know that exhibition will be called TV. I am my energy now. The maximum value of tension is 18 million. The mass is four K. G minus 9.81 So this is equal to 20 minus 9.81 This is the acceleration. So if we saw this, this is a quarter 20 minus 9.81 It is a quarter 10.2 or if we write geophysical to 10. So it is around 10 winter per second squared. So option B will be correct, right?

Averell Hause · Answer

party. We have the lower block and the free body diagram for the lower block. We have the force of gravity going down and then going straight up would be tension. So too. So we can say that the force the net force acting on the lower block f sub l that would be the net force on the lower block It I would be equal in t&#x27;s up to minus EMS G. And so we know that this is gonna be equaling a and we can say But then the mass times acceleration would be equal in attention minus the mass times acceleration due to gravity. Therefore, the tension is gonna be the mass multiplied by the sum of the acceleration due to gravity plus the acceleration of the system. So T&#x27;s up to is gonna equal 3.50 kilograms multiplied by 9.8 plus 1.6 meters per second squared. This is gonna equal 39.9 Newton&#x27;s. So this would be for a detention. So too, and then we could say for part a we have the upper block and here we have going up tensions of one going down, tensions up to and then going down the force of gravity. Now here B of the mass and we can say that the force on the upper block f sub you is equaling the mass times acceleration once again. But this would be equal to t sub one minus t sub two minus m g. And so we could say that Then we know that I am a well equal t sub one minus thi sep two plus mg and this is going to be equal to rather we can say that this becomes t sub one equaling t 72 plus um G plus m a and then we can solve So t&#x27;s of one would be equaling m times G plus a plus mg plus m A. And this is gonna be equaling to, um, times G plus A. And so essentially, t someone is gonna equal two times attention. So too, and so attentions of one equaling two times 39.9. Newton&#x27;s attention. Someone is then equaling 79.8 Newtons. This would be a rancher Answer for the tension sub one. This would be attention acting upwards on the upper block and the tension sub to this would be tensions up to for the lower block for be. Then we want to find the acceleration of the elevator before the upper string breaks. So for part B, we can say that some of forces in the UAE direction is equaling the mass times acceleration in the UAE direction. And this is gonna equal t someone minus t&#x27;s up to plus mg. And so we can say again, t someone is equaling t sub two plus m times G plus A. And we can then say that t someone is equaling two brother m g plus a plus m times g plus I. And at this point, we can say that the acceleration is going to be equaling t someone minus two mg divided by two m and so this is gonna be equaling T&#x27;s of 1 70 Rather, In this case, teeth of one becomes the maximum tension of 85 Newtons. So this would be 85 0.0 Newton&#x27;s. This would be minus two times 3.50 kilograms multiplied by 9.80 meters per second squared, and this is be divided by two times 3.50 kilograms and we find that the acceleration of the system right before the upper string breaks is 2.34 meters per second squared. This would be our final answer for part B. That is the end of the solution. Thank you for watching.

Jayashree Behera · Answer

so the object is five kilograms. It weighs five kilograms. No for the weight off the Arctic, we have force off gravity. Pointing downwards on the object is on the surface. Therefore, there should be some normal reaction to do the contact forces. If the contact falls off, the object on the surface subjects the normal force. Then the tension in the string is 30. Newton&#x27;s we&#x27;re pulling object to the stream and attention in the string is 30 Newton&#x27;s and the surface on which the object is placed is Ah, hi, some friction. So we must have some fictional force in the opposite direction off the applied force. So here. So there&#x27;s a free body diagram. Let me just did I go the values. So that&#x27;s the tension. Then we have moss five kilograms and then the coefficient of friction. And that&#x27;s the coefficient off. Kinetic friction makes sense because the object is moving and we have to find it. Find the acceleration off the moving object. If the object wasn&#x27;t moving, then you would have needed the coefficient of static friction. So here we have to go fishin off kinetic friction. No, we&#x27;re going to use Newton&#x27;s law Newton&#x27;s third law, and we will apply it on the horizontal and vertical direction and saw them to find the acceleration off project. So let&#x27;s soul for the wide election first, because that&#x27;s a simpler one that is simply than the ex direction. So I&#x27;m assuming my vertical direction to be Why, obviously and the upward is positive. I access. And to the right, we have wanted Eric Sexes now in the water girl direction. So I have Newton&#x27;s losses that the net force is equal. Do mass times accidentally shipped. And since we&#x27;re talking about the article election here, therefore the acceleration should be vertical direction. And so you should have coefficient y in the forces and excellent issue. No, let&#x27;s write down all the forces. So since I have taken my positive by axes to be upward, so all the forces that wind upwards should be positive. So I have end here than F G. That is the force, or due to gravity off the gravitational force, which is equal to mass times acceleration due to gravity, and that there&#x27;s pointing downward. Therefore, a negative sign before that. And this is equal to mass times acceleration off. Dante now again, since the since the object is not moving in the vertical direction, it means that the acceleration in the water collection is zero and therefore both the forces balance each other. So now we have the normal force on the object. Now let&#x27;s all for the horizontal direction. So we have attention in the direction off the applied force minus the coefficient of friction minus the frictional force f in the opposite direction and the negative sign. And this is equal do mass times X. It extends for the acceleration in the horizontal direction and eggs is the one that we need to find that destroyed the gender, Uh, question So that is there. And here I have substituted the values for the forces. Now we know that the fictional force with the equal to since we&#x27;re talking about kind of extraction, therefore that frictional force will be given by the kinetic coefficient of friction times. The normal force on normal forces formed the biggest equal. Tow the gravitational force here so we can substitute that in place off normal force. And now we have the fictional force. So they used this expression over here. So we have a pension minus fictional force to be quite do Mustangs acceleration. And from here we find that X is equal to do Linus Yuki n g over a over m savi move to the next place. We just write down the expression once again. Oops! And now we can substitute history values. We know the value so we can substitute them along with gravity that is basically 10 meters per second squared approximately or you can use 9.8 meters was sick and scared. So that&#x27;s the acceleration due to gravity was substituted. Those three on dhe find the acceleration to be five meters per second squared. So that&#x27;s that acceleration off. The object in the horizontal direction on the object is obviously not moving in the vertical direction. So a lie is Siegel on DDE. If you match the options, the answer is going to be option. Thank you

Meghan Miholics · Answer

in this question, we are told that a 50 kilogramme box hangs from a broker and were asked to find the tension in the rope in e Siris of different situations. So before we go into any calculations, let&#x27;s go ahead and do everybody diagram. So for the box, we&#x27;ve got the tension force from the rope is going up and the gravitational force is pulling down. Let&#x27;s call the up direction to be positive in this case. So what this means is that the sum of forces, the net force on this object will always equal tension minus the gravitational force. So let&#x27;s go ahead and see how we can use this in the different situations to calculate the tension. So in part, a were told that the box is at rest. So what that means is that the velocity is equal to zero, but also the acceleration is equal to zero, right. The box is not moving and the velocity is not changing, So the acceleration is there for a zero. If the acceleration is equal to zero than the net force, the overall force has to be equal to zero. And so, since we have two forces here tension and gravity. If the Net Force is going to equal zero, it means that those two forces completely balance each other out. Right. So another way to see that is, you know, mathematically the net force is equal to T minus F g. The past equals zero. If we pull f g over to the other side, that shows us that tension. The magnitude of the tension is equal to the magnitude of the gravitational force. They completely balanced each other up so we can calculate the tension in this case as mass times G. The mass is 50 kilograms and the acceleration due to gravity is 9.81 meters per second squared. And so the tension in the rope when the box is at rest will be 490 Newtons. So that is our first tension force. Okay. Ah, let&#x27;s do part B. So in part B, we&#x27;re told that the box moves at a study five mirrors per second. But that motion, even though the box is moving that motion, is study. It means that&#x27;s constant, right? The velocity is not changing. And so that means that the acceleration is again zero This, of course, implies via Newton&#x27;s second law that the net force, the total force, is equal to zero. And so we get the exact same situation that we had in part a. That tension is equal to the gravitational force and so attention is equal to 490 Newton&#x27;s no need Teoh do that calculation again. So this is the answer for part B. Okay, let&#x27;s take a look at part C. So in part C, things get a little bit more interesting. So we&#x27;re told that the box is moving at five meters per second again, but this time it&#x27;s not moving Study. It&#x27;s not at a constant speed. The acceleration is equal to five meters per second squared. We&#x27;re told that it is speeding up, so that means that it&#x27;s going to have an acceleration of positive five meters per second squared. Okay, so that means that the Net Force is not zero right, so we can calculate what the net force is. The net force will be the mass times the acceleration and that will give me a value of 250 Newtons. Well, I&#x27;ve already created an equation that I can use this and write. I created this equation earlier. The net force is equal to the tension minus the gravitational force. So I know the net force has to be 250 Newtons. I know what the gravitational forces. I&#x27;ve already calculated that I&#x27;m times G 4 90 and so I can use this equation. Lips two very simply solve for the tension. I just need to pull the 4 90 over to the other side. So I&#x27;m just adding for 90 to both sides and so I&#x27;ll get t is equal to 7 40 Nunes So this is the tension in the case that the accelerator but in the case that the box is accelerating upward, which makes sense, we need to provide more than the gravitational force in order to accelerate the box upwards. Okay, and then we just have one more part. And in this part again, we&#x27;re told that the box is moving five meters per seconds. But again, it&#x27;s not a study motion. We have an acceleration. It is slowing down at a rate of five meters per second. So if something is slowing down, then in the positive direction, then the acceleration is negative. So let me say that again. It&#x27;s moving up so it&#x27;s moving in the positive direction. But it is slowing down, so it will have a negative acceleration. A negative acceleration implies that the initial of Aussie is bigger than the final velocity. Right? So we are slowing down. Our initial are velocities are getting smaller and smaller. So our acceleration is negative five meters per second squared so we can calculate our net force is going to be 50 times negative. Five. So this time our net force will be negative 250 news. And I think that makes sense because if something is, you know, slowing down, it needs to have an overall force that is dragging it in the opposite direction of its motion. So the net force here is negative 2 50 and again, we can use that same equation. The net force is equal to tension minus, uh, the forced you to gravity. We know what the net forces we know, the gravitational forces and we can just go ahead and solve for attention. So what will ah, again ad for 90 symbol signs. And so on one side we get tension. On the other side, we get negative to 50 plus for 90. That will give us a total of 2 40 So in the case where the box is slowing down, the gravitational force is more than the tension writes the gravitational forces, overpowering the tension, slowing the elevator down overall for the box.

Ren Jie Tuieng · Answer

Now this question. We have an Eddie veto, right? A two blocks are actually hung from the ceiling off this elevator. Now, given that it accelerates upwards at magnitude off 1.6 just for a second puts again, we want to find what its detention in these drinks. So very simple to simplify this to find t one, I just had to consider this ISS a body, a single object or party shinko see to Mrs. And we can see that this and new body, including the two messes, experiences t one as well as his weight Total weight. And we know that nets acceleration must be following d uh, following the elevator. So the nets f nets this t one mine smg she must be close to. And it, in this case, the M is the total mess off both blocks, which would give us seven kilograms, right, Because each of the 3.5 so t one is basically and she pussy. So seven times, uh, 9.1 plus 1.6 about 77 79.9. You tits and ah t two I t to again, we just consider this block and this experiences again. M g n t true, But I am, of course, is just 3.5. So t two just be 3.5 to rest time snipe 0.1 small 0.6. So they still half off t two spoke to Unite Point night you tents? Considering that the strings county, we&#x27;ve stead maximum tension off. Uh, 85 mutants? No, the finalized. He makes me acceleration. Well, we know that the tension in the first string is actually higher One. So estan s, this is below 85 mutants, then the other string t to will definitely be bill. If I and it will not break, we just have to close to the wife&#x27;s t one and consider t one being equals 2 85 mutants. In this case, this is goes to the total mess off the blocks, which is seven kilograms stry plus Hey and you want to find what I saw? Eight. We just fight seven kilograms over. Subject away. De gravitational acceleration. Just 9.1 Suspect two point tree treat. Is this the maximum acceleration before the a string will break

Ze-Han Lee · Answer

So in this problem we have Ah, we have a book whose masses to kilograms. And we have a fool&#x27;s acting on this book so that at the speed increases from full meters, a second to eight meters per second and that the time is, uh, 0.5 seconds. Okay, so we can determine the average of force f. So we know that times t equal m times v two minus sometimes be want, right? So we too is eight meters a second. Everyone&#x27;s full meters a second, so we can see that after here equals 16 Newton&#x27;s all right.

Problem

A worker moves a 30 $\mathrm{kg}$ box by pulling …

Problem 14 Easy Difficulty

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