a-student-attaches-a-15-kg-mass-to-a-40-cm-long-string-the-student-then-begins-to-whirl-the-mass-in-

Question

A student attaches a 1.5 kg mass to a 40 cm long string. The student then begins to whirl the mass in a horizontal circle with increasing speed. If the string snaps when the mass has a speed of 6 m/s, what is the breaking strength of the string?
(A) 60 N
(B) 120 N
(C) 180 N
(D) 240 N

Vishal Gupta · Accepted Answer

what, in the given problem, length off the spring is given. S l is equal to 40 centimeter. Or we can say this is 0.40 m moss. A fetched with each is MM is equal to 1.5 kilograms. Mhm. The string snaps when their speed achieved by the mask is 6 m for a second. So the centrifugal force acting on this mask under which the spring will snap that centrifugal force is given by the expression and we squared by our and for our means, the radius of a circle. This is the land of the strength, So plugging in all known values here Mm mass is the 1.5 kg. For we. This is 6 m per second to the square for Lent. This is 0.40 meter. So this force at which the string Biggs comes out Toby 135 Newton. So out off the four options given eight. Option says the breaking strength of the string is 60 Newton BC&#x27;s. This is 1 20. Newton C. Says this is 1 80 Newton and D. C is this is 240 Newton. But as the string is breaking. Have one. 35 Newton so we can see the breaking strength. Offspring is 120 Newtons. So the answer off the question is option B, which says the breaking strength is 120 Newton. Thank you.

Vishal Gupta · Answer

you know, given problem. There is an object moving over a vertical circular track. This is the work because circular crack the object is attached with a spring whose land every is 1.5 meter, and the muscled object is we&#x27;ll get you. So when the object will be at the bottom, most point we have to find the tension in the string. At the bottom was point. The two forces will be acting on objects. One is X wait 15 vertically downward and a lever is tension D A. As this is the point. A the lower most white, the bottom, OTA said. Your track. So in this condition, as in order to complete a circle, there should be always a centripetal force acting on the object. Centripetal force should act the words the center only. So if I use free body diagram off the object at this point a using free body diagram F B d off the object at e. We know that force will be de a minus Angie, and it should be equal. Do M B squared by L. M. Is the length of the string, which will serve as the radios off the circular track speed, constant speed off The object has been even as well. I need a poor second. So here the tension in the string will become MGI Bless em. The square by l or I can write em You can cheap less Reese whereby L So when I put these values, I get it. That&#x27;s a two for mass 9.8 to meet a possibly swear for acceleration due to gravity, this is well square for velocity. This is length one point fight. So it will come out to be do in 29 point it plus 1 44 on one point fight. So finally uncertainly 211 or six Mutant This is the tension in the screen in the object is at the bottom off the circular part No, we have to find the tension in the spring when the object may be at the top off the But in that kiss, both the forces acting on the object we being downward direction Vacate mg intention db displaying to be so using free body diagram again this time db less mg boat combined together Maybe the necessary centripetal force again So baby will be given us m into G plus the square by l. So when I put these values No, Sorry, it will become you re making an s d B equals Tow em. These were by l minus energy so and can be taking a common out will become be scarred by l minus G on putting these values do at the mosque. Me sweat means 1 44 1.5 minus 9.8 And then I calculate it. I get the answer. 1 72 point four new done. So this is the answer for the second part. So these are the two tensions in the string and the bottom most point and and that top most point. Thank you.

Averell Hause · Answer

Silver Party. We can say that we&#x27;re the intuitive concludes that the intuitive conclusion is that, of course, the tension is greatest at bottom of swinging, and this will be supported if we apply Newton&#x27;s second law, where T minus mg. The tension minus the weight equals the mass times, the in centripetal acceleration and so solving for tea. This would be equaling to the mass multiplied by G plus the squared over R. And so here we can see that increasing the speed eventually leads to the tension at the bottom of the circle, reaching the breaking value of 40 Newton&#x27;s. So the so the answer for part a simply that the tension is greatest at the bottom of the swing. This would be your answer. Four parts should be solving the above equation here for the Velocity V. We can see that the Velocity V is gonna be equaling square root of our multiplied by the tension over the mass minus G, the acceleration due to gravity. So this would be a square root of 0.91 meters, multiplied by 40 Newton&#x27;s divided by 0.37 kilograms, minus 9.8 meters per second squared extend the square root and we find that the velocity V is equaling 9.5 meters per second. This would be our final answer for part B. That is the end of the solution. Thank you for watching.

Ryan Hood · Answer

and part of this problem, whereas to draw free body diagram of this stone, we know that the radius of the string is zero point nine meters. The mass of the stone is you&#x27;re appointing eight kilograms and a maximum attention. We could have been a string of sixteen Newton&#x27;s before it, Ricks. So if you imagine our stone here, we know there&#x27;s going to be gravity pulling it down of magnitude and times G. Now it doesn&#x27;t actually move down because there&#x27;s gonna be a normal force by the table. And then there&#x27;s also going to be a radio force because of the string. And so I&#x27;m assuming that&#x27;s being world around by, say this point here. And so that&#x27;s the three body diagram. There&#x27;s only three forces here, and that completes party for Part B. We need to find the maximum velocity that we can have before the string breaks. And so for this, we&#x27;re just going to notice that the radio force is equal to the tension. Now we know the radio acceleration. It&#x27;s given in the book as b squared over R, and we also know that the radio force is equal to mass times the radio acceleration. And so it&#x27;s equal to the mass times velocity squared over R. And so we can take this expression here and plug it in for the left side. Here we do that. We get that the tension is equal to don&#x27;t be squared over R. Now we can plug in our max tension to figure out what our Maxwell Long City is. We also have deployed under Massena rays, which are givens in the problem. We knew that we were sixteen unions is equal to zero point eight kilograms. It&#x27;s on speed squared over zero point nine meters. Now we consult us for B and when we do that we have is eight point two three meters per second and that&#x27;s the final answer.

Abhishek Jana · Answer

the storm that is traveling in a circular part on DA. It&#x27;s connected through a string on one end of the strain is fixed. So we consider this as our fixed end and the other end is where the stone is attached and that is creating a circular part. Now we can immediately see that intense a circular part, and the stone is following that path. That means there is some acceleration that is directing towards the center. And we can call that a radial exploration which is equal to V squared over are these the velocity and with the stone is spinning and then our is the radius of the cell feel apart. And the only force that we have here is the tensile force during the stream and no other forces there on this direction. Now, if we actually try to, uh, draw the free body diagram the stone. So here we&#x27;re considering that the stone is on the right side of the circular part. So that means let&#x27;s say the stone is around here and, ah, if we consider the radial direction as our ex direction, so this will be the ex direction for us, and if we consider the line, or let&#x27;s say the point that it&#x27;s coming towards us Assauer positive wide direction and we can mark that like so So this direction right here is coming towards us. So that&#x27;s the positive direction. That&#x27;s the convention that we&#x27;re choosing to choosing for this problem now if we look at the forces that&#x27;s acting on the stone here. So for this ex direction, we have being, Ah, radial acceleration. And then for the forces, the only force that&#x27;s acting on this direction on the ex direction is thie tense. I&#x27;ll force of the attention due to the strength so I can call that tea. And for the right direction, we have the gravity that&#x27;s acting downwards. So that means the force will be masters gravity and to counterbalance. That will have the, um, equal and opposite normal force, which should be which mean save n right. And this is from Newton&#x27;s third Law that these two forces should be equal if the system is stable. So by stable, I mean that there&#x27;s no movement in why direction for that stone, so that means these two forces air balancing each other. So that&#x27;s why there&#x27;s no movement. But of course there&#x27;s movement in extraction that Andi That&#x27;s why we have to figure out the tensile force that&#x27;s been acted. Ah, on the stone. So yeah, so that&#x27;s the everybody diagram that we want and then to solve for the Mac or maximum velocity that the stone can have before it breaks. We need the extraction of force because that&#x27;s where the 10 side forces, so for extraction will have f X equals M times X s o. That&#x27;s the radial exploration for us. And from here we can write effects as the tensile force which should be equal to mass times a eggs. And for our case, it is. The X is a radio and that is be spread over our And from there, if we solve for C will get square root off tee times are over m. And now we can put all the values and salt for V. So that&#x27;s for tension. We have 16 mutants. Then we have for radius on for the length of the string, we have 0.9 meter. Then for the mass, we have a kilogram 0.8 kilograms. Sorry. And ah, from here we can get the velocity as eight find to do meters per second. Now, we actually we should actually cross check whether our units are correct or not. And to do so, we can actually expand this unique Newton on. We know from this force aggression that Newton should be the unit of mass times the unit of acceleration. So that&#x27;s that should be ah, so for our case, that should be kilograms. We&#x27;re using the unit s o. That&#x27;s ridiculous. Grand times. Ah, leader for a second squared and then we have a meter over here, and then we divide that a kilogram, So then we can get rid of the kilograms. And since we have a square root over here, so let&#x27;s not forget the square root. So we have meet us great over second spread. And if we take the square of that, we get meters per second. So that means our units are consistent. So we should actually do this little calculation on the site. So to check whether the the units they were using is actually consistent with the units that were given. Thank you

Luis Mendoza · Answer

this problem we&#x27;re dealing with, We have a table. Where is in a stream? We&#x27;ll run around in a circle. A lot of people. So initially we are fixing the tension of the street at 50 Newtons such that the rock Ha ha ha has a speed of 20.4 meters per second. We call that the one I moved in a trajectory with a radius off 2.5 meters that we call that first radio are want their own. We pull into stream to make it shorter eventually breaks the instant before it breaks. We see that the rug has a had a velocity off 51 minutes per second was it&#x27;s running circle or particular trajectory with radios over one leader. So now the question is what is the danger of the spring? What it breaks. So in order to pursue, like we find a massive Dirac. So he&#x27;s presented the case, Huh? Used to be a dribble of force before we would cost velocity. So detention on the street has to equal to this criminal force From here were sold for the mass of the wrong. You don&#x27;t have the mask off 0.3003 kilograms. Now we can get here. Good luck. Later, case. We&#x27;re just breaks on. We do the same thing with centripetal forces of T two for two, equal to Mask Beach was squared or R squared. Or now we know Mass up the rock. Have fun in the previous case so we can use blowing the numbers for for the two or two on em. I would find that the station detention of which the stream breaks us out to be 781 humans.

Problem

A planetoid orbits a planet in a circular path at…

Problem 27 Easy Difficulty

A student attaches a 1.5 kg mass to a 40 cm long string. The student then begins to whirl the mass in a horizontal circle with increasing speed. If the string snaps when the mass has a speed of 6 m/s, what is the breaking strength of the string?
(A) 60 N
(B) 120 N
(C) 180 N
(D) 240 N

Answer

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