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A 1.80-kg object is held 1.15 m above relaxed, massless vertical spring with a force constant of 335 N/m. The object is dropped onto the spring: (a) How far does th...

Question

A 1.80-kg object is held 1.15 m above relaxed, massless vertical spring with a force constant of 335 N/m. The object is dropped onto the spring: (a) How far does the object compress the spring?(b) Repeat part (a), but this time assume constant air-resistance force of 0.600 N acts on the object during its motion_(c) How far does the object compress the spring if the same experiment is performed on the Moon, where 9 = 1.63 m/s? and air resistance is neglected?

A 1.80-kg object is held 1.15 m above relaxed, massless vertical spring with a force constant of 335 N/m. The object is dropped onto the spring: (a) How far does the object compress the spring? (b) Repeat part (a), but this time assume constant air-resistance force of 0.600 N acts on the object during its motion_ (c) How far does the object compress the spring if the same experiment is performed on the Moon, where 9 = 1.63 m/s? and air resistance is neglected?



Answers

A 1.50 -kg object is held 1.20 $\mathrm{m}$ above a relaxed massless, vertical spring with a force constant of $320 \mathrm{N} / \mathrm{m} .$ The object is dropped onto the spring. (a) How far does the object compress the spring? (b) What If? Repeat part (a), but this time assume a constant air-resistance force of 0.700 $\mathrm{N}$ acts on the object during its motion. (c) What If? How far does
the object compress the spring if the same experiment is performed on the Moon, where $g=1.63 \mathrm{m} / \mathrm{s}^{2}$ and air resistance is neglected?

Question 1 25. When the object will fall from a height, etch on the spring, then it will compress that spring by on distance. X So we this is their spring and there is that object. This is high When it will fall on thy spring, it will compress that spring our distance x So taking this as the difference level. We will said that the initial gravitational potential energy all the objectives. MGI one day mg edge Bless X and they should be called the finally spring potential energy hockey excess square. So now it's obstructing the values. Mass is 1.5 KD in 29.8 into 1.2 which is the height last eggs. This is equal to have in tow state on base into excess fire. This will give us a quarter ticket question solving this quarter ticket question. We will get the value off X equal to zero point the A one made up. Now in the second part of the question five me When the same experiment is conducted on moon, then we can die. The mass 1.5 x election due to gravity on moon is 1.63 in tow. Height. 1.2 less new genuine length. X is equal to 1. 60 exists where so again, this will give us supported ticket question. Who's road saves? Executed? Judo point 14 So seven maker so on the moon that spring will become pressed by 0.14 to 7 m. Five c. Venda retarding force is also taken into consideration. Then we convert any shelter additional potential energy mg at x x His name for Parsi. My next of Urban by the frictional force have been toe edge. That's sex. They should be equal to the final spring potential energy. Uh, yeah, excess five. So substituting the values 115 into 9.8 in tow. 1.2 less eggs, one bite. See my nest force, you know Born seven in tow 1.2 That's X. This is equal to half times 20 exits flat. Simplifying this, we will get 16.8 less What? Dean X It would do 1 60 Exit squad. This is a quarter ticket question on its solution. It's X equal to zero point 371 m

Okay, so here we have a pretty complicated problem. Well it's at least complicated in the way that Rmg Y is gonna be a D plus X since it falls the lowest level after the spring compresses. So what we have is MG times D plus X. The distance that it follows, plus the distance that the spring compresses is equal to one F k X squared. So then we distribute the M G D plus mg X. Like that. And then we subtract both sides and then we're gonna have a quadratic form which is a little messy. But we get one half K X squared minus M G. E. Plus sex. Let's just keep it like that. And then I went ahead and graph this thing in does most to get the zeros which were at negative 0.1 and 0.1892 So we're gonna take the positive version of that. So that would be 0.1892 And that is going to be the distance that the spring compresses just about 19 centimeters. So moving on a party, uh we see that the spring is going to be done on the moon. So we're going to have a slightly different gravity of 1.63 Yeah. So what that means is that we're looking at the same equation, but this time instead of 9.81 just gonna write 1.63 And Rasmus look at the positive roots, which is going to be 0.634 So like that. Okay, Okay, okay, meters. So it's about three times less. And that makes about sense since the gravity of the moon is about 1/6 of Earth's and then repeat part A. But this time assume that there is some air resistance during the motion, so that air resistance is gonna be throughout here, the distance that it travels. So the work done by the their resistance is going to be the force which is 7.7 newtons times the distance D. Plus X. And this is gonna be on Earth again. So it's gonna be easier upon 29 2 plus 1.2. Yeah. Mhm. Mhm. And when we calculate that, we got 0.7 times 1.2 plus 0.1892 we've got 0.11 point oh three jules. And we're just going to subtract that from the last part over here. So what we're going to have is uh we're gonna add right here those 13 jewels. Mhm. And then put that back into gizmos. It's Mhm. And the graph completely disappears. Uh huh. That's strange. Yeah. And I fell for the classic dez most trick of zooming out. Um We actually don't cross the X. Axis at any point. Um So it doesn't actually cross it again. So that's kind of weird. We're gonna have to look into that a little bit further. Oh, and now I see my she was supposed to be a minus sign right here. So in Desmond's, that's gonna be a result of 0.1. So the spring compresses about 10 centimetres, sir, 11 70.1 of five m. So artsy. We get the compression of 0.1 old five meters.

This question cover the concept of the total energy for solution and the energy is equivalent to the sum of the potential energy. First, the current treatment had an instinct and that's what energies are secure into half of the spring content K into the square. So first we need to find that out energy. So the total energy is half of the mass M into the square plus half of the supreme constant K. And do the squad of cooperation. So he is half of the mass of the object and that is accumulated. Sorry, uh, must we have to find out. So look at right. The bus is a curently two and 2 E minus K. X square before peace. But let's say this is a question and the total energy easy to get into half of the spring constant K. And that is 23.51 newtons per meter Into the squad to amplitude. And that is if you're 2011979 notice square. So the total energy Is 0.46, Now let's obscured the value after and then the question one. So the mass of the object is to into 0.4604 Jews 23.51 Newton Bomb Eater into the position. And the position is Uh 7.4 71 centimeter, therefore minus tenders minus two uh, m square upon the square of the speed. And the speed is 0.72 86. We just for a second square. So the mass of the object is 1.487. Look

In the first part of this problem, we are going to calculate the amount by which the spring is straight. What this amount is, uh, why the extension in the spring is calculated through the Hook's Law, which can be written as, uh, physicals, too. Kyi here, the surface, the Applied Force and this case, the spring constant. So we can write this question for why, which is the extension in the spring as y Z equals two afterward. It by care. In this case, this forces equals to the weight of the object, and this can be written as mg. So we can wire this question as y Z equals two mg divided by care. Let's put the values into this gradient so it will be wise equals two into 1.1 kg into 9.0 metro parts can square divided by K, which is given by 120 Newton parameter. So from here we can write the value for this. Why is 9.0 multiply Waiter is four minus 2 m. So this is the answer to the first part of this problem. In second part of this problem, we have to calculate the dispute with which the object passes through its original position and the web. So the dispute is, uh, we in order to kill clear dispute, we need to apply the law of conservation of energy. Uh, so we can write to the total energy at the point of release, as, uh only wanted by to care. Why not square? This is the elastic potential energy at the point of release and they should be calls to the total energy. When the object passes through its original position, that should be equals two unworried by two m, we square. So this is the kind of energy at the point, plus the gravitational potential energy which is equal to MGH, plus the elastic potential energy which can be written as one divided by two Ky Square Now, by inserting values into the square. And we can right here when you worried by two into 120 Newton parameter into 0.29 made her whole square as equals to one divided by two into 1.1 kg into with square. So this week is unknown, plus into 1.1 kg into the gravitational acceleration, which is equal to 9.80 Metropolis can square into the video for edge, which is equals two zero point 20 m. Plus the last two potential energy which is written as one he wanted by two. Yeah, into 120 newtons per meter into 0.90 m hole square. So from here, we can write the value for this we as well as equals two 2.1 m per second. So this is the answer to the second part of this problem. End of the question. Thank you.


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