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Problem No: I: (a) force Of magnitude 5.0 N applied to each end ofa spring cause the spring tor sretch 3.5 cmt from ils relaxed length, how far do forces of magnitu...

Question

Problem No: I: (a) force Of magnitude 5.0 N applied to each end ofa spring cause the spring tor sretch 3.5 cmt from ils relaxed length, how far do forces of magnitude ZO N cause the same spring t0 streteh? What is the spring constani of this spring? (c) How much work Is done by the applied forces in stretching the spring 3.5 cm from its relaxed length? W (akxxatsktxtf W6Answer: (a) 4.9 cm;, (b) 1.4 Nicm; and (c) 88 ml

Problem No: I: (a) force Of magnitude 5.0 N applied to each end ofa spring cause the spring tor sretch 3.5 cmt from ils relaxed length, how far do forces of magnitude ZO N cause the same spring t0 streteh? What is the spring constani of this spring? (c) How much work Is done by the applied forces in stretching the spring 3.5 cm from its relaxed length? W (akxxatsktxtf W6 Answer: (a) 4.9 cm;, (b) 1.4 Nicm; and (c) 88 ml



Answers

(a) If forces of magnitude $5.0 \mathrm{N}$ applied to each end of a spring cause the spring to stretch $3.5 \mathrm{cm}$ from its relaxed length, how far do forces of magnitude $7.0 \mathrm{N}$ cause the same spring to stretch? (b) What is the spring constant of this spring? (c) How much work is done by the applied forces in stretching the spring $3.5 \mathrm{cm}$ from its relaxed length?

So we're told we have a spring with the spring constant Of 3, 5 times 10 to the 4th newtons per meter. And they ask us how much energy is stored or how much work does it take to compress the spring? Um 0.50 point 05 m. Or whether it be five centimeters. So to compress it or to stretch it that much. So the 1st 1st question is about stretching it. Um and so we know that the formula for um kind of this uh energy stored in the spring, which would be equal to the change in the energy and the system, which would be equal to the work that we had to do on the on the spring is one half K, delta X squared. And in this case delta X is five centimeters. We know what K is. So plugging all of this in, we get 43.8 jewels. Now then they ask us now if we, instead of stretching at 5cm we compress it 5cm. Well that basically gives us the same result because again we have delta X. Squared here. So um compressing it or stretching it by the same amount creates, you know um The requires the exact same amount of energy. Again assuming it's a linear spring such that the force is proportional to displacement. So that would be changed. The force is equal to the spring, concert times changing displacement. Or this change in length of this spring. Um So again uh it doesn't matter stretching it or compressing it. If we do buy the same amount, we're going to have to require to do the same amount of work to do that.

Well, and this problem, we have a block and the spring arranged here. And, uh, the block is free to slide without friction on the surface. And in doing so, it stretches or compresses the spring. Let's see here we're told when the block is pulled to four centimeters, we must apply a magnet force of magnitude 360 Newtons. So, um, assuming that the, um that the force that the spring is linear than the force is going to be have a proportionality constant, Mhm, um, K Delta X. So with Delta X is the change in length from its unstructured on stretched length. And so if we take the in stretched length to be zero, we'll measure zero from there. Then at four, we have this 360 Newton, so we can figure out that. Then we have a spring constant here of 9000 Newtons per meter. So again, just plugging in, you know, four centimeters here, 360 Newtons here and then solving for cape. So then they ask us for a bunch of stuff about how much work is done between different states. Um, we have the the spring compressed being compressed. So, um, let's see, here we have. Uhh. So we have that we pull the block 11 centimeters and release it. How much work does the spring due? On the block as the block moves from four from five centimeters to three centimeters. Um, it turns out we don't actually need this 11 centimeters. That's just, um, I guess, um they said we stretched it to that much. So you can obviously it's gonna go. You know, it's going to get back to these these positions that they give us. Um, so we we basically start our measurement from five centimeters, so the spring is stretched five centimeters, and we know the work. Um, the energy stored in the in the spring is one half the spring. Constant times the change in length squared. So again, we're measuring. Um, the the UN compressed length or the UN stretched length is X equals zero. So Delta X is just gonna be X so we can plug in our values. We know that the spring constant and we know we have five centimeters here, so just making the unit conversion. And when you get 11 to 5 jewels, so that's our That's our nominal base value because we're measuring everything from five centimeters with respect to that. And so then they said between five centimeters and three centimeters. Well, we can figure out the energy stored in the spring at three centimeters again just plugging in three centimeters here to the expression for the work energy stored in the spring. And we get four point oh, five tools. So that means that the work and going from our initial state to our first state, um, or the energy the or the basically the work that's done on the by the spring on the mass as we go from in this state here to this state. Here is then the difference here. And that's 7.2 jewels. So, um, we lost this much energy in the spring, so that means that we did what we lost. We lost this much energy in the spring going from here to here, So that must have been put into energy in the block, and that would be in to turn into kinetic energy in the block. Now they ask us when the spring is at minus three centimeters. Um, how much between our initial state. How much energy change in the spring was between our initial state? That's five. And now it's minus three. Again, we can put minus three into our equation here and again that winds up being, um, squared so we can actually get the same value. Um, as when the spring. I stretched three centimeters. Now it's compressed three centimeters. So we wind up with If we look at the If we look at the change in energy, why did I put a zero? There should be a two. The energy, the change of energy from going from five to minus three. Is this minus? Um, this minus this, which is the same thing. So we got again. The spring did an amount of work. 7.2 jewels of work on the mask for going from five to minus three. Um, if we go from five to minus five, we see that again because this is squared in here. This value is the same as this value and the working going from five. When the spring is stretched to five to win, it's minus five. The spring has not done any work on the on the block in that in that range? Um, the different, um, in that region in that range of motion, because the the energy compressed in the spring is the same in each case. And now if we go to, um, minus nine centimeters, we can see that now we have a energy stored in the spring is 36.45 jewels, um, taking, subtracting that from this, we get minus 25.2 jewels. So in this case, when the spring has gone past, you know, five minus five centimeters and always compressing the spring, I'm even more. We're actually doing work on the spring at that point. So the amount of work that the spring does on the block is a negative because we're actually the work is being done from the block on the spring. So what's happening here is that the kinetic energy, um, that was in the block is now being changed into potential energy. Uh, is being changed into potential energy in the spring. So those are the change in the changing? Basically, we're just looking for the change in work. Uh, the change in energy stored in the spring because then that that energy must have gone into the block since we're assuming that this is a closed system and energy is not going anywhere else, Yeah.

So first it wants us to find the spring constants that will use books law F equals K X and put me in the values that were given. So 45 k times five on by both sides by five. We'll end up getting k equal seven Now for part B, we'll go ahead and integrate our force so we'll get one half que x squared and we're going. We're having a displacement from 0 to 3 here, so we'll go ahead blood those in And so we'll have one half time seven and I like Teoh and put into brackets C B squared minus a square. Once again, this is just gonna be three squared. So this will be nine, I guess closed brackets not really necessary. And then they will end up giving us an answer of 31.5. And now, for part, C will be doing something just a little more interesting. So we'll have it. Going from work equals one half K X squared. But this time we'll be going from 5 to 10. We have a displacement from five going from 15. 20 and 15 25. So now the close brackets gonna make sense when we put that in there. So I have one half times seven and we'll ever closed. Brackets of 10 squared, just 100 minus five squared. Just 25. We'll have one half time. Seven. I'm 75. That's gonna end up Equalling 262.5, right? Yes. So go ahead and circle the other stuff too.

And this problem, We're gonna talk about the elastic force. So you need to remember that, For example, if we have a spring that has a new elastic, constant spring constant K and then we and then this year that is shown on the screen is the unscratched length of the string zero. And then the string is compressed. Yeah, by a distance, X, then the force that's exerted on the string is equal to minus K X. Okay, So if X is positive, then the string is compressed and the force acts in the opposite direction to the compression. Okay, So if the object is compressed in the force acts in this direction while if the string is stretched, then the force acts in the opposite direction. Okay. And X would be negative in this case. Uh, also, we need to remember that the potential energy associate ID with a spring or any elastic forces equal to case k times X squared, divided by two. Okay, So what we have in our problem is a spring that has Oh, that is compressed. Actually, it's stretched stretched. I'm sorry. By a distance, acts off 0.5 m and for that of force off. 160 movements is required. Mhm. Um, And from this, we can already obtain what is the value of K? Because we know that the magnitude of the forces able to K X okay, this is the magnitude of the force. That's why I didn't right the minus here. So okay. Is af divided by f X? Uh, f is 160. Newton's, uh, ax is 0.5 m. So Okay. Is 3200 newtons per meter? Yeah, in question A. We have to find what's the force necessary to stretch? Um, the spring by 0.0 15 m. And the answer is that the force, it was K X. It's equal to 3200 newtons per meter Time, 0.0 15 m. So f is equal to ah, 48 Newtons. Yeah, and then we have to find what is the force necessary to compress the spring by 0.2 m. So the force eyes equal to, um que x Again, this is the magnitude of the force. Ofcourse the forces are gonna be opposite direction off compressions, for example, with the string. The spring is compressed. Then the forces pointing in the opposite direction the same as if the spring had bean. If the spring is, uh, stretched okay, it's always pointing to the opposite direction of the motion. Yeah, but the magnitude is just 32 100 Newtons. Commuter time, 0.2 m. So F is 64 newts. Then in question be we have to find how much work must be done. And the work is just the variation and energy. So it's equal to the potential energy you in the first case. Okay, it's always key. Expert over to in the first case K is 3200 Newtons Cream eater X is, um zero point 0 15 m. So the work must be able to 48 Newtons. Okay? Yeah. Then in the second case again, K is 3200, uh, Newton's commuter that and multiplied by 0.2 m square, uh, divided by two which is equal to is actually, I'm sorry. Made a mistake up here. The work is actually a zero point 36 years. Yeah, on in the second case, the work is 0.64 Jews. Okay? And this includes the actual


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