how-much-work-is-done-in-stretching-the-spring-of-exercise-3-from-a-length-of-100-in-to-120-in-2

Name: how much work is done in stretching the spring of exercise 3 from a length of 100 in to 120 in 2
Uploaded: 2021-11-26
Duration: 59 s
Channel: Numerade
Description: How much work is done in stretching the spring of Exercise 3 from a length of 10.0 in. to 12.0 in.?

Question

How much work is done in stretching the spring of Exercise 3 from a length of 10.0 in. to 12.0 in.?

Lucas Finney · Accepted Answer

This problem we have a continuation of the previous problem where we are asked how much work is done in stretching the spring of exercise three from a length of 10 inches to 12 inches. Where we found that for the spring in exercise three, our spring constant was equal to four. So in this case we&#x27;d have that the work done will be the integral from 10.8 or 10 up to 12 of four X. Dx. Which will then be four x squared over two, evaluated from 10 to 12 or two X squared, evaluated from 10 to 12. It will then be two times 12 squared, So two times 1 44 minus 10 squared, so minus 100 Which will then clearly B2 times 44 or 88 pound inches.

Ashwin Narayan · Answer

fly tells us that before so required stretches scream distinct. This is Kate case. That&#x27;s what we&#x27;re trying to catch. A late the worst part to step in. We&#x27;re dealing with a variable force, so we have to take an interview. Starting position The ending position Force. Okay. Changing. Hey, so the first step that we need to do it actually calculate what cave so that we can plug it in here, do that Rehab of information. We know that a force of 250 Newtons stretches this spring 30 centimeters or 0.3 meters. But this fall for Kay just dividing we get that K is equal to 833. Really? Yeah. Newtons per meter. We&#x27;re here. And that gives us all the information we need. We want a couple the work. When we start X one meters, 2.5 meters. Our screens positive is 833.33 meeting her meter and are variable of integration here. Okay, so we&#x27;re basically taking the invisible of some 10. And we can do that taking any derivatives and full do you 833 and now the integral of ex just expert too, that we can pluck in 0.5 meters where it over to might quit too neither you. I want all this together. You get an amp yours. 87. Wait. Five mutant new meters can make sure you never work out. We have units meters straight. Here. 10 This is unit for me too. They get Newton meters, also known as

Adnan Gill · Answer

so hooks Losses F sequel to K X. We are told that a force of 215 news compresses the spring by 30 centimeters, or 0.3 meters, so that gives K is equal to 25 divided by 0.3 or 2500 divided by three mutant R meter work is equal to the integral x one to x two fdx. So since Casey quote to 207,500 over three so we can write F is equal to 250 0/3 times X. So now we confined work which would be equal to begin integral to 50 0/3 x from 0.2 meters, 20.5 meters and DX. So work is equal to 25 year 0/3 and times the integral, which would be X squared over two over the limit from 0.2 to 0.5. So this would be to 500 earth three times you don&#x27;t 30.5 squared over two minus 0.2 squared over two, and when solving this comes out to be equal to 87.5 ju

Adnan Gill · Answer

so if you stop the question off, we start with F is equal to K X, and we&#x27;re told that a force off 90 Newton stretches the spring by one meter. So that means that key is equal to 19. You can per meter and then it says What is the work done and stretching it when it is stretched by an and multi 4 to 5 meters. So we have work is equal to 1/2 key X squared. We just calculated that K is equal to 90 and then access file. So this repeat 1/2 times 90 times five square, and that comes out to people to 1125 Jews. So in order to stretch that spring by five mutant five meters, you need to do it work off 1125 Jews

Mike Gaerlan · Answer

All right. So the force, uh, done by a spring is equal to K X. Like so. Okay, so, um, the work done, it&#x27;s going to be the integral of K x d x. Like So, um, for the first part of the problem, we know that it took to Jules to stretch the spring to 15 centimeters. So that&#x27;s gonna be, um, from zero, huh? Zero 25 0.5 Actually, 0.5 Like so, uh, que ex the ex. Okay, so, uh, this is now going to be k X squared, divided by two, like so. And this is from 0 to 0.5 is the same as 1/20. So I&#x27;m gonna use 1/20 like that. Okay, so if you plug in 1/20 we get one over 800. Okay. Um, like so. And then if you like, in zero, just get zero. So K is going to be 1600 here. Okay. 1600. Now, um, we need to find the work done to stretch a spring from 5 to 20. So we&#x27;re going to take our 1600. Okay. Uh, ex d x, We&#x27;re trying to find the work. And they were trying to work. Find the work from 5 to 10 or 0.5 Right, Um, 0.5 20.10 Okay, so that&#x27;s all in meters. So then we get, uh, 1600 x squared. So we get 800 or the integral 800 x squared from this is from 1/20 to 1/10. Okay, that&#x27;s puggle those in. So we get 800 over 100 minus 800 over 400. Okay, so this becomes eight. Minus 800. 400 is too. So a total answer is going to be six, Jules.

Yuou Sun · Answer

come in Steaks. You can&#x27;t steal yourself. Hooker slaw. Right? So 90 is equal to okay times why, Right? So pay is just 90. And the problem lies. Gusto. So how much work does it take to strike the spring fire meters? Five meters we can use W is equal to your growth after. Yeah, right. And I was able to pay X so w is equal to on half times k times X square actives fire in case 90 Just compute 45 times. Don&#x27;t you fight, right? Oh, that&#x27;s sorry. 45 25 90. That&#x27;s 1125. Sure, that&#x27;s great.

Problem

A 160 -lb person compresses a bathroom scale 0.08…

Problem 4 Easy Difficulty

How much work is done in stretching the spring of Exercise 3 from a length of 10.0 in. to 12.0 in.?

Answer

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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