 Let's take a look at the solution for question three of the practice midterm exam for math 12-20 here. Suppose that 16 foot pounds of work is needed to stretch a spring from its natural length of 12 inches to 18 inches. How much work is required to stretch the spring from 18 inches to 24 inches? So we get a spring question going on here and at some point we're gonna have to use Hooke's law, right? The force to hold a spring extended beyond its equilibrium is going to be k times x here where x equals the distance beyond equilibrium. So some things we notice here is that the natural length of the spring is 12 inches so that's our equilibrium. That's what we mean here by x equals zero. Some other things we should point out is that work here is being measured in foot pounds and force will be measured in pounds, right? And so because we're measuring things in foot pounds we do need to convert things into feet. So by 12 inches we really mean it's a foot. And here 18 inches, right? 18 inches would be one and a half feet but we have to think of beyond equilibrium. 18 inches is actually six inches beyond equilibrium so it's a half foot. So this represents x equals 0.5. That's probably one of the most confusing parts I think about this problem. And so later on we're gonna stretch the spring from 18, right, which was x equals 0.5 feet to 24 inches which would be x equals one. So that's something to keep track of here. Now notice the other thing we have here is that we were told to suppose that 16 foot pounds of work is necessary to stretch. You have to be very careful here, right? Force is required to hold the spring stretched but work is required to extend the spring. So we're actually told the work here and you can see this from the units here. Foot pounds describes work, pounds describes force here. So what we're seeing here is the following information. It takes 16 foot pounds of work to extend it from zero to 0.5 and this will be kx dx. That's what we have right here. So our first goal is to identify the spring constant for which when you take the antiderivative of kx squared over two as you go from zero to one half or 0.5, whichever you prefer. Plugging that in there, you're gonna k over two times one half squared. This is gonna give us k over eight. This is equal to 16 times in both sides by eight you're gonna get k equals 16 times eight which that's gonna be 32, 64, 128. That's our spring constant. This k equals 128. So now what we have to answer is the work. The work associated from integrating from one half to one this 128x dx. For which case we're gonna see that when you take the antiderivative you're gonna get 64x squared as you go from one half to one. Plugging the numbers in there 64 times you're gonna get a one squared which is one minus a one half squared which is one fourth. One takeaway of fourth is three fourths. Four goes into 64 16 times so we get 64 times three which should then give us a 48 foot pounds which then gives us answer F right here. So on this example, like all spring examples using Hook's law over here we need to identify what is the spring constant. On some of the story problems we saw in the homework they tell you the force to hold it extended beyond equilibrium. So then you would use the force function to solve for k. In this example though we were told how much work was required to stretch it so we actually started with the integral of kx. So do pay very close attention to the distinction. I knew this was work one because it told me it was work but also because the measurements was in foot pounds that's a measurement of work. If we use scientific units force would be measured in newtons but work would be measured in joules which of course is a newton meter.