 In this video we provide the solution to question number two from practice exam number one for math 1220, in which case, suppose that 16 foot pounds of work is needed to stretch a spring from its natural length of 12 inches to the distance of 18 inches. And then we're asked how much work is required to stretch the spring from 18 inches where currently it's at to 24 inches. Okay, some things to note here. Well, we measured we're measuring work in foot pounds, but we're measuring distance in feet. So at some point we probably want to change from inches to feet. So I'm going to try to do that right now. So 12 inches, of course, is the same thing as one foot 18 inches, which is a 12 inches plus six. That's actually going to be 1.5 feet, like so. Similarly, this 18 inches right here becomes 1.5 feet. And then 24 inches is two times 12. So that then becomes two feet. So that unit conversion is going to be very useful in this situation. I also want to point out this one that many of the many of these spring problems we've seen helps the tell us the force required to hold the spring. It's six inches past its equilibrium, but this one doesn't tell us the force to hold it. It tells us the work to stretch it. Okay. And so what that means is we actually have this integral 16 here. That's the work required to stretch it from equilibrium, which will be 0x to 1.5. And then kx dx. So the force required to hold a spring extended is going to be force equals kx where k is the spring constant and x is the distance beyond equilibrium. We don't know the force. We know the work. So we actually have an integral in play right here. So if you find the antiderivative, you're going to get kx squared over two as you evaluate from 0 and 1.5. So when you plug it in 0, just make everything disappear. When you plug in the 1.5, you square that and so you end up with k over 8. Remember, this is equal to 16. So when you solve for k in this situation, which is what we're going for right now, k is going to equal 8 times 16, which would give you 120, excuse me, 128, which would be your value there for for k. If you have that k value in play, you can then solve the problem that we're trying to do here. So we need to find the work required to extend it from a half inch to one foot beyond equilibrium. We now know that k is 128. We didn't calculate this x dx here. So the 128 is long for the journey right now. The antiderivative is going to look like it did a moment ago, x squared over two. We evaluate from 1.5 to 1 this time. Now, notice the 2 goes into the 128 evenly. I actually used 64 times there. So you're left with the x squared as you go from 1.5 and 1. So plug in the 1, 1 squared is a 1. Plug in the 1.5, 1.5 squared is a 1.4. So then we have 1 minus 1.4. We can think of 1 as 4.4. So that would simplify to be 64 times 3.4. 4 goes into 64 16 times. So we end up with 16 times 3. Or in other words, we end up with 48 foot pounds, which then turns out to be option F.