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A particle moves from the origin to the point $x=3 \mathrm{m}, y=6 \mathrm{m}$ along the curve $y=a x^{2}-b x,$ where $a=2 \mathrm{m}^{-1}$ and $b=4 .$ It's subject to a force $c x y \hat{\imath}+d \hat{\jmath},$ where $c=10 \mathrm{N} / \mathrm{m}^{2}$ and $d=15 \mathrm{N}$.Calculate the work done by the force.

135J

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University of Sheffield

in this problem, we are asked about a particle that moves in a curve. Um, and it's being moved by a force that varies in the X direction, but it actually does not very in the why direction Ah, with X. And so, um, what I'm going to do is I'm going to separate the work into an X component of the work and the UAE components of the work. So work in the X direction is the integral of the force in the X Direction D X, which is going to give us the definite integral integral from 0 to 3 because it send, it says that's it moves the particle moves from X equals zero to x equals three. Um, and then it tells us that the force is C X y. So see, X, Why is the force in the X direction D X? But now we're also given that why is a function of X so see, axe were given that why equals X squared minus B X groups DX. All right, um, I'm gonna choose to put the numbers in here for a B and C. I think it might make it a little easier to understand. We're given that, see is 10 and these are all s I BCE units. I'm gonna go ahead and distribute the actually, I'm going to distribute the never mind. I'm not gonna distribute it yet. I'll do it in another step. Okay? A is to and B is four. Okay, Now, before I do anything else, I'm going to distribute the 10 X. So that's going to give me 20 X to the third power minus 40 acts to the second power T X and I could do this integral. That's going to give me 20/4 x to the fourth power minus 40/3 X to the third power from 0 to 3. But I noticed that they're all multiplied by X. So at zero, this is, uh, the expression is gonna be eat going to evaluate to zero. So I'm just going to put in the three, which is, um, 20/4 times three to the fourth power, minus 40/3 times three to the third power, which gives us, if you put that in a calculator. 40 five. Jules. So this is part of our answer. But we also have to work in the Y direction. So I'm gonna go to a new page. Work in the Y direction is the integral of the force in the Y direction de y. This one's easier because the force in the Y direction is constant. It's a constant, but DE which and we're told the D is 15. Well, I'm going to write de That's the as the constant were given, um de y um uh, and were also given that it goes from 0 to 6 meters in the Y direction. So we're told that the constant D is equal to 15. If we integral integrate that, we just get whoops. We just get 15. Why? From 0 to 6 and again, that one's gonna evaluate 20 So our answer is that the work in the Y direction is 15 times six, which is 90 Jules. So our total work is 90 plus 45 jewels, which is 135 Jules

Pennsylvania State University