CALC An object moving in the xy-plane is subjected to the force F...

Question

CALC An object moving in the xy-plane is subjected to the force F(arrow on top) =(2xy î+x² ĵ) N, where x and y are in m.
a. The particle moves from the origin to the point with coordinates (a, b) by moving first along the x -axis to (a, 0) , then parallel to the y -axis. How much work does the force do?

Accepted Answer

Hi, everyone in this practice problem, we are asked to determine the work done by the force. We will have a box subjected to a two dimensional force with a force equation of F equals X Y squared, I plus three X Y J Newton, depending on the X and Y positions. The X and A Y coordinates are measured in beta. And the box actually starts at the origin 30.0 comma zero, move straight 2.0 comma M and then takes a right angle turn, moving straight to N comma M. We're asked to determine the work done by the force. And the options are going to be a half M squared and squared B two M squared N C two M squared and squared and D A third M cube N. So uh in order for us to actually determine the total work done, we wanna um first decompose our factor equation of the force factor. So F here equals to X Y squared, I Plus 3 XYJN. And from here, we can determine that our FX is going to be X Y squared and our F Y is then going to be three X Y. Next, we want to identify the uh the movement or the displacement of the box. So it will start at 0.0 comma zero, moving to 00.0 comma M. So that is going to be the first uh movement of the box. And throughout this position, the uh the thing that change is actually the Y. And throughout this whole position, the X actually remains at zero next from 0.0 comma M the box moves to point N comma M. And throughout this, this movement, the X is actually the thing that's changing and the Y actually equals to M or stays at point M. So now we can actually start tackling the problem. So we wanna uh recall that work will equals to the end the girl of F dot or F Ds where the DS is actually the displacement. So work because we have the force in the factor component, we have to actually decompose this um from one movement to the other. So I am going to start off with the first movement which is going from 10.0, comma 0 to 00.0 comma M. So um throughout this movement, the AX will equals to zero. And the thing that's changing is the Y. So therefore, we have to actually look at the F Y component. So this is going to be Y I to Y F F Y D Y. Next we have the second movement which is uh changing in the X direction. So F uh this is going to be the integral from X I to X F FX D X just like. So OK, so now we are ready to actually substitute the values that we know. So substituting things work will then equals due to integral from the Y I is going to be zero from 00.0 comma zero. And the Y F is going to be M from 00.0 comma M F Y. We have three X Y dot D Y and then plus the integral from X I X I is going to be um zero from 00.0 comma M and X F is going to be N from point N comma M and the FX is going to be X Y squared and we still will have the D X. Next, we want to substitute our X equals to zero and Y equals to M into our equation. So the X equals zero is applied to the first part of the movement. So therefore, the first part of the movement will then be uh integral from zero to M three multiplied by zero, multiplied by Y D Y. So this whole thing of the first term will then equals to zero. So we can essentially neglect that the second term will then be the integral from zero to N of X Y squared. And the Y is M, so we know that this is going to be M squared X D X. So simplifying things, the work is essentially just the integral from zero to N of M squared multiplied by X D X M squared is just a constant. And uh according to the integral rule where we have a constant inside of the integral, we can't actually uh move it outwards outside the integral. So this will then be M squared integral from zero to N of X D X. Now we want to recall that the integral of X D X is going to be equals to half X squared plus C. And in this case, because our integral is bounded, so we can determine that essentially the work will equals to M squared multiplied bye half X squared, evaluated from 0 to N. This will then equals to M squared multiplied by half and squared minus zero. So this will then be equals or the work will then be equals to half M squared and squared, which will actually be the answer to our problem. So the work done should equals to half M squared and squared, which will correspond to option A in our answer choices. So answer A is going to be the answer to this problem. So with the work done by the force equaling to half M squared M squared option A is going to be the answer to this problem. And that will essentially be it for this problem. Practice this problem if you guys still have any sort of confusion please make sure to check out our other lesson videos on similar topics and that will be all for this one. Thank you.

Key Concepts

Work Done by a Force

Line Integrals

Vector Fields

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