A force acting on a particle moving in the xy plane is given by Fx = (2yi +x^2j)N, where x and y are in meters. The particle moves from the origin to a final position having coordinates x=5.00m and y=5.00m, as in Fig. Calculate the work done by F along (a) OAC, (b) OBC, (c) OC.(d) Is F conservative or nonconservative? Explain.
The plane is a square like below but with connected line on every side. I don't know how to solve for c) OC.Please help.
Bl---------C(5,5)
l
l
Ol----------A
I know the answer is 66.7J but how do I get there?
To calculate the work done by a force along a specific path, we need to use the formula W = ∫ F · dr, where F is the force vector and dr is a displacement vector along the path.
(a) OAC:
To calculate the work done along path OAC, we need to break it up into two components: OA and AC.
1. OA: The force vector, F, is given as Fx = 2yi + x^2j.
Along OA, both x and y are changing, so we need to express the force vector in terms of x only:
F = 2y(i) + x^2(j) = 2x(i) + x^2(j).
The displacement vector, dr, along OA is given by dr = dx(i) + dy(j).
Applying the dot product:
W(OA) = ∫ F · dr = ∫ (2x(i) + x^2(j)) · (dx(i) + dy(j)).
Integrating along the path OA from O to A:
W(OA) = ∫[0 to 5] (2x(dx) + x^2(dy)).
2. AC: Since AC is a straight vertical line, the force vector F is given by Fx = 0, and Fy = x^2.
The displacement vector, dr, along AC is given by dr = dy(j).
Applying the dot product:
W(AC) = ∫ F · dr = ∫ (0(i) + x^2(j)) · dy(j).
Integrating along the path AC from A to C:
W(AC) = ∫[0 to 5] (x^2(dy)).
To calculate the work done along path OBC (b), we similarly break it up into two components: OB and BC.
(b) OBC:
1. OB: Similarly to (a), F = 2x(i) + x^2(j). The displacement vector, dr, is given by dr = dx(i) + dy(j).
Applying the dot product:
W(OB) = ∫ F · dr = ∫ (2x(dx) + x^2(dy)).
Integrating along the path OB from O to B:
W(OB) = ∫[0 to 5] (2x(dx) + x^2(dy)).
2. BC: Since BC is a straight horizontal line, the force vector F is given by Fx = 2y and Fy = 0.
The displacement vector, dr, along BC is given by dr = dx(i).
Applying the dot product:
W(BC) = ∫ F · dr = ∫ (2y(i) + 0(j)) · dx(i).
Integrating along the path BC from B to C:
W(BC) = ∫[0 to 5] (2y(dx)).
Now, for c) OC, OC is a diagonal path of the square. To calculate the work done along OC, we will break it up into two components: OA and AC and sum up the work done along those paths.
1. OA: Since we have already calculated the work done along OA in (a), we can use the result from W(OA) for this component.
2. AC: Similarly to (a), F = 2x(i) + x^2(j), and the displacement vector, dr, is given by dr = dx(i) + dy(j).
Applying the dot product:
W(AC) = ∫ F · dr = ∫ (2x(dx) + x^2(dy)).
To integrate along path AC from A to C, we need to find the range of x that corresponds to this path.
The equation of the diagonal line AC can be expressed as y = x.
So, substituting this into the range of y in path AC (0 to 5), we get x = y = 0 to x = y = 5.
Therefore, we need to integrate from x = 0 to x = 5.
W(AC) = ∫[0 to 5] (2x(dx) + x^2(dy)).
Finally, we can find the total work done along OC by adding the work done along OA and AC: W(OC) = W(OA) + W(AC).
(d) To determine if the force F is conservative or nonconservative, we need to check if the work done by the force along a closed path is zero. If the work done is always zero for any closed path, the force is conservative; otherwise, it is nonconservative.