Please help. If you need the figure, let know how I can show you and I will do it. A particle moves in the xy plane (see the figure below) from the origin to a point having coordinates
x = 7.00 m and y = 4.00 m
under the influence of a force given by F = 4y^2i− 2xj.
(a) What is the work done on the particle by the force F if it moves along path 1 (shown in red)?
J
(b) What is the work done on the particle by the force F if it moves along path 2 (shown in blue)?
J
(c) What is the work done on the particle by the force F if it moves along path 3 (shown in green)?
J
(d) Is the force conservative or nonconservative?
if ∇ × F = 0, the force is conservative
i j k
d/dx d/dy d/dz
4y^2 -2x 0
[d/dx (-2x) -d/dy(4y^2)]k
=-2 - 8y
sorry, not a conservative field. The result is path dependent and I can not help without the drawing.
However I can make a guess that one of them is along the x axis from 0 to 7, then up 4 to (7,4)
from 0,0 to 0,7 motion only in x so
work = Fx dx = 4 y^2 dx = 0
then at x = 7, straight up 4 units
work = Fy dy = -2(7)(4) = -56 Joules
The figure is a rectangle on an xy plane. Line 1(red) starts from the origin (0,0) and goes right on the x-axis until it reaches the end of the rectangle, then it goes up until it reach the end of the rectangle and stop there. Line 2(blue) also starts from the origin but it goes up on the y-axis until it reach the end of the rectangle then it turn right and goes until it stop where line 1 has stopped at. Line 3 (green) is a diagonal line that starts from the origin and goes from bottom left to upper right until it stop at where line 1 and 2 have stopped at. On top of the rectangle, where are the lines have stropped is a written ( 7.00,4.00).
The figure is a rectangle on an xy plane. Line 1(red) starts from the origin (0,0) and goes right on the x-axis until it reaches the end of the rectangle, then it goes up until it reach the end of the rectangle and stop there.
==========================================
That is the one I just did for you.
Part B
up to (0,4) then right to (7,4)
first y motion
Fx dx = 0
Fy dy -2*0 dy = 0
now x motion at y = 4 from x 0 to x = 7
Fx dx = 4 y^2 dx = = 4 y^2 dx = 4*16 (7-0) = 64
Part C
y = (4/7) x
dy/dx = 4/7
now integrate along that line
4 y^2 dx - 2 x dy
but
dx = 7 dy/4 and x = 7 y/4
so
4 y^2 (7/4)dy - 2(7y/4) dy
integrate from y = 0 to y = 4
7 y^3/3 - (7/2)y^2/2
7(64)/3 - (7/4)(16)
To find the work done on the particle by the force F along each path, we can use the formula for work:
W = ∫F · dr
where W is the work done, F is the force vector, and dr is the differential displacement vector along the path.
(a) For path 1 (shown in red):
To find the work done along this path, we need to determine the differential displacement vector dr along the path. Since the path is a straight line from the origin to (7, 4), the displacement vector can be written as dr = dx i + dy j, where dx and dy are the differentials of x and y, respectively.
The differential displacements dx and dy can be found as:
dx = 7 m
dy = 4 m
Substituting the values of dx and dy into the equation for the force F = 4y^2i - 2xj, we get:
F = 4(4)^2i - 2(7)j
F = 64i - 14j
Now we can find the work done:
W = ∫(64i - 14j) · (dx i + dy j)
W = ∫(64dx - 14dy)
W = 64∫dx - 14∫dy
W = 64x - 14y
Substituting the final coordinates of the particle into the equation, we get:
W = 64(7) - 14(4)
W = 448 - 56
W = 392 J
Therefore, the work done on the particle by the force F along path 1 is 392 J.
(b) For path 2 (shown in blue):
To find the work done along this path, we need to determine the differential displacement vector dr along the path. Since the path is a straight line from the origin to (7, 4), the displacement vector can still be written as dr = dx i + dy j.
Using the same values of dx and dy as in the previous part:
dx = 7 m
dy = 4 m
Substituting the values of dx and dy into the equation for the force F = 4y^2i - 2xj, we get the same force vector as before:
F = 64i - 14j
Using the same formula for work, we get:
W = ∫(64i - 14j) · (dx i + dy j)
W = ∫(64dx - 14dy)
W = 64∫dx - 14∫dy
W = 64x - 14y
Substituting the final coordinates of the particle into the equation, we get:
W = 64(7) - 14(4)
W = 448 - 56
W = 392 J
Therefore, the work done on the particle by the force F along path 2 is also 392 J.
(c) For path 3 (shown in green):
To find the work done along this path, we once again need to determine the differential displacement vector dr along the path. Since the path is a straight line from the origin to (7, 4), the displacement vector remains as dr = dx i + dy j.
Using the same values of dx and dy as before:
dx = 7 m
dy = 4 m
Substituting the values of dx and dy into the equation for the force F = 4y^2i - 2xj, we once again get the same force vector:
F = 64i - 14j
Using the same formula for work, we have:
W = ∫(64i - 14j) · (dx i + dy j)
W = ∫(64dx - 14dy)
W = 64∫dx - 14∫dy
W = 64x - 14y
Substituting the final coordinates of the particle into the equation, we get:
W = 64(7) - 14(4)
W = 448 - 56
W = 392 J
Therefore, the work done on the particle by the force F along path 3 is also 392 J.
(d) To determine if the force is conservative or nonconservative, we can test for the conservative property. A force is conservative if the work done by it only depends on the initial and final positions and not on the path taken.
From the previous calculations, we can see that the work done by the force F is the same for all three paths 1, 2, and 3. Therefore, the force is conservative, as the work done only depends on the initial and final positions of the particle and not the path taken.