# physics

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^2iā 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?

1. if ā Ć 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

posted by Damon
2. 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).

posted by joy
3. 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.

posted by Damon
4. 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)

posted by Damon

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