A particle moves along line segments from the origin to the points (3, 0, 0), (3, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field
F(x, y, z) = z2i + 2xyj + 3y2k.
Find the work done.
I'm very confused on this?
To find the work done by the force field, we need to calculate the line integral of the force field along the path taken by the particle.
The line integral is defined as follows:
W = ∫F · dr
where F is the force field and dr is the differential displacement vector along the path.
In this case, the force field is given by F(x, y, z) = z^2i + 2xyj + 3y^2k.
First, let's calculate the differential displacement vector dr. Since the particle is moving along line segments, dr can be expressed as:
dr = dx i + dy j + dz k
where dx, dy, and dz are the differentials of x, y, and z coordinates, respectively.
Next, let's consider each segment of the path and calculate the line integral along each segment.
1) From the origin to (3, 0, 0):
On this segment, only x is changing, so dx = 3 and dy = dz = 0. Therefore, dr = 3i.
W1 = ∫F · dr
= ∫(z^2i + 2xyj + 3y^2k) · (3i)
= ∫(3z^2) dx
= 3 ∫z^2 dx
= 3 ∫(0^2) dx
= 3 ∫0 dx
= 0
2) From (3, 0, 0) to (3, 3, 1):
On this segment, both x and y are changing. Therefore, we need to calculate dx, dy, and dz.
dx = 0, dy = 3, dz = 1. Therefore, dr = 3i + 3j + k.
W2 = ∫F · dr
= ∫(z^2i + 2xyj + 3y^2k) · (3i + 3j + k)
= ∫(3z^2) dx + ∫(6xy) dy + ∫(9y^2) dz
= 3 ∫z^2 dx + 6 ∫xy dy + 9 ∫y^2 dz
3) From (3, 3, 1) to (0, 3, 1):
On this segment, only x is changing. So dx = -3 and dy = dz = 0. Thus, dr = -3i.
W3 = ∫F · dr
= ∫(z^2i + 2xyj + 3y^2k) · (-3i)
= ∫(-3z^2) dx
= -3 ∫z^2 dx
= -3 ∫(1^2) dx
= -3 ∫1 dx
= -3x
4) From (0, 3, 1) back to the origin:
On this segment, both x and y are changing. Therefore, we need to calculate dx, dy, and dz.
dx = 0, dy = -3, dz = -1. Therefore, dr = -3j - k.
W4 = ∫F · dr
= ∫(z^2i + 2xyj + 3y^2k) · (-3j - k)
= ∫(-6xy) dy + ∫(-9y^2) dz
= -6 ∫xy dy - 9 ∫y^2 dz
Finally, the total work done is the sum of the line integrals along each segment:
W = W1 + W2 + W3 + W4
Calculate each line integral and substitute the values of dx, dy, and dz that we've determined for each segment to find the total work done.
This is a closed path
check first if field is potential in which case integral is zero
F = z^2 i + 2xy j + 3y^2k
del cross F = 0 ????
nope no way
so first line =
3 i + 0 j + 0 k
F dot line = Fx *3 = 0*3 = 0
second line
in x from 3 to 3 = 0
in y from 0 to 3 = 3
in z from 0 to 1 = 1
Fy = 2 x y = 6 y
y work = 6 y dy from 0 to 3 = 6*9/2 = 27
z work = 3 y^2 from z =0 to z = 1
but y = 3 z so 27 z^2 from 0 to 1
27 z^3/3 0 to 1 = 9
so total work in line 2 = 27+9 = 36
so far 27 + 36 Joules
now do lines three and four
and get the total for all four
and I am thinking you are in MITx as well :)