Asked by Anon
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.
F(x, y, z) = z2i + 2xyj + 3y2k.
Find the work done.
Answers
Answered by
Damon
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 :)
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 :)
Answered by
Anon
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