A particle moves along line segments from the origin to the points (2, 0, 0), (2, 5, 1), (0, 5, 1), and back to the origin under

Question

1 year ago

12

the influence of the force field
$F(x,y,z) = z^2i+5xyj+2y^2k$
Use Stokes Theorem to find the work done:

1 answer:

kondaur [170] · Answer 1 · 2023-09-20T12:07:19+03:00

A particle moves along line segments from the origin to the points, work done is mathematically given as

W=184.5units

<h3>What is the solution of an equation that matches the model.?</h3>

Considering line segments from (0,0,0)origin to (2,0,0) , (2,5,1) and (0,5,1) under the infulence of force

Generally, the equation for force is mathematically given as

F = z2i + 5xyj + 2y2k

Therefore, Considering u*v

$u * v = (u_1j+u_2j+u_3k) * (v_1i+v_2j+v_3k)$

$u* v = u_1v_1(i * i) + u_1v_2(i * j)+u_1v_3(i * k) + u_2v_1(j * i) + u_2v_2(j * j)+u_2_3(j* k) + u_3v_1(k * i) + u_3v_2(k * j)+u_3v_3(k * k)$

Where

$i * i = j *j=k * k=0$

Hence

$u* v = u_1v_2k-u_1v_3j-u_2v_1k+u_2v_3i +u_3v_1j - u_3v_2i$

$u = (u_1,u_2,u_3) = (0,5,1)\\\\v = (v_1,v_2,v_3) = (-3,0,0)$

The normal equation formed

-2y + 15z = 0

z= (1/5)y

Considering the level surface and differential surface area

h(x,y,z) = -y + 5z =0

$dS = |grad(h)| dA$

In terms of the x and y coordinates of (2,0,0) and (2,5,1) and (0,5,1), we can state that the ranges are 0 to 3 and 5 respectively we have

$0 \leq x \leq 2 \ and \ 0 \leq y \leq 5$

Using strokes theorem to evaluate

$F = z2i + 5xyj + 2y2k$

$curl \ F = 4yi+2zj+5yk = (4y,2z,5y)$

$curl \ F * nds= \frac{1}{5}(-2z + 25y)\ dy \ dx$

In conclusion, The work done is

$W=\int _CF *dr$

$\int _C F * dr = \int \int curl \ F * n ds$

$\int _C F *dr = \frac{123}{2}\int_0^3 \ dx$

$\int _C F * dr = \frac{123*3}{2}$

W= 184.5units