Question #0fc26

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Answer 1 · 2017-12-09T16:12:22

$W_1 = \int_0^4x^2dx=64/3$ .
$W_2 = \int_{x=0}^{x=4}(x^3-8x^2+12x)dx=-32/3$ .
$W_3 = 0$

$vecF(x,y)=x(x+y)\hati + xy^2\hatj;$

Section 1: $\quad (0,0)\rightarrow(4,0)$

$y=0; \qquad dy=0; \qquad vecF(x,y)=x^2\hati;\qquad dvecr = dx\hati;$

$W_1=\int_0^4vecF(x,y).dvecr = \int_0^4x^2dx=64/3$ .

Section 2: $\quad(4,0)\rightarrow(0,4)$
Equation of the line segment is: $\quad y=4-x;$
$dvecr = dx\hati + dy\hatj; y=4-x; \qquad dy=-dx$

$vecF(x,y).dvecr = x(x+y)dx + xy^2dy,$
$\qquad \qquad \qquad \qquad = 4xdx-x(4-x)^2dx = -(x^3-8x^2+12x)dx$

$W_2=\int_{(4,0)}^{(0,4)}vecF(x,y).dvecr,$
$= -\int_{x=4}^{x=0}(x^3-8x^2+12x)dx =\int_{x=0}^{x=4}(x^3-8x^2+12x)dx$
$=[x^4/4-8/3x^3+6x^2]_0^4=-32/3$ .

Section 3: $\quad (0,4)\rightarrow(0,0)$
$x=0; dx=0; \qquad vecF(x,y)=vec0;$

$W_3=0$

Net Work Done:
$W_{"net"} = W_1 + W_2 + W_3 = 64/3-32/3+0=32/3$ .