Question #0fc26

1 Answer
Dec 9, 2017

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

Explanation:

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.