Question

Question #0fc26

Answer 1

`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`.