F=(2xy +z³)i + x³j + 3xz²k find a scalar potential and work done in moving an object in the field from (1,-2,1) to (3,1,4)​

Question

F=(2xy +z³)i + x³j + 3xz²k find a scalar potential and work done in moving an object in the field from (1,-2,1) to (3,1,4)​

in progress 0
Acacia 4 years 2021-08-15T09:42:18+00:00 1 Answers 75 views 0

Answers ( )

  1. Nem
    0
    2021-08-15T09:43:25+00:00
    Reply

    Step-by-step explanation:

    Given:

    \textbf{F} = (2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}

    This field will have a scalar potential \varphi if it satisfies the condition \nabla \times \textbf{F}=0. While the first x- and y- components of \nabla \times \textbf{F} are satisfied, the z-component doesn’t.

    (\nabla \times \textbf{F})_z = \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right)

    \:\:\:\:\:\:\:\:\: = 3x^2 - 2x \ne 0

    Therefore the field is nonconservative so it has no scalar potential. We can still calculate the work done by defining the position vector \vec{\textbf{r}} as

    \vec{\textbf{r}} = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}

    and its differential is

    \textbf{d} \vec{\textbf{r}} = dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}}

    The work done then is given by

    \displaystyle \oint_c \vec{\textbf{F}} • \textbf{d} \vec{\textbf{r}} = \int ((2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}) • (dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}})

    \displaystyle = (x^2y + xz^3) + x^3y + xz^3|_{(1, -2, 1)}^{(3, 1, 4)}

    = 422

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )