Question

A force defined by the law $\overrightarrow{F}=(3x^2i+7xy^2j+3k)N$ acts on a particle that is shifted from the origin of the coordinate system to a point P (2 m, 8 m, 0 m).

(a) Calculate the work done by the force when the particle follows path OAP.

(b) Calculate the work done by the force, when the particle follows a curvilinear path defined by the law $y=2x^2$.

Answer 1

(a) The work done by the force when the particle follows path OAP is $\frac{7192}{3}$ J.

(b) The work done by the force when the particle follows a curvilinear path defined by the law $y=2x^2$ is $2056$ J.

Answer 2

The work done by a variable force $\overrightarrow{F}$ acting on a particle displaced along a path is given by the line integral: $W = \int \overrightarrow{F} \cdot d\overrightarrow{r}$

Given $\overrightarrow{F}=(3x^2i+7xy^2j+3k)N$ and $d\overrightarrow{r} = dxi + dyj + dzk$. So, $\overrightarrow{F} \cdot d\overrightarrow{r} = (3x^2)dx + (7xy^2)dy + (3)dz$.

First, let's check if the force is conservative by calculating its curl: $\nabla \times \overrightarrow{F} = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3x^2 & 7xy^2 & 3 \end{vmatrix} = i \left( \frac{\partial(3)}{\partial y} - \frac{\partial(7xy^2)}{\partial z} \right) - j \left( \frac{\partial(3)}{\partial x} - \frac{\partial(3x^2)}{\partial z} \right) + k \left( \frac{\partial(7xy^2)}{\partial x} - \frac{\partial(3x^2)}{\partial y} \right)$ $= i (0 - 0) - j (0 - 0) + k (7y^2 - 0) = 7y^2 k$

Since $\nabla \times \overrightarrow{F} \neq \overrightarrow{0}$, the force is non-conservative, which means the work done depends on the path taken. The particle moves from O (0, 0, 0) to P (2 m, 8 m, 0 m). Notice that the z-coordinate is 0 for both start and end points. Also, the paths described are in the xy-plane, implying $z=0$ and $dz=0$ along the entire path. Therefore, the term $3dz$ in the work integral will always contribute zero. So, $W = \int (3x^2dx + 7xy^2dy)$.

(a) Work done by the force when the particle follows path OAP.

The path OAP consists of two segments:

1. Segment OA: From O (0, 0, 0) to A (2, 0, 0). Along this segment, $y=0 \implies dy=0$ and $z=0 \implies dz=0$. $x$ varies from 0 to 2. $W_{OA} = \int_{x=0}^{2} 3x^2dx + \int_{y=0}^{0} 7x(0)^2(0) = \int_{0}^{2} 3x^2dx = [x^3]_{0}^{2} = 2^3 - 0^3 = 8 \text{ J}$.

2. Segment AP: From A (2, 0, 0) to P (2, 8, 0). Along this segment, $x=2 \implies dx=0$ and $z=0 \implies dz=0$. $y$ varies from 0 to 8. $W_{AP} = \int_{x=2}^{2} 3(2)^2(0) + \int_{y=0}^{8} 7(2)y^2dy = \int_{0}^{8} 14y^2dy = 14 \left[ \frac{y^3}{3} \right]_{0}^{8} = 14 \left( \frac{8^3}{3} - \frac{0^3}{3} \right) = 14 \times \frac{512}{3} = \frac{7168}{3} \text{ J}$.

Total work done along path OAP: $W_{OAP} = W_{OA} + W_{AP} = 8 + \frac{7168}{3} = \frac{24 + 7168}{3} = \frac{7192}{3} \text{ J}$.

(b) Work done by the force when the particle follows a curvilinear path defined by the law $y=2x^2$.

The path is from O (0, 0, 0) to P (2, 8, 0). Along this path, $y=2x^2 \implies dy = 4xdx$. Also $z=0 \implies dz=0$. $x$ varies from 0 to 2. $W_{curvilinear} = \int_{x=0}^{2} (3x^2dx + 7xy^2dy)$ Substitute $y=2x^2$ and $dy=4xdx$: $W_{curvilinear} = \int_{0}^{2} (3x^2dx + 7x(2x^2)^2(4xdx))$ $W_{curvilinear} = \int_{0}^{2} (3x^2 + 7x(4x^4)(4x))dx$ $W_{curvilinear} = \int_{0}^{2} (3x^2 + 112x^6)dx$ $W_{curvilinear} = \left[ \frac{3x^3}{3} + \frac{112x^7}{7} \right]_{0}^{2}$ $W_{curvilinear} = \left[ x^3 + 16x^7 \right]_{0}^{2}$ $W_{curvilinear} = (2^3 + 16 \cdot 2^7) - (0^3 + 16 \cdot 0^7)$ $W_{curvilinear} = 8 + 16 \cdot 128 = 8 + 2048 = 2056 \text{ J}$.

The work done for the two paths is different, which is expected for a non-conservative force.