To find the work done by a force $\mathbf{F}$ over a straight line, we use the line integral of the force along the path of motion. The work done $W$ is given by:

$W = \int_C \mathbf{F} \cdot d\mathbf{r}$

where:

• $\mathbf{F} = xyi + (y-x)j$ is the force vector.
• $d\mathbf{r} = dx\,i + dy\,j$ is the differential displacement vector.
• $C$ is the path of motion from the point $(1,1)$ to the point $(2,3)$.

Step 1: Parameterize the Path

The path from $(1,1)$ to $(2,3)$ can be parameterized as:

$x(t) = 1 + t, \quad y(t) = 1 + 2t$

where $t$ varies from 0 to 1.

Step 2: Express the Force and Displacement in Terms of $t$

The force vector in terms of $t$ becomes:

$\mathbf{F}(t) = (x(t)y(t))i + (y(t) - x(t))j$ Substituting the parameterized coordinates: [ \mathbf{F}(t) = \left[(1+t)(1+2t)\right]i + \left[(1+2t) - (1+t)\right]j ] Simplify: $\mathbf{F}(t) = \left(1 + 3t + 2t^2\right)i + \left(t\right)j$

The differential displacement vector is: $d\mathbf{r} = \frac{dx}{dt}dt\,i + \frac{dy}{dt}dt\,j$ $d\mathbf{r} = (1)dt\,i + (2)dt\,j = dt\,i + 2dt\,j$

Step 3: Compute the Dot Product $\mathbf{F}(t) \cdot d\mathbf{r}$

$\mathbf{F}(t) \cdot d\mathbf{r} = \left(1 + 3t + 2t^2\right)(dt) + \left(t\right)(2dt)$ $\mathbf{F}(t) \cdot d\mathbf{r} = \left(1 + 3t + 2t^2 + 2t\right)dt$ Simplify: $\mathbf{F}(t) \cdot d\mathbf{r} = \left(1 + 5t + 2t^2\right)dt$

Step 4: Integrate to Find the Work Done

Integrate from $t = 0$ to $t = 1$: $W = \int_0^1 \left(1 + 5t + 2t^2\right)dt$

Compute the integral: $W = \left[t + \frac{5t^2}{2} + \frac{2t^3}{3}\right]_0^1$ Substitute the limits: $W = \left(1 + \frac{5}{2} + \frac{2}{3}\right) - \left(0 + 0 + 0\right)$ $W = 1 + 2.5 + 0.6667 = 4.1667$

Final Answer

The work done by the force $\mathbf{F}$ over the straight line from $(1,1)$ to $(2,3)$ is approximately 4.17 units.

Would you like more details or have any other questions?

Here are some related questions:

1. What is the significance of the work done by a force in physics?
2. How do you parameterize a straight line path between two points?
3. Can work done by a force be negative? If so, what does it signify?
4. How do you interpret the dot product in the context of work?
5. What is the physical meaning of the force field given in this problem?

Tip: When integrating a polynomial, carefully check each term's power and coefficient to ensure accuracy in the antiderivative.