Question

Find the work done by the force field \(\mathbf{F}\) on a particle moving along the given path. \(\mathbf{F}(x, y)=2 x \mathbf{i}+y \mathbf{j}\) \(C:\) counterclockwise around the triangle with vertices \((0,0),\) \((1,0),\) and (1,1)

Short answer

The work done by the force field \( \mathbf{F}(x, y) \) on a particle moving along the path \( C \) is 1.5 joules (or units of work).

Step-by-step solution

Parameterize the path

Split the path into three line segments corresponding to the sides of the triangle: \[ C_1: \mathbf{r}(t) = (t,0), 0 \leq t \leq 1 \], \[ C_2: \mathbf{r}(t) = (1,t), 0 \leq t \leq 1 \], and finally \[ C_3: \mathbf{r}(t) = (1-t,1-t), 0 \leq t \leq 1 \]. These defines the path of the particle using a parameter \(t\) that varies from 0 to 1.

Calculate the line integral over each segment

The line integral of \( \mathbf{F} \) over each segment \( C_i \) of the path \( C \) is given by: \[ \int_{C_i} \mathbf{F} \cdot d\mathbf{r} \] where \( d\mathbf{r} \) is the differential displacement along \( C_i \). We calculate this for each segment and add up the results.

Evaluate the integrals

Integrate for each segment: in case of \( C_1 \): \[ \int_{0}^{1}(2t \mathbf{i} + 0 \mathbf{j}) \cdot (dt \mathbf{i}+ 0 \mathbf{j}) = \int_{0}^{1} 2t dt = [t^2]_0^1 = 1 \], for \( C_2 \): \[ \int_{0}^{1}(2 \mathbf{i} + t \mathbf{j}) \cdot (0 \mathbf{i} + dt \mathbf{j}) = \int_{0}^{1} t dt = [0.5 t^2]_0^1 = 0.5 \], and in case of \( C_3 \) it's: \[ \int_{0}^{1} ((2-2t) \mathbf{i} + (1-t) \mathbf{j}) \cdot (-dt \mathbf{i} - dt \mathbf{j}) = \int_{0}^{1} 2t dt + (1-t) dt = [-t^2 + t]_0^1 = 0 \]. Putting it all together, we get \( 1 + 0.5 + 0 = 1.5 \).

Result and interpretation

The work done by the force field \( \mathbf{F}(x, y) \) on a particle moving along the path \( C \) is the sum of the work done over each segment: 1.5 joules (or units of work), which is also the direction-independent scalar product (the dot product) of the force field and the displacement vectors. This implies that the force field \( \mathbf{F}(x, y) \) has done a net positive work on the particle along the path, meaning it has added energy to the particle.