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Find the work done when a force moves a particle in the xy plane from to along the parabola . Is the work done different when the path is the straight line
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Q. Find the work done when a force moves a particle in the xy plane from to along the parabola . Is the work done different when the path is the straight line
- Parameterize force vector: To find the work done, we need to calculate the line integral of the force vector along the path. For the parabola , we'll parameterize the path using as the parameter, so . Then we'll substitute into the force vector and integrate from to .
- Find displacement vector: First, let's parameterize the force vector for the parabola. Since , we have $\vec{F} = (x^\(2\) - (\sqrt{x})^\(2\) + \(2\)x)\mathbf{i} - (\(2\)x\sqrt{x} + \sqrt{x})\mathbf{j} = (x^\(2\) - x + \(2\)x)\mathbf{i} - (\(2\)x^{\(3\)/\(2\)} + \sqrt{x})\mathbf{j} = (x^\(2\) + x)\mathbf{i} - (\(2\)x^{\(3\)/\(2\)} + \sqrt{x})\mathbf{j}.
- Calculate dot product: Now we'll find the differential displacement vector \(d\mathbf{r}\) along the parabola. Since \(y = \sqrt{x}\), \(dy = (1/2)x^{(-1/2)}dx\). The displacement vector is \(d\mathbf{r} = dx\mathbf{i} + dy\mathbf{j} = dx\mathbf{i} + (1/2)x^{(-1/2)}dx\mathbf{j}\).
- Integrate dot product: The work done is the integral of the dot product of the force vector and the displacement vector. So, we need to calculate \(\int_{0}^{1} \mathbf{F} \cdot d\mathbf{r}\). This is \(\int_{0}^{1} [(x^2 + x)\mathbf{i} - (2x^{\frac{3}{2}} + \sqrt{x})\mathbf{j}] \cdot [dx\mathbf{i} + (\frac{1}{2})x^{-\frac{1}{2}}dx\mathbf{j}]\) from \(0\) to \(1\).
- Integrate dot product: The work done is the integral of the dot product of the force vector and the displacement vector. So, we need to calculate \(\int_{0}^{1} \vec{F} \cdot d\vec{r}\). This is \(\int_{0}^{1}[(x^2 + x)\mathbf{i} - (2x^{\frac{3}{2}} + \sqrt{x})\mathbf{j}] \cdot [dx\mathbf{i} + (\frac{1}{2})x^{-\frac{1}{2}}dx\mathbf{j}]\) from \(0\) to \(1\). Performing the dot product, we get \(\int_{0}^{1}[(x^2 + x)dx - (2x^{\frac{3}{2}} + \sqrt{x})(\frac{1}{2})x^{-\frac{1}{2}}dx]\) from \(0\) to \(1\). This simplifies to \(\int_{0}^{1}(x^2 + x - x - \frac{1}{2})dx\) from \(0\) to \(1\).
