Question

The work done by the force \( \mathbf{F} = (x + y) \hat{i} - (x^2 + y^2) \hat{j} \),
where \( \hat{i} \) and \( \hat{j} \) are unit vectors in \( \mathbf{O X} \) and \( \mathbf{O Y} \) directions, respectively, along the upper half of the circle \( x^2 + y^2 = 1 \) from \( (1, 0) \) to \( (-1, 0) \) in the \( xy \)-plane is

• A. \( -\pi \)
• B. \( -\frac{\pi}{2} \)
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

When calculating work using a force and displacement vector, break the problem into a line integral and parametrize the path. Remember to simplify integrals where possible.

Answer 1

The Correct Option is B

We are given the force \( \mathbf{F} = (x + y) \hat{i} - (x^2 + y^2) \hat{j} \) and need to calculate the work done by this force along the upper half of the circle \( x^2 + y^2 = 1 \) from \( (1, 0) \) to \( (-1, 0) \) in the \( xy \)-plane. The work done by a force along a path is given by the line integral: \[ W = \int_C \mathbf{F} \cdot d\mathbf{r}. \] For this problem, the path \( C \) is the upper half of the circle \( x^2 + y^2 = 1 \), so we parametrize the path as: \[ x = \cos(t), \quad y = \sin(t), \quad t \in [0, \pi]. \] The differential displacement vector \( d\mathbf{r} \) is given by: \[ d\mathbf{r} = \frac{d}{dt}(\cos(t), \sin(t)) \, dt = (-\sin(t), \cos(t)) \, dt. \] Now, we compute the dot product \( \mathbf{F} \cdot d\mathbf{r} \): \[ \mathbf{F} = (x + y) \hat{i} - (x^2 + y^2) \hat{j} = (\cos(t) + \sin(t)) \hat{i} - (1) \hat{j}. \] Thus, the dot product is: \[ \mathbf{F} \cdot d\mathbf{r} = (\cos(t) + \sin(t)) (-\sin(t)) + (-1)(\cos(t)) = -\cos(t)\sin(t) - \sin^2(t) - \cos(t). \] Simplifying the integrand: \[ \mathbf{F} \cdot d\mathbf{r} = -\cos(t)\sin(t) - (\sin^2(t) + \cos(t)). \] Now, the work done is: \[ W = \int_0^\pi \left[ -\cos(t)\sin(t) - (\sin^2(t) + \cos(t)) \right] dt. \] We can break the integral into parts: \[ W = \int_0^\pi -\cos(t)\sin(t) \, dt - \int_0^\pi (\sin^2(t) + \cos(t)) \, dt. \] Evaluating each integral: - The first term: \[ \int_0^\pi -\cos(t)\sin(t) \, dt = 0 \quad \text{(since it's an odd function over a symmetric interval)}. \] - The second term: \[ \int_0^\pi (\sin^2(t) + \cos(t)) \, dt = \int_0^\pi \sin^2(t) \, dt + \int_0^\pi \cos(t) \, dt. \] The integral of \( \cos(t) \) over \( [0, \pi] \) is zero, and the integral of \( \sin^2(t) \) is \( \frac{\pi}{2} \), so we have: \[ \int_0^\pi (\sin^2(t) + \cos(t)) \, dt = \frac{\pi}{2}. \] Thus, the total work is: \[ W = -\frac{\pi}{2}. \] Hence, the work done is \( -\frac{\pi}{2} \), and the correct answer is (B).