CALC An object moving in the xy-plane is subjected to the force $\vec{F} = (2 x y \hat{i} + x^{2} \hat{j}) N$ , where x and y are in m. The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis. How much work does the force do?

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Step 1: Recall the formula for work done by a force: $ W = \int \mathbf{F} \cdot d\mathbf{r} $, where $ \mathbf{F} $ is the force vector and $ d\mathbf{r} $ is the infinitesimal displacement vector. In this case, $ \mathbf{F} = (2xy \hat{i} + x^2 \hat{j}) $ and the path is piecewise (first along the x-axis, then along the y-axis).

Step 2: For the first segment of the path (from $ (0, 0) $ to $ (a, 0) $ along the x-axis), the displacement vector is $ d\mathbf{r} = dx \hat{i} $. Substitute $ y = 0 $ into $ \mathbf{F} $, which simplifies to $ \mathbf{F} = x^2 \hat{j} $. Since $ \mathbf{F} \cdot d\mathbf{r} = 0 $ (no $ \hat{i} $ component in $ \mathbf{F} $), the work done along this segment is zero.

Step 3: For the second segment of the path (from $ (a, 0) $ to $ (a, b) $ along the y-axis), the displacement vector is $ d\mathbf{r} = dy \hat{j} $. Here, $ x = a $ is constant, so $ \mathbf{F} = (2ay \hat{i} + a^2 \hat{j}) $. The dot product $ \mathbf{F} \cdot d\mathbf{r} = a^2 dy $.

Step 4: Integrate $ \mathbf{F} \cdot d\mathbf{r} $ over the second segment of the path. The limits of integration for $ y $ are from $ 0 $ to $ b $. Thus, $ W = \int_0^b a^2 dy $. Since $ a^2 $ is constant, the integral simplifies to $ W = a^2 \int_0^b dy = a^2 [y]_0^b $.

Step 5: Evaluate the integral result. Substitute the limits of integration into $ [y]_0^b $, which gives $ W = a^2 (b - 0) = a^2 b $. This is the total work done by the force along the given path.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Work Done by a Force

Work is defined as the integral of force along a path. In this case, the work done by the force F on the object as it moves from one point to another can be calculated by integrating the force vector along the specified path. The work done is dependent on both the magnitude of the force and the displacement in the direction of the force.

Line Integral

A line integral is used to calculate the work done by a force field along a specific path. It involves integrating the force vector along the trajectory of the object, which can be broken down into segments if the path is not straight. In this problem, the path consists of two segments: first along the x-axis and then parallel to the y-axis.

Vector Fields

A vector field assigns a vector to every point in space, representing quantities like force or velocity. In this question, the force F is a vector field defined in the xy-plane, where its components depend on the coordinates x and y. Understanding how to evaluate the force at different points along the path is crucial for calculating the work done.