| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the work done by the force F on the particle, we need to calculate the dot product of the force vector and the displacement vector of the particle. The force vector F is given as (-1, 0, 1), and the displacement vector of the particle can be found by taking the derivative of the position vector (t, t^2, t^3) with respect to time t: dr/dt = (1, 2t, 3t^2) The work done by the force F on the particle is given by the integral of the dot product of the force vector and the displacement vector over the time interval [0, 1]: W = ∫[0,1] F · dr/dt dt Substituting the values, we get: W = ∫[0,1] (-1, 0, 1) · (1, 2t, 3t^2) dt Expanding the dot product, we get: W = ∫[0,1] (-1 + 3t^2) dt Evaluating the integral, we get: W = [-t + t^3] from 0 to 1 W = (-1 + 1) - (0 - 0) W = 0 Therefore, the work done by the force F on the particle is 0. The correct answer is C. 0. |
C |
| deepseekr1 | The work done by a constant force \( \mathbf{F} \) on a particle moving along a path is given by the line integral \( \int \mathbf{F} \cdot d\mathbf{r} \). Since \( \mathbf{F} \) is constant, this simplifies to \( \mathbf{F} \cdot (\mathbf{r}(1) - \mathbf{r}(0)) \), where \( \mathbf{r}(t) \) is the position vector of the particle at time \( t \). 1. **Force Vector**: \( \mathbf{F} \) is a unit force parallel to \( (-1, 0, 1) \). Normalizing this vector gives \( \mathbf{F} = \left( -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right) \). 2. **Displacement Vector**: The particle moves from \( t = 0 \) to \( t = 1 \) along the path \( \mathbf{r}(t) = (t, t^2, t^3) \). The displacement is \( \mathbf{r}(1) - \mathbf{r}(0) = (1, 1, 1) - (0, 0, 0) = (1, 1, 1) \). 3. **Dot Product**: Compute \( \mathbf{F} \cdot (\mathbf{r}(1) - \mathbf{r}(0)) \): \[ \left( -\frac{1}{\sqrt{2}} \right) \cdot 1 + 0 \cdot 1 + \left( \frac{1}{\sqrt{2}} \right) \cdot 1 = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 0. \] The work done is \( 0 \), so the correct answer is **C**. **Answer: C** |
C |
| gpto1 | C | C |
| GPT-4o | To determine the work done by the force \( F \) on the particle, we need to compute the line integral of the force along the given path. Given: - The force vector \( F \) is parallel to \( (-1, 0, 1) \) and is a unit vector. - The path of the particle is \( \vec{r}(t) = (t, t^2, t^3) \) for \( t \) in \([0, 1]\). First, we normalize the force vector to make it a unit vector: \[ \vec{F} = \frac{(-1, 0, 1)}{\sqrt{(-1)^2 + 0^2 + 1^2}} = \frac{(-1, 0, 1)}{\sqrt{2}} = \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \] Next, we find the derivative of the path with respect to \( t \): \[ \vec{r}'(t) = \frac{d}{dt}(t, t^2, t^3) = (1, 2t, 3t^2) \] The work done by the force along the path is given by the line integral of the dot product of \( \vec{F} \) and \( \vec{r}'(t) \) from \( t = 0 \) to \( t = 1 \): \[ W = \int_{0}^{1} \vec{F} \cdot \vec{r}'(t) \, dt \] Calculate the dot product \( \vec{F} \cdot \vec{r}'(t) \): \[ \vec{F} \cdot \vec{r}'(t) = \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \cdot (1, 2t, 3t^2) = -\frac{1}{\sqrt{2}} \cdot 1 + 0 \cdot 2t + \frac{1}{\sqrt{2}} \cdot 3t^2 \] \[ = -\frac{1}{\sqrt{2}} + \frac{3t^2}{\sqrt{2}} \] \[ = \frac{3t^2 - 1}{\sqrt{2}} \] Now, we integrate this expression from \( t = 0 \) to \( t = 1 \): \[ W = \int_{0}^{1} \frac{3t^2 - 1}{\sqrt{2}} \, dt = \frac{1}{\sqrt{2}} \int_{0}^{1} (3t^2 - 1) \, dt \] Evaluate the integral: \[ \int_{0}^{1} (3t^2 - 1) \, dt = \left[ t^3 - t \right]_{0}^{1} = (1^3 - 1) - (0^3 - 0) = 0 \] Thus, \[ W = \frac{1}{\sqrt{2}} \cdot 0 = 0 \] Therefore, the work done by the force \( F \) on the particle is: \[ \boxed{C} \] |
C |
| llama3.1-70B-Instruct | A |
C |
| llama3.1-405B-Instruct | To find the work done by the force F on the particle, we need to calculate the dot product of the force vector and the displacement vector of the particle. The force vector F is given as (-1, 0, 1), and the displacement vector of the particle can be found by taking the derivative of the position vector (t, t^2, t^3) with respect to time t: dr/dt = (1, 2t, 3t^2) The work done by the force F on the particle is given by the integral of the dot product of the force vector and the displacement vector over the time interval [0, 1]: W = ∫[0,1] F \* dr/dt dt Substituting the values, we get: W = ∫[0,1] (-1, 0, 1) \* (1, 2t, 3t^2) dt Expanding the dot product, we get: W = ∫[0,1] (-1 + 3t^2) dt Evaluating the integral, we get: W = [-t + t^3] from 0 to 1 W = (-1 + 1) - (0 - 0) W = 0 Therefore, the work done by the force F on the particle is 0. The correct answer is C. 0. |
C |
| DeepSeek-Coder-V2 | B. -1/(4sqrt(2)) | C |
| Qwen2-72B-Instruct | B. -1/(4sqrt(2)) | C |
| baichuan4 | C | C |
| doubao | B The force vector F can be taken as (-1/sqrt(2), 0, 1/sqrt(2)). The velocity vector of the particle is (1, 2t, 3t^2). The dot product of F and the velocity vector is (-1/sqrt(2))*1 + 0*2t + (1/sqrt(2))*3t^2 = (-1/sqrt(2)) + (3t^2/sqrt(2)). Integrating this from 0 to 1 gives the work done: [-t/sqrt(2) + t^3/sqrt(2)] from 0 to 1 = (-1/sqrt(2) + 1/sqrt(2)) - (0) = -1/(4sqrt(2)). |
C |
| minimax | To find the work done by a force along a path, we use the formula: \[ W = \int_C \vec{F} \cdot d\vec{r} \] where \( W \) is the work, \( \vec{F} \) is the force vector, and \( d\vec{r} \) is the differential displacement vector along the path \( C \). Given that the force \( \vec{F} \) is parallel to the vector \( (-1, 0, 1) \), we can represent it as \( \vec{F} = k(-1, 0, 1) \), where \( k \) is a constant. Since it's a unit force, \( k \) must be such that the magnitude of \( \vec{F} \) is 1. However, we don't need to know the exact value of \( k \) because the magnitude of \( \vec{F} \) will cancel out with the magnitude of \( d\vec{r} \) in the dot product. The path of the particle is given by \( \vec{r}(t) = (t, t^2, t^3) \). The differential displacement vector \( d\vec{r} \) is the derivative of \( \vec{r}(t) \) with respect to \( t \), which is \( d\vec{r} = (1, 2t, 3t^2)dt \). Now we calculate the dot product \( \vec{F} \cdot d\vec{r} \): \[ \vec{F} \cdot d\vec{r} = k(-1, 0, 1) \cdot (1, 2t, 3t^2)dt = k(-1 \cdot 1 + 0 \cdot 2t + 1 \cdot 3t^2)dt \] \[ = k(-1 + 3t^2)dt \] The work done by the force from \( t = 0 \) to \( t = 1 \) is: \[ W = \int_0^1 k(-1 + 3t^2)dt \] Since \( k \) is a constant, we can take it out of the integral: \[ W = k \int_0^1 (-1 + 3t^2)dt \] \[ W = k \left[ -t + t^3 \right]_0^1 \] \[ W = k \left[ (-1 + 1^3) - (-0 + 0^3) \right] \] \[ W = k \left[ -1 + 1 \right] \] \[ W = k \cdot 0 \] \[ W = 0 \] So the work done by the force along the given path is 0, regardless of the value of \( k \), because the path is symmetric about the \( t \)-axis and the force is constant. The negative and positive contributions to the work cancel each other out. The correct answer is: C. 0 |
C |
| kimi | To solve this problem, we need to calculate the work done by the force F on the particle. The work done by a force is given by the dot product of the force vector and the displacement vector over the path taken by the particle. The force F is a constant unit force parallel to the vector (-1, 0, 1), so we can write it as: \[ F = (-1, 0, 1) \] The path of the particle is given by the parametric equations: \[ x(t) = t \] \[ y(t) = t^2 \] \[ z(t) = t^3 \] The displacement vector \( \Delta \mathbf{r} \) between time \( t = 0 \) and \( t = 1 \) is the difference between the position vectors at these times: \[ \Delta \mathbf{r} = \mathbf{r}(1) - \mathbf{r}(0) = (1, 1^2, 1^3) - (0, 0^2, 0^3) = (1, 1, 1) \] Now, we calculate the dot product of the force vector and the displacement vector: \[ \mathbf{F} \cdot \Delta \mathbf{r} = (-1, 0, 1) \cdot (1, 1, 1) \] \[ \mathbf{F} \cdot \Delta \mathbf{r} = (-1)(1) + (0)(1) + (1)(1) \] \[ \mathbf{F} \cdot \Delta \mathbf{r} = -1 + 0 + 1 \] \[ \mathbf{F} \cdot \Delta \mathbf{r} = 0 \] The work done by the force is the magnitude of the dot product, which is 0. Therefore, the work done by the force F on the particle is 0. Answer: C. 0 |
C |
| wenxinyiyan | B | C |
| zhipu | To determine the work done by the force \( \mathbf{F} \) on a particle moving along a specified path, we use the formula for work: \[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \] where \( \mathbf{F} \) is the force vector and \( d\mathbf{r} \) is the differential displacement vector along the path. Given: - The force \( \mathbf{F} \) is a constant unit force parallel to the vector \( (-1, 0, 1) \). To make it a unit vector, we normalize \( (-1, 0, 1) \): \[ \mathbf{F} = \frac{(-1, 0, 1)}{\sqrt{(-1)^2 + 0^2 + 1^2}} = \frac{(-1, 0, 1)}{\sqrt{2}} = \left( -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right) \] - The path of the particle is given by \( \mathbf{r}(t) = (t, t^2, t^3) \) for \( t \) from 0 to 1. - The differential displacement vector \( d\mathbf{r} \) is: \[ d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) dt \] where \( \frac{dx}{dt} = 1 \), \( \frac{dy}{dt} = 2t \), \( \frac{dz}{dt} = 3t^2 \). So, \[ d\mathbf{r} = (1, 2t, 3t^2) dt \] Now, the work done \( W \) is: \[ W = \int_{0}^{1} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{1} \left( -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right) \cdot (1, 2t, 3t^2) dt \] Calculating the dot product: \[ \mathbf{F} \cdot d\mathbf{r} = \left( -\frac{1}{\sqrt{2}} \cdot 1 \right) + \left( 0 \cdot 2t \right) + \left( \frac{1}{\sqrt{2}} \cdot 3t^2 \right) = -\frac{1}{\sqrt{2}} + \frac{3t^2}{\sqrt{2}} \] So, the integral for work becomes: \[ W = \int_{0}^{1} \left( -\frac{1}{\sqrt{2}} + \frac{3t^2}{\sqrt{2}} \right) dt \] \[ W = \int_{0}^{1} \left( \frac{-1 + 3t^2}{\sqrt{2}} \right) dt \] \[ W = \frac{1}{\sqrt{2}} \int_{0}^{1} (-1 + 3t^2) dt \] Evaluate the integral: \[ \int_{0}^{1} (-1 + 3t^2) dt = \left[ -t + t^3 \right]_{0}^{1} = (-1 + 1^3) - (-0 + 0^3) = -1 + 1 = 0 \] Therefore: \[ W = \frac{1}{\sqrt{2}} \cdot 0 = 0 \] The work done by the force \( \mathbf{F} \) on the particle is \( 0 \). So, the correct answer is: \[ \boxed{C} \] |
C |
| gpto1 | C | C |
| deepseekv3 | To solve this problem, we need to calculate the work done by the force \( F \) on the particle moving along the given path. The work \( W \) done by a constant force \( F \) along a path \( \mathbf{r}(t) \) is given by the dot product of the force and the displacement vector: \[ W = \mathbf{F} \cdot \Delta \mathbf{r} \] where \( \Delta \mathbf{r} \) is the displacement of the particle from the initial to the final position. ### Step 1: Determine the Force Vector \( \mathbf{F} \) The force \( F \) is a constant unit force parallel to the vector \((-1, 0, 1)\). To find the unit vector in the direction of \((-1, 0, 1)\), we normalize it: \[ \mathbf{u} = \frac{(-1, 0, 1)}{\sqrt{(-1)^2 + 0^2 + 1^2}} = \frac{(-1, 0, 1)}{\sqrt{2}} = \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \] Since \( F \) is a unit force, \( \mathbf{F} = \mathbf{u} = \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \). ### Step 2: Determine the Displacement Vector \( \Delta \mathbf{r} \) The path of the particle is given by \( \mathbf{r}(t) = (t, t^2, t^3) \). We need to find the displacement vector from \( t = 0 \) to \( t = 1 \): \[ \Delta \mathbf{r} = \mathbf{r}(1) - \mathbf{r}(0) = (1, 1, 1) - (0, 0, 0) = (1, 1, 1) \] ### Step 3: Calculate the Work Done Now, we calculate the work done by taking the dot product of \( \mathbf{F} \) and \( \Delta \mathbf{r} \): \[ W = \mathbf{F} \cdot \Delta \mathbf{r} = \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \cdot (1, 1, 1) \] \[ W = \left(-\frac{1}{\sqrt{2}}\right) \cdot 1 + 0 \cdot 1 + \left(\frac{1}{\sqrt{2}}\right) \cdot 1 \] \[ W = -\frac{1}{\sqrt{2}} + 0 + \frac{1}{\sqrt{2}} = 0 \] ### Final Answer The work done by the force \( F \) on the particle is \( 0 \). **Answer:** C. 0 |
C |