Subtopic: Tsiolkovsky Rocket Equation & Specific Impulse
Question Statement:
A particle acted upon by a constant force 4î + ĵ - 3k̂ N is displaced from point A
with position vector î + 2ĵ + 3k̂ m to point B with position vector 5î + 4ĵ + k̂ m.
The work done by this force is _______ J (answer in integer).
Official Answer Key:
24.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2024 Q.30)**
**1. Given Force and Position Vectors**:
- Constant applied force vector:
$$\vec{F} = 4\hat{i} + 1\hat{j} - 3\hat{k}\text{ N}$$
- Initial position vector $\vec{r}_A$:
$$\vec{r}_A = 1\hat{i} + 2\hat{j} + 3\hat{k}\text{ m}$$
- Final position vector $\vec{r}_B$:
$$\vec{r}_B = 5\hat{i} + 4\hat{j} + 1\hat{k}\text{ m}$$
**2. Displacement Vector $\Delta\vec{r}$**:
$$\Delta\vec{r} = \vec{r}_B - \vec{r}_A = (5 - 1)\hat{i} + (4 - 2)\hat{j} + (1 - 3)\hat{k} = 4\hat{i} + 2\hat{j} - 2\hat{k}\text{ m}$$
**3. Work Done Calculation via Dot Product**:
$$W = \vec{F} \cdot \Delta\vec{r} = (4)(4) + (1)(2) + (-3)(-2)$$
$$W = 16 + 2 + 6 = 24\text{ J}$$
**4. Official Master Key**:
- **Official Answer Key**: **24.0** (Acceptable Range: `[24, 24]`)
- **Exam Takeaway**: For a constant force, work done is the scalar product of force and displacement: $W = \vec{F} \cdot \Delta\vec{r} = 24\text{ J}$.