🚀 Work Done by a Variable Force
When the force applied changes with position, we calculate work using integration (adding up tiny bits of work):
- For a small displacement \( \Delta x \), work done is \( \Delta W = F(x) \cdot \Delta x \).
- Total work = area under the \( F(x) \) vs. \( x \) graph: \[ W = \int_{x_i}^{x_f} F(x) \, dx \]
📊 Example: Pushing a Trunk
A woman pushes a trunk with:
- 100 N force for first 10 m.
- Force decreases linearly to 50 N over the next 10 m (total distance = 20 m).
- Frictional force = 50 N (opposite direction).
Work done by the woman:
\[ W_{\text{woman}} = \text{Area of rectangle ABCD} + \text{Area of trapezoid CEID} \] \[ = (100 \, \text{N} \times 10 \, \text{m}) + \frac{1}{2} \times (100 \, \text{N} + 50 \, \text{N}) \times 10 \, \text{m} = 1000 \, \text{J} + 750 \, \text{J} = 1750 \, \text{J} \]Work done by friction:
\[ W_{\text{friction}} = (-50 \, \text{N}) \times 20 \, \text{m} = -1000 \, \text{J} \] Key takeaway: Opposing forces do negative work! 🌟💥 Work-Energy Theorem for Variable Forces
The change in kinetic energy (\( \Delta K \)) equals the total work done:
\[ K_{\text{final}} – K_{\text{initial}} = W_{\text{total}} \]Derivation:
- Start with \( \frac{dK}{dt} = F \cdot v \) (from Newton’s 2nd law).
- Integrate over displacement: \[ \int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x) \, dx \quad \Rightarrow \quad \Delta K = W \]
🎯 Important Concepts for NEET
- 🔧 Work by variable force: Calculate using integration or area under the \( F(x) \)-\( x \) graph.
- ⚡ Work-energy theorem: \( \Delta K = W_{\text{net}} \). Applies even for changing forces!
- 📉 Negative work: Forces opposing motion (like friction) reduce kinetic energy.
- 📐 Graphical methods: Break complex forces into shapes (rectangles, triangles) for easy area calculation.
- 💡 Real-world examples: Bullets losing kinetic energy, pushing objects with decreasing force.
