In real-life situations, a force is rarely constant. Most forces change with position, time, or displacement. Such a force is called a variable force.
1. Work Done for a Small Displacement
Consider a force \(F(x)\) that varies with position \(x\).
If the displacement \(\Delta x\) is very small, the force can be assumed to be approximately constant over that small interval.
Therefore, the small amount of work done is:
\[ \Delta W = F(x)\,\Delta x \]
where:
- \(F(x)\) = force at position \(x\)
- \(\Delta x\) = small displacement
2. Total Work Done
To find the total work done from the initial position \(x_i\) to the final position \(x_f\), we add the work done over all small intervals:
\[ W = \sum F(x)\,\Delta x \]
The summation extends from \(x_i\) to \(x_f\).
3. Limiting Case
As the displacement intervals become smaller and smaller,
\[ \Delta x \rightarrow 0 \]
the number of intervals increases without limit. In this limit, the summation becomes a definite integral.
Thus,
\[ W=\lim_{\Delta x \to 0}\sum F(x)\,\Delta x \]
\[ \boxed{W=\int_{x_i}^{x_f} F(x)\,dx} \]
This is the general expression for the work done by a variable force.
4. Graphical Interpretation
If a graph of force \(F\) versus displacement \(x\) is plotted, then:
\[ \boxed{\text{Work Done = Area under the } F\text{-}x \text{ graph}} \]
between the limits \(x_i\) and \(x_f\).
The rectangular strips used in the summation method combine to form the area under the curve as \(\Delta x\) approaches zero.
5. Key Points
- A force that changes with position is called a variable force.
- For a very small displacement, the force can be treated as constant.
- Total work is obtained by adding the work done over many small intervals.
- In the limit of infinitesimal displacement, summation becomes integration.
- The work done by a variable force is:
\[ \boxed{W=\int_{x_i}^{x_f} F(x)\,dx} \]
- The area under the force-displacement graph represents the work done.
6. Formula Summary
\[ \Delta W = F(x)\,\Delta x \]
\[ W = \sum F(x)\,\Delta x \]
\[ W = \lim_{\Delta x \to 0}\sum F(x)\,\Delta x \]
\[ \boxed{W=\int_{x_i}^{x_f} F(x)\,dx} \]
7. Conclusion
For a variable force, the work done cannot be calculated using \(W = Fs\) because the force changes throughout the motion. Instead, the work done is found by integrating the force over the displacement, which is equal to the area under the force-displacement graph.
Frequently Asked Questions (FAQs)
Q1. What is a variable force?
Answer: A variable force is a force whose magnitude or direction changes with position, time, or displacement.
Q2. Why can't we use \(W = Fs\) for a variable force?
Answer: The formula \(W = Fs\) is valid only when the force remains constant throughout the displacement. For a variable force, the force changes continuously, so integration is required.
Q3. What is the formula for work done by a variable force?
\[ W=\int_{x_i}^{x_f} F(x)\,dx \]
Q4. What does the area under the Force-Displacement graph represent?
Answer: The area under the Force-Displacement graph represents the work done by the force.
Q5. What is the SI unit of work done?
Answer: The SI unit of work is Joule (J).
\[ 1~\text{Joule} = 1~\text{Newton} \times 1~\text{metre} \]
Q6. Give an example of a variable force.
Answer: The restoring force of a spring is a variable force.
\[ F=-kx \]
Multiple Choice Questions (MCQs)
1. A force that changes with displacement is called:
A) Constant Force
B) Variable Force
C) Balanced Force
D) Net Force
Answer: B) Variable Force
2. For a variable force, the work done is given by:
A) \(W=Fs\)
B) \(W=mv\)
C) \(W=\int F(x)\,dx\)
D) \(W=ma\)
Answer: C) \(W=\int F(x)\,dx\)
3. The work done by a variable force is equal to:
A) Slope of the F-x graph
B) Area under the F-x graph
C) Intercept of the graph
D) Velocity of the object
Answer: B) Area under the F-x graph
4. The SI unit of work is:
A) Newton
B) Watt
C) Joule
D) Pascal
Answer: C) Joule
5. When \(\Delta x \to 0\), the summation \(\sum F(x)\Delta x\) becomes:
A) Differentiation
B) Multiplication
C) Division
D) Integration
Answer: D) Integration
6. The expression \(W=\lim_{\Delta x\to0}\sum F(x)\Delta x\) represents:
A) Momentum
B) Power
C) Work done by a variable force
D) Acceleration
Answer: C) Work done by a variable force
7. Which of the following is a variable force?
A) Weight near Earth's surface
B) Constant applied force
C) Spring force
D) None of these
Answer: C) Spring force
8. The area under a Force-Displacement graph has the unit:
A) Newton
B) Joule
C) Metre
D) Watt
Answer: B) Joule
9. If force remains constant throughout motion, the work done is:
A) \(W=Fs\)
B) \(W=mv\)
C) \(W=ma\)
D) \(W=F/s\)
Answer: A) \(W=Fs\)
10. In the formula \(W=\int_{x_i}^{x_f} F(x)\,dx\), \(dx\) represents:
A) Force
B) Velocity
C) Infinitesimal displacement
D) Acceleration
Answer: C) Infinitesimal displacement
Quick Revision
\[ \Delta W = F(x)\,\Delta x \]
\[ W=\sum F(x)\,\Delta x \]
\[ W=\lim_{\Delta x\to0}\sum F(x)\,\Delta x \]
\[ \boxed{W=\int_{x_i}^{x_f}F(x)\,dx} \]
Remember: For a variable force, the work done is equal to the area under the Force–Displacement (F-x) graph.
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