The Work-Energy Theorem states that the net work done on a particle is equal to the change in its kinetic energy.
\[ W=\Delta K=K_f-K_i \]
This theorem is valid for both constant and variable forces.
Work Done by a Variable Force
For a variable force \(F(x)\), the force changes with position. Therefore, the work done cannot be calculated using
\[ W=F \cdot s \]
Instead, work is calculated by integration:
\[ W=\int_{x_i}^{x_f}F(x)\,dx \]
where:
- \(F(x)\) = Variable force
- \(x_i\) = Initial position
- \(x_f\) = Final position
Derivation of Work-Energy Theorem for a Variable Force
Consider a particle of mass \(m\) moving along the x-axis under the action of a variable force \(F(x)\).
Step 1: Write the Expression for Kinetic Energy
Kinetic energy of the particle is:
\[ K=\frac{1}{2}mv^2 \]
Differentiating both sides with respect to time:
\[ \frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2}mv^2 \right) \]
Since \(m\) and \(\frac{1}{2}\) are constants:
\[ \frac{dK}{dt} = \frac{1}{2}m \frac{d}{dt}(v^2) \]
Using chain rule:
\[ \frac{d}{dt}(v^2) = 2v\frac{dv}{dt} \]
Substituting:
\[ \frac{dK}{dt} = \frac{1}{2}m \left( 2v\frac{dv}{dt} \right) \]
\[ \frac{dK}{dt} = mv\frac{dv}{dt} \]
Step 2: Apply Newton's Second Law
According to Newton's Second Law:
\[ F=m\frac{dv}{dt} \]
Substituting into the above equation:
\[ \frac{dK}{dt} = Fv \]
Step 3: Express Velocity as Rate of Change of Position
We know:
\[ v=\frac{dx}{dt} \]
Substituting:
\[ \frac{dK}{dt} = F\left(\frac{dx}{dt}\right) \]
Multiplying both sides by \(dt\):
\[ \frac{dK}{dt}dt = F\left(\frac{dx}{dt}\right)dt \]
Since:
\[ \frac{dK}{dt}dt=dK \]
and
\[ \frac{dx}{dt}dt=dx \]
Therefore:
\[ dK=F\,dx \]
Step 4: Integrate Both Sides
Integrating from initial kinetic energy \(K_i\) to final kinetic energy \(K_f\), and from initial position \(x_i\) to final position \(x_f\):
\[ \int_{K_i}^{K_f} dK = \int_{x_i}^{x_f}F(x)\,dx \]
Since:
\[ \int dK=K \]
Applying limits:
\[ [K]_{K_i}^{K_f} = \int_{x_i}^{x_f}F(x)\,dx \]
\[ K_f-K_i = \int_{x_i}^{x_f}F(x)\,dx \]
Step 5: Recognize Work Done
The right-hand side represents work done:
\[ W=\int_{x_i}^{x_f}F(x)\,dx \]
Therefore:
\[ W=K_f-K_i \]
This proves the Work-Energy Theorem.
Final Formula
\[ \boxed{W=K_f-K_i} \]
or
\[ \boxed{W=\Delta K} \]
Work-Energy Theorem in Terms of Velocity
Since:
\[ K_i=\frac{1}{2}mv_i^2 \]
and
\[ K_f=\frac{1}{2}mv_f^2 \]
Substituting:
\[ W= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \]
\[ \boxed{ W= \frac{1}{2}m \left( v_f^2-v_i^2 \right) } \]
Physical Meaning of the Theorem
- If \(W > 0\), kinetic energy increases.
- If \(W < 0\), kinetic energy decreases.
- If \(W = 0\), kinetic energy remains constant.
Important Points
- The theorem is valid for both constant and variable forces.
- Work and kinetic energy are scalar quantities.
- The theorem relates work directly to change in kinetic energy.
- It is an integral form of Newton's Second Law.
- It is very useful when time is not required.
Limitations of the Work-Energy Theorem
1. Loss of Time Information
The theorem gives information only about the initial and final states. It does not provide information about the time taken during motion.
2. Scalar Nature
Newton's Second Law is a vector equation:
\[ \vec{F}=m\vec{a} \]
The Work-Energy Theorem is a scalar equation:
\[ W=\Delta K \]
Therefore, directional information is lost.
FAQs
Q1. What is the Work-Energy Theorem?
The Work-Energy Theorem states that the net work done on a body equals the change in its kinetic energy.
Q2. Is the theorem valid for a variable force?
Yes, the theorem is valid for both constant and variable forces.
Q3. Why is integration used in the theorem?
Integration is used because the force varies with position and cannot be treated as constant.
Q4. Is work a scalar quantity?
Yes, work is a scalar quantity.
Q5. What happens if net work done is zero?
The kinetic energy of the particle remains constant.
MCQs
1. The Work-Energy Theorem states that:
A) Work done = Momentum
B) Work done = Change in kinetic energy
C) Work done = Force × Time
D) Work done = Acceleration
Answer: B) Work done = Change in kinetic energy
2. The SI unit of work is:
A) Newton
B) Joule
C) Watt
D) Pascal
Answer: B) Joule
3. If net work done is negative, kinetic energy:
A) Increases
B) Decreases
C) Remains constant
D) Becomes infinite
Answer: B) Decreases
4. Work and kinetic energy are:
A) Vector quantities
B) Scalar quantities
C) Tensor quantities
D) None of these
Answer: B) Scalar quantities
5. Which law is used to derive the Work-Energy Theorem?
A) Kepler's Law
B) Newton's Second Law
C) Hooke's Law
D) Boyle's Law
Answer: B) Newton's Second Law
Summary
For a variable force, work done is given by:
\[ W=\int_{x_i}^{x_f}F(x)\,dx \]
Using Newton's Second Law and the definition of kinetic energy, we obtain:
\[ W=K_f-K_i \]
Thus, the net work done on a particle is equal to the change in its kinetic energy. This result is known as the Work-Energy Theorem.
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